Solutions to
Mathematics PULLOUT WORKSHEETS FOR CLASS X First Term By
Surender Verma M.Sc. (Mathematics), B.Ed. Delhi Public School, Dwarka, New Delhi
Since 1950
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CONTENTS 1. Real Numbers ............................................................................................... 7-15 > Worksheets (1 to 6) >Assessment Sheets (1 and 2)>Chapter Test 2. Polynomials ................................................................................................ 16-28 > Worksheets (10 to 15) >Assessment Sheets (3 and 4)>Chapter Test 3. Pair of Linear Equations in Two Variables ...................................... 29-50 > Worksheets (18 to 29) >Assessment Sheets (5 and 6) >Chapter Test 4. Triangles ...................................................................................................... 51-72 > Worksheets (33 to 45) >Assessment Sheets (7 and 8)>Chapter Test 5. Introduction to Trigonometry ............................................................... 73-87 > Worksheets (50 to 58) >Assessment Sheets (9 and 10)>Chapter Test 6. Statistics ...................................................................................................... 88-97 > Worksheets (62 to 65) >Assessment Sheets (11 and 12)>Chapter Test PRACTICE PAPERS (1 to 5) ......................................................................... 99-132
–3–
Solutions to
PULLOUT WORKSHEETS [Summative Assessments] [FIRST TERM]
Chapter
1
REAL NUMBERS WORKSHEET–1
125 53 1 = = = 0.0625 4. 3 3 16 2 5 16 × 5 125 Clearly, the decimal form of 4 3 termi2 .5 nates after four places.
1. (B)
2. (C) We know that the factors of a prime are 1 and the prime itself only. Therefore, the common factor of p and q will be 1 only. Hence, HCF (p, q) = 1. 3. (A) As prime factors of 1005 are: 1005 = 5 × 3 × 67. ∴ 7 is not a prime factor of 1005. 4. Hint:
24 192 = = 0.192 . 125 1000
5. LCM =
First number × Second number HCF
96 × 404 = 24 × 404 = 9696. 4 6. (i) 660; (ii) 330 Hint: Going in opposite direction to the factor tree, we obtain 2 × 165 = 330 (ii) and 2 × 330 = 660 (i).
=
7. HCF = 3; LCM = 420 Hint: 12 = 22 × 3; 15 = 3 × 5; 21 = 3 × 7. 8. (i) Terminating
(ii) Non-terminating repeating. 9 1 1 = = . 108 12 22 × 31
9. Hint: Let 5 – 2 3 = ⇒ R E A L
3 =
10. Let a be any odd positive integer and b = 4. By Euclid’s lemma there exist integers q and r such that a = 4q + r, 0 ≤ r < 4 . .. a = 4q or 4q + 1 or 4q + 2 or 4q + 3. Therefore, for a to be odd, we have to take a = 4q +1 or 4q +3. 11. The maximum capacity of a bag will be the HCF of 490, 588 and 882. Let us find out the required HCF by prime factorisation method. 490 = 2 × 5 × 72 588 = 22 × 3 × 72 882 = 2 × 32 × 72 ∴ HCF = 2 × 72 = 98 Thus, the maximum capacity of a bag is 98 kg.
WORKSHEET–2 1. (A) HCF (p, q) = 1 ⇒ p and q are coprime. If p and q are coprime with q ≠ 0 and
p is q
a rational number, then q has only 2 and 5 as prime factors. i.e., q = 2m × 5n where, m and n are nonnegative integers.
543 543 Hint: = 1 3. 250 2 × 5
Hint:
As RHS of this equation is rational, but LHS is an irrational so a contradiction.
a ;b≠0 b 5b – a 2b
N U M B E R S
2. (B) Going to opposite direction to the factor tree, we obtain 3 × 7 = 21 (ii) and 2 × 21 = 42 (i). 3. (A) Required number = =
23×1449 161 1449 = 207. 7
7
4. 2 = 1.414... and
3 = 1.732...
3 Therefore, we can take 1.5 = 2 3 as 2 < < 3. 2
5. Hint: As 12576 > 4052 .. . 12576 = 4052 × 3 + 420 Further 4052 = 420 × 9 + 272 Further 420 = 272 × 1 + 148 Further 272 = 148 × 1 + 124 Further 148 = 124 × 1 + 24 Further 124 = 24 × 5 + 4 Further 24 = 4 × 6 + 0. In the last equation, remainder is zero. Hence, the required HCF = 4. 6. First given number is composite as 5 × 3 × 11 + 11 = 11 (15 + 1) = 11 × 16 = 11 × 2 × 8 But second given number is prime as 5 × 7 + 7 × 3 + 3 = 35 + 21 + 3 = 59. 7. No. Prime factors of 6n will be of type 2n × 3n. As it doesn't have 5 as a prime factor, so 6n can't end with the digit 5. 8. Hint: Let a be any positive integer .. . a = 3q or 3q + 1 or 3q + 2 . .. a2 = 9q2 = 3m; m = 3q2 or a2 = (3q +1)2 = 3m + 1, m = q (3q + 2) or a2 = (3q + 2)2 = 3m + 1, m = 3q2 + 4q + 1. 9. We represent 6, 72 and 120 in their prime factors. 6 =2 × 3 72 = 23 × 32 120 = 23 × 3 × 5 Now, HCF = 2 × 3 = 6 And LCM = 23 × 32 × 5 = 360. 10. Hint: Let
2 − 5 = x, a rational number ⇒ 2 =x+ 5 Squaring both sides, we get
2 = x2 + 5 + 2x 5 – x2 – 3 ⇒ 5 = 2x RHS of this last equation is rational, but LHS is an irrational which is a contradiction.
8
11. Length = 6 m 30 cm = 630 cm Breadth = 5 m 85 cm = 585 cm Height = 3 m 60 cm = 360 cm The required length of the tape will be the Highest Common Factor (HCF) of the numbers 630, 585 and 360. Let us find out the HCF. 630 = 2 × 32 × 5 × 7 585 = 32 × 5 × 13 360 = 23 × 32 × 5 ∴ HCF = 32 × 5 = 45 Hence, the length of the tape will be 45 cm.
WORKSHEET–3 43 43 × 5 215 = = 4 = 0.0215 3 4 2 ×5 (2 × 5) 10 Hence, the number terminates after four places of decimal.
1. (C)
2. (A)
4
(
2− 3
)(
) ( 2) −( 3)
2+ 3 =
2
2
= 2 – 3 = – 1. – 1 is a rational number. 3. (C) 128 = 27; 240 = 24 × 3 × 5. Now, HCF (128, 240) = 24 = 16. 4. Hint: First number =
HCF × LCM = 232. Second number
5. No. Hint: Prime factors of 15 n will not be of type 2n × 5n. 6. Rational number = 0.27 Irrational number = 0.26010010001... . 7. (i) (ii)
145 29 8 232 = × = = 0.232 . 625 125 8 1000
7 125 875 × = = 0.0875 . 80 125 10000
8. Let us assume, to the contrary that 2 is rational. We can take integers a and b ≠ 0 such that a 3 = , where a and b are coprime. b ⇒ 3b2 = a2 ⇒ a2 is divisible by 3 ⇒ a is divisible by 3 ...(i) We can write a = 3c for some integer c M A T H E M A T
I C S – X
⇒ a2 = 9c2 ∴ ⇒ 3b2 = 9c2 [ a2 = 3b2] ⇒ b2 = 3c2 2 ⇒ b is divisible by 3 ⇒ b is divisible by 3 ...(ii) From (i) and (ii) we observe that a and b have atleast 3 as a common factor. But this contradicts the fact that a and b are coprime. This means that our assumption is not correct. Hence,
3 is an irrational number.
9. As:
1032 = 408 × 2 + 216 ...(i) 408 = 216 × 1 + 192 ...(ii) 216 = 192 × 1 + 24 ...(iii) 192 = 24 × 8 + 0 ...(iv) ⇒ HCF = 24 ∴ From (iii) ⇒ 24 = 216 – 192 = 216 – [408 – 216] {... Use (ii)} = 2 × 216 – 408 = 2[1032 – 2 × 408] – 408 {... Use (i)} 24 = 1032 × 2 – 5 × 408 ⇒ m = 2.
10. Hint: Let x be any positive integer. Then it is of the form 3q or 3q + 1 or 3q + 2. If x = 3q, then x3 = (3q)3 = 9m; m = 3q3 If x = 3q + 1, then x3 = (3q + 1)3 = 9m + 1; m = q(3q2 + 3q + 1). If x = 3q + 2, then x3 = (3q + 2)3 = 9m + 8; m = q (3q2 + 6q + 4). 11. The maximum number of columns must be the highest common factor (HCF) of 616 and 32. Let us find out the HCF by the method of Euclid's division lemma. Since 616 > 32, we apply division lemma to 616 and 32, to get 616 = 32 × 19 + 8 Since the remainder 8 ≠ 0, we apply the division lemma to 32 and 8, to get 32 = 8 × 4 + 0 R E A L
N U M B E R S
The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 8, the HCF of 616 and 32 is 8 Hence, the maximum number of columns is 8.
WORKSHEET– 4 1. (B)
(
6− 5
)(
) ( 6) − ( 5)
6+ 5 =
2
2
= 6 – 5 = 1 = Rational number.
2. (B) Hint: Denominator is not in the exact form of 2m × 5n, where m, n are non-negative integers. 3. (C) 0 ≤ r < b. 4. Hint: 107 = 4 × 26 + 3. 5. Hint: 7 × 13 = (ii) and (ii) × 11 = (i). 6. Let us represent each of the numbers 30, 72 and 432 as a product of primes. 30 = 2 × 3 × 5 72 = 23 × 32 432 = 24 × 33 Now, HCF = 2 × 3 = 6 And LCM = 24 × 33 × 5 = 2160. 7. Here, 396 > 82. ∴ 396 = 82 × 4 + 68 Further 82 = 68 × 1 + 14 Further 68 = 14 × 4 + 12 Further 14 = 12 × 1 + 2 Further 12 = 2 × 6 + 0 In the last equation, the remainder is zero and the divisor is 2. Hence, the required HCF = 2. a 8. Hint: Let 3 + 2 5 = ; b ≠ 0 b a – 3b ⇒ = 5 = Rational 2b Which is a contradiction as 5 is an irrational number. Hence, 3 + 2 5 is an irrational number. 9. (i) The given fraction can be written as 43 43 × 5 = = 0.0215 24 ·53 104 Hence, the given number terminates after four places of decimal.
9
(ii) The given fraction can be written as 4
359 2 × 359 5744 = 5 = 5 2×5 2 × 55 100000 = 0.05744 Hence, the given number terminates after five places of decimal. 10. The required number of students will be the highest common factor (HCF) of 312, 260 and 156. Let us find out the HCF by the method of prime factorisation. 312 = 23 × 3 × 13 260 = 22 × 5 × 13 156 = 22 × 3 × 13 ∴ HCF = 22 × 13 = 52 Number of buses required Total number of students = Number of students in one bus 312 + 260 + 156 = = 14 52 Thus, the maximum number of students in a bus and number of buses required are 52 and 14 respectively.
11. Hint: Let x = any positive integer x = 5m, 5m + 1, 5m + 2, 5m + 3 or 5m + 4 Now take square of all form.
WORKSHEET–5 1. (C) Hint: LCM of 18, 24, 30, 42 = 2520 ∴ Required number = 2520 + 1 = 2521. 2. (C) Let the quotient is m when n2 – 1 is divided by 8. ∴ n2 – 1 = 8 × m ⇒ n2 – 1 = An even integer. ⇒ ∴
n2
= An even integer + 1 = Odd integer n = An odd integer.
3. (B) Hint: HCF (65, 117) = 13 Now, 65m – 117 = 13. ∴ m = 2 will satisfy this equation. 4. Prime factors of numbers 1 to 10 are: 1 = 1; 2 = 1 × 2; 3 = 1 × 3; 4 = 1 × 22
10
5 = 1 × 5; 6 = 1 × 2 × 3; 7 = 1 × 7; 8 = 1 × 23; 9 = 1 × 32; 10 = 1 × 2 × 5 Now, LCM = 1 × 23 × 32 × 5 × 7 = 8 × 9 × 5 × 7 = 2520 is required number. 5. Hint: ⇒
5−
3
5+
3
= 2x – 15
4 – 15 = 2x – 15
⇒ x = 2, which is a rational number. 6. Hint: Any odd positive integer will be type of 4q + 1 or 4q + 3 ∴
(4q + 1)2 = 16q2 + 8q + 1 = 8 (2q2 + q) + 1
Also, (4q +
3)2
= 8n + 1 = 16q2 + 24q + 9 = 8 (2q2 + 3q + 1) + 1 = 8n + 1.
7. 35 cm Hint: Find HCF.
a b where a, b are integers and b ≠ 0 Squaring on both sides,
8. Hint: Let
5−3 2 =
5 + 18 − 6 10 = ⇒ ⇒
23 −
a2 b2
a2 = 6 10 b2
23b 2 − a 2
6b 2 9. (i) Terminating.
= 10 ... a contradiction. (ii) Terminating.
10. The required number of burfis will be the highest common factor of 420 and 130. Let us find out the HCF using Euclid's division lemma. It is clear that 420 > 130. We apply Division lemma to 420 and 130, to get 420 = 130 × 3 + 30 Since the remainder 30 ≠ 0, so we apply Division lemma to 130 and 30, to get 130 = 30 × 4 + 10 M A T H E M A T
I C S – X
Again the remainder 10 ≠ 0, so we apply Division lemma to 30 and 10, to get 30 = 10 × 3 + 0 Now, the remainder is zero. So the HCF of 420 and 130 is the divisor at the last stage that is 10. Hence, the required number of burfis is 10. 11. Let n = 3q, 3q + 1 or 3q + 2. Case I: If n = 3q, then n = 3q divisible by 3 n + 2 = 3q + 2 ⇒ Not divisible by 3 n + 4 = 3q + 4 = 3(q + 1) + 1 ⇒ Not divisible by 3. Case II: If n = 3q + 1 then only n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + 1) is divisible by 3. and if n = 3q + 2 then only n + 4 = 3q + 6 = 3(q + 2) is divisible by 3.
WORKSHEET– 6 1. (C) 3825 = 52 × 32 × 17 So, 11 is not a prime factor of 3825. 2. (C) As p and p + 1 are two consecutive natural numbers, HCF = 1 and LCM = p (p + 1). 3. (A) Hint: The given number is
51 17 or 1500 500
∴ Denominator = 500 = 22 × 53 Clearly, the denominator is exactly in the form 2m × 5n, where m and n are nonnegative integers; so the given number is a terminating decimal expansion. 4. Hint: ... 8 = 23; 9 = 32; 25 = 52 ∴
HCF (8, 9, 25) = 1 LCM (8, 9, 25) = 1800.
5. Hint: ∴ ⇒ ⇒ R E A L
HCF (210, 55) = 5 210 × 5 + 55y = 5 55y = 5 – 1050 −1045 y= = – 19. 55 N U M B E R S
6. Irrational Hint:
2− 3 2+ 3
=
x 3
x 3 ⇒ 7 3 – 12 = x = Irrational. 7. Rational Number = 0.55 Irrational number = 0.5477477747... . 8. 15 Hint: HCF (1380, 1455, 1620) = 15. 9. (i) 0.052. (ii) 5.8352. 10. We know that any positive integer is either of the form 3q, 3q + 1 or 3q + 2 for some integer q. Now, three cases arise. Case I. When p = 3q, ⇒
7–4 3 =
p + 2 = 3q + 2 and p + 4 = 3q + 4 Here, p = 3q is exactly divisible by 3 p + 2 = 3q + 2 leaves 2 as remainder when it is divided by 3 p + 4 = 3q + 4 or 3 (q + 1) + 1 leaves 1 as remainder when it is divided by 3. Case II. When p = 3q + 1, p + 2 = 3q + 3 and p + 4 = 3q + 5 Here, p = 3q + 1 leaves 1 as remainder when it is divided by 3 p + 2 = 3q + 3 or 3 (q + 1) is exactly divisible by 3 p + 4 = 3q + 5 or 3(q + 1) + 2 leaves 2 as remainder when it is divided by 3. Case III. When p = 3q + 2, p + 2 = 3q + 4 and p + 4 = 3q + 6 Here, p = 3q + 2 leaves 2 as remainder when it is divided by 3. p + 2 = 3q + 4 or 3(q + 1) + 1 leaves 1 as remainder when it is divided by 3 p + 4 = 3q + 6 or 3(q + 2) is exactly divisible by 3. Hence, in all the cases, one and one number out of p, p + 2 and p + 4 is divisible by 3, where p is any positive integer.
11
OR Any positive odd integer is type of 2q + 1 where q is a whole number. ∴ (2q + 1)2 = 4q2 + 4q + 1 = 4q (q + 1) + 1 ...(i) Now, q(q + 1) is either 0 or even So it is 2m, where m is a whole number. ∴ from (i) ⇒ (2q + 1)2 = 8m + 1. 11. Since, height of each stack is the same, therefore, the number of books in each stack is equal to the HCF of 96, 240 and 336. Let us find their HCF 96 = 24 × 2 × 3 240 = 24 × 3 × 5 336 = 24 × 3 × 7 So, HCF = 24 × 3 = 48. Now, number of stacks of English books
5. The required number would be the HCF of 967 – 7 = 960 and 2060 – 12 = 2048. Let us find the HCF of 960 and 2048 by using Euclid’s algorithm. Since 2048 > 960 ∴ 2048 = 960 × 2 + 128 960 = 128 × 7 + 64 128 = 64 × 2 + 0 Since the remainder becomes zero and the divisor at this stage is 64, the HCF of 960 and 2048 is 64. Hence, the required number is 64. 6.
96 =2 48 Number of stacks of Hindi books =
240 =5 48 Number of stacks of Mathematics books =
=
336 = 7. 48
ASSESSMENT SHEET–1 1. (D) The denominator of each fraction in the options (A), (B) and (C) can be expressed in the form 2n 5m, where m, n being whole numbers. 2. (A) Let x be any positive integer then it is of the form 3q or 3q + 1 or 3q + 2. So, x2 can be written in the form 3m or 3m + 1. 3. HCF × LCM = Product of the two numbers ⇒ 40 × 252 × p = 2520 × 6600 ⇒
p=
2520 × 6600 = 1650. 40 × 252
4. True, If the number 3n ends with the digit 0, then its prime factorisation contains the prime 5. But by the Fundamental Theorem of Arithmetic, there is no prime other than 3 in the factorisation of 3n.
12
Clearly, and ∴ HCF and LCM
456 = 23 360 = 23 = 23 = 23
× × × ×
3 × 19 32 × 5 3 = 24 32 × 5 × 19 = 6840.
7. Let us assume the contrary that
3 is a
rational number. a So, 3 = , where a and b are coprime. b
a2 (Squaring both sides) b2 ⇒ 3b2 = a2 ⇒ a2 is divisible by 3 ⇒ a is divisible by 3 because 3 is a prime. We can write a = 3c for some integer c Substituting a =3c in 3b2 = a2, we get 3b2 = 9c2 ⇒ b2 = 3c2 ⇒ b2 is divisible by 3 ⇒ b is divisible by 3. Therefore, both a and b are divisible by 3. But this contradicts the fact that a and b are coprime that is, no common factor other than 1. ∴
3=
M A T H E M A T
I C S – X
5000 = 5 × 103 = 5 × (2 × 5)3 = 23 × 54.
Consequently, we arrive at the result that our assumption that
3 is rational, is wrong. Hence, 3 is an irrational number. 8. Let a be any odd positive integer. Then, it is of the form 6p + 1, 6p + 3 or 6p + 5. Here, three cases arise. Case I. When a = 6p + 1, ∴ a2 = 36p2 + 12p + 1 = 6p(6p + 2) + 1 = 6q + 1, where q = p(6p + 2). Case II. When a = 6p + 3, ∴ a2 = 36p2 + 36p + 9 = 36p2 + 36p + 6 + 3 = 6(6p2 + 6p + 1) + 3 = 6q + 3, where q = 6p2 + 6p + 1. Case III. When a = 6p + 5, ∴ a2 = 36p2 + 60p + 25 = 36p2 + 60p + 24 + 1 = 6(6p2 + 10p + 4) + 1 = 6q + 1, where q = 6p2 + 10p + 4. Hence, a is of the form 6q + 1 or 6q + 3.
ASSESSMENT SHEET–2
Clearly, the decimal expansion terminates after four decimal places. 2. (C) LCM (p, q) = x3 y2 z3. 3. HCF × LCM = Product of the two numbers. 9 × LCM = 306 × 657
306 × 657 = 22338. 9 4. The maximum number out of 3, 5, 15, 25, 75 is 75. Therefore, the HCF of 525 and 3000 is 75.
⇒
LCM =
5. The denominator of R E A L
= 6. Let ∴
514 = 0.0514. 10 4
x = 2p + 1 and y = 2q + 1 x2
+ y2 = (2p + 1)2 + (2q + 1)2 = 4p2 + 4p + 1 + 4q2 + 4q + 1 = 4(p2 + p + q2 + q) + 2 =S+T S = 4(p2 + p + q2 + q) and T = 2
where
S is divisible by 4 and so an even integer. T is not divisible by 4 but an even integer. Therefore, S + T is even, as sum of any two evens is even, and not divisible by 4. 7. Let us assume the contrary that rational number.
5 is a
We can take coprime a and b (say) such that a 5 = ;b≠0 b ⇒
b 5=a
Square both the sides to get 5b2 = a2 ⇒ a2 is divisible by 5
14587 1. (D) = 11.6696. 1250
⇒
257 257 × 2 257 = = 3 5 × 10 5 × 10 3 × 2 5000
Further,
257 is 5000. 5000
N U M B E R S
⇒ a is divisible by 5 because if square of a number is divisible by a prime, then the number is divisible by the prime. Let us take some integer c such that a = 5c Square both the sides to get a2 = 25c2 Substitute
a2 = 25c2 in 5b2 = a2 to get 5b2 = 25c2 b2 = 5c2
⇒ b2 is divisible by 5 ⇒ b is divisible by 5 Therefore, both a and b are divisible by 5.
13
This contradicts the fact that a and b are coprime that is a and b have no common factor. ∴ Our assumption is false.
3. (C) As, 8q is even and 6 is even, 8q + 6 is even.
56125 449 4. ... 0.56125 = = 100000 800
5 is an irrational
So, we conclude that
=
number. 8. Any positive integer n can be written in the form 3q, 3q + 1 or 3q + 2. Here, three cases arise as follows: Case I. When n = 3q, ∴ n3 = (3q)3 = 27q3 ∴
n3 + 1 = 27q3 + 1 = 9 × 3q3 + 1 = 9m + 1, where m = 3q3.
Case II. When n = 3q + 1, n3 = (3q + 1)3
∴
= 27q3 + 1 + 3(3q + 1) × 3q = 27q3 + 27q2 + 9q + 1 ∴
n3 + 1 = 27q3 + 27q2 + 9q + 2 = 9 (3q3 + 3q2 + q) + 2
∴ 2n × 5 m = 2 5 × 5 2 n = 5, m = 2. 5.
n3 = (3q + 2)3 = 27q3 + 8 +
2– 9
∴
n3
+1=
27q3
=9
+ 8 + 54q2 + 36q +
54q2
+ 36q + 9
(3q3
6q2
+ 4q + 1)
+
= 9m, where m = 3q3 + 6q2 + 4q + 1. Hence, n3 + 1 can be expressed in the form 9m, 9m + 1 or 9m + 2 for some integer m.
CHAPTER TEST 1. (D) Since 32844 = 2 × 2 × 3 × 7 × 17 × 23 So, 11 is not prime factor of 32844. 2. (A) .. .
LCM =
306 × 1314 18
= 22338.
14
2
= 2 – 2 18 + 9
7. No. Hint: Prime factors of 9n will be type of 32n, i.e., 3 × 3 × ... 3 Even no. of times. 8.
120 105 150 HCF LCM
3 × 6q (3q + 2) = 27q3
)
= irrational. 6. Yes. 2 × 3 × 5 × 13 × 17 + 13 = 13 × (2 × 3 × 5 × 17 + 1) = 13 × 511 = a composite number.
where m = 3q3 + 3q2 + q ∴
(
= 11 – 2 18
= 9m + 2, Case III. When n = 3q + 2,
449 449 = 5 32 × 25 2 × 52
∴ And
= 23 × 3 × 5 =3× 5 × 7 = 2 × 3 × 52 = 3 × 5 = 15 = 23 × 3 × 52 × 7 = 8 × 3 × 25 × 7 = 4200.
9. Hint: Let ⇒
2 – 3 3 = x where x is rational.
(
2–3 3
)
2
= x2
⇒ 2 + 27 – 6 6 = x2 ⇒
29 – x2 = 6 6
⇒
29 – x 2 = 6 6. 6
Since 6 is not a perfect square. So always irrational.
6 is
∴ It's a contradiction. M A T H E M A T
I C S – X
10. We know that any positive integer is of the form 3q or 3q + 1 or 3q + 2. Case I:
n = 3q
⇒
n3 = (3q)3 = 9 × 3q3 = 9m
⇒
n3 + 1 = 9m + 1, where m = 3q3.
Case II:
n = 3q + 1 n3 = (3q + 1)3
⇒
= 27q3 + 1 + 27q2 + 9q
⇒
= 9q (3q2 + 3q + 1) + 1 = 9m + 1 n3 + 1 = 9m + 2, where m = q(3q2 + 3q + 1).
Case III: ⇒
n = 3q + 2 n3 = (3q + 2)3 = 27q3 + 8 + 54q2 + 36q
R E A L
N U M B E R S
n3 + 1 = 27q3 + 54q2 + 36q + 9 = 9(3q3 + 6q2 + 4q + 1) = 9m, where m = 3q3 + 6q2 + 4q + 1. Hence, n3 + 1 can be expressed in the form 9m, 9m + 1 or 9m + 2, for some integer m. 11. Length = 8.25 m = 825 cm Breadth = 6.75 m = 675 cm Height = 4.50 m = 450 cm The required length of the rod will be the highest common factor of 825 cm, 675 cm and 450 cm. Now, 825 = 3 × 52 × 11 675 = 33 × 52 450 = 2 × 32 × 52 So, HCF (825, 675, 450) = 3 × 52 = 75 Hence, length of the rod is 75 cm. ❑❑
15
Chapter
2
POLYNOMIALS WORKSHEET–10
1. (C) Hint: put x2 + 2x + 1 = 0 and solve for x. 2. (C) Since the given graph of y = p(x) cuts x-axis at three points, so the number of zeroes of p(x) are 3. 3. (A) Hint:
1 1 α+β + = . αβ α β
4. Let one zero be α, then the other one will 1 be . α k 1 ∴ α. =– α 15 ⇒ 5.
⇒
p(x) = x2 – 3x + 2.
9. According to the division algorithm, p(x) = g(x) × q(x) + r(x) ⇒ x3 – 3x2 + x + 2 = g (x) × (x – 2) + (– 2x + 4) (As given in question) x3 – 3x2 + 3x – 2 x–2 To find g(x), we proceed as given below.
⇒
g(x) =
k = – 15. Sum of zeroes (S) = – =
2 3 + 4 3
3–8 5 = – 4 3 4 3
1 2 3 =– × 2 4 3 Now, required polynomial will be 1 5 x2 – Sx + P, i.e., x2 + x – 2 4 3
Product of zeroes (P) = –
or 4 3 x2 + 5x – 2 3 . 6. Let f (x) = 2x2 + 2ax + 5x + 10 If x + a is a factor of f (x), then f (– a) = 0 Therefore, 2a2 – 2a2 – 5a + 10 = 0 ⇒ a = 2. 7. x3 – 4x2 + x + 6 Hint: If the roots are α, β and γ of a cubical polynomial, then the polynomial will be (x – α) (x – β) (x – γ) = (x – 3) (x – 2) (x + 1) = x3 – 4x2 + x + 6.
16
8. Solving α + β = 3 and α – β = –1, we get α = 1, β = 2 ... Polynomial is x2 – (α + β) x + αβ
Thus, g(x) = x2 – x + 1. 1 3 10. − ; 3 2
Hint: 6x2 – 7x – 3 = 0 ⇒ 6x2 – 9x + 2x – 3 = 0 ⇒ ⇒
3x (2x – 3) + 1 (2x – 3) = 0 (2x – 3) (3x + 1) = 0
...
x=
...
α+β=
3 7 –b −1 + = = 6 a 2 3
...
α.β=
–1 3 –1 c . = = . a 3 2 2
1 3 or − 3 2
M A T H E M A T
I C S – X
11. Let p(x) = x4 + x3 – 34x2 – 4x + 120 Given zeroes of p(x) are 2 and – 2 ... (x – 2) (x + 2) = x4 – 4 is a factor of p(x). We divide p(x) by x2 – 4, x2 + x – 30 x2 – 4 x4 + x3 – 34x2 – 4x + 120 x4 – 4x2 – + x3 – 30x2 – 4x +120 x3 – 4x + – – 30x2 + 120 – 30x2 + 120 – + 0
... p(x) = (x2 – 4) (x2 + x – 30) . . . Other zeroes of p(x) are given by x2 + x – 30 = 0 2 ⇒ x + 6x – 5x – 30 = 0 ⇒ x(x + 6) – 5(x + 6) = 0 ⇒ (x – 5) (x + 6) = 0 x = 5, – 6 Hence, all the zeroes are 2, – 2, 5 and – 6.
WORKSHEET– 11
5. p = 2 Hint: (2)3 – 3(2)2 + 3(2) – p = 0 ⇒ ⇒ ∴
8 – 12 + 6 – p = 0 2 – p =0 p = 2.
6. Let α and β be the two zeroes of f(x) = ax2 + 2x + 3a
3a 2 and αβ = =3 a a According to the question, Then, α + β = –
–2 =3 a ⇒
a=
–2 . 3
7. Let the third zero be α, then sum of the zeroes = – ⇒
2 + 3 + α= –
coefficient of x 2 coefficient of x 3 –6 1
⇒ α= 1 Hence, the third zero is 1. 8. Let us divide 6x4 + 8x3 + 17x2 + 21x + 7 by 3x2 + 4x + 1.
1. (A) ∵ p(x) = 2x2 – 2x + 1 ∴ Sum of zeroes = 1 1 Product of zeroes = . 2 2. (A) Let α = 5 and β = – 5, then the quadratic polynomial will be x2 – (α + β)x + αβ or x2 – 25. 3. (D) Let us take option (D) p(x) = (x2 – 2) – (x2 + 3x) = – 3x – 2 This is a linear polynomial. 4. For zeroes of p(x), put p(x) = ⇒ 4x2 – 4x + 1 = 2 ⇒ 4x – 2x – 2x + 1 = ⇒ 2x (2x – 1) – 1(2x – 1) = ⇒ (2x – 1) (2x – 1) = ∴ 2x – 1 =
0 0 0 0 0 0 1 1 x= , . 2 2
∴ P O L Y N O M
I
A L S
Clearly, the remainder is x + 2. Now, ax + b = x + 2 Comparing like powers of x both the sides, we obtain a = 1, b = 2. 9. We know that, Dividend = (Divisor × Quotient) + Remainder ⇒ 4x3 – 8x2 + 8x + 1 = g(x) × (2x – 1) + x + 3 ⇒ g(x) × (2x – 1) = 4x3 – 8x2 + 7x – 2 g(x) =
4x 3 – 8x 2 + 7 x – 2 2x – 1
17
Now,
2. (C) 2
2x – 3x + 2 2x – 1 4x3 – 8x + 7x – 2 2
4x3 – 2x2 – + – 6 x2 + 7 x – 2 – 6x2 + 3x – + 4x – 2 4x – 2 – + 0
Now,
3+ 1
=– And product of zeroes = =
Coefficient of x Coefficient of x 2
3 Constant term . Coefficient of x 2
(
11. p(x) = x 2 − 2x − 12; − 2 2, 3 2
)
2
Hint: For zeroes: x – 2x – 12 = 0 ⇒
x 2 + 2 2 x − 3 2 x − 12 = 0
⇒
(x – 3 2 ) (x + 2 2 ) = 0
⇒
x = – 2 2 or x = 3 2 .
WORKSHEET–12 1. (C)
Sum of zeroes =
– (– 5) = 15 1 3
3 9 2 Product of zeroes = = . 2 1 3
18
k=
– 3k 1
6 = 2. 3
1 . α
1 4a = α a2 + 4 a2 – 4a + 4 = 0 ⇒ (a – 2)2 = 0 ⇒ a = 2.
3, 1
sum of zeroes =
∴
So,
Hint: x2 – 3 x – x + 3 = 0 x=
6= –
will be
3 and 1
⇒
⇒
3. (D) Let one zero be α, then the other one
Hence, g(x) = 2x2 – 3x + 2. 10.
Sum of zeroes = 6
α.
4. (A) Let the zeroes be α, β, γ. If γ = – 1, then c αβγ = – 1 If γ = – 1, then αβ = c ...(i) Further, (– 1)3 + a (– 1)2 + b (– 1) + c = 0 ⇒ –1 + a–b +c = 0 ⇒ c=b–a+1 ...(ii) From equations (i) and (ii), we have αβ = b – a + 1. 5. Given polynomial is: f (x) = x2 – px – 2p – c . .. α + β= p and α . β = – 2p – c ... (α + 2) (β + 2) = αβ + 2 (α + β) + 4 = – 2p – c + 2p + 4 = (4 – c). 6. λ = 6 Hint: (α + β)2 = (α − β)2 + 4αβ. 7. x = –1 or 3; f(x) = Hint: x= ∴ Sum of zeroes = Product of zeroes = ∴ p(x) = =
x2 – 2x – 3 – 1 or 3, 2 –3 x2 – (α + β)x + αβ x2 – 2x – 3.
47 4 Hint: f (x) = {x2 – (sum of roots) x + (product of roots)}
8. x2 – x –
M A T H E M A T
I C S – X
9. The number which to be subtracted is the remainder when 4x4 + 2x3 – 8x2 + 3x – 7 is divided by 2x2 + x – 2. To find the remainder, we proceed as following. 2x2 – 2 2x2 + x – 2 4x4 + 2x3 – 8x2 + 3x – 7 –
4x4 + 2x3 – 4x2 – + –
4x2
+ 3x – 7
4 – 4x – 2x + + – + 5x – 11
Hence, 5x – 11 must be subtracted from 4x4 + 2x3 – 8x2 + 3x – 7 so that it becomes exactly divisible by 2x2 + x – 2. 10. g(x) = x2 + 2x + 1 Hint:
p(x) = g(x) × q(x) + r(x)
⇒
g(x) =
where,
p(x) = 3x3 + x2 + 2x + 5 q(x) = 3x – 5 r(x) = 9x + 10.
p ( x ) – r( x ) q( x )
5 5 and x = – are zeroes of 3 3 p(x) = 3x4 + 6x3 – 2x2 – 10x – 5, so p(x) is
11. Since x =
divisible by x –
Hence, all the zeroes of p(x) are –
5, 3
5 , – 1 and – 1. 3
WORKSHEET–13
2
and
3(x + 1)2 = 0 x = – 1 and x = – 1
⇒ ⇒
5 5 5 x + , i.e., x2 – 3 . 3 3
1. (A) Hint: Given polynomial can be written as: p(x) = 2x2 + 3x – 11
–b a c Product of zeroes = . a Sum of zeroes =
2. (B) Sum of zeroes = – 99 = –ve Product of zeroes = 127 = +ve If the sum of both zeroes is negative, then the zeroes would be either both negative or one negative and other one positive. If the product of both the zeroes is positive, then the zeroes would be either both positive or both negative. Consequently, we obtain that both the zeroes are negative. 3. (D) We know that the degree of the remainder is less than the degree of divisor. Here, degree of the divisor is 3, therefore, the possible degree of the remainder can be any out of 0, 1 and 2. 4. Hint: Substitute x = – 2 in x2 + 2x + k = 0. 5. Since α, β are the zeroes of x2 + px + q, then α + β = –p; αβ = q p 1 1 α +β Now, + = =– q α β αβ And
1 1 1 1 × = = α β αβ q
So the polynomial having zeroes will be Here, other two zeroes of p(x) are the other two zeroes of quotient 3x2 + 6x + 3 Put 3x2 + 6x + 3 = 0 P O L Y N O M
I
A L S
1 1 and β α
p 1 1 1 1 1 p(x) = x2 – + x + × = x 2 + x + α β α β q q
or p(x) = qx2 + px + 1.
19
6. g(x) = x2 + 2x + 7. Hint: Divide x3 + 3x – 14 by x – 2. 7. ...
p(x) = g(x) = q(x) = r(x) =
3x2 – 3x + 12. x2 – x + 4 3 0.
=
(α + β )2 − 2αβ = b2 – 2ac . 2 2 2 αβ
c
OR Let us divide x4 + 2x3 + 8x2 + 12x + 18 by x2 + 5.
1 1 8. ,– 7 7 Hint: For zeroes: 21x2 – 3 = 0 1 x2 = 7 ...
α 2 + β2 1 1 = + α 2β2 α2 β2
11. Hint:
x= ±
1 . 7
9. Since a = 2 is a zero of a3 – 3a2 – 10a + 24, therefore a3 – 3a2 – 10a + 24 is divisible by a – 2. Further the obtained quotient will provide the other two zeroes.
Clearly, the remainder is 2x + 3. Now, px + q = 2x + 3 Comparing like powers of x both the sides, we get p = 2, q = 3.
WORKSHEET–14
Put a2 – a – 12 = 0 for other zeroes. ⇒ (a – 4) (a + 3) = 0 ⇒ a = –3, 4 Thus, the other two zeroes are – 3 and 4. 10. g(x) = x + 1. Hint: Applying division algorithm, we get x4 + 1 = g(x) × (x3 – x2 + x – 1) + 2 ⇒
g(x) = =
4
x –1 x – x2 + x – 1
3. (A) α + β =
3 1 , αβ = 2 2
... (α – β)2 = (α + β)2 – 4αβ
3
( x + 1 )( x – 1 )( x 2 + 1 )
9 1 –2= 4 4 1 α–β= ± 2 =
( x – 1 )( x 2 + 1 )
= x + 1.
20
1. (C) If a quadratic polynomial has equal roots , then its discriminant must be zero. So, b2 – 4ac = 0 ⇒ b2 = 4ac This last equation holds if a and c have same sign. 2. (D) Sum of zeroes = – 3 + 7 = 4, Product of zeroes = (– 3) × 7 = – 21 A polynomial may be k(x2 – 4x – 21) where k has infinitely many real values. Hence, infinitely many number of polynomials can be.
⇒
M A T H E M A T
I C S – X
...
1 1 α = , β = 1 or α = 1, β = 2 2
5 ... α + 2 = , β + 2 = 3 or α + 2 = 3, 2 5 . 2 Hence, the required polynomial can be β+2=
15 5 11 5 x2 – + 3 x + × 3, i.e., x2 – x+ . 2 2 2 2 4. Let zeroes be α and β. α + β = 6, αβ = 4 Using (α – β)2 = (α + β)2 – 4αβ, we get (α − β)2 = 62 – 4 × 4 = 20 ⇒ α – β, = ± 2 5 Thus, the difference of zeroes is ± 2 5 .
(α + β )2 − 2αβ
αβ 13 25 − 12 = = . 6 6
6. Hint: x2 – 1 = (x + 1) (x – 1) ... x = – 1 or 1, both will satisfy with the given polynomial. ... we get, p + q + r + s + t = 0 ...(i) and p–q+r–s+t=0 ...(ii) From (ii), p+r+t=q+s From (i), 2 (q + s) = 0 ⇒ q + s = 0 .. . p + r + t = q + s = 0. 7. No. Hint: Divide q(x) by g(x). If the remainder obtained is zero, then the g(x) is a factor of q(x) otherwise not. 8. a = 1, b = 7 Hint: Put remainder = 0 and equate coefficient of x in the remainder and constant term with zero. 9. According to division algorithm, p(x) = g(x) × q(x) + r(x) P O L Y N O M
I
6x2 + 3x + 2, g(x) = 3 2x2 – x, r(x) = – 2 8x3 + 6x2 – x + 7, g(x) = 2x2 + 1 4x + 3, r(x) = – 5x + 4 9x2 + 6x + 5, g(x) = 3x + 2, 3x, r(x) = 5.
10. Given quadratic polynomial is 5 5 x2 + 30x + 8 5 To find its zeroes, put 5 5 x2 + 30x + 8 5 = 0
5 5 x2 + 20x + 10x + 8 5 = 0
⇒ ⇒ 5x ⇒
(
)
A L S
i.e.,
( 5) (
) 5x + 4) = 0
5x + 4 + 2 5
( 5x + 2
⇒
α 2 + β2 α β 5. Hint: = + β α αβ =
(i) p(x) = q(x) = (ii) p(x) = q(x) = (iii) p(x) = q(x) =
x=– x=–
5x + 4 = 0
2 5
or x =
–4 5
2 5 4 5 or x = – 5 5
–2 5 4 5 6 5 – =– 5 5 5 And product of zeroes
So, sum of zeroes =
2 5 4 5 8 = – = . × – 5 5 5
Also,
sum of zeroes = – =–
And product of zeroes = =
Coefficient of x Coefficient of x 2
30 6 5 =– 5 5 5
Constant term Coefficient of x 2 8 8 5 = . 5 5 5 Hence verified.
OR Hint:
Let S =
α−1 β−1 + α+1 β+1
α − 1 β − 1 P= α + 1 β + 1
21
... Required polynomial p(x) = x2 – Sx + P. 11. As
3 and – 3 are the zeroes of the given 2 2
3 quadratic polynomial, so x – and 2 3 x + will be the factors of that, Conse2
3 3 quently, x – × x + , i.e., 2 2
2 3 x – 2
must be the factor of that. Let us divide 3 2x4 – 10x3 + 5x2 + 15x – 12 by x2 – . 2
2x2 – 10x + 8
x2 –
3 2x4 – 10x3 + 5x2 + 15x – 12 2 2x4 – 3x2 –– + – 10x3 + 8x2 + 15x – 12 – 10x3 +
+ 15x – 8x2 8x2 –
0
– 12 – 12 +
Now, 2x4 – 10x3 + 5x2 + 15x – 12
= x2 –
3 2 (2x – 10x + 8) 2
By splitting –10x, we factorise 2x2 – 10x + 8 as (x – 4) (2x – 2). So, its zeroes are given by x = 4 and x = 1. Therefore, the zeroes of the given polynomial are
3, – 2
3 , 1 and 4. 2
WORKSHEET–15 1. (B) Hint: f (x) = x2 – px – ( p + c) (α + 1) (β + 1) = αβ + (α + β) + 1 .
22
2. (A) Hint:
1 1 1 α+β+ γ + + = . βγ γα αβγ αβ
3. (D) Let zeroes be α and β, then (α – β)2 = 144 ⇒ α – β = + 12 α + β= – p αβ = 45 Also, we have (α – β)2 = (α + β)2 – 4αβ 144 = p2 – 180 ⇒ p = ± 18.
...(i) ...(ii) ...(iii)
4. Let the given linear polynomial be y = ax + b ....(i) This passes through points (1, –1), (2, 1) and
3 , 2
0
∴ –1= a+ b 1 = 2a + b
...(ii) ...(iii)
3 a+b ...(iv) 2 Solving equations (ii) and (iii), we get a = 2, b = – 3 which satisfy to equation (iv). Consequently, using equation (i), we get y = 2x – 3 ∴ Polynomial is p(x) = 2x – 3 3 Since p(x) = 0 if x = 2 3 ⇒ x= is zero of p(x). 2 0=
5. Let us divide ax3 + bx – c by x2 + bx + c by the long division method.
x2
ax – ab 3 ax + bx – c + bx + c 3 ax + abx2 + acx – – – – abx2 + (b – ac) x – c – abx2 – ab2 x – abc + + + (ab2 + b – ac)x + abc – c
Put remainder = 0 M A T H E M A T
I C S – X
⇒ (ab2 + b – ac)x + (abc – c) = 0 ⇒ ab2 + b – ac = 0 and abc – c = 0 Consider abc – c = 0 ⇒ (ab – 1) c = 0 ⇒ ab = 1 or c = 0. Hence, ab = 1. 6. Hint: Let p(x) = x3 – mx2 – 2npx + np2 (x – p) is a factor of p(x) ⇒ p(x) = 0 at x = p. ⇒ p3 – p2m – p2n = 0 ⇒ p2 [(p – (m + n)] = 0 ⇒ p = m + n where p ≠ 0. 7. x3 – 4x2 + x + 6 Hint: The required cubic polynomial is given by (x – 3) (x – 2) (x + 1) or x3 – 4x2 + x + 6 This is the required polynomial. 8. – 2, 3, 4 Hint: α+β+γ=5 αβ + βγ + αγ = –2 αβγ = –24 Let αβ = 12 ... γ = –2 ... α+β =7 ⇒ (α – β)2 = 1 ⇒ α–β =±1 ... α + β = 7 and α – β = 1 ⇒ α =4 β =3 or α + β = 7 and α – β = –1 ⇒ α =3 β = 4. 9. f (x) would become exactly divisible by g(x) if the remainder is subtracted from f(x). Let us divide f(x) by g(x) to get the remainder. x2 + 6x + 8 x2 – 4x + 3 x4 + 2x3 – 13x2 – 12x + 21 x4 – 4x3 + 3x2 – – + 6x3 – 16x2 – 12x + 21 6x3 – 24x2 + 18x – + – 2 8x – 30x + 21 8x2 – 32x + 24 – – + 2x – 3
Hence, we should subtract 2x – 3 from f(x). P O L Y N O M
I
A L S
10. If 2 ±
(
3 are zeroes of p(x), then x – 2 + 3
(
)
)
and x – 2 – 3 are factors of p(x). Consequently
{x – (2 + 3 )} {x – (2 – 3 )}
i.e., (x – 2)2 – 3, i.e., x2 – 4x + 1 is factor of p(x). Further,
x2 – 2x – 35 x2 – 4x + 1 x4 – 6x3 – 26x2 + 138x – 35 x4 – 4x3 + x2 – – + – 2x3 – 27x2 + 138x – 35 – 2x3 + 8x2 – 2x + – + – 35x2 + 140x – 35 – 35x2 + 140x – 35 – + + 0 Clearly x2 – 2x – 35 is a factor of p(x) ⇒ (x – 7)(x – 5) is a factor of p(x) ⇒ x – 7 and x + 5 are factors of p(x) ⇒ x – 7 = 0 and x + 5 = 0 give other zeroes of p(x) ⇒ x = 7 and x = – 5 are other zeroes of p(x). Hence, 7 and – 5 are required zeroes. 11. Hint:
α2 β 2 α 4 + β4 + 2 = 2 β α α 2β 2 =
{(α + β)2 − 2αβ}2 − 2α 2β2 . α 2 β2
OR Given polynomial is: f (x) = pqx2 + (q2 – pr)x – qr Put f (x) = 0 to find roots. pqx2 + (q2 – pr) x – qr = 0 ⇒ pqx2 + q2x – prx – qr = 0 ⇒ qx(px + q) – r(px + q) = 0 ⇒ (px + q)(qx – r) = 0 ⇒
x=–
q r or x = q p
23
Sum of roots
=– =–
(
q r pr – q 2 + = q p pq
Coefficient of x Coefficient of x
2
6x2 + 7 2 x + 4 x – 2
q r qr r Product of roots = – p × q = – p = – pq
=
6x3 + 2x2 – 10x – 4 2 6x 3 – 6 2 x 2 – + 7 2x2 – 10x – 4 2 7 2x2 – 14x + – 4x – 4 2 4x – 4 2 – + 0
Constant term . Coefficient of x 2
ASSESSMENT SHEET– 3 1. (C) ⇒
Discriminant = 0 b2 – 4ac = 0 b2
⇒
= 4ac
2. (D) Required quadratic polynomial = x2 – (sum of zeroes)x + product of zeroes = x2 – 2 3x – 5 3 .
) (6x2 + 3
2x + 4 2x + 4
= x– 2
(
) (3
)(
(
)(
Hence, 3 2 x + 4
f(x) = x2 – p(x +1) – c = ∴ α + β = p; αβ = Now, (α + 1) (β + 1) = = = =
x2 – px – (c + p) – (c + p) αβ + (α + β) + 1 – (c + p) + p + 1 –c–p+p+1 1 – c.
2 x2 – 10x – 4 2
2 is a zero of f (x), (x – 2 ) is a factor of f(x).
2x + 4
2x + 4
)
)
2x + 1
)
2 x + 1 gives x = –
Therefore, other two zeroes are – –
)
2 2 3
1 2 2 2 and 3
2 . 2
6. p(y) = y2 +
3 5 y–5 2
Here, a = 1, b =
4. Correct,
24
(
or x = –
p(x)= x2 – ax – (a + 1) At x = – 1, p(x) = (–1)2 – a( –1) – (a + 1) = 1 + a – a –1 = 0 q(x) = ax2 – x – (a + 1) at x = – 1, q(x) = a( –1)2 – ( –1) – (a + 1) = a + 1 – a –1 = 0 Therefore, x + 1 is the common factor of p(x) and q(x).
As
) (6 x 2 + 7
= x– 2
LHS = = positive sign ⇒ RHS = 4ac must be positive sign. ⇒ c and a have same signs.
5. Let f(x) = 6x3 +
(
∴ f (x) = x – 2
b2
3.
)
Let us divide f(x) by x – 2 .
3 5 ,c=–5 2
Discriminant 2
3 5 D = b2 – 4ac = – 4 × 1 × (– 5) 2
=
125 45 + 20 = 4 4
–3 5 125 ± –b± D 2 4 Now, y = = 2a 2×1 M A T H E M A T
I C S – X
Here, remainder is (20 + a)x + b + 25. If the polynomial p(x) is exactly divisible by 2x2 – 5, the remainder must be zero. ∴ (20 + a)x + (b + 25) = 0 Comparing the coefficients of like powers of x between both the sides, we have 20 + a = 0 and 25 + b = 0 ⇒ a = – 20 and b = – 25.
–3 5 5 5 ± 2 2 = 2 –8 5 2 5 2 2 = or 2 2
=
5 or – 2 5 2
ASSESSMENT SHEET–4
5 and – 2 5 . 2
Hence, the zeroes are
1. (C) Sum of zeroes = –
7. α and β are zeroes of f (x) = x2 – x – 2 Sum of roots = α + β = –
–1 =1 1
–2 =–2 1 ∴ (2α + 1) + (2β + 1) = 2(α + β) + 2
Product of roots = αβ =
1 . 3 2. (B) At x = 2, p(x) = 0, i.e., p(2) = 0 ∴ a (2)2 – 3 × 2 (a – 1) – 1 = 0 ⇒ 4a – 6a + 6 – 1 = 0
...(i) ...(ii)
...(iii)
⇒
= 4αβ + 2 (α + β) + 1
Now,
= 4 (– 2) + 2 (1) + 1 [Using (i) and (ii)] = –5 ...(iv) Now, required polynomial can be given by x2 – {(2α + 1) + (2β + 1)}x + (2α + 1)(2β + 1) [Using (iii) and (iv)] i.e., x2 – 4x – 5. 8. Let us divide p(x) by
– 5.
3x2 + 4x + 5 2x2 – 5 6x4 + 8x3 – 5x2 + ax + b 6x4 –
– 15x2 + 8x3 + 10x2 + ax + b 8x3 – 20x + –
P O L Y N O M
I
A L S
5 . 2
α +β 1 1 + – 2αβ = – 2αβ αβ β α
=
5 –2×4 4
= –
27 . 4
4. Using division algorithm, we have g(x) × (x – 2) – 2x + 4 = x3 – 3x2 + x + 2 ⇒
g(x) =
x 3 – 3x 2 + 3x – 2 x–2
Here, at x = 2, x3 – 3x2 + 3x – 2 = 8 – 12 + 6 – 2 = 0 ∴
= x3 – 3x2 + 3x – 2 = (x – 2) (x2 – x + 1)
2
10x + (20 + a)x + b 10x2 – 25 – + (20 + a)x + b + 25
a= Sum of zeroes = α + β = 5 Product of zeroes = αβ = 4
3.
And (2α + 1) (2β + 1) = 4αβ + 2α + 2β + 1
2x2
2
Product of zeroes =
= 2(1) + 2 [Using (i)] =4
–3 2 = 3
= ⇒
( x – 2)( x2 – x + 1) ( x – 2)
g(x) = x2 – x + 1.
25
3 2 The required polynomial is given by k [x2 – sx + p]
5. Given s = 2 and p = –
Let us draw the graph of p(x) using this table. Y (0,6) 6–
3 i.e., k x 2 – 2x – , where k is any real 2 number.
5– (–1, 4)
3–
2
= 4 3 x + 8 x – 3x – 2 3
2–
= 4x( 3x + 2) – 3 ( 3x + 2)
1–
or x =
Thus, the zeroes are α = –
–
–
–
–
–
1
2
3
= –
X
p(x) = – x2 + x + 6
Y
3 2 3 and β = 4 3
5 5 3 =– 4 3 4× 3
Coefficient of x =– Coefficient of x 2
Product of zeroes = αβ = –
0 –1 –
–3 –
3 4
3 –5 3 2 3 =– + = 4 12 3
From the graph, it is clear that the zeroes of p(x) are – 2 and 3.
(ii)
Let y = p(x) y = x3 – 4x
∴
The table for some values of x and their corresponding values of y is given by x
–2
–1
0
1
2
y
0
3
0
–3
0
Let us draw the graph of p(x) by using this table.
2 3 . 3 4 3
(–1, 3)
Constant term 2 3 = . Coefficient of x 2 4 3
3– 2– 1– (2, 0)
–
–
–
–
X
(–2, 0)
–
y = p(x) y = – x2 + x + 6
Y
–
Hence verified.
–3
– 2 –1 0 –1 –
1
2
3
X
–2 –
The table for some values of x and their corresponding values of y is given by
26
– 2 –1
–2 –
Sum of zeroes = α + β
7. (i) Let ∴
–3
or 4 x – 3 = 0
–2 3 = 3 3
=–
–
X
To find zeroes of f (x), put f(x) = 0
x=
(3, 0)
(–2, 0)
= ( 3x + 2)(4x – 3)
–2
(2, 4)
4–
6. Let f(x) = 4 3 x 2 + 5 x – 2 3
3x + 2 = 0
(1,6)
x
–2
–1
0
1
2
3
y
0
4
6
6
4
0
p(x) = x3 – 4x – – 3
(1, –3)
Y
From the graph, it is clear that the zeroes of p(x) are – 2, 0 and 2 . M A T H E M A T
I C S – X
8. Let f (x) should be added to p(x) so that the resulting polynomial is exactly divisible by g(x). Since the degree of f (x) is less than that of g(x). So, f(x) may be ax + b i.e., f(x) = ax + b. Therefore, the new dividend would be x4 + 2x3 – 2x2 – 5x + 7 + ax + b, i.e., x4 + 2x3 – 2x2 + (a – 5)x + b + 7 Let us divide this new dividend by g(x).
7 . 2 So, option (B) is correct. ⇒ x = 0 or –
4. Hint: g(x) = =
p( x ) – (7 – 5 x ) 2x
2x 3 + 4 x 2 + 5x + 7 – 7 + 5x 2x
= x2 + 2x + 5.
2
x +1 x2 + 2x – 3 x4 + 2x3 – 2x2 + (a – 5)x + b + 7 x4 + 2x3 – 3x2 + – – x2 + (a – 5)x + b + 7 –
2x – 3 – + (a – 7)x + b + 10
CHAPTER TEST
∴
–5 1 , αβ = 2 2
α + β + αβ = –2
So, option (A) is correct. 2. (B) p(x) = x2 – (α + β)x + αβ = x2 + x – 2 So, option (B) is correct. 3. (B) Hint: p(x) = x x + 7 2 ∴ zeroes are given by 7 xx+ = 0 2
P O L Y N O M
αβ =
x2 +
Thus, the remainder obtained must be zero. ∴ (a – 7)x + b + 10 = 0 ⇒ (a – 7)x + (b + 10) = 0 . x + 0 ⇒ a – 7 = 0 and b + 10 = 0 ⇒ a = 7 and b = – 10 Hence, f (x) = 7x – 10.
1. (A) ∵ α + β =
5. Hint: α + β =
I
A L S
6. Hint:
⇒
24 6 = 4 5 5 –9 5 4 5
=
–9 . 4
α= –β
α+β= 0
⇒
–b =0 a
3k + 1 =0 2
⇒
k=
–1 . 3
7. Answer may vary. 8. If α, β and γ are the zeroes of a cubic polynomial f(x), then f(x) = x3 – (α + β + γ) x2 + (αβ + βγ + γα) x – αβγ Here, α + β + γ = 4, αβ + βγ + γα = 1 and αβγ = – 6 ∴ f(x) = x3 – 4x2 + x + 6. 9. We know that Dividend = Quotient × Divisor + Remainder ⇒ Dividend – Remainder = Quotient × Divisor Clearly, RHS of the above result is divisible by the divisor. Thus, if we subtract remainder from the dividend, then it will be exactly divisible by the divisor. Dividing x4 + 2x3 – 13x2 – 11x + 10 by x2 – 4x + 3, we get
27
∴ Other zeroes are given by
x2 + 6x + 8 x2 – 4x + 3 x4 + 2x3 – 13x2 – 11x + 10 x4 – 4x3 + 3x2 – – + 3 6x – 16x2 – 11x +10 6x3 – 24x2 + 18x – – + 2 8x – 29x + 10 8x2 –
–
32x + 24 – + 3x – 14
Quotient = x2 – 6x + 8 and remainder = 3x – 14. Thus, if we subtract the remainder 3x – 14 from x4 + 2x3 – 13x2 – 11x + 10, it will be divisible by x2 – 4x + 3. 10. Hint: Since x =
–1 1 and x = are zeroes. 3 3
1 1 Therefore, x – x + will be a 3 3
1 factor of p(x), i.e., x – is a factor of p(x). 3 2
3x2 – 15x + 18
x2
1 3x4 – 15x3 + 17x2 + 5x – 6 – 3 3x4 – x2 + – – 15x3 + 18x2 + 5x – 6 – 15x3 + 5x – + 2 18x – 6 18x2 – 6 – + 0
28
3x2 – 15x + 18 = 0 ⇒
x2 – 5x + 6 = 0
⇒
(x – 3) (x – 2) = 0
∴
x = 3, 2.
11. We have 4 3 x 2 + 5x – 2 3 =
(
)(
3 x + 2 4x – 3
)
So, the value of 4 3 x 2 + 5 x – 2 3 is zero when, 3 x + 2 = 0 or 4 x – 3 = 0, i.e., when x =
or x =
–2 3
3 . Therefore, the zeroes of 4
4 3x2 + 5x – 2 3 are
–2 3
and
3 . 4
Now, sum of zeroes
–2 3 =
+
–5 3 = 4 4 3 – Coefficient of x Coefficient of x 2
–2 3 –2 3 × Product of zeroes = = 3 4 4 3
=
Constant term . Coefficient of x 2
❑❑
M A T H E M A T
I C S – X
Chapter
3
PAIR OF LINEAR EQUATIONS IN TWO VARIABLES WORKSHEET–18
1. (B) Since (3, a) lies on the equation 2x – 3y = 5, therefore, (3, a) must satisfy this equation. ∴ 2 (3) – 3 (a) = 5 ⇒
⇒ a=
3a = 1
1 . 3
–3 2 = k –9
2. (A) Hint: ⇒
k = 6.
3. (C) Hint: The condition of inconsistency of two equations a1x + b1y = c1 and a2x + b2y = c2 is given by a1 = b1 ≠ c1 . a2 b2 c2 Hint: Joining the given equations, we get 80x + 80y = 240 or x + y= 3 ...(i) Subtracting given first equation from other one, we get 6x – 6y = – 6 or x – y = –1 ...(ii) Solving equations (i) and (ii), we obtain x = 1, y = 2. 5. x = 3, y = 2
1 1 = u, = v. x−y x+y
6. False Let us substitute c = 40, The given equations become x – 2y = 8 or 5x – 10y = 40 R
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7. The given equations are 4(2x + 3y) = 9 + 7y and 3x + 2y = 4 or 8x + 5y –9 = 0 3x + 2y – 4 = 0 By cross-multiplication, we have
–y 1 x = = 1 –5 –2
⇒
x = – 2 and y = 5 Hence, x = – 2, y = 5 is the solution of the given system of equations. 8. To draw a line, we need atleast two solutions of its corresponding equations. x + 3y = 6; at x = 0, y = 2 and x = 3, y = 1. So, two solutions of x + 3y = 6 are: x
0
3
y
2
1
2x – 3y = 12; at x = 0, y = –4 and at x = 6, y = 0 So, two solutions of 2x – 3y = 12 are:
∴ Given equations become 10u + 2v = 4 and 15u – 5v = – 2.
P A I
⇒ The equations represent a pair of coincident lines. ⇒ The equations have infinitely many solutions.
−y 1 x = = − 16 15 32 + 27 − − 20 + 18
4. x = 1, y = 2
Hint: Let
1 –2 8 = = 5 –10 40
Here,
x
0
6
y
–4
0
Now, we draw the graph of given system of equations by using their corresponding solutions given in the above tables.
E Q U A T
I O N S
...
29
2. (A) Condition for parallel lines is
b a1 c = 1 ≠ 1 b2 c2 a2 ⇒
–3 –2 1 = ≠ –1 k 3
⇒ k = – 6.
3. (D) As y = 2 and y = 7, both represent straight lines parallel to x-axis ∴ y = 2 and y = 7 are parallel lines. Hence, the given pair of equations has no solution. 4. The given lines to be coincident, if From the, graph the two lines intersect the y-axis at (0, 2) and (0, – 4). 9. Let the fixed charges and change per km be Rs. x and Rs. y respectively. x + 10y = 105 ...(i) x + 25y = 255 ...(ii) Subtracting equation (i) from equation (ii), we get 15y = 150 y = 10 ...(iii) From equations (i) and (iii), we get x= 5 Now, the fare for travelling a distance of 35 km = x + 35y = 5 + 35 × 10 = Rs. 355. Fixed charge = Rs. 5 Charge per km = Rs. 10 Total charge for 35 km = Rs. 355.
WORKSHEET–19 1. (C) x – 5y = 5. (2, k) lies on it.
30
∴
2 – 5(k) = 5
⇒
k= –
⇒ 3 . 5
5(k) = – 3
– ( k – 3) 3 k = = –k k 12 III II I
Taking I and II, we have k2 = 36 ⇒ k = ± 6. Taking II and III, we have k2 – 3k = 3k ⇒ k(k – 6) = 0 ⇒ k = 0 or 6 Using (i) and (ii), we obtain k = 6. 5. x = 5, y = 2 Hint: Adding the given equations, we get 2x + y = 12 Subtracting the given equations, we get 3x + y = 17 Now, (i) – (ii) ⇒ – x = – 5 x=5 ∴ from (ii ) ⇒ 3(5) + y = 17 y = 2.
...(i)
... (ii)
...(i) ...(ii)
6. Yes. Applying the condition c1 a1 b1 = = c a2 b2 2 We have –3 2 1 = = –9 6 3 That is true. Therefore, the pair of equations is consistent with infinitely many solutions. M A T H E M A T
I C S – X
Equations (i) and (ii) from the required pair of linear equations. On solving these two equations, we will find starting salary of x = Rs. 13000 and fixed increment of y = Rs. 500.
615 255 , y= 104 52 Hint: The system: 9x – 10y + 15 = 0 5x + 6y – 60 = 0 By cross-multiplication, we have −y x 1 = = . 600 − 90 − 540 − 75 54 + 50
7. x =
WORKSHEET–20 1. (B) As the lines are intersecting each other, 3 –3 2 ≠ ⇒ a≠ . a –1 2
8. For equation 3x + y – 2 = 0, x
0
1
y
2
–1
2. (D) Line x = a is parallel to y-axis and the line y = b is parallel to x-axis. These lines intersect each other at (a, b). 3. 3x – y – 5 = 0 and 6x – 2y – k = 0 have no solution. ⇒ These equations represent a piar of parallel lines. 3 –5 –1 = ≠ ⇒ k 6 –2 ⇒ k ≠ 10.
For equation 2x – 3y – 5 = 0, x
–2
1
y
–3
–1
4. No. For infinitely many solutions, the following condition must be satisfied. λ 3 7 = = 2 6 –14 1 –7 1 3 But, here ≠ as ≠ – 2 14 2 6 Hence, no value of ‘λ’ provides the pair of infinitely many solutions.
As the lines corresponding to the given equations intersect each other at (1, –1), the required solution is x = 1, y = –1. 9. Let the man's starting salary and fixed increment be x and y respectively. According to the question, x + 4 y = 15000 x + 10 y = 18000 P A I
R
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...(i) ...(ii)
5. The given system of equations can be written as ax + by – (a – b) = 0 bx – ay – (a + b) = 0 By cross-multiplication, –y x = – a( a + b) + b ( a – b) – b( a + b) – a ( a – b) (ii ) (i)
1 – a2 – b2 (iii ) ⇒ Taking (i) and (iii) simultaneously, we get x = 1 and y = – 1
E Q U A T
I O N S
...
31
Hence x = 1, y = – 1 is the solution of the given system of equations. 6. x = 6, y = – 4, m = 0 1 1 Hint: Take = u and = v. x y 7. No; (6, 0) , (4, 0) Hint: For x + 3y = 6 x
0
3
y
2
1
WORKSHEET–21 1. (B) Hint: According to the condition of infinitely many solutions, we reaches at 2a – b 21 a+b = = . 2 3 7
2. (C) Hint: Simplifying the given linear equations, we have
7 2 8 7 – = 5, + = 15 y x y x Now take
For 3x + 9y = 12 x
1
4
y
1
0
Let us draw the graph of lines using the tables obtained above.
1 1 = u, = v; and solve. y x
3. (D) Let unit's and ten's digit be x and y respectively. x + y= 9 ...(i) 10y + x + 27 = 10x + y ...(ii) Solving equations (i) and (ii), we have x = 6, y = 3 Hence, the required number is 3 × 10 + 6, that is 36. 4. False ... Equations are 5x – 5y = 3 and 10x – 10y – 3 = 0 a1 b1 5 1 –5 1 = = ; = = – a2 10 2 b2 10 2 a1 c1 b1 c1 3 = =1 ∴ = ≠ a2 b2 c2 c2 3
∴
In the graph, lines are parallel. So, the pair of equations is not consistent. The lines intersect the x–axis at (4, 0) and (6,0). 8. Let the initial length be x and breadth be y. Then according to question, xy – 9 = (x – 5) (y + 3) ...(i) and xy + 67 = (x + 3) (y + 2) ...(ii) Simplifying equations (i) and (ii), we have 3x – 5y = 6 ...(iii) 2x + 3y = 61 ...(iv) On solving (iii) and (iv), x = 17, y = 9 Hence, length of rectangle is 17 units and that of breadth is 9 units.
32
⇒
lines are parallel.
5. p ≠ 6 Hint:
3 5 ≠ p 10
⇒ p≠6
∴ p can take any value but not 6. 6. x =
− a + 4b 4a–b , y= 5b 5a
Hint:
x – 3b (2 a + b) + 2b ( a + 2b )
=
–y – 2 a (2 a + b ) + 3 a (a + 2b )
=
1 2 a × 2b – 3 a × 3b
M A T H E M A T
I C S – X
Take first and third terms as well as second and third terms and solve. 7. a = 7, b = 3 Hint: For infinitely many solutions,
2b + 1 2 − (a − 4) = = 5b − 1 − ( a − 1) 4 Take
1 a−4 = 2 a−1
⇒
a=7
and
1 2b + 1 = 2 5b − 1
⇒
b = 3.
8. Table for values of x and y as regarding equation 3x + y – 5 = 0 is x
0
1
y
5
2
Similarly table for equation 2x – y – 5 = 0 is x
0
1
y
–5
–3
Let us draw the graph of lines using the tables obtained above.
9. Let son's present age be x years and father's present age be y years. Since, the father's age is 3 years more than 3 times the son's age. ... y = 3x + 3 ⇒ 3x – y = 0 ...(i) 3 years hence, father's age = (y + 3) years and son's age = (x + 3) years. Since, 3 years hence, the father's age will be 10 years more than twice the son's age. ... y + 3 = 2 × (x + 3) + 10 ⇒ 2x + 6 + 10 – y – 3 = 0 ⇒ 2x – y + 13 = 0 ...(ii) Subtracting equation (ii) from equation (i), we have x = 10 Substituting x = 10 in equation (i), we have y = 3 × 10 + 3 = 33. Son's age = 10 years; father's age = 33 years. OR Let the speed of train and bus be x km and y km respectively. According to the given conditions, 4=
60 240 + x y
...(i)
Distance Using Time = Speed And 4 +
10 100 200 = + 60 x y
...(ii)
1 1 = u and y = v in (i) and (ii) x and simplifying, we get 15u + 60v = 1 ...(iii) 24u + 48v = 1 ...(iv) On solving equations (iii) and (iv), we obtain
Putting
u= i.e., The lines intersect y-axis at (0, 5) and (0, – 5). P A I
R
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I
N E A R
1 1 and v = 80 60
x = 60 and y = 80
Hence, the speed of train is 60 km/hr and that of bus 80 km/hr.
E Q U A T
I O N S
...
33
WORKSHEET–22
On solving
1. (D) The condition to be coincident for lines ax + by + c = 0 and dx + ey + f = 0 is given by c a b = = f e d ⇒
ae = bd ; bf = ce.
Note: Two lines are coincident if both the equations follow the condition of infinitely many solutions. 2. (C) Let the required equation be ax + by + c = 0. Then,
a c 6 = ≠ 2 –5 – 3
⇒
a 6 = = k (say) 2 – 3
⇒a=
2 k, b = – 3 k, c any real number
Then, 2 k – 3 k + c = 0 Putting c = –1, we have ⇒ ⇒
2 kx –
3 ky – k = 0 2x –
3 y = 1.
3. (A) For no solutions,
–(k – 2) k 3 = ≠ 12 k –k ⇒
k= ± 6
if
k= 6
∴ if
6 3 6–2 4 2 = ≠ = = 12 6 6 6 3
True
Required value of k, can be 6 or – 6.
4. For infinite number of solutions, we have –3 –7 2 = = p+q – ( p + q – 3) – (4 p + q)
34
–7 –3 = , – (4p + q) – ( p + q – 3) we obtain p = – 5, q = – 1. 5. x = 1, y = 2 Hint: Adding and subtracting the given two equations, we have x+y=3 ...(i) and x–y=–1 ...(ii) Now, solve equations (i) and (ii). 6. x = a2, y = b2 Hint: Given system of linear equations may be written as bx + ay – ab (a + b) = 0 b 2x + a2y – 2 a2 b 2 = 0 Solve these two equations by the method of cross-multiplication. 7. Let the two digits number be 10x + y. Since ten's digit exceeds twice the unit's digit by 2 ∴ x = 2y + 2 ⇒ x – 2y – 2 = 0 ...(i) Since the number obtained by interchanging the digits, i.e., 10y + x is 5 more than three times the sum of the digits. ... 10y + x = 3 (x + y) + 5 ⇒ 2x – 7y + 5 = 0 ...(ii) On solving equations (i) and (ii), we obtain x = 8 and y = 3 ... 10x + y = 83 Hence, the required two digits number is 83. 8. Tables for equations 3x + y – 11 = 0 and x – y – 1 = 0 are respectively.
k= –6 –6 –8 3 4 = ≠ = 12 –6 –6 3
∴
True
2 –3 = and p+q – ( p + q – 3)
x
3
4
y
2
–1
and
Let us draw the graph. x
0
1
y
–1
3
M A T H E M A T
I C S – X
Put x – y = u, x + y = v and solve further find x and y. OR Let each boy receive Rs. x and the number of boys be y. Then sum of money which is distributed is Rs. xy. Had there been 10 boys more, each would have received a rupee less, ... (y + 10) (x – 1) = xy
From the graph, it is clear that the lines intersect each other at a point A(3, 2). So the solution is x = 3, y = 2. The line 3x + y – 11 = 0 intersects the y-axis at B(0, 11) and the line x – y – 1 = 0 intersects the y-axis at C (0, –1). Draw the perpendicular AM from A on the y-axis intersect it at M. Now, in ∆ABC, base BC = 11 + 1 = 12 units, height AM = 3 units. 1 × base × height ... ar(∆ABC) = 2 1 = × 12 × 3 = 18 sq. units 2 x = 3, y = 2 ; Area = 18 sq. units. 9. Speed of boat = 6 km/hr, Speed of stream = 2 km/hr Hint: Let the speed of boat in still water = x km/h and the speed of stream = y km/h 12 40 + = 8 x–y x+y
...(i)
Distance Using Time = Speed 16 32 + = 16 x–y x+y
P A I
R
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...(ii) I
N E A R
⇒ 10x – y = 10 ...(i) Had there been 15 boys fewer, each would have received Rs 3 more, ... (y – 15) (x + 3) = xy ⇒ 5x – y = –15 ...(ii) Solving (i) and (ii), we get x = 5 and y = 40 ... xy = 200 Hence, sum of money = Rs. 200 And number of boys = 40.
WORKSHEET–23 1. (A) Here,
–3 2 9 = ≠ 6 –9 –5
... Lines are parallel. 2. (B) As the given equation are homogeneous so only solution will be x = 0 and y = 0. 3. (C) Hint: Put
1 = v and solve. y
1 =u, x
4. The given equations have a unique solution ⇒
a b ≠ ⇒ am ≠ bl. m l
5. The given equation can be written as 6ax + 6by = 3a + 2ab ...(i) and 6bx – 6ay = 3b – 2a ...(ii) Multiplying equation (i) by a and (ii) by b and adding the results, we have 6(a2 + b2) = 3(a2 + b2) 1 ⇒ x= 2
E Q U A T
I O N S
...
35
Substituting x =
1 in equation (i), we have 2
6a + 6by = 3a + 2b 2
1 3 1 1 Thus, the solution is x = , y = . 3 2 6. a = 5, b = 1 Hint: Two linear equations a1 x + b1 y + c1 = 0 and a2 x + b2 y + c2 = 0 have infinite number of solutions if a1 b c = 1 = 1. a2 b2 c2
⇒
6by = 2b
⇒
y=
7. Given system of linear equations can be written as (a + 2b) x + (2a – b) y – 2 = 0 (a – 2b) x + (2a + b) y – 3 = 0 By cross-multiplication, x – 3 (2 a – b) + 2 (2 a + b) (i ) –y = – 3 ( a + 2b) + 2 ( a – 2b) (ii) 1 = ( a + 2b) (2a + b) – (2a – b) ( a – 2b) (iii )
OR Let fare from A to B and from A to C be Rs. x and Rs. y respectively. According to the given conditions, 2x + 3y = 795 ...(i) 3x + 5y = 1300 ...(ii) Solving eqn. (i) and (ii), we obtain x = 75, y = 215 Hence, fares from A to B is Rs. 75 and from A to C is Rs. 215. 9. Let us make the table for the values of x and corresponding to the equation 2x + y – 8 = 0 x
2
4
y
4
0
Similarly, for the equation x – y – 1 = 0 x
4
3
y
3
2
Let us draw the graph.
5b – 2a 10 ab a + 10 b Again taking (ii) and (iii), y = 10 ab
Taking (i) and (iii), x =
5b – 2a a + 10 b , y = is the 10 ab 10 ab solution of the given system of equations.
Thus, x =
8. Speed of rowing = 6 km/hr, Speed of current = 4 km/hr Hint: 2 × (x + y) = 20 [Time × Speed = Distance] 2 × (x – y) = 4 Where, x = speed of rowing and, y = speed of current.
36
M A T H E M A T
I C S – X
From the graph, the lines intersect each other at the point A(3,2). Therefore, the solution is x = 3, y = 2. The lines intersect the y-axis at B(0,8) and C (0, – 1). To find the area of the shaded portion, that is, ∆ABC, draw perpendicular AM from A on the y-axis to intersect it in M. Now, AM = 3 units and BC = 8 + 1 = 9 units.
1 × BC × AM 2 1 = ×9×3 2 27 sq. units = 2 x = 3, y = 2; Area = 13.5 sq.units. ...
ar(∆ABC) =
– 12p – 12q = – 15p + 3 ⇒ 2p – 10q = 9 and 3p – 12q = 3 ⇒ p = 5, q = 1. 6. x = 1, y = 1, Hint: Take
∴ given equation can be written as: 3 u+v= ⇒ 4u + 4v = 3 4 1 1 1 u– v=– ⇒ 4u – 4v = – 1. 2 2 8
and
1 −1 ,y= 2 3
7. x =
Hint: Put
WORKSHEET–24
1 1 = u, =v 3x + y 3x − y
1 1 = u and = v. x y
8. Table for values of x and y corresponding to equation 4x – 5y – 20 = 0 is
1. (C) For unique solution,
p 4 ≠ ⇒ p ≠ 4. 2 2 2. (A) In the case of no solution,
x
5
0
y
0
–4
Similarly for the equation 3x + 5y – 15 = 0
–5 –1 3 = ≠ ⇒ k ≠ 10. –k 6 –2
3. (D) x = 80, y = 30 Hint: x + 2y = 140, 3x + 4y = 360. 4. True According to the conditions of consistency, either
x
5
0
y
0
3
Let us draw the graphs for the two equations.
2 2 3 ≠ – 5 or 3 = – 5 = 1 3 –5 3 –5 3 2 2
Clearly, the first condition holds. Hence, the system of equations is consistent with a unique solution. 5. For infinitely many solutions,
⇒ P A I
– (5p – 1) p + q 2( p – q) = = – 12 3 4 4p + 4q = 6p – 6q and R
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L
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N E A R
E Q U A T
I O N S
...
37
As the graphs of the two lines intersect each other at the point A(5, 0), the required solution is x = 5, y = 0. The graphs intersect the y-axis at B (0, 3) and C(0, – 4). Therefore, the coordinates of vertices of the triangle ABC are A(5, 0), B(0, 3) and C(0, – 4). x = 5, y = 0 and (5, 0), (0, 3), (0, – 4). 9. Let speeds of two cars that start from places A and B be x km/hr and y km/hr respectively. Case I: When two cars travel in same direction. Let the cars meet At C
Distance travelled by the car that starts from A AC = 5 × x Similarly distance for other car BC = 5 × y . .. AC – BC = 5x – 5y ⇒ 5x – 5y = 100 ⇒ x – y = 20 ...(i) Case II: When two cars travel in opposite directions. Let the cars meet at D
WORKSHEET–25 1. (C) The given equations represent to be parallel lines if 2(k – 1) 1 –1 = ≠ 3 –1 –1
k –1= –
⇒
3 2
1 . 2 2. (D) For the point of intersection of any line with x-axis, put y = 0 ... – 3x + 7 (0) = 3 ⇒ x =–1 So the required point is (– 1, 0). 3. (B) x – y = 0 ...(i) 2x – y = 2 ...(ii) – + – – x = –2 (Subtracting) . .. x = 2. k= –
⇒
Further y = x = 2. 4. For inconsistency, k+2 6 –(3k + 2)2 = ≠ 3 2 –4
k + 2 = 4 and (3k + 2)2 ≠ 8
⇒ ⇒
k = 2 and k ≠
Hence,
k = 2.
1 ± 2 2 –2 3
(
)
5. m ≠ 4 Distance travelled by the car that starts from A AD = 1 × x Similarly distance for other car BD = 1 × y ... AD + BD = x + y ⇒ x + y = 100 ...(ii) Solving equations (i) and (ii), we get x = 60 and y = 40 Hence, speeds of two cars that start from place A and B are 60 km/h and 40 km/h respectively.
38
Hint:
m –2 . ≠ 2 –1
6. Given system of equations can be written as 2x + 3y – 18 = 0 ...(i) x – 2y – 2 = 0 ...(ii) Now,
2 3 ≠ 1 –2
Hence the system has unique solution. Now, by cross-multiplication on (i) and (ii), we get M A T H E M A T
I C S – X
To draw the graphs of the equations (i) and (ii), we need atleast two solutions of each of the equations. These solutions are given below:
–y x 1 = = – 6 – 36 – 4 + 18 – 4 – 3 ⇒ x = 6, y = 2 Thus, the solution of given system is x = 6, y = 2.
x y = 4x – 8
7. x = 5, y = –1 Hint: Take
1 1 = u, = v and solve. x+y x–y
8. Let Meena received x notes of Rs. 50 and y notes of Rs. 100 Since total number of notes is 25 ... x + y = 25 ...(i) Since the value of both types of notes is Rs. 2000. ... 50x + 100y = 2000 and x + 2y = 40 ...(ii) Solving equations (i) and (ii), we get x = 10, y = 15 Hence, Meena received 10 notes of Rs. 50 and 15 notes of Rs. 100. OR Let the length and breadth of rectangle be x units and y units respectively. Then area of rectangle = xy sq. units Case I: The length is increased and breadth is reduced by 2 units. ... (x + 2) (y – 2) = xy – 28 ⇒ xy – 4 – 2 x + 2 y = xy – 28 ⇒ x – y = 12 ...(i) Case II: The length is reduced by 1 unit and breadth increased by 2 units. ∴ (x – 1) (y + 2) = xy + 33 ⇒ xy – 2 – y + 2x = xy + 33 2x – y = 35 ...(ii) Solving equations (i) and (ii), we get x = 23 and y = 11 Hence, the length of the rectangle is 23 units and the breadth is 11 units. 9. The given linear equations are 4x – y – 8 = 0 and 2x – 3y + 6 = 0 P A I
R
O F
L
I
...(i) ...(ii)
N E A R
x 2x + 6 y 3
0
2
–8
0
0
–3
2
0
Plot the points A(0, – 8), B(2, 0), C(0, 2) and D(– 3, 0) on graph paper and join them to form the lines AB and CD as shown in figure.
The graphs of these lines intersect each other at P(3, 4). So, unique solution: x = 3, y = 4. Also, the graphs meet the x-axis at D(– 3, 0) and B(2, 0). Hence, the triangle formed by the lines is PBD with vertices P(3, 4), B(2, 0) and D(– 3, 0).
E Q U A T
I O N S
...
39
7. x = – 2, y = 5 and m = – 1
WORKSHEET–26
Hint: 2x + 3y = 11 ⇒ y =
1. (C) For coincident lines, 7 1 2 = = 14 2 k k = 4.
⇒
Substitute this value of y in 2x – 4y = – 24 and solve for x.
2. (A) The given system of equations can be written as x + 2y = 140, 3x + 4y = 360 Solving this system, we obtain x = 80, y = 30. 3. Adding the given equations, we have 3x = 0 ⇒ x = 0 Substituting x = 0 in any of the given equations, we get y = 0 Hence, the required solution is x = 0, y = 0. 4. False a1 2 b1 5 c1 = , = , =6 a2 4 b2 10 c2 1 1 ... = ≠6 2 2 ⇒ They are parallel.
Hint: 6. Put
8. The given system of linear equations is 2x – y – 5 = 0 ...(i) 3x + y – 5 = 0 ...(ii) To draw the graph of equations (i) and (ii), we need atleast two solutions of each of the equations, which are given below: x y = 2x – 5 x
–3x+ 5 y= 3
0
4
–5
3
0
–3
5
–4
Using these points, we are drawing the graphs of lines as shown below:
Hint: As
5. a = –1, b =
11 – 2x 3
5 2
2 – (2a + 5) 5 . = = 2b + 1 –9 15
1 1 = u and = v in given system of y x
equations. u + v – 7= 0 2u + 3v – 17 = 0 By cross-multiplication,
...(i) ...(ii)
u –v 1 = = – 17 + 21 – 17 + 14 3–2
⇒ u = 4, v = 3 1 1 ⇒x= ,y= 4 3
1 1 ,y= is the solution of the 3 4 given system of equations. Hence, x =
40
From the graph, the lines intersect each other at the point P(2, –1). Therefore, the solutions is x = 2, y = –1. The lines meet the y-axis at the points Q(0, 5) and R(0, –5). 9. Let the fixed charge and additional charge for each day be Rs. x and Rs. y respectively. M A T H E M A T
I C S – X
Since saritha paid Rs. 27 for a book kept for 7 days ... x + 4y = 27 ...(i) Also, Susy paid Rs. 21 for the book kept for 5 days ... x + 2y = 21 ...(ii) Subtracting equation (ii) from (i), we get 2y = 6 ⇒ y = 3 Again substituting y = 3 in equation (ii), we get x = 15 Hence, the fixed charge is Rs. 15 and additional charge for each day is Rs 3. OR Son's age = 10 years, father's age = 40 years. Hint: Let the present age of father and son be x and y years respectively. ... x + 5 = 3 (y + 5) And x – 5 = 7 (y – 5).
WORKSHEET–27 1. (A) Hint: In a cyclic quadrilateral,
To solve the system for x and y, using the method of cross-multiplication, we have –y x = – q( p + q) – p( p – q ) – p( p + q) + q( p – q) =
x 1 –y = = 2 2 2 2 – p – q2 –p –q –p – q
⇒
2
⇒
x = 1, y = –1.
8. The given system of equations can be written as 3x – 4y – 1 = 0 ...(i) 6x – 15y = 0
3. (A) For infinite number of solutions,
...(ii)
To draw the graph, we need atleast two solutions for each of the equations (i) and (ii), which are given below: x
3
7
– 3x – 1 y= 4
2
5
x
∠A + ∠C = 180° and ∠B + ∠D = 180°. 2. (C) Hint: Both lines are passing through the origin.
1 – p2 – q 2
y=
6x + 15 8
3 2
11 2
3
6
Let us draw the graph to use these points.
2 p – q –21 p+q = = 3 2 –7 p + q = 6 and 2p – q = 9 p = 5 , q = 1.
⇒
4. False, as a + 5b = – 10. 5. False, x = 4, y = 1 does not satisfy the second equation. 6. No solution Hint: 2x + 3y = 7, 6x + 9y = 11 Here,
2 3 7 Parallel lines. = ≠ 6 9 11
7. The given system of linear equations can be written as px + qy – (p – q) = 0 qx – py – (p + q) = 0 P A I
R
O F
L
I
N E A R
From the graph, it is clear that the lines are parallel. Hence, the given system of equations is inconsistent.
E Q U A T
I O N S
...
41
9. Let the fraction be
x y
On adding 1 to each of numerator and 6 denominator, the fraction becomes 5 x +1 6 ... = y +1 5 ⇒ 5x + 5 = 6y + 6 ⇒ 5x – 6y = 1 ...(i) Further, on subtracting 5 from each of numerator and denominator, the fraction becomes ...
3 2
x–5 3 = y–5 2
⇒ 2x – 10 = 3y – 15 ⇒ 2x – 3y = – 5 ...(ii) Solving equations (i) and (ii), we get x = 11, y = 9 Hence, the required fraction is
11 . 9
OR Rs. 6000, Rs. 5250 Hint: Let incomes of x and y be 8x and 7x respectively; and expenditures of them be 19y and 16y respectively. ∴ 8x – 19y = 1250 ...(i) 7x – 16y = 1250 ...(ii)
WORKSHEET–28 1. (A) Hint:
k–3 k 3 . = = k k 12
2. (C) The condition that the given system of equations has unique solution is represented by a1 b1 ≠ . a2 b2 3. (B) The condition that the given system of equations represents parallel lines is
42
p–2 p2 + 1 5 = ≠ 2 3 3p + 1 ⇒
5p = – 5 ⇒ p = – 1.
4. (A) Multiplying first equation by 2 and the other one by 3 and adding, we get 1 21.8x = 10.9 ⇒ x = 2 1 Substituting x = in any of the given 2 1 equations, we have y = . 3 1 1 ∴ x= , y = . 3 2 5. True The condition for parallel lines is
–2 2 –3 = ≠ 6 –6 5 1 1 –3 ⇒ = ≠ 3 3 5 The condition holds. The lines are parallel. 6. x = a2 , y = b2 1 1 Hint: Put = u and = v. y x 7. Given system of linear equations can be written as: (a – b) x + (a + b) y – (a2 – 2ab – b2) = 0 (a + b) x + (a + b) y – (a2 + b2) = 0 By cross - multiplication, x – ( a + b )(a 2 + b 2 ) + ( a + b) ( a2 – 2 ab – b 2 ) =
–y – ( a – b)( a + b ) + ( a + b) ( a2 – 2ab – b2 )
=
1 ( a – b ) ( a + b) – ( a + b ) ( a + b)
⇒
2
2
–y 1 x = = 2 2 – 2 ( b a + b) – 2b ( a + b) – 4 ab
Hence, the solution of given system of equations is x = a + b, y = –
2ab . a+b
M A T H E M A T
I C S – X
8. To draw graph of the equation, we need atleast two solutions. Two solutions of the equation 4x + 3y – 24 = 0 are mentioned in the following table: x
0
6
y
8
0
area of ∆ ABC = =
0
1
y
–2
0
x
0
4
y
–4
–4
and
1 × 10 × 8 = 40 sq. unit. 2
9. Rs. 600, Rs. 700 Hint: Let cost price of trouser be Rs. x and that of shirt Rs. y. Then 125 x+ 100 110 x+ 100
Similarly, two solutions of each of the equations 2x – y – 2 = 0 and y + 4 = 0 are respectively. x
1 × base × height 2
⇒
110 y = 1520 100 125 y = 1535 100
25 x + 22 y = 30400 22 x + 25 y = 30700 OR
Using the tables obtained above, let us draw the graph.
6 l of 50% and 4 l of 25%. Hint: Let x litres of 50% acid and y litres of 25% acid should be mixed.
50 25 40 y= x+ ( x + y) 100 100 100 x + y = 10 ⇒
2x = 3y x + y = 10
WORKSHEET–29 1. (C) x = 9, y = 6 Hint: x – y = 3 and 2x + 3y = 36. 2. (D)
3p 18
p
⇒
2
∴ Observing the graph, the lines meet each of her pairwise in A(3, 4), B(– 1, – 4) and C(9, – 4). Hence, the vertices of the triangle ABC so obtained are A(3, 4), B(–1, – 4) and C(9, – 4). P A I
R
O F
L
I
N E A R
=
6 ≠ 24
=
3 ≠ 2
p=
50 75 2 3
3.
3. (A) Solving 3x – 2y = 4 and 2x + y = 5, we have x = 2, y = 1. Now, substituting these values of x and y in y = 2x + m, we have 1 = 2 × 2 + m ∴ m = – 3.
E Q U A T
I O N S
...
43
4. x = b, y = – a a2
b2
Hint: x – y = ab (a + b), ax – by = 2ab Solving the equation we get x = b1, y = – a. 5. For inconsistency,
9. Two solutions of 6y = 5x + 10 are:
α–3 3 α = ≠ ⇒ α2 = 36 and α α 12
3α ≠ α2 – 3α ⇒ α = ± 6 and α ≠ 0 or α ≠ 6 ⇒ α = –6. 6. x =
−26b 22a ,y= 5 5
Hint:
4bx + 3ay – 2ab = 0 3bx + ay – 8ab = 0.
7. 3x + 2y = 800. 12x + 8y = 3000; Not possible Hint: Let cost of 1 chair be Rs. x and that of 1 table be Rs. y. ∴3x + 2y = 800, 12x + 8y = 3000. 8. Let the actual prices of tea-set and lemonset be Rs. x and Rs. y respectively According to the question, Case I: Selling price – cost price = Profit ⇒ 0.95x + 1.15y – (x + y) = 7 ⇒ – 0.05x + 0.15y = 7 ...(i) Case II: Selling price – Cost price = Profit ⇒ 1.05x + 1.10y – (x + y) = 13 ⇒ 0.05x + 0.10y = 13 ...(ii) Solving equations (i) and (ii), we get x = 100 , y = 80 Hence, actual prices of tea-set and lemonset are Rs. 100 and Rs. 80 respectively. OR The person invested Rs. 500 at the rate of 12% per year and Rs. 700 at the rate of 10% per year. Hint: Let the person invested Rs. x at the rate of 12% per year and Rs. y at the rate of 10 % per year ∴ ⇒ and
44
12x 10 y + = 130 100 100 6x + 5y = 6500 12y 10x + = 134 100 100
⇒ 5x + 6y = 6700 ...(ii) Adding and subtracting (i) and (ii), we get x + y = 1200 ...(iii) x – y = – 200 ...(iv)
...(i)
x
–2
4
y
0
5
Two solutions of y = 5x – 15 are x
3
2
y
0
–5
Now, we draw the graph of the system on the same coordinate areas.
(i) From the graph, we look that the two lines intersect each other at A(4, 5). (ii) The vertices of the triangle : A (4, 5); B (–2, 0); C (3, 0). Height of ∆ABC corresponding to the base BC, h = 5 units and base, b = BC = 5 units Now, ar(∆ABC) =
1 ×b×h 2
=
1 ×5×5 2
= 12.5 square units. M A T H E M A T
I C S – X
Solving equations (iii) and (v), we have
ASSESSMENT SHEET–5
x=
1. (C) Let us check option (C). 2 5 2 5 x + y = (– 3) + (– 2) = – 2 – 5 = – 7 3 2 3 2
3x –
1 1 y = 3 (– 3) – (– 2) = – 9 + 1 = – 8. 2 2
2. (D) 4x – y = 42
∴
⇒ x– y=2 x – 2y = 8 – + – y= –6
Therefore, xy =
x= – Therefore, xy = – Hence,
∴ from (i) ⇒ x = – 4.
7 3 2 = = 4a + b a+b–3 a+b
a – 5b = 0.
4. False, because the given pair of equations has infinitely many solutions at k = 40 only. 5. Given equations are 2y – x. (x + y) = 1
1
x + y=
...(i)
y–x
2 and (x + y)x – y = 2 ...(ii) Substituting the value of x + y from equation (i) in equation (ii), we get 1 y–x 2
P A I
=2
1 2
x+y= R
O F
L
3 3 or – . 16 4
Let us apply cross-multiplication method to solve these equations. –y 1 x = = 2 1 b 1 1 2 1 a – 2 – + + 2 – + 2 + 2 a a a b b b ab ba
⇒
a2 b2 – a2 y b2 x = = a–b –b+a –a+b
Taking
a2 b2 b2 x = a– b –b+a
x–y
I
a2 b2 – a2 y = a–b –a + b
and
⇒ (2x – y)x – y = 21 ⇒ (x – y)2 = 1 ⇒ x–y=±1 ⇒ x–y=1 or x – y = –1 Substituting x – y = 1 and x – y equation (ii), we get respectively x+y=2 and
3 . 16
y x + – (a + b) 2 = 0 b a x y + 2 – 2= 0 a2 b
b a1 c = 1 = 1 c a2 b2 2
⇒
xy =
3 1 ;y= 4 4
6. Given equations can be written as
3. For coincident lines
⇒
3 4
Solving equations (iv) and (vi), we have
...(i) ...(ii)
x + y = – 10
⇒
1 3 ;y= . 2 2
⇒ ...(iii) ...(iv) = –1 in ...(v) ...(vi)
N E A R
⇒
x=
a 2 b 2 ( a – b) b 2 ( a – b)
and y =
( a – b) a2 b 2 a2 ( a – b)
x = a2 and y = b2.
7. Given equations of lines are: 3x + y + 4 = 0 ...(i) and 6x – 2y + 4 = 0 ...(ii) To draw the graphs of lines (i) and (ii), we need atleast two solutions of each equation.
E Q U A T
I O N S
...
45
For equation (i), two solutions are: x y
0 –4
Sale price of the sweater = y – y ×
–3 =
5
For equation (ii), two solutions are: x
0
2
y
2
8
Let us draw the graphs of the lines (i) and (ii) y x + – 2= 0 b a
90 y 100
10 100
108 90 x+ y = 1008 100 100
∴ ⇒ Case II:
108x + 90y = 100800
Sale price of the saree = x + x × =
110 x 100
Sale price of the sweater = y – y × =
92 y 100
...(i) 10 100
8 100
92 110 x+ y = 1028 100 100 ⇒ 110x + 92y = 102800 ...(ii) Adding equations (i) and (ii), we get 218x + 182y = 203600 ...(iii) Subtracting equation (i) from (ii), we get 2x + 2y = 2000 or 218x + 218y = 218000 ...(iv) (Multiplying by 109) Solving equation (iii) and (iv), we have x = 600 and y = 400 Hence, the cost price of the saree is Rs. 600 and the list price of the sweater is Rs. 400.
∴
From the graph it is clear that the two lines intersect each other at a point, P (–1, – 1), therefore, the pair of equations consistent. The solution is x = –1, y = – 1. 8. Let the cost price of the saree and the list price of the sweater be Rs. x and Rs. y respectively. Now two cases arise. Case I: 8 Sale price of the saree = x + x × 100 108 = x 100
46
ASSESSMENT SHEET–6 1. (B) For no solution, k 7 3 = ≠ 4k + 1 2k 12 7 3 ∴ = ⇒ k = 14. 2k 12 2. (B) For infinitely many solutions, k k 13 = = 6 k+4 39 Taking
13 k = ⇒ k = 2. 39 6
M A T H E M A T
I C S – X
3. x = – a and y = k must satisfy both the given equations. Let us substitute these values of x and y in bx – ay + 2ab = 0 b (–a) – ak + 2ab = 0 ⇒ –ak + ab = 0 ⇒ k = b. 4. Yes, because –3 b c 1 a1 2 = 1 = 1 , i.e., = = . c b2 –9 3 2 6 a2 5. Given equations are: 4x + 5y = 2 (5p + 2q)x + (4p – 2q)y = p + q + 3 Here, a1 = 4, b1 = 5, c1 = – 2, a2 = 5p + 2q, b2 = 4p – 2q, c2 = –p –q –3 For infinitely many solutions, c b a1 = 1 = 1 c2 b2 a2 ⇒
–2 5 4 = = –p – q – 3 4 p – 2q 5 p + 2q
Taking
5 4 = 4 p + 2q 5 p + 2q
⇒ 25p + 10q = 16p – 8q ⇒ 9p = – 18q ⇒ p = – 2q Also, taking
and x – y= 2 ...(iv) Adding equations (iii) and (iv); and subtracting equation (iv) from (iii), we get respectively. x = 1 and y = – 1. 7. One of the given equation is 2x + y = 14 ...(i) Here, at x = 0, y = 4 and at x = 2, y = 0 Two solutions of equation(i) are given in the following table: x
0
2
y
4
0
Another given equation is 2x – y = 4 ...(ii) Two solutions of equation (ii) are given by the following table: x
0
2
y
–4
0
Let us draw the graph of the two equations (i) and (ii) by using their corresponding tables. ...(i)
–2 4 = –p – q – 3 5 p + 2q ⇒ –10p – 4q = – 4p – 4q –12 ⇒ p= 2 Substitute p = 2 in equation p = – 2q to get q = – 1. Hence, p = 2, q = – 1. 6. Given system of equations is 43x + 67y = –24 ...(i) 67x + 43y = 24 ...(ii) Adding (i) and (ii); and subtracting (i) from (ii), we get respectively. 110x + 110y = 0 and 24x – 24y = 48 i.e. x + y= 0 ...(iii) P A I
R
O F
L
I
N E A R
E Q U A T
I O N S
...
47
From the graph, vertices of the triangle ABC are A(0,4), B(0, – 4) and C(2, 0).
1 × base × height 2 1 = × AB × OC 2
ar(∆ABC) =
=
1 × (4 + 4) × 2 2
= 8 square units. Thus, area of the triangle is 8 square units. 8. Let Anand had a total of x oranges; and he makes lot A of p oranges and lot B of remaining x – p oranges. There are two cases now. Case I: 2 Selling price of lot A = Rs. p 3 Selling price of lot B = Rs. (x – p) ∴
2 p + x – p = 400 3 3x – p = 1200
⇒ ...(i) Case II: Selling price of lot A = Rs. p 4 Selling price of lot B = Rs. (x – p) 5 4 ∴ p + (x – p) = 460 5 4x + p = 2300 ...(ii) Add equations (i) and (ii) to get 7x = 3500 ⇒ x = 500 Hence Anand had a total number of 500 oranges.
CHAPTER TEST 1. (C) A pair of linear equations is said to consistent, if the lines intersect each other at a point or coincide. 2. (C) 6, 36 Hint: Let the son's age = x, father's age = y ∵ y = 6x
48
and Solve yourself.
y + 4 = 4(x + 4)
3. (C) The lines are coincident –1 3 8 = = –k 6 16
⇒
⇒ k = 2.
4. Yes, For consistency, Either
2a b 2a –a b ≠ or = = 4a 2b 4a –2b 2b
Here the relation
–a b 2a = = , –2 a 2b 4a
1 1 1 = = holds. 2 2 2 ⇒ The pair is consistent. i.e.,
5.
21x + 47y = 47x + 21y = + + 68x + 68y = ⇒
110 162 + 272
x + y= 4
...(i)
Subtracting the given equation, we get – 26x + 26y = – 52 ⇒ x – y= 2 ...(ii) Solve equation (i) and (ii) and we get x = 3, y = 1. 6. We are given
and
2xy 3 x+y = 2
...(i)
xy –3 2 x – y = 10
...(ii)
Taking equation (i)
2xy 3 x+y = 2 ⇒ 3x + 3y = 4xy Now, Taking equation (ii)
...(iii)
xy –3 2 x – y = 10 M A T H E M A T
I C S – X
⇒
– 6x + 3y = 10xy
...(iv)
Multiplying equation (iii) by 2 and adding its with (iv), we get
∴ Selling price of a sweater = y –
9y = 18xy ∴
x=
Putting x = ⇒
=
1 2
108 x 90 y + = 1008 100 100
1 in equation (iv), we get 2
∴
y=
–3 2
This,
x=
1 –3 and y = . 2 2
108x + 90y = 100800 6x + 5y = 5600 ...(i) If profit on a saree is 10% and discount on a table is 8%, then total selling price is Rs. 1028.
7. The given system of equations will have infinite number of solutions if
1 2 = a+b a+b–2
a + b – 2 = a – b and 2a + 2b – 4 = a + b
⇒ a + b – a + b = 2 and a + b = 4 b = 1 and a = 3
Hence, the given system of equations will have infinite number of solutions, if a = 3, b = 1. 8. Let the cost price of saree be Rs. x and list price of sweater be Rs. y. Profit on a saree = 8% ∴
Selling price of a saree = x + =
8 x 100
108 x 100
Discount on a sweater = 10% P A I
R
O F
L
I
110 x 92 y = 1028 + 100 100
110x + 92y = 102800 55x + 46y = 51400 ...(ii) Multiplying eqn. (i) by 55 and eqn. (ii) by 6, we get
1 1 = and a–b a+b–2
⇒
8y 10 x + y – ⇒ x + = 1028 100 100 ⇒
1 2 1 = = a+b–2 a+b a–b
⇒
90 y. 100
According to condition, we have
– 3 + 3y = 5xy
⇒
10 y 100
N E A R
330x + 275y = 308000 ...(iii) 330x + 276y = 308400 ...(iv) Subtract eqn. (iv) from eqn. (iii), we get y = 400 the value of y, put in eqn. (i), we have x = 600 Cost price of saree = Rs. 600 List price of sweater = Rs. 400. 9. To draw the graph of a line, we are required atleast two solutions of its corresponding equation. At x = 0, 5x – y = 5 gives y = –5 At x = 1, 5x – y = 5 gives y = 0 Thus, two solutions of 5x – y = 5 are given in the following table: x
0
1
y
–5
0
I O N S
...
E Q U A T
49
x
9
1
y
–4
0
x + 2y = 1 x
2
4
y
5
–7
6x + y = 17 Now, we will draw the graphs of the three lines on the same coordinate axes.
Similarly, we can find the solutions of each remaining equation as given in the following tables:
50
From the graph, it is clear that the lines form a triangle ABC with vertices A(1, 0), B(3, –1) and C(2, 5). ❑❑
M A T H E M A T
I C S – X
Chapter
4
TRIANGLES WORKSHEET–33
1. (A) ∵ ∆ABC ~ ∆DEF
ar (∆ABC) BC2 AC2 AB2 = = 2 2 = 2 ar (∆DEF) EF DF DE
∴
Taking
ar (∆ABC) BC2 = 2 ar (∆DEF) EF
⇒
64 BC2 = 2 121 (15.4)
⇒
BC =
∴
Hint: ar(∆DEF) = ⇒
[Using (i)]
and ∠1 = ∠1 ∆PQS ~ ∆TQR
6. 1 : 4
64 × 15.4 × 15.4 121 BC = 11.2 cm.
⇒
QT QR = PQ QS
⇒
(Common) (SAS criterion) Hence proved.
1 ar ∆ABC 4
A
D
ar(∆ DEF) 1 = . 4 ar(∆ ABC) B
E 7. 13 m Hint: Distance between tops = AD
2. (D) Observing the figure, we obtain ∠A = ∠R, ∠B = ∠Q, ∠C = ∠P ∴ ∆ABC ~ ∆RQP.
∴ AD =
AE2 + DE2
=
(5)2 + (12)2
F
C
= 13 m.
3. Yes. Here, 262 = 242 + 102 = 676 ⇒ AC2 = AB2 + BC2 ∴ ∆ABC is a right triangle.
8. Hint: Use Pythagoras Theorem.
4. In ∆ABC, LM || CB AM AL …(i) = AB AC (Basic Proportionality Theorem) Similarly in ∆ADC,
∴
9. Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio. Proof: ABC is a given triangle in which DE || BC. DE intersects AB and AC at D and E respectively.
∴ AN = AL …(ii) AD AC Comparing equations (i) and (ii), we have
AM AN = . AB AD 5.
As ∠1 = ∠2 ⇒ PQ = PR
QT QR = PR QS T R
I
A N G L E S
Hence proved. We have to prove …(i) (Given)
AE AD = CE BD Let us draw EM ⊥ AB and DN ⊥ AC. Join BE and CD.
51
Now,
Also,
ar(∆ADE) =
1 × base × height 2
=
1 × AD × EM 2
...(i)
ar(∆ADE) =
1 × AE × DN 2
...(ii)
ar(∆BDE) =
1 × BD × EM ...(iii) 2
1 × CE × DN ...(iv) 2 Dividing equation (i) by equation (iii) and equation (ii) by equation (iv), we have ar(∆CDE) =
and But
ar (∆ADE) AD = ar(∆BDE) BD
...(v)
AE ar(∆ADE) = CE ar(∆CDE)
...(vi)
ar(∆BDE) = ar(∆CDE)
...(vii)
(Triangles are on the same base and between the same parallels BC and DE) Comparing equations (v), (vi) and (vii), we have
AD = BD 2nd Part: As ∠B = ⇒ AB = ⇒ AD + DB = ⇒ AD = ∴
AE . CE
∠C AC AE + EC AE
(... BD = EC)
AE AD = EC DB
∴ By converse of Basic Proportionality Theorem, DE || BC. Hence proved.
WORKSHEET– 34 1. (A) ∆BAC ~ ∆ADC ⇒
52
BC AC = ⇒ y2 = 16 × 4 ⇒ y = 8 cm. AC DC
AB 2. (D) DC2 = BC2 + BD2 = BC2 + 2
= BC2 + =9+ ⇒
DC =
2
1 (AC2 – BC2) 4
1 (25 – 9) = 9 + 4 = 13 4
13 cm.
3. (C) Hint: As DE || BC AE AD = EC DB
∴ ⇒
2x − 1 2x + 5 = . x − 3 x −1
4. DE || BC and DB is transversal ⇒
∠EDA = ∠ABC (Alternate interior angles)
Similarly, ∠AED = ∠ACB Consequently, ∆ADE ~ ∠ACB (AA similarity) ∴
AD 2 ar (∆ADE) = ar (∆ABC) AB 2
⇒
ar (∆ADE) AD 2 = 2 153 9 AD
⇒
ar(∆ADE) = 17cm2.
5. No. Here,
5 DP 1 = = 10 2 PE
And
6 1 DQ = = QF 18 3
∵
DQ DP ≠ QF PE
Therefore, PQ is not parallel to EF. 6. Hint: Use Basic Proportionality Theorem. M A T H E M A T
I C S – X
7. As AB = BC = AC
1 ∴ AD ⊥ BC ⇒ BD = BC 2 ∴ Using Pythagoras Theorem AB2 = AD2 + BD2 B 1 AD2 = AB2 − BC 2
⇒
3AB2 4 4AD2 = 3AB2.
2
Let us construct a ∆PQR with ∠Q = 90° such that PQ =A′B′ and QR + B′C′ ...(ii)
A
D
C
=
⇒
Hence proved
OR Let ABCD be a rhombus Since, diagonals of a rhombus bisect each other at right angles, ∴ AO = CO, BO = DO, ∠AOD = ∠DOC ∠COB = ∠BOA = 90° Now, in ∆AOD AD2 = AO2 + OD2 ...(i) Similarly, DC2 = DO2 + OC2 ...(ii) CB2 = CO2 + BO2 ...(iii) and BA2 = BO2 + AO2 ...(iv) Adding equations (i), (ii), (iii) and (iv), we have AD2 + DC2 + CB2 + BA2 = 2(DO2 + CO2 + BO2 + AO2) BD 2 AC2 BD 2 CA 2 = 2 4 + 4 + 4 + 4
= BD2 + CA2.
Hence proved
1 BC 3 Using Pythagoras Theorem.
8. Hint: BD = DE = EC =
9. Statement: In a triangle, if square of the largest side is equal to the sum of the squares of the other two sides, then the angle opposite to the largest side is a right angle. Proof: We are given a triangle ABC with A′C′2 =A′B′2 + B′C′2 ...(i) We have to prove that ∠B = 90° T R
I
A N G L E S
In ∆PQR, PR2 = PQ2 + QR2 (Pythagoras Theorem) =A′B′2 + B′C′2 ...(iii) [From (ii)] 2 2 2 But A′C′ =A′B′ + B′C′ ...(iv) [From (i)] From equations (iii) and (iv), we have PR2 = A′C′2 ⇒ PR = A′C′ ...(v) Now, in ∆ABC and ∆PQR, A′B′ = PQ [From (ii)] B′C′ = QR [From (ii)] A′C′ = PR [From (v)] Therefore, ∆A′B′C′ ≅ ∆PQR (SSS congruence rule) ⇒ ∠B′ = ∠Q (CPCT) But ∠Q = 90° ... ∠B′ = 90°. Hence proved. 2nd Part In ∆ADC, ∠D = 90° ... AC2 = AD2 + DC2 = 62 + 82 = 36 + 64 = 100 In ∆ABC, AB2 + AC2 = 242 + 100 = 676 and BC2 = 262 = 676 Clearly, BC2 = AB2 + AC2 Hence, by converse of Pythagoras Theorem, in ∆ABC, ∠BAC = 90° ⇒ ∆ABC is a right triangle.
53
7. In ∆ABC and ∆AMP ∠A = ∠A (Common) ∠ABC = ∠AMP = 90°
WORKSHEET– 35 1. (B)
⇒
ar(∆ ADE) DE2 = ar (∆ ABC) BC2 2 BC 3 ar(∆ADE) = BC2
(i) ∴ 2
(ii) ∴ × 81 = 36 cm2.
2. (A) ∆OAB ~ ∆OCD ⇒
OA OB = OC OD
⇒
OB = 4 ×
3 = 6 cm. 2
⇒
3x + 4 3 x + 19 AD BE = ⇒ = x x+3 DC EC
⇒ 3x2 + 19x = 3x2 + 4x + 9x + 12 ⇒ 6x = 12 ⇒ x = 2.
EF DE = ST TU
∴
EF DE ≠ . TU ST
…(i)
AC || PR ⇒
AP CR = AO CO
…(ii)
BQ CR = BO CO (By converse of BPT)
6. 1 : 2. Hint: Let AB = BC = a ... AC = 2 a ...
54
AB2 ar( ∆ABE) = . ar( ∆ACD) AC2
∴
2 ar (∆ABC) = ar (∆ AXY) 1
Y
X
∆ABC ~ ∆ AXY
B
C
2 ar ( ∆ABC) 2 AB = ar( ∆AXY) = 1 AX
2 2 −1 AB BX = ⇒ = . 1 AX AB 2
9. Hint: Prove converse of Pythagoras Theorem.
WORKSHEET– 36 A
AD 2 6 = = DC 7 21
AP BQ = AO BO
⇒ BC || QR.
BC CA = . MP PA (∴ Corresponding sides of similar triangles are proportional.)
1. (A) In triangle ABC,
5. AB || PQ ⇒
From (i) and (ii)
As
⇒
4. No, ∆FED ~ ∆STU ∵ Corresponding sides of the similar triangles are in equal ratio. ∴
(AA criterion)
8. Hint: ar(∆ AXY) = ar(BXYC) + ar(BXYC) ⇒ 2. ar(∆AXY) = ar(BXYC) + ar(∆ AXY) A = ar(∆ABC)
3. In ∆ABC, To make DE || AB, we have to take ⇒
∆ABC ∼ ∆AMP,
2 18 – 14 BE = = 7 EC 14 ∴ 2. (A) ⇒
BE AD = EC DC
D
C
B
E
⇒ DE || AB .
EF 2 ar (∆DEF) = ar (∆ABC) BC2 ar(∆DEF) = 54 ×
16 = 96 cm2. 9
3. (B) ∆ABD ~ ∆BCD ⇒
BD AB = CD BC
M A T H E M A T
I C S – X
⇒ ⇒
3.6 5.4 = 5.2 BC
So, by AA rule of similarity, ∆AQO ~ ∆BPO
5.4×5.2 BC = 3.6 = 7.8 cm.
⇒
∠BAC = ∠DAE ∠ACB = ∠AED
(Common angle) (Each 90°)
∆ABC ~ ∆AED
(AA criterion)
AB =
10 × 9 10 AQ = ⇒ AQ = 6 6 9 ⇒ AQ = 15 cm. OR Let the height of the tower be h metres
⇒
4. In ∆ABC and ∆ADE,
∴
AO AQ = BO BP
AC 2 + BC 2 =
P
25 + 144 = 13 cm
AC AB BC = = Now, AE DE AD
A x
12m B
⇒
5 12 13 = = AE DE 3
⇒
36 15 DE = cm and AE = cm. 13 13
C
8m
8 AB BC 12 = ⇒ = 40 h PQ QR
⇒
12 × 40 = 60 metres. 8 8. Hint: As ∆AOB ~ ∆COD
⇒
h=
A
B
2
9 6 3 = = ≠ . 25 5 5 Hence, it is not correct to say that ratio of
6.
…(i)
BD2
…(ii)
=
+
Adding (i) and (ii), we get AE2 + BD2 = AC2 + EC2 + DC2 + BC2 = (AC2 + BC2) + (EC2 + DC2) AE2 + BD2 = AB2 + DE2. Hence proved. 7. In ∆AQO and ∆BPO, ∠QAO = ∠PBO (Each 90°) ∠AOQ = ∠BOP (Vertical opposite angles) T R
I
A N G L E S
D
C
( 2 CD ) ar(∆AOB) AB = = 2 ar(∆COD) CD 2 CD
6 . 5
BC2
O
2
AE2 = AC2 + EC2 DC2
R
40m
∆ABC ~ ∆PQR.
5. No. Ratio of areas of two similar triangles = square of ratio of their corresponding altitudes.
areas of the triangles is
Q
2
=
4 . 1
9. Hint: Prove Pythagoras Theorem. For 2nd Part: …(i) ∴ AB2 = AD2 + BD2 Also AC2 = AD2 + CD2 …(ii) From (i) and (ii), ⇒ AB2 − AC2 = BD2 − CD2 ⇒ AB2 + CD2 = AC2 + BD2. Hence proved.
WORKSHEET–37 1. (A) ∵ ∴
∠M = 180° – (∠L + ∠N) (ASP) = 180° – (50° + 60°) = 70° ∆LMN ~ ∆PQR ∠ M = ∠Q ⇒ ∠Q = 70°.
55
2. (C) In ∆KMN, as PQ || MN, KP KQ = PM QN KP KQ = PM KN – KQ
⇒ ⇒
KN PM –1 = KQ KP
⇒
20.4 13 –1 = KQ 4
In right ∆AOE, AO2 = AE2 + OE2 From equations (i) and (ii), we have
13 17 20.4 =1+ = 4 KQ 4
⇒
20.4× 4 KQ = 17 KQ = 4.8 cm.
⇒ ⇒
AF2 + OF2 = AE2 + OE2 Similarly, we can find out that
...(iii)
BD2 + OD2 = BF2 + OF2
...(iv)
CE2
...(v)
and
+
OE2
=
CD2
+
OD2
Adding equations (iii), (iv) and (v), we arrive
3. (A) ∆ABC ~ ∆DEF.
AF2 + BD2 + CE2 = AE2 + CD2 + CD2 + BF2.
4. ∵ ∆ABC ~ ∆PQR
Hence the result. 2
2
9 3 = = = 9 : 1. 1 1
ar (∆ PRQ) QR = 2 ar (∆ BCA) BC 5. True Hint: Use Basic Proportionality Theorem 6. Hint: A B Use: ∠1 = ∠2 3 ∠3 = ∠4. 2
∴
...(ii)
8. ∆ABC ~ ∆PQR ⇒
BC AB = and ∠B = ∠Q PQ QR
⇒
1 BC AB = 2 and ∠B = ∠Q 1 PQ QR 2
⇒
AB BP = and ∠B = ∠Q PQ QM
7. Draw EOF || AD A E
D
D
4 F
C
F
O
B
C
E
1
OB2 = EO2 + EB2 OD2 = OF2 + DF2 ∴ OB2 + OD2 = EO2 + EB2 + OF2 + DF2 = EO2 + CF2 + OF2 + AE2 [∵ DF = AE, EB = CF] = (EO2 + AE2) + (CF2 + OF2) OB2 + OD2 = OA2 + OC2. OR Join OA, OB and OC ∴
In right ∆AOF, AO2 = AF2 + OF2
56
...(i)
(∵ BD = DC and QM = MR) ⇒
∆ABD ~ ∆PQM
⇒
AD AB = . PM PQ
Hence proved.
OR Hint: Show that ∆ABD ~ ∆PQM M A T H E M A T
I C S – X
Similarly, we can prove that
ar (∆ABC) AB = ar (∆PQR) PQ 9. Let the two given triangles be ABC and PQR such that ∆ABC ~ ∆PQR BC AB = QR PQ
∴
...(i)
and
2
...(vi)
AC ar(∆ABC) = ar(∆PQR) PR
2
...(vii)
From equations (v), (vi) and (vii), we obtain 2
AB BC ar(∆ABC) = = ar(∆PQR) PQ QR
2
2
AC = . PR
Let us draw perpendiculars AD and PM from A and P to BC and QR respectively. ∴ ∠ADB = ∠AMQ = 90° ...(ii) Now, in ∆ABD and ∆PQM, ∠B = ∠Q (∵∆ABC~∆PQR) ∠ADB = ∠AMQ [From (ii)] So, by AA rule of similarity, we have ∆ABD ~ ∆PQM AD AB = PM PQ
⇒
Hence, the theorem. Further, in the question, 2 ar(∆ABC) BC = ar(∆DEF) EF
⇒
⇒
BC AD = QR PM
=
...(iv)
1. (B) Ratio of areas of two similar triangles = Ratio of squares of their corresponding sides. = 42 : 92 = 16 : 81.
⇒
5 AB AD = ⇒ AB = 21 × = 15 cm. 7 AC AE
3. (C) ∠M = ∠Q = 35° (Corresponding angles) QR PQ = MN ML (Ratio of corresponding sides)
[Using (iv)]
A N G L E S
8 × 15.4 = 11.2 cm. 11
2. (A) In ∆ABC, DE || BC
1 × BC × BC = 2 1 × QR × QR 2
I
64 × 15.4 × 15.4 121
WORKSHEET–38
1 × BC × AD ar (∆ABC) Now, = 2 1 ar (∆PQR) × QR × PM 2
T R
BC =
...(iii)
From equations (i) and (iii), we get
BC = QR
BC 2 64 = 15.4 × 15.4 121
2
...(v)
⇒
MN = 5 ×
12 = 10 cm. 6
57
4 . Yes.
the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.
5 2 AP = = 7.5 3 AQ 2 4 BP = = 3 6 BR AP BP = AQ BR
Here,
Hence, due to the converse of Basic Proportionality Theorem, AB || QR. Proof: We are given a ∆ABC in which
5. ∵ DB ⊥ BC and AC ⊥ BC
AC2 = AB2 + BC2
∴ DB || AC Now, ∠DBA = ∠BAC
(Alternate angles)
And, ∠DEB = ∠ACB
(Each 90°)
∴
∆BDE ~ ∆ABC
(AA similarity)
DE BE = (Corresponding sides) BC AC AC BE = . BC DE
⇒
Hence proved.
We need to prove ∠ABC = 90°. Let us construct a ∆PQR such that ∠PQR = 90° and
...(ii)
QR = BC
...(iii)
PR2 = PQ2 + QR2 PR2 = AB2 + BC2
...(iv)
[Using equations (ii) and (iii)] From equations (i) and (iv), we have
Hint: See worksheet-35 Sol. 8. 7. Hint: AM =
PQ = AB
Using Pythagoras Theorem in ∆PQR, we have ∴
AX 2 − 2 6. = 2 AB
...(i)
AC = PR Now, in ∆ABC and ∆PQR,
1 AB; 2
1 AC 2 Using Pythagoras Theorem.
AL =
8. Hint: Join AC and use Basic proportionality theorem. A
...(v)
B
AB = PQ
(From (ii)]
BC = QR
[From (iii)]
AC = PR
[(From (v)]
So,
∆ABC ≅ ∆PQR
(SSS congruence)
∴
∠ABC = ∠PQR
(CPCT)
But
∠PQR = 90°
(By construction)
∴
∠ABC = 90°.
Hence proved.
2nd Part: E
X
As AB2 = 2AC2
F
A
= AC2 + AC2 D
C
9. Converse of Pythagoras Theorem: In a triangle, if square of one side is equal to
58
⇒
AB2 = BC2 + AC2 [... BC = AC]
⇒
∠C = 90°.
B
C
M A T H E M A T
I C S – X
WORKSHEET–39 1. (A) Let the length of shadow is x metres. BE = 1.2 × 4 = 4.8 m
⇒ 5AB – 5AD = 4AD ⇒
∆ABC ~ ∆DEC AB BC = DE EC 4.8 + x 3.6 ⇒ = x 0.9 3.6x = 4.32 + 0.9x.
(
3a
∆ADE ~ ∆ABC
5 DE AD = = 9 AB BC
∴
)
2
∴
∠EDC ~ ∠BCD (Alternate interior angles)
= a2 + 3a2
i.e.,
...(iii)
∠DEF = ∠CBF
... (iv)
From equations (iii) and (iv), we have ∆DEF ~ ∆CBF
(AA similarity) 2
5 3 AB AB –1 = ⇒ = 2 2 AD AD DE || BC ⇒ ∆ABC~∆ADE 5 AB BC = = . 2 DE AD
∴
∠EDF = ∠BCF
Similarly,
AD 3 AB – AD 2 = ⇒ = 3 DB 2 AD
⇒
...(ii)
[Using (i)] ∵ DE || BC and DC is a transversal
= 4a2 = (2a)2 According to the converse of Pythagoras Theorem, the angle opposite to longest side is of measure 90°. 3. (A)
5 AD = ...(i) 9 AB
As DE || BC,
4.32 x= = 1.6 m. 2.7 2. (B) Here,(a)2 +
AD 5 5 AD = ⇒ = AB – AD 4 4 DB
6.
DE 25 ar (∆DEF) = . = 81 ar (∆CFB) BC
⇒
[Using equation (ii)] 7.
A D
4. No. In ∆PQD and ∆RPD, ∠PDQ = ∠PDR = 90° B
L
C
E
M
F
AB = AC; DE = DF
But neither ∠PQD = ∠RPD nor ∠PQD = ∠PRD Therefore, ∆PQD is not similar to ∆RPD. A
5. Hint: ∆BAC ~ ∆ADC BC BA AC ⇒ = = AC DC AD ⇒ CA2 = BC × CD. I
AB DE = =1 AC DF
⇒
AB AC = also ∠A = ∠D DE DF
⇒
∆ABC ∼ ∆DEF
∴
AL2 ar(∆ABC) = DM 2 ar( ∆DEF)
1
⇒ 2 B
T R
∴
A N G L E S
D
C
4 AL = 5 DM
∴ Ratio of corresponding heights is 4 : 5.
59
OR Proof: Draw a ray DZ parallel to the ray XY. In ∆ADZ, XY || DZ ∴
AY AX 2 = = YZ XD 3
⇒ 2YZ = 3AY
... (i)
In ∆YBC, BY || DZ YZ BD 1 ∴ (... BD = DC) = = ZC DC 1 ⇒ 2YZ = 2ZC ... (ii) From (i) and (ii), 2ZC = 3AY ... (iii) Now, AC = AY + YZ + ZC 8 3 3 = AY + AY + AY = AY 2 2 2 = 4AY
AC(AD + CD) = AB2 + BC2 AC.AC = AB2 + BC2 AC2 = AB2 + BC2. Hence proved. 2nd part: AB2 = = = = =
AD2 + BD2 AD2 + (3CD)2 AD2 + 9CD2 AD2 + CD2 + 8CD2 AC2 + 8CD2
1 = AC2 + 8 BC 4
WORKSHEET– 40
(∵ ∆ABC ~ ∆PQR)
BD =
9. Statement : In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two sides. Proof: We are given a right triangle ABC right angled at B. We need to prove that AC2 = AB2 + BC2 Let us draw BD ⊥ AC. Now, ∆ADB ~ ∆ABC AB AD So, = (sides are proportional) AC AB or AD .AC = AB2 ...(i) Also, ∆BDC ~ ∆ABC BC CD So, = AC BC or CD .AC = BC2 ...(ii) Adding equations (i) and (ii) we get AD . AC + CD . AC = AB2 + BC2
60
AB AB AC = = QR PQ PR
1. (A)
1 BC; 2 1 AE = EB = AB 2 Using Pythagoras Theorem. Hint:
1 ∵ CD = 4 BC Hence proved.
2AB2 = 2AC2 + BC2.
Therefore, AC : AY = 4 : 1. Hence Proved. 8. 2 5 cm
2
12 10 7 = = 9 y x
⇒
and
x=
7×9 21 = 12 4
y=
9×10 15 = . 12 2
2. (C) Required ratio =
4 16 = = 4 : 5. 5 25
3. 17 m Hint: N P 8m W
15 m
O
E
S
Use Pythagoras Theorem and find OP. M A T H E M A T
I C S – X
4. Hint: Let AB = AC = BC = ... a2 =
c b a b 2 + c2
3 2 Also, ar(∆ABE) = a 4 3 2 ar(∆BCF) = c 4 3 2 b . ar(∆ACD) = 4
7. Hint: A
D E
B
C
Prove that ∆AEB ~ ∆DEC. 8. See worksheet–33, sol. 9 (Ist part). 2nd part Join EF and join BD to intersect EF at O.
5. See worksheet − 36 sol. 6. 6. Let ABCD be a quadrilateral of which diagonals intersect each other at O. It is given that AO BO = CO DO CO AO = ...(i) DO BO In ∆AOB and ∆COD, ∠AOB = ∠COD (Vertical opposite angles)
∵ AB || DC, and EF || AB, ∴ AB || DC || EF In ∆ABD, EO || AB, DO DE = BO AE
or
CO AO = [From (i)] DO BO Hence, by SAS rule of similarity, we obtain ∆AOB ~ ∆COD ⇒ ∠BAO = ∠DCO i.e. ∠BAC = ∠DCA These are alternate angles. Therefore, AB || CD and AC is transversal ⇒ ABCD is a trapezium. Hence proved
OR Hint: As ∠BAC = ∠EFG ; ∠ABC = ∠FEG And ∠ACB = ∠FGE
1 1 ∠ACB = ∠FGE 2 2 ∴ ∠ACD = ∠FGH and ∠DCB = ∠HGE ∴ ∆DCA ~ ∆HGF Similarly ∆DCB ~ ∆HGE.
(Basic Proportionality Theorem) Similarly, in ∆BCD, DO CF = BO BF
I
A N G L E S
...(ix)
Using equations (viii) and (ix), we obtain the required result. BF AE = . FC ED
WORKSHEET– 41 1. (B)
EF DF DE = = BC AC AB (i) (ii) (iii)
=
DE + EF + DF AB + BC + CA (iv)
∴
T R
...(viii)
⇒
Perimeter of ∆DEF 4 = 3 + 2 + 2.5 2 [Taking (ii) and (iv)]
⇒ Perimeter of ∆DEF = 15 cm.
61
2. (A) DE || BC ⇒
x+2 x = x–2 x –1
⇒ ⇒
x2 – 4 = x2 – x x = 4. We need to prove
3. ∆KNP ~ ∆KML ⇒
c ac x = ∴x= . b + c b +c a
ar (∆APB) = In right ∆ABC,
4. Hint: A
B
AC =
L
1 ar (∆AQC) 2
AB2 + BC2
= AB
...(i) (∵ AB = BC)
2
P D
C
Now,
ar(∆APB) =
3 AB2 4
And
ar(∆AQC) =
3 AC2 4
Prove that ∆ADL ~ ∆CPD. 1 5. Hint: 2AP = PC ⇒ AP = AC 3
=
1 Similarly, BQ = BC 3 Use Pythagoras Theorem.
6. PQ || BC ⇒
AQ AP 1 = = QC PB 2
ar (∆APB) = ar (∆AQC)
ar (∆ABC) AB = = ( 3 )2 = 9 AP ar (∆APQ) AB ∵ AP = 3
⇒
ar (∆ABC) −1=8 ar (∆APQ)
)
2
⇒
ar(∆APB) =
3 AB 2 1 4 = 2 3 AB 2 2
1 ar(AQC). 2 Hence proved.
7. Hint:
ar (∆APQ) 1 ar ( BPQC) 8 = ⇒ = . ar ( BPQC) 8 ar (∆APQ) 1
∴ Ratio of area of ∆APQ and trapezium BPQC is 1 : 8. OR Let the given square be ABCD. Let us draw an equilateral triangle APB and another equilateral triangle AQC on the side AB and on the diagonal AC respectively.
62
(
3 AB2 ...(iii) 2 Divide equation (ii) by equation (iii), we obtain
2
⇒
3 AB 2 4
=
∴ Also ∆APQ ~ ∆ABC ⇒
...(ii)
Extend AD till E such that AD = DE and similarly, PM = MN M A T H E M A T
I C S – X
Prove that ∆ACE ~ ∆PRN ∠ 1 = ∠2 ...(i) ∠ 3 = ∠5 But (CPCT) ∠ 3 = ∠4 ∠ 4 = ∠6 ... ∠ ∠5 = ∠6 ...(ii) Adding (i) and (ii), ∠1 + ∠5 = ∠2 + ∠6 ∠BAC = ∠QPR ⇒ ... ∆ABC ~ ∆PQR. (By SAS) 8. Let us take two similar triangles ABC and PQR such that ∆ABC ~ ∆PQR. AB BC CA = = PQ QR RP
...
...(i)
Therefore,
BC × AM ar (∆ ABC) = QR × PN ar (∆ PQR) =
BC 2 QR 2
...(v)
[Using (iv)] From results (i) and (v), we arrive BC 2 AB 2 ar (∆ ABC) CA 2 = = = . QR 2 ar (∆ PQR) PQ 2 RP 2
Hence the result. Further, consider the question in the following figure.
We need to prove
ar (∆ABC) AB 2 BC 2 CA 2 = = = 2 2 ar(∆PQR) PQ QR RP 2 ∠ABO = ∠CDO and ∠BAO = ∠DCO (Alternate angles) ⇒ ∆AOB ~ ∆COD (AA rule)
ar (∆AOB) AB 2 ⇒ ar (∆COD) = CD 2 Let us draw AM ⊥ BC and PN ⊥ QR. ∵ ∆ABC and ∆PQR ∴ ∠B = ∠Q ...(ii) In ∆ABM and ∆PQN, ∠B = ∠Q [From (ii)] and ∠M = ∠N (Each 90°) ∴ ∆ABM ~ ∆PQN (AA criterion) AB AM = PQ PN
∴
1 ⇒ ar(∆COD) = 84 × 2 = 21 cm2.
2
CD 1 = ∵ AB 2
WORKSHEET– 42 1. (A)
∆OBC ~ ∆ODA
(AA criterion)
...(iii)
From equations (i) and (iii), we have BC AM = QR PN
Now,
And T R
I
...(iv)
1 × base × height ar(∆ABC) = 2 1 = × BC × AM 2 1 ar(∆PQR) = × QR × PN 2 A N G L E S
⇒ ⇒
OB OC BC = = =2 OD OA DA BC = 2DA = 2 × 4 = 8 cm.
2. (D) Let the given areas be 2x and 3x. Required ratio =
2x : 3x =
2 : 3.
3. (C) x = 11 or 8 OD OC Hint: Use = . OB OA
63
4. True Geometrical figures which are equiangular i.e., if corresponding angles in two geometrical figure are same, are similar. 5. In right ∆ADC, AD2 = AC2 – CD2 = (2CD)2 – CD2 [∵ AC = BC = 2 CD] = 3 CD2.
∴ Using Pythagoras Theorem AD2 = AE2 + DE2 = AE2 + (BE − BD)2 ⇒ AD2 = AE2 + BE2 + BD2 − 2.BE.BD ⇒
1 1 a a 2 3
6. Hint: BMDN is a rectangle. ∆BMD ~ ∆DMC DM DN ⇒ = ⇒ DM2 = DN × MC MC DM Also, ∆ ∆BND ~ ∆DNA. ⇒
=
⇒
DN DM = ⇒ DN2 = DM × AN. AN DN
⇒
= AD2 + (3BD)2 = AD2 + 9BD2 = AD2 + BD2 + 8BD2
GE 3x = 2AB 3x + 4x
(∴ DC = 2 AB)
GE =
...(i)
4 FG 4 = ⇒ FG = AB ...(ii) 7 AB 7 Adding equations (i) and (ii), we get
⇒
6 4 AB + AB 7 7
10 AB 7 7 EF = 10AB.
⇒
EF =
Hence proved.
and
64
⇒
WORKSHEET– 43 1 . (C) 294 cm2 Hint: Prove that ∆OBP ~ ∆OAQ.
A
3. Hint: Draw AM ⊥ BC and DN ⊥ BC As ∆AOM ~ ∆DON
A
⇒
BE = EC =
B
D
2
2AC2 = 2AB2 + BC2 .
E
C
M
1 × BC × AM ar (∆ ABC) 2 = 1 ar (∆ DBC) × BC × ON 2 =
C
N
B
AE ⊥ BC
1 a 2 1 1 BD = BC = a 3 3
1 = AB2 + 8 BC 4
2. (A) 6 cm Hint: Use AA-similarity to prove ∆AOB ~ ∆COD.
8. Hint: Let AB = BC = AC = a ∴
9AD2 = 7AB2.
AC2 = AD2 + DC2
⇒
⇒
7 a2 a2 a2 = − 9 9 3
9. See worksheet-39 sol. 9 (I part).
GE BE = DC BC
GE + FG =
a2 a2 − + 9 3
Hint: 2nd Part:
6 AB 7 Similarly, ∆DGF ~ ∆DBA ⇒
AC2
= a2 +
7. Let BE = 3x and EC = 4x. In ∆BCD, GE || DC ∆BGE ~ ∆BDC ∴ ∴
2
1 AD2 = AE2 + EC2 + a − 2 3
O D
AM AO = . DN OD
M A T H E M A T
I C S – X
4. Hint: Use concept of similarity.
⇒
5. Draw AP ⊥ BC ∴
A
⇒
B
2
1 D
P
⇒ C
⇒
AB2 = AP2 + BP2
∴
∠ALE = ∠CLB (Vertically opposite angles) ∆LAE ~ ∆LCB (AA criterion) AE LE = (Corresponding sides) BC BL 2BC EL = BC BL
EL = 2BL.
[Using equation (i)] Hence proved.
OR
= AP2 + (BD + DP)2
Hint:
AB2 = AP2 + BD2 + DP2 + 2BD. DP
⇒
= AD2 + BD (BD + 2DP) [... BP = PC] ⇒ AB2 − AD2 = BD × CD. Hence proved. 6. Hint: As AD is median a
P
b
AB2 + AC2 = 2(AD2 + BD2)
⇒
BC2 2 AB2 + AC2 = 2 AD + 4
x p
From figure, show x =
ab . a+b
7. ∆MDE and ∆MCB, ∠MDE = ∠MCB MD = MC
so,
(Alternate angles) (M is mid-point of CD)
∠DME = ∠CMB (Vertically opposite angles)
⇒2(AB2 + AC2) = 4AD2 + BC2 ...(i) ⇒ Similarly, 2(AB2 + BC2) = 4BE2 + AC2 ...(ii) 2 2 2 2 2(AC + BC ) = 4CF + AB ...(iii) Add (i), (ii) and (iii), 3 (AB2 + AC2 + BC2) = 4 (AD2 + BE2 + CF2). 8. See worksheet-33 sol. 9 (Ist part). 2nd Part: Draw EM || AB M is a point on CB ∴ EM || AB
∴ ∆MDE ≅ ∆MCB, (ASA criterion) ⇒ DE = CB (CPCT) ⇒ AE – AD = BC ⇒ AE = 2BC ...(i) (∵ BC = AD) Now, in ∆LAE and ∆LCB, ⇒ ∠LAE = ∠LCB (Alternate angles) T R
I
A N G L E S
⇒
CM CE = MB AE
CM DE = MB EB From (i) and (ii),
Also
...(i) ...(ii)
CE DE = . AE EB
65
WORKSHEET– 44 4 3 BE a ⇒ a= 3 4 ∴ AB2 + BC2 + CA2 = a2 + a2 + a2 = 3a2
1. (D) BE =
4 =3 × BE 3
2
= 16 BE2. 2. (B) Hint: Use pythagoras theorem 3. (B) Hint: Using basic proportionality theorem we get, x = 4. 4. Hint: In ∆ACD and ∆ABC, ∠A = ∠A ∠ACD = ∠ACB = 90° ⇒ ∆ACD ~ ∆ABC …(i) ⇒ AC2 = AB. AD ∆BCD ~ ∆BAC ⇒ BC2 = BA. BD …(ii) ∴ Applying (ii) ÷ (i) gives the result. 5. Let the given parallelogram be ABCD We need to prove that AC2 + BD2 = AB2 + BC2 + CD2 + DA2 Let us draw perpendiculars DN on AB and CM on AB produced as shown in figure.
In ∆BMC and ∆AND, BC = AD (Opposite sides of a ||gm) ∠BMC = ∠AND (Each 90°) CM = DN (Distance between same parallels)
66
∴ ∆BMC ≅ ∆AND (RHS criterion) ⇒ BM = AN ...(i) (CPCT) Using result on obtuse triangle ABC (∠ABC > 90°), AC2 = AB2 + BC2 + 2AB.BM ...(ii) Using result on acute triangle ABD (∠BAD < 90°), BD2 = AB2 + DA2 – 2AB .AN ⇒ BD2 = CD2 + DA2 – 2AB . BM ...(iii) [Using (i) and AB = CD] Adding equations (ii) and (iii), we have AC2 + BD2 = AB2 + BC2 + CD2 + DA2. Hence proved. 6. Hint: AP || QB || RC Use Basic proportionality theorem. 7. See worksheet 38 sol. 9 (Ist part). 8. Let us produce AD to J and PM to K so that DJ = AD and MK = PM. Join CJ and QK.
In ∆ADB and ∆JDC, AD = JD, ∠ADB = ∠JDC, BD = CD ⇒ ∆ADB ≅ ∆JDC (SAS criterion of congruence) ⇒ AB = JC ...(i) (CPCT) Similarly, we can prove that PQ = KR ...(ii) According to the given conditions, we have AB AC AD = = PQ PM PR ⇒
AJ JC AC = 2 = PK KR PR 2 [Using (i) and (ii)]
M A T H E M A T
I C S – X
JC AJ AC = = KR PK PR
⇒ ⇒
∴
∆AJC ~ ∆PKR
(SSS criterion of similarity) ⇒ ∠JAC = ∠KPR (Corresponding angles) i.e., ∠DAC = ∠MPR ...(iii) Similarly, we can prove that ∠DAB = ∠MPQ ...(iv) Adding equations (iii) and (iv), we obtain ∠BAC = ∠QPR ...(v) Thus, in ∆ABC and ∆PQR, we have
AC AB = PQ PR
⇒
h=
Add (i) and (ii), to get the result.
WORKSHEET– 45 1. (D) ∆ABC ~ ∆PQR 20 10 = h 50
5×7 25 × 49 ⇒ h2 = = 3.5 cm. 10 100
3. (B) Altitude AM divides base BC in two equal parts. That is BM = MC = 7 cm using Pythagoras Theorem in right ∆ABM,
(Given)
and ∠BAC = ∠QPR [From (v)] Therefore, ∆ABC ~ ∆PQR. (SAS criterion of similarity) Hence proved. OR Hint: As ∆ADC is obtuse angled triangle, AC2 = AD2 + CD2 + 2 . CD . DE …(i) and ∆ADB is acute-angled triangle So, AB2 = AD2 + BD2 − 2 . DB . DE …(ii)
⇒
25 × 49 100 52 = 2 ⇒ h2 = 49 100 h
AM = =
252 – 7 2 =
(25 + 7) (25 – 7)
32 × 18 = 24 cm.
4. (i) We know that diagonal of a square 2 × side
= In square AEFG, AF =
2 AG
...(i)
In square ABCD,AC =
...(ii) 2 AD Using equations (i) and (ii), we obtain AF AC = . ...(iii) AG AD (ii) ∠GAF = ∠DAC (Each 45°) ⇒ ∠GAF – ∠GAC = ∠DAC – ∠GAC ⇒ ∠CAF = ∠DAG ...(iv) From equations (iii) and (iv), we have ∆ACF ~ ∆ADG. (SAS criterion) 5 . Hint: ∵ ∠1 = ∠2 ∴ PQ = PR QT QR = . ∴ PQ QS
6. Hint: Draw AM ⊥ BC and DN ⊥ BC. A 7. Hint: Fig. ⇒
h=
50 × 20 = 100 m. 10
2. (A) The ratio of similar triangles is equal to the ratio of squares of their corresponding altitudes. T R
I
A N G L E S
b
C
c p a
B
67
8. Hint for 1st part: Prove Pythagoras Theorem. Hint for 2nd part: Let the given triangle is ABC with ∠A = 90° Draw AD ⊥ BC ∆ABC ~ ∆DAC and ∆ABC ~ ∆DBA. 9. Hint: Let the DC = AB = x 3 4 Then QC = x and AP = x 5 5 ∆QRC ~ ∆PRA. OR Let the given parallelogram be ABCD. We need to prove AC2 + BD2 = AB2 + BC2 + CD2 + DA2 Let us draw perpendiculars DN on AB and CM on AB produced as shown in figure. In ∆BMC and ∆AND
BC = AD
(opposite sides of a parallelogram) ∠BMC = ∠AND (Each 90°) CM = DN (Distance between same parallels) ∴ ∆BMC ≅ ∆AND (RHS criterion) ⇒ BM = AN ...(i) CPCT In obtuse triangle ABC (∠ABC > 90°), AC2 = AB2 + BC2 + 2AB . BM ...(ii) In acute triangle ABD (∠BAD < 90°), BD2 = AB2 + DA2 – 2AB . AN ⇒ BD2 = CD2 + DA2 – 2AB . BM ...(iii) [Using (i) and AB = CD] Adding equations (ii) and (iii), we have AC2 + BD2 = AB2 + BC2 + CD2 + DA2. Hence proved.
68
ASSESSMENT SHEET–7 1. (C) In ∆ABC, PQ || BC ∴ ∴ ∴ 2. (B) ⇒
AQ AP = QC BP
2.4 2 = ⇒ BP = 3.6 cm BP 3 AB = AP + BP = 2.4 + 3.6 = 6 cm.
9 BC BC2 ar(∆ABC) = ⇒ = 2 EF 4 ar(∆DEF) EF
2
3 BC = . 2 EF
3. Let draw AM ⊥ BC and DN ⊥ BC ∵ ∠AMO = ∠DNO = 90° and ∠AOM = ∠DON
∴ ∴
∆AMO ~ ∆DNO (AA similarity) AM AO = ...(i) DO DN
1 × BC × AM ar(∆ABC) 2 Now, ar(∆DBC) = 1 × BC × DN 2
=
AO . DO
[Using (i)]
4. True, because ∆BCD ~ ∆CAD ⇒ CD2 = BD . AD. 5. PQ || BC and AB is transversal ∴ ∠APQ = ∠ABC ...(i) (Corresponding angles) In ∆ABC and ∆APQ, ∠BAC = ∠PAQ (Common) M A T H E M A T
I C S – X
∠ABC = ∠APQ [From (i)] so, by AA criterion of similarity,
We need to prove
AD 2 ar(∆ABC) = ar(∆PQR) PM 2
∆ABC ~ ∆APQ
ar( ∆ABC) AB2 ∴ ar( ∆APQ) = AP2 Subtracting unity from both the sides, we have. ⇒
2 2 ar( ∆ ABC) – ar( ∆ APQ) = AB – AP 2 AP ar(∆ APQ) 2
AB ar(trapezium BPQC) = –1 AP ar(∆APQ) ...(ii) It is given that
⇒
AP 1 PB = ∴ =2 AB 2 AP
In ∆ABD and ∆PQM, ∠ADB = ∠PMQ = 90° ∠ABD = ∠PQM (∴∆ABC ~ ∆PQR) ∴ ∆ABD ~ ∆PQM (AA criterion of similarity) AB AD = ...(i) ⇒ PQ PM (Corresponding sides) We know that the ratio of areas of two similar triangles is equal to ratio of squares of their corresponding sides AB 2 ar(∆ABC) = ...(ii) ar(∆PQR) PQ 2 From equations (i) and (ii), we have
∴
ar(∆ABC) AD 2 = . ar(∆PQR) PM 2
PB + 1= 2 + 1 AP
∴
ASSESSMENT SHEET–8
PB + AP AB =3⇒ =3 AP AP
⇒
2
AB ∴ ...(iii) =9 AP From equations (ii) and (iii), we have
Hence proved.
1. (B)
30 6 x 15 = ⇒x= ,⇒x= cm. 8 5 4 6+2
2. (A) ∆PQR ~ ∆CAB,
ar(∆APQ) 1 = ar(trapezium BPQC) 8
⇒
PR PQ QR = = CA BC AB
⇒ ar(∆APQ) : ar(trapezium BPQC)
or
AB BC CA = ⇒ . QR PQ PR
= 1 : 8. 6. See worksheet-36 sol. 6. 7. See worksheet-33 sol. 9 (Ist part). 8. We are given two triangles ABC and PQR such that ∆ABC ~ ∆PQR. Draw perpendiculars AD and PM on BC and QR respectively.
3. In ∆ABC, ∠M || AB. Using Basic Proportionality Theorem, we have 2x 2x + 3 BC AC = ⇒ = x – 3 BM AL x–2 ⇒ ⇒
2x2 – 4x = 2x2 + 3x – 6x – 9 – x – 9 ⇒ x = 9.
4. False, because ∆PQD and ∆PRD don’t obey any condition of similarity. 5. Let the given right angled triangle be ABC with ∠C = 90° such that AC = b, BC = a and AB = c. T R
I
A N G L E S
69
∵
∆ABC ~ ∆PQR
∴
1 BC BC AB 2 = = 1 QR PQ QR 2
⇒
BD AB = QM PQ
...(i)
(a) In ∆ABD and ∆PQM, Using Pythagoras Theorem, we have AB2 = AC2 + BC2 ⇒ c2 = a2 + b2 ...(i) Area of equilateral triangle drawn on side BC
3 2 a ...(ii) 4 Similarly, areas of equilateral triangles drawn on side BC and side AB are respectively =
=
3 2 b 4
...(iii)
3 2 c ...(iv) 4 Sum of areas of equilateral triangles drawn on the sides BC and AC And
=
3 2 3 2 a + b 4 4 [Adding (ii) and (iii)] =
=
3 2 (a + b2) 4
3 2 c [Using (i)] 4 = Area of equilateral triangle drawn on hypotenuse AB. Hence proved. =
6.
⇒
BD AB = QM PQ
[From (ii)]
∠ABD ~ ∠PQM (∵∆ABC ~ ∆PQR) So, by SAS criterion of similarity, we have ∆ABD ~ ∆PQM ⇒
AD AB = PM PQ
(b) ∵ ∆ABD ~ ∆PQM, [From part (a)] ⇒ ∠ADB = ∠PMQ ⇒ 180° – ∠ADC = 180° – ∠PMR [From figure] ⇒ ∠ADC = ∠PMR. Hence proved. 7. See worksheet-38 sol. 9 (Ist part). 8. BD = BE (Given) In ∆OBD, AF ⊥ OB and BD ⊥ OB ∴ AF || BD ∴ ∆OAF ~ ∆OBD ⇒
OA AF = OB BD
⇒
AF OA = BE OB
...(i) [Using (i)]
In ∆AFC and ∆BEC, ∠FAC = ∠EBC (Each 90°) ∠FCA = ∠ECB (Vertically opposite angles) So by, AA criterion of similarity, ∆AFC ~ ∆BEC AC AF = ...(ii) BC BE Comparing equations (i) and (ii), we have
⇒
OA AC = OB BC
70
M A T H E M A T
I C S – X
OC – OA OA = OB – OC OB
⇒
⇒ OA × OB – OA × OC = OB × OC – OB × OA ⇒ (OA + OB) × OC = 2OA × OB Dividing both sides by OA × OB × OC, we get 1 2 1 + = . OB OC OA
Hence proved.
CHAPTER TEST 1. (B) BC =
52 +122 = 13 cm
∆ABD ~ ∆CBA AD AB = AC BC
⇒
⇒ AD =
5 × 12 60 = cm. 13 13
∆ 16 40 2 P2 2. (D) 1 = 12 = = 2 25 ∆2 50 P2
Clearly,
PN PM = NR MQ
⇒
MN || QR. ∆AOB ~ ∆COD
5. ⇒
AO BO = CO DO
⇒
7x – 9 9x – 8 = 2x – 1 3x
⇒ 3x2 – 2x – 8 = 0 ⇒ (x – 2) (3x + 4) 3 ⇒ x = 2 or x = – 4 ⇒ x = 2. (Negative value rejected) 6. ∴ ∆ABE ≅ ∆ACD ∴ AB = AC and AE = AD (CPCT) Consider AB = AC ⇒ AD + DB = AE = EC ⇒ DB = EC ...(i) (∴ AE = AD) Also AD = AE ...(ii) (Proved above) Dividing equation (ii) by equation (i), we have AE AD = EC DB Hence, in ∆ABC
1 AE 1.5 AD = ⇒ = EC EC 3 DB
⇒
3 = 2 cm. 1.5
⇒ ⇒
⇒
EC =
4. Yes. = 15.2 – 5.7 = 9.5 cm NR = PR – PN = 12.8 – 4.8 = 8 cm Now,
T R
I
...(iii)
AE AD = EC DB DE || BC (Converse of Basic Proportionality Theorem) ∠ADE = ∠ABC and ∠AED = ∠ACB ∆ADE ~ ∆ABC.
7. Hint:
MQ = PQ – PM
and
(Corresponding sides)
⇒ 21x2 – 27x = 18x2 – 16x – 9x + 8
⇒ ∆1 : ∆2 = 16 : 25. 3. (C)
(AAA criterion of similarity)
PM 5.7 = = 0.6 9.5 MQ PN 4.8 = = 0.6 NR 8
A N G L E S
x AC = z BC y AC ∆RCA ~ ∆QBA ⇒ = . z AB ∆PAC ~ ∆QBC ⇒
8. Hint: Draw MN || AD, passing through O to intersect AB at M and DC at N.
71
∆ABC ~ ∆PQR ⇒
Use Pythagoras Theorem for ∆AOM, ∆BOM, ∆CON and ∆DON. 9. Hint :
BC AB = and ∠B = ∠Q QR PQ
⇒
AB 2BD = and ∠B = ∠Q PQ 2QM
⇒
BD AB = and ∠B = ∠Q QM PQ
⇒
∆ABD ~ ∆PQM
⇒
AB AD = . PQ PM
72
M A T H E M A T
I C S – X
Chapter
5
INTRODUCTION TO TRIGONOMETRY WORKSHEET–50
⇒ 3A = 90° – A + 26° ⇒ A =
116° 4
=1+
⇒ sin
cos 59° =1+1=2 cos 59°
Hence, the given equation is valid.
⇒ A = 29°. 2. (D) 2 sin
sin 10° + cos 59° sec 59° sin 10°
=
1. (B) sin 3A = cos (A – 26°) ⇒ sin 3A = sin {90°– (A – 26°)}
1 x x = 1 ⇒ sin = 2 2 2
7.
4 2
cot 30°
x π x π = sin ⇒ = 2 2 6 6
=
π ⇒x= ⇒ x = 60°. 3
3. (B) sin θ = ⇒1–
sin2
⇒ cos θ =
24 24 ⇒ sin2 θ = 25 25
2
2 242 2θ = 7 θ=1– ⇒ cos 252 252
7 25 1 sin θ + cos θ co s θ
1 − sin 2 θ
( 3)
2
+
1
1 – 2 3 2 2
=
4 1 8+8−3 4 + – = 3 6 2 3
=
13 . 6
2
3 2 ⇒ sin (x + y) = sin 90° and cos (x – y) = cos 30° ⇒ x + y = 90° and x – y = 30°
Adding and subtracting, we get respectively
9. cosec A = sin A =
.
cos A =
6. True,
=
4
– cos2 45°
i.e., x = 60° and y = 30°.
5. cot 25° + tan 41°.
LHS =
sin 60°
2x = 120° and 2y = 60°
24 25 24 1 49 = 25 + = + = = 7. 7 7 7 7 7 25 25 sin θ
1 2
8. sin (x + y) = 1 and cos (x – y) =
Now, tan θ + sec θ =
4. tan θ =
+
cos 80° + cos 59° cosec 31° sin 10°
I O N
T O
1 = cosec A
sin A 1 = cos A 3
cot A =
1 =3 tan A
T R
1 10
1 – sin 2 A =
tan A =
cos(90° – 10°) + cos 59° cosec (90° – 59°) sin 10°
I N T R O D U C T
10
1–
1 = 10
I G O N O M E T R Y
3 10
73
sec A =
10 . 3
1 = cos A
10. Hint: RHS =
sin C =
sin 2 A 1 – cos2 A = 1 – cos A 1 – cos A
7.
(1 – cos A) (1 + cos A) 1 – cos A
=
sin θ (1 – 2 sin 2 θ) (2cos 2 θ – 1 )
Hint: LHS = =
cos 60° + sin 30° − cot 30° tan 60° + sec 45° − cosec 45°
=
OR
=
sin θ sin θ × cos 2 θ = . cos θ × cos 2 θ cos θ
= =
1. (A) sin (θ + 36°) = cos θ ⇒ sin (θ + 36°) = sin (90°– θ) ⇒ θ + 36° = 90° – θ ⇒ 2θ = 54° ⇒ θ = 27°.
=
25 5 ⇒ sec2 θ = 16 4
⇒ sec2 θ – 1 =
25 –1 16
⇒
9 16
tan2 θ =
⇒ tan θ =
3 . 4
4. (A) Hint: ∠A = 30º, ∠B = 90º, ∠C = 60º. 5. sec θ + tan θ =
=
=
6. sin A =
74
1 sin θ (1 + sin θ) + = cos θ cos θ cos θ
1 + sin θ 1 − sin 2 θ
=
b+a b+a b − a
a b = a2 1− 2 b
1+
=
24 7 , cos A = , 25 25
b+a . b−a
b+a b2 − a2
3 3
3 –3 . 3 cot θ tan (90° − θ) − sec (90° − θ) cosec θ sin θ cos (90° − θ) + cos θ sin (90° − θ) cot θ cot θ − cosec θ cosec θ sin θ sin θ + cos θ cos θ
cot 2 θ − cosec 2 θ sin 2 θ + cos 2 θ
= cosec2 θ – 1 – cosec2 θ [∵ sin2 θ + cos2 θ = 1] = – 1.
2. (C) Hint: Divide numerator and denominator by cos θ. sec θ =
1 1 + – 3 1– 3 2 2 = × 3 3+ 2– 2
8. Given expression
WORKSHEET–51
3. (D)
7 24 and cos C = . 25 25
9.
1 2 Hint: Draw ∆ ABC with AB = BC = AC = a (say) Draw AD ⊥ BC ∴ ∠BAD = ∠DAC = θ = 30° and BD = DC = a/2
∴
sin θ =
⇒ sin 30° =
1 BD a/2 = = AB a 2 1 . 2
1 tan θ tan θ 10. LHS = + 1 1 − tan θ 1− tan θ =
tan 2 θ 1 + tan θ − 1 tan θ (1 − tan θ)
=
1 ( tan θ − 1)
1 2 tan θ − tan θ
M A T H E M A T
I C S – X
=
⇒
tan 3 θ − 1 1 tan θ − 1 tan θ
...
= tan θ + cot θ + 1 =
4. (D) Hint: (1 + sin θ) (1 – sin θ) = cos2 θ 1 = . sec 2 θ 5. (A) Hint: sec 4A = cosec (90º – 4A).
tan 2 θ + 1 sec2 θ +1 = +1 tan θ tan θ
= 1 + sec θ cosec θ = RHS OR
LHS
cos θ = sin 60° ⇒ θ = 30° 1 3 3 2cos2 30°– 1 = 2× − 1 = − 1 = 2 2 4
cos A − cos A cot A – cos A sin A = = cos A cot A + cos A + cos A sin A
1 cos A – 1 sin A = 1 cos A + 1 sin A
6. Hint: cos (90º – θ) = sin θ, sin (90º – θ) = cos θ. 7.
1 – sin 2 θ (1 + sin θ)(1 – sin θ) = (1 + cos θ)(1 – cos θ) 1 – cos2 θ 2
49 cos2 θ 7 = = cot2 θ = = . 2 8 64 sin θ
OR
1 −1 cosec A – 1 sin A = = = RHS 1 + 1 cosec A + 1 sin A
cosec 2 θ + cot 2 θ cosec 2 θ – sec 2 θ =
WORKSHEET–52 =
1. (A) Hint: tan 5º = cot 85º; tan 25º = cot 65º. 2. (C)
=
8 tan x = 15
⇒
tan2 x =
225 64
⇒
sec2x – 1 =
225 64
⇒
sec x =
17 8
sec2 x =
289 64
1 + 2 cot 2 θ cot 2 θ – tan 2 θ 1+ 2×3 7 21 = = . 1 8 8 3– 3 3 2
8 17
⇒ cos x =
15 – 8 64 8 = 1– = – 289 17 17
=
1 1 3 + + +1 4 2 4
=
5 1+ 2 + 3 + 4 10 = . = 2 4 4
LHS = =
sec θ = cosec 60° I O N
T O
2
9. Hint:
7 . 17
I N T R O D U C T
)
2 3 1 1 2 = + + ( 1) + 2 2 2
Now, sin x – cos x = 1 – cos x – cos x
3. (A) Hint:
(
1 + cot 2 θ − 1 + tan 2 θ
8. sin2 30° + sin2 45° + sin2 60° + sin2 90°
2
=
1 + cot 2 θ + cot 2 θ
T R
1 − sin θ 1 − sin θ × 1 + sin θ 1 − sin θ (sec θ − tan θ)2
I G O N O M E T R Y
75
OR
Hint:
7. Hint: LHS =
cos A sin 2 A + LHS = 1 − tan A sin A − cos A
8. LHS =
cos 2 A sin 2 A – = cos A − sin A cos A – sin A
=
= cos A + sin A.
.
=
WORKSHEET–53 1. (A) tan θ =
= cos θ =
∴
32 + 42
3
25 = 5
A
4
2. (C) Hint: 1 + tan θ + sec θ 1 + cot θ + cosec θ 1 + sec θ = cot θ cot θ
1 − cos2 θ
.
6. True. Hint: A6 + B6 = (A2 + B2) [(A2 + B2)2 – 3A2 B2].
76
B
=
2 cot 15° 2 sin 68° – 5 tan 75° cos 22°
–
3 tan 45° tan 20° tan 40° tan50° tan70° 5
=
2 cot (90° − 75°) 2 sin (90° − 22°) – 5 tan 75° cos 22°
3 × 1 × tan ( 90° – 70°) tan (90° – 50°) tan 50° tan 70°
–
3. (B) Hint: A + B = 90º; A – B = 30º. 4. (A) tan 2θ = cot (θ + 9°) ⇒ tan 2θ = tan [90° – (θ + 9°)] ⇒ 2θ = 90° – θ – 9° ⇒ 3θ = 81° ⇒ θ = 27°. 5. cot θ =
cot A + 1 − cosec A
9. Given expression
5
4 Base = 5 Hypotenuse
cos θ
(cot A + cosec A ) − (cosec2A – cot2 A )
(cosec A + cot A) [1 − cosec A + cot A] cot A − cosec A + 1 = cosec A + cot A = RHS.
4 1− 1 − cos θ 5 = 1. = 4 1 + cos θ 9 1+ 5
=1+
cot A − 1 + cosecA cot A + 1 − cosecA
=
Perpendicular 3 = C Base 4
BC =
cos A − sin A + 1 cos A + sin A − 1
Dividing Numerator and Denominator by sin A, we get
cos 2 A − sin 2 A = cos A − sin A
10. Hint: 1 – 2sin2 θ = 2cos2 θ – 1 = cos2 θ – sin2 θ
1 − 2sin θ cos θ + 1 + 2sinθ cos θ . (sin θ + cos θ)(sin θ − cos θ)
5
=
2 tan 75° 2 cos 22° – 5 tan 75° cos 22°
–
3 cot 70° cot 50° tan 50° tan 70° 5
=2 –
–
3×
=2–
2 5 1 1 × tan 50° tan 70° tan 70° tan 50° 5 2 3 10 − 2 − 3 5 – = = = 1. 5 5 5 5
M A T H E M A T
I C S – X
10. Given expression = 8 3 cosec2 30°. sin 60°. cos 60°. cos2 45°. sin 45°. tan 30°. cosec3 45°. 1
= 8 3 ×
. sin (90° – 30°). sin 30° cos (90° – 30°) cos2 (90° – 45°). sin 45°. 2
1
Hint: tan x =
sin 2 30°
× cos 30°. sin 30°.
sin2
45°.
⇒ sin x =
4. (A)
cos 30° sin 30°. sin 30° = 8 3 × × cos 30° sin2 30°
=8 3 × 1 × 1 × 1 = 8 3 . OR Hint: sec2 θ = x 2 +
1 1 + 2 2 16 x
... sec2 θ – 1 = x 2 +
1 1 − 2 2 16 x
⇒
1 tan2 θ = x − 4x
15 8
15 8 , cos x = 17 17
161 64 225 – = . 289 289 289
sin 30° + tan 45° – cosec 60° sec 30°+ cos 60° + cot 45°
=
3 3 –4 3 3 –4 × 3 3+4 3 3 –4
=
43 – 24 3 27 + 16 – 24 3 = . 11 27 – 16
5. ∵ A + B + C = 180º ∴
C+A 180º – B = cot 2 2 B B = cot (90º – ) = tan 2 2 = RHS.
LHS = cot
2
1 tan θ = ± x − . 4 x
6. Yes.
7 . 25 7. LHS = (cosec A – sin A) (sec A – cos A) Hint: Both sides =
WORKSHEET–54
1 1 – sin A – cos A = sin A cos A
3 , 4 let BC = 3x and CA = 4x
1. (B) As sin A =
∴ AB = ( 4 x )2 – (3 x )2 = 7 x 3x BC 3 Now, tan A = = = . AB 7x 7
I N T R O D U C T
2
2 3 = 3 = . 4 2 3
1 2 +1– 2 3 = 3 +2 3 –4 = 2 1 4+ 3 +2 3 + +1 2 3
sin 2 45° sin 45° sin 3 45°
...
1 1+ 3
∴ sin2 x – cos2 x =
sin 30° 1 sin 45°. × 3 cos 30° sin 45°
×
1 3
3. (D)
sin 30° 1 3 cos 30° sin 45°
= 8 3×
2 tan 30° 2. (C) = 1 + tan 2 30°
2×
I O N
=
1 – sin2 A 1 – cos2 A × sin A cos A
=
cos 2 A × sin 2 A sin A cos A
= sin A cos A T O
T R
I G O N O M E T R Y
.... (i)
77
RHS =
1 1 = sin A cos A tan A + cot A + cos A sin A
sin A cos A sin 2 A + cos 2 A = sin A cos A ....(ii) 2 2 (∵ sin A + cos A = 1) From equations (i) and (ii) we obtain LHS = RHS. 8. 7 sin2 θ + 3(1 – sin2 θ) = 4 Let sin θ = x ∴ 7x2 + 3 – 3x2 = 4 1 1 ⇒ x=± ⇒ 4x2 = 1 ⇒ x2 = 2 4 1 −1 ∴ sin θ = or sin θ = 2 2 1 sin θ = – is not possible as θ is acute. 2
∴ sec θ + cosec θ =
∴ cos θ =
Now, 5 sin θ – 3 tan θ = 5 × 0.8 – 3 ×
Now, cosec B = cosec 60° =
sin 2 θ – 1 = 4 sin θ sec2 θ –1 cos2 θ = 4 sin θ tan θ .... (ii) From (i) and (ii), LHS = RHS.
=
2+2–
( 3)
2
sec θ – cosec θ sec θ + cosec θ
7.
=
36 8 64 + 2× – 1. (A) Required value = 25 100 6 100
78
( )
3+2+4 = 9. 4–3 6. False. Hint: ∠A = 30º, ∠B = 60º.
=
1 (192 + 216 – 400) 300 1 2 = ×8= . 12 3
2
2
2 2 1 3 +4 + 5× 0 + 3 2 3
=
WORKSHEET–55
= 25 ×
2 3
5. (A) Given expression
2 + 2 . Hence proved 3
= 4 mn = 4
4 =0 3
3. (D) Hint: Divide numerator and denominator by sin A. 3 1+ 1 + cot A 4 . = 3 1 – cot A 1– 4 2 4. (C) sec A = ⇒ sec A = sec 30° ⇒ A = 30° 3 ⇒ A + B = 90° ⇒ B = 90°– 30° = 60°
3 2
9. 1 Hint: cos (40º + θ) = sin {90° – (40° + θ)} = sin (50° – θ) and cos 40° = sin 50°. 10. LHS = m2 – n2 = (tan θ + sin θ)2 – (tan θ – sin θ)2 = 4sin θ tan θ ...(i) RHS.
1 − 0.36 = 0.8
0.8 4 sin θ = = 0.6 3 cos θ
And tan θ =
=
⇒ cosec θ = 2
1 – cos 2 θ =
2. (C) sin θ =
=
1 1 – cos θ sin θ = sin θ − cos θ 1 1 sin θ + cos θ + cos θ sin θ 1 sin θ = 1 (sin θ + cos θ) × sin θ
(sin θ
− cos θ ) ×
3 4 = 3 1+ 4 1–
1 . 7
M A T H E M A T
1 – cot θ 1 + cot θ
Hence proved.
I C S – X
cos θ − sin θ cos θ − sin θ = cos θ + sin θ cos θ + sin θ
8.
cot θ – 1 . cot θ + 1 (Dividing numerator and denominator by sin θ)
=
p –1 q p–q = . = p p+q +1 q 2
sin 35° cos 55° = + – 2cos 60° cos 55° sin 35° 2
2
sin 35° cos 55° = + cos 90° − 35° sin 90° − 55° ( ) ( ) – 2cos 60° 2
sin 35° cos 55° = + cos 55° sin 35°
2
– 2cos 60°
=
–
⇒
⇒
1 = 2−2 × = 2 – 1 = 1 2 OR Given expression
1 2 3 –1 = × 3 3
6– 3 3 = . 3 3
m2 n2 =1 − sin 2 A tan 2 A
m2 − n2 cos2 A =1 sin 2 A m2 – 1 = (n2 – 1) cos2 A m2 − 1 = cos2 A. n2 – 1 Hence proved
⇒
OR
cos 58° sin 22° + sin 32° cos 68°
–
1 tan 60°
10. tan A = n tan B n and sin A = m sin B ⇒ cot B = tan A m 1 ⇒ sin B = sin A ⇒ cosec B = sin A m .. . cosec2 B – cot2 B = 1
⇒
= 1 + 1 – 2 cos 60°
=
=2– =2–
9. Given expression 2
1 cos 38° = 2– tan 18° tan 35° tan 60° × 1 1 × tan 18° tan 35° cos 38° ×
cos 38°cosec 52° tan 18° tan 35° tan 60° tan 72° tan 55° sin 22° cos 58° + cos (90° – 22° ) sin (90° – 58° ) cos 38°cosec ( 90° – 38° )
tan 18° tan 35° tan 60° tan ( 90° – 18° ) tan (90° – 35° )
=
cos 58° sin 22° + cos 58° sin 22°
–
cos 38° sec 36° tan 18° tan 35° tan 60° cot 18° cot 35°
I N T R O D U C T
Consider an equilateral triangle PQR in which PS ⊥ QR. Since PS ⊥ QR so PS bisects ∠P as well as base QR. We observe that ∆PQS is a right triangle, right-angled at S with ∠QPS = 30° and ∠PQS = 60°. For finding the trigonometric ratios, we need to know the length of the sides of the triangle. So, let us suppose PQ = x 1 x Then , QS = QR = 2 2 and (PS)2 = (PQ)2 – (QS)2 = x2 –
I O N
T O
∴ PS = T R
x2 3x 2 = 4 4
3x 4
I G O N O M E T R Y
79
x 1 QS (i) cos 60° = = 2 = 2 PQ x
(ii) sin 60° =
2
cos 27° sin 29° = + –4 cos (90° − 29°) sin ( 90° − 29°) 1 × 2
3x 2 = 3 2 x
PS = PQ
QS (iii) tan 30° = = PS
= 1 + 12 –
x 2 = 1 . 3x 3 2
4 = 0. 2
5. (B) Given expression 2 2 4 2 1 1 1 2 3 = 4 + –3 – − 1 ( ) 2 2 2 2
WORKSHEET–56
1 1 1 3 + – 3 – 1 – =4 16 4 2 4
1. (B) b2x2 + a2y2 = b2a2 cos2 θ + a2b2 sin2 θ = a2b2. 2. (A) A = 90° – 60° = 30° ∴ cosec A = cosec 30° = 2.
12 122 3. (C) tan θ = ⇒ 1 + tan2 θ = 1 + 2 5 5 6.
13 ⇒ sec θ = 5
=
1 3 3 +1– +3– 4 2 4
=
9 17 8 – = = 2. 4 4 4
1 cos A sin A sin A + = + tan A sin A 1 + cos A 1 + cos A
1 + sin θ 1 + sin θ cos θ sec θ + tan θ Now, = = 1 − sin θ 1 – sin θ sec θ – tan θ cos θ 13 12 25 + 5 5 = = 5 = 25 1 13 12 – 5 5 5 12 tan θ = 5
sin θ = =
1 = cosec θ 1 5 1+ 12
2
1 1 + cos2 θ
12 = . 13
80
=
cos A + cos 2 A + sin 2 A sin A (1 + cos A )
=
1 + cos A = cosec A sin A (1 + cos A )
= 2. 7. ∵
sin θ =
4 3 ∴ cosec θ = 3 4
7 9 = 4 16
∵ cos θ =
1 – sin2 θ =
∴
7 4 and cot θ = 3 7
sec θ =
1–
Now, LHS =
cosec 2 θ – cot 2 θ = sec 2 θ – 1
=
7 7 = = RHS. 9 3
2
sin 29° cos 27° 2 4. (A) + – 4 cos 45° cos 61° sin 63°
2
16 7 9 – 9 9 = 9 9 16 –1 7 7
M A T H E M A T
Hence Proved. I C S – X
8. Hint: LHS =
1 1 cos A – sin A sin A
–
2
1 sin A
1 sin A = – sin A 1 – cos A
=
sin A (1 – cos A )
2
sin2 A + cos2 A = 1.
Hint: (x + 1)2 = x2 + 52
= cot A.
1 = tan 30° 3
3. (A) tan A =
⇒ ∴ ∠A = 30°
OR Hint: Using a3 + b3 = (a2 + b2 – ab) (a + b) we get
sin 3 θ+ cos3 θ + sin θ . cos θ sin θ+ cos θ (sin θ + cos θ)(sin 2 θ + cos2 θ − sin θ . cos θ) = sin θ + cos θ + sin θ ⋅ cos θ = 1 − sin θ . cos θ + sin θ . cos θ = 1 .
( 2 + 2 sin θ )(1 – sin θ ) 9. (1 + cos θ)( 2 – 2 cos θ) 2 (1 + sin θ )(1 – sin θ ) = 2 (1 + cos θ )(1 – cos θ )
∴
∠C = 180° – A – B = 180° – 120°
= 60° Now, sin A cos C + cos A sin C = sin 30° cos 60° + cos 30° sin 60° =
3 1 1 3 × + × = 1. 2 2 2 2
1 2
4. (C) cos α = ⇒
⇒
⇒ cos α = cos 60°
α = 60° 1 3
tan β =
⇒ tan β = tan 30°
β = 30°.
Now, sin (α + β) = sin (60° + 30°) = sin 90° = 1.
cos 2 θ 1 – sin 2 θ = = = cot2 θ sin 2 θ 1 – cos 2 θ
5. (B) tan 1° tan 2°.... tan 43° tan 44° tan 45° tan 46° tan 47°..... tan 88° tan 89°
2 225 = 15 = . 8 64
= (tan 1° tan 89°)(tan 2° tan 88°)....(tan 43° tan 47°)(tan 44° tan 46°) tan 45°
10. Hint: p2 – 1 = sec2 θ + tan2 θ + 2 sec θ tan θ – 1 = 1 + tan2 θ + tan2 θ + 2 sec θ tan θ – 1 = 2 tan θ (tan θ + sec θ) Similarly p2 + 1 = 2 sec θ (tan θ + sec θ).
= (tan 1° cot 1°)(tan 2° cot 2°)....(tan 43° cot 43°)(tan 44° cot 44°) tan 45° = (1) × (1) × .... × (1) × (1) × tan 45° = (1 × 1 ×....× 1 × 1) × tan 45° = 1 × 1 = 1.
WORKSHEET–57 1. (B) Hint:
x − y = cos2 A 2
2. (A)
1 – cos 2 A – 1 + cos A sin A (1 – cos A ) cos A (1 – cos A )
and ∴
sin 2 A – 1 + cos A = sin A (1 – cos A )
=
∴
x−y 2 = sin A x+y
6. Given expression x + y = 2 cot A x – y = 2 cos A
I N T R O D U C T
= I O N
T O
T R
tan 50° + sec 50° + cos 40° cosec 50° cot 40° + cosec 40°
I G O N O M E T R Y
81
tan 50° + sec 50° = + cot (90° − 50°) + cosec (90° − 50°)
cos 40°cosec (90° – 40°) 1 tan 50° + sec 50° = + cos 40°. cos 40° tan 50° + sec 50° = 1 + 1= 2.
7. LHS = tan (A – B) = tan (60° – 30°) = tan 30° 1 . 3 RHS
=
=
1 3–1 3 3 = = = 1 1+1 1+ 3 . 3 3–
1 = = LHS. 3
2 3 2
Hence verified.
sin6 θ 3 sin 2 θ 1 . + +1 8. RHS = 6 2 cos θ cos θ cos2
=
sin 6 θ + cos6 θ + 3 sin 2 θ cos 2 θ cos6 θ
( sin
2
3
θ + cos θ ) 2
cos 6 θ
= sec6 θ = LHS. [∵ sin 2 θ +cos2 θ = 1] OR Hint: Numerator of LHS = tan θ + sec θ – (sec2 θ – tan2 θ) = (tan θ + sec θ) – (tan θ + sec θ) (sec θ – tan θ) = (tan θ + sec θ) (1 – sec θ + tan θ). 9. cos θ + sin θ = 2 cos θ Squaring both sides, we get cos2 θ + 2 cos θ sin θ + sin2 θ = 2 cos2 θ ⇒ 2 cos2 θ – cos2 θ – 2 cos θ sin θ = sin2 θ ⇒ cos2 θ – 2 cos θ sin θ = sin2 θ Adding sin2 θ to both sides , we have
82
⇒ cos θ – sin θ =
2 sin θ Hence proved.
l tan θ + m sec θ = n ...(i) × l′ l′ tan θ – m′ sec θ = n′ ...(ii) × l ⇒ ll′ tan θ + ml′ sec θ = nl′ l′ l tan θ – m′ l sec θ = n′ l – + – (m′ l + ml′) sec θ = nl′ – n′ l
10. Hint:
⇒
tan A − tan B tan 60° − tan 30° = 1 + tan A ⋅ tan B 1 + tan 60° ⋅ tan 30°
=
⇒ sin2 θ + cos2 θ – 2 cos θ sin θ = sin2 θ + sin2 θ ⇒ (cos θ – sin θ)2 = 2 sin2 θ
sec θ =
Similarly, tan θ =
nl' − n'l m'l + ml' nm' + mn' . lm' + ml'
WORKSHEET–58 1. (D) Given expression =
cos2 (90 − 70°) + cos2 70° sec 2 (90° − 40°) − cot 2 40°
+ 2 {cosec2 58° – cot 58° tan (90° – 58°)} =
sin 2 70° + cos2 70° cosec 2 40° – cot 2 40° + 2 (cosec2 58° – cot2 58°)
1 + 2(1) = 1 + 2 = 3. 1 2. (A) sec 5A = cosec (A – 36°) ⇒ sec 5A = sec {90° – (A – 36°)} 5A = – A + 126° ⇒ A = 21°. =
3. (D) Given expression = sin2 5° + sin2 10° ... + sin2 40° + sin2 45° + sin2 50° + ... + sin2 80° + sin2 85° + sin2 90° 2
1 = cos2 85° + cos2 80° + .... + cos2 50° + 2 + sin2 50° + .... + sin2 80° + sin2 85° + (1)2 = (cos2 85° + sin2 85°) + (cos2 80° + sin2 80°) 1 + .... + (cos2 50° + sin2 50°) + +1 2 1 = (1 + 1 + .... 8 terms) + +1 2
M A T H E M A T
I C S – X
=8+
[(a2 − b2)2 − (a2 + b2)2] tan2 θ + 4 a2b2 + 4 ab (a2 − b2) tan θ − (a2 + b2)2 = 0 2 2 2 ⇒ – 4a b tan θ + 4ab (a2 – b2) tan θ − a4 − b4 + 2a2b2 = 0 2 2 2 2 2 − 4a b tan θ + 4ab (a − b ) tan θ − (a2 − b2)2 = 0 2 2 2 2 2 4a b tan θ − 4ab (a − b ) tan θ + (a2 − b2)2 = 0 ⇒ [2ab tan θ − (a2 − b2)]2 = 0 ⇒ 2ab tan θ = a2 − b2
1 1 +1= 9 . 2 2
4. (A) tan 3x =
1 1 1 1 1 . + = + =1 2 2 2 2 2
⇒ tan 3x = tan 45° ⇒ x = 5. (B) cosec A = cos A =
45° = 15°. 3
2 ⇒ sin A =
1–
1 – sin 2 A =
1 2
1 1 = 2 2
⇒
tan A = 1, cot A = 1 Now,
(
4 tan 2 A – cos 2 A
)
=
1 +3×1 2 1 41 − 2
2×
= 6. True Hint:
a cos θ + b sin θ = 4 a sin θ – b cos θ = 3 ⇒ a cos θ sin θ + b sin2 θ = 4 a sin θ cos θ – b cos2 θ = 3 – + –
4 = 2. 2
...(i) × sin θ ...(ii) × cos θ sin θ cos θ
b = 4 sin θ – 3 cos θ a = 4 cos θ + 3 sin θ
∴ a2 + b 2 = 16 sin2 θ + 9 cos2 θ – 12 sin θ cos θ + 16 cos2 θ + 9 sin2 θ + 12 sin θ cos θ = 16 + 9 = 25.
9. LHS =
⇒
−
tan θ + 2ab =
(a2
+
b2)
= (sin2 A + cos2 A + sin A cos A) = sin2 A . cos2 A = RHS. 10. m = cosec θ – sin θ = = n = sec θ – cos θ = =
. sec θ
= (a2 + b2 . 1 + tan 2 θ Squaring both sides: (a2 − b2)2 tan2 θ + 4 a2b2 + 4 ab (a2 − b2) tan θ = (a2 + b2)2 (1 + tan2 θ) = (a2 + b2)2 + (a2 + b2)2 tan2 θ I N T R O D U C T
sec 3 A – cosec 3 A
1 (sin A cos A + cos 2 A + sin 2 A) sin A cos A (sin A − cos A) = 1 (sin A − cos A) sin 3 A cos 3 A
a2 + b2 (a2 − b2) tan θ + 2ab = cos θ b2)
(1+ cot A + tan A )( sin A – cos A )
cos A sin A + 1 + (sin A – cos A ) sin A cos A = 1 1 – 3 cos A sin 3 A
7. (a2 − b2) sin θ + 2ab . cos θ = a2 + b2 Divide by cos θ
(a2
a2 − b2 . 2 ab
8. Hint: Use (a2 + b2)3 = a6 + b6 + 3a2b2 (a2 + b2).
2 sin 2 A + 3 cot 2 A
Similarly,
tan θ =
I O N
T O
(
Now, LHS = m 2 n
2 3
Hence proved.
1 – sin θ sin θ 2
cos θ 1 – sin2 θ = sin θ sin θ 1 – cos θ cosθ sin 2 θ 1 – cos2 θ = cos θ cos θ 2 2 3
) + (mn ) 2
2
cos 4 θ sin 2 θ 3 cos 2 θ sin 4 θ 3 × × = + 2 2 sin θ cos θ sin θ cos θ
T R
I G O N O M E T R Y
83
(
= cos 3 θ
)
2 3
(
+ sin 3 θ
)
4. True, because LHS = tan 60° =
2 3
= cos2 θ + sin2 θ = 1 = RHS. OR LHS = (1 + cot A – cosec A)(1 + tan A + sec A) cos A 1 = 1 + − sin A sin A
= =
(sin A + cos A )2 – 12 sin A cos A 2
=
2 sin A cos A =2 sin A cos A
6.
= RHS.
Hence proved.
ASSESSMENT SHEET–9 1. (B) cos 30° =
sin 30° = ⇒
y=
⇒
10 3 BC ⇒ = x 2 AC
2
2
cos 55° + cos(90° − 35°) – 2 cos 60° 2
= 1 + 1 – 1 = 1.
=4
(Given)
=4
)
1 4
1 2 1 1 When sin x = – , cos y = 2 2 1 1 When sin x = , cos y = – 2 2
⇒
cos 55° cos 55° 1 = + –2× ° ° cos 55 cos 55 2
... (i)
we get 2(sin2 x + sin2 x) = 1 ⇒ sin2 x =
3. We know: sin θ = cos (90° – θ) so, given expression
}
2
x + cos y 2 sin
(
2. (D) tan 2θ = cot (θ + 15°) ⇒ tan 2θ = tan [90°– (θ + 15°)].
{
(4 )
2
2 sin 2 x + cos 2 y
10 3 cm. 3
2
sin 2 x + cos 2 y
(Given)
⇒ 4 = 41 ⇒ 2(sin2 x + cos2 y) = 1 ... (ii) Substituting cos y = – sin x from (i) in (ii),
y×3 AB 1 ⇒ = AC 2 20 3
cos(90° − 35°) cos 55°
⇒ ⇒
2sin x + cos y = 1 2sin x + cos y = 20 sin x + cos y = 0 16
20 20 3 x= ⇒x= cm 3 3
⇒
84
= 2{(sin2 A + cos2 A)3 – 3sin2 A cos2A (sin2 A + cos2 A)} 2 2 2 – 3{(sin A + cos A) – 2sin2 A cos2 A} + 1 [∵ (a + b)3 = a3 + b3 + 3ab(a + b) and (a + b)2 = a2 + b2 + 2ab] = 2(1 – 3sin2 A cos2 A) – 3(1 – 2sin2 A cos2 A) + 1 2 [∵ sin A + cos2 A = 1] = 2 – 6 sin2 A cos2 A – 3 + 6 sin2 A cos2 A +1=0
sin A + cos A –1 cos A + sin A + 1 × sin A cos A
sin A + 2 sin A cos A + cos 2 A – 1 sin A cos A
=
2 2 tan 30° 3 = RHS = = 3. 1 1 – tan 2 30° 1+ 3 6 6 5. 2 (sin A + cos A) – 3 (cos4 A + sin4 A) + 1
sin A 1 + 1+ cos A cos A
=
3 and
sin x = ±
Hence, sin x = –
1 1 1 , cos y = or sin x = , 2 2 2
1 . 2 7. We know that
cos y = –
sin (90°– θ) = cos θ, tan (90°– θ) = cot θ, sec(90°– θ) = cosec θ M A T H E M A T
I C S – X
Now,
+ =
+
=
{
ASSESSMENT SHEET– 10
sec 2 (90° – θ) – cot 2 θ 2(sin 2 25° + sin 2 65°) 2
2
1. (A) Given expression = sin 25° cos (90° – 25°) + cos 25° sin (90°– 25°) 2 2 = sin 25° + cos 25° = 1.
2
2 cos 60° tan 28° tan 62° 3(sec 2 43° – cot 2 47°)
sec 2 (90° − θ) − cot 2 θ
}
2 sin 2 25° + cos 2 ( 90° − 25°)
2 cos2 60° + tan 2 28° tan 2 (90° − 28°)
{
}
3 sec 2 43° − cot 2 ( 90° − 43°)
(
cosec 2 θ − cot 2 θ
2 sin 2 25° + cos 2 25°
+
)
2cos2 60° + tan 2 28° . cot 2 28°
1 + = 2×1
3(sec 2 43° − tan 2 43°)
1 ×1 4 3×1
2×
1 1 2 = + = . 2 6 3 8. Given equation are: sin θ + cos θ = p ... (i) sec θ + cosec θ = q ... (ii) Squaring both the sides of equation (i), we get sin2 θ + cos2 θ + 2sin θ cos θ = p2 Subtract unity from both the sides to get p2 – 1 = 2sin θ cos θ ... (iii) Equation (ii) can be written as
q=
2 sin θ cos θ – 2 sin θ – cos θ cos θ cos θ 2. (B) = cos θ 2 sin θ 2 sin θ + cos θ + cos θ cos θ 4 2× – 1 2 tan θ – 1 5 3 = = = . 4 2 tan θ + 1 11 2× + 1 3 1 cos 45° 2 3. = 2 sec 30° + cosec 30° +2 3 1 1 2 3 = = × 2 2 1+ 3 2 1+ 3
(
= =
3×
(
(
3 –1
)(
3 +1
)
3 –1
)
)
3– 3 × 4 2
=
2 2
3 2– 6 . 8
4. False, because cos2 23° – sin2 67° = 0, 0 is not a positive value. 5. LHS =
1 1 + cos θ sin θ
sin θ + cos θ q= sin θ cos θ
3
2 2
(
)
=
cos A 1 + sin A + cos A 1+ sin A cos2 A + (1 + sinA)
2
(1 + sin A) cos A
... (iv)
=
cos2 A + 1 + sin 2 A + 2sin A (1 + sin A) cos A
sin θ + cos θ q (p2 – 1) = × 2sin θ cos θ sin θ cos θ
=
2 (1 + sin A ) 2 + 2 sin A = (1 + sin A ) cos A (1 + sin A) cos A
⇒ q (p2 – 1) = 2(sin θ + cos θ) ⇒ q (p2 – 1) = 2p. Hence proved.
=
⇒
From equations (iii) and (iv), we get
I N T R O D U C T
I O N
T O
2 = 2 sec A cos A = RHS.
T R
Hence proved.
I G O N O M E T R Y
85
6. Let us construct a triangle ABC in which AB = BC = AC = a (say). Draw AD ⊥ BC. AD bisects BC a ⇒ BD = DC = 2 AD bisects ∠BAC ⇒ θ = 30° In right angled ∆ABD. a AD2 = AB2 – BD2 = a2 – 2 3a2 a2 2 =a – = 4 4
3 a 2 Now, in ∆ ABD,
⇒
2
AD =
a 2
BD tan θ = ⇒ tan 30° = AD
3
1 . 3
⇒ tan 30° =
2∠A = 135° ⇒ ∠A = 67 a 2
Thus, ∠A = 67
7. ( a2 – b2) sin θ + 2ab cos θ = a2 + b2 (Given) Divide both sides by cosθ to get ( a2 – b2) tan θ + 2ab = (a2 + b2) sec θ Squaring both sides, we get (a2 – b2)2 tan2 θ + 4a2 b2 + 4ab(a2 – b2) tan θ = (a2 + b2)2 sec2 θ ⇒ (a2 – b2)2 tan2 θ – (a2 + b2)2 tan2 θ + 4ab (a2 – b2) tan θ – (a2 + b2)2 + 4a2b2 = 0 ∵ sec2 θ = 1 + tan2 θ) (∵ 2 2 2 ⇒ – 4a b tan θ + 4ab (a2 – b2) tan θ – (a2 – b2)2 = 0 2 2 2 2 2 ⇒ – 4a b x + 4ab (a – b ) x – (a2 – b2)2 = 0 where x = tan θ This is a quadratic equation in x. Here, discriminant,
(
D = 16a2b 2 a2 − b 2
)
2
(
− 4 × 4 a 2b 2 a 2 − b 2
)
2
=0 ∴x=
− 4 ab( a2 − b2 ) − 0
⇒ tan θ =
86
2 × ( − 4 a2 b 2 ) a2 − b2 . 2 ab
=
8. Since ABC is a acute angled triangle so, ∠A < 90°, ∠B < 90° and ∠C < 90°. Also ∠A + ∠B + ∠C = 180° ....(i ) 1° (Given) sin (A + B – C) = 2 ⇒ sin (A + B – C) = sin 30° ⇒ ∠A + ∠B – ∠C = 30° .... (ii ) Similarly, ∠B + ∠C – ∠A = 45° .... (iii ) Add equations (ii ) and (iii ) to get 1° 2∠B = 75° ⇒ ∠B = 37 2 Subtract equation (ii) from equation (i) to get 2∠C = 150° ⇒ ∠C = 75° Subtract equation (iii) from equation (i) to get
a2 − b2 2 ab
Hence proved.
1° 2
1° 1° , ∠B = 37 and ∠C = 75°. 2 2
CHAPTER TEST 1. (A) x =
sec θ 1 tan θ and = 2 x 2
∴
1 2 x2 – 2 = 2 x
sec 2 θ tan 2 θ – 4 4
sec 2 θ – tan 2 θ 1 = 2 = . 2 4 2. (C) Hint.
cos 2 20° + cos 2 70°
(
2
2
2 sin 59 + sin 31° 2
)
2
sin 70° + cos 70°
(
2
2
)
2 sin 59° + cos 59°
= =
2 k
2 k
1 2 ⇒ k = 4. = 2 k
3. (C) sin4 θ + cos4 θ = 1 + 4k sin2 θ cos2 θ ⇒ (sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ = 1 + 4k sin2 θ cos2 θ ⇒ 2 sin2 θ cos2 θ (– 1 – 2k) = 0 ⇒ – 1 – 2k = 0 ⇒ k = –
1 . 2
M A T H E M A T
I C S – X
tan θ = 4 θ + 1 = 42 + 1
4. tan2
⇒
sec2
⇒
θ=
[Using equation (i)] 2 4 – 3 (a 2 – 1)2 3 2 . a –1 = 4 4 Hence proved.
(
= 1–
17
1 1 (tan2 θ + 2 sec2 θ) = (16 + 2 × 17) 10 10 = 5. 5. False. Suppose A = 30° and B = 60° Then, LHS = tan (A + B) = tan (30° + 60°) = tan 90° ⇒ LHS = undefined .... (i) and RHS = tan A + tan B = tan 30° + tan 60° ∴
9. LHS =
=
=
1 4 1+ 3 + 3 = = = 3 3 3 ⇒ RHS = a real number .... (ii) From results (i) and (ii), it is clear that the given identity is false.
6.
–1 7 Hint:
=
cos 55° = cos (90° – 35°) = sin 35° cos 70° = sin 20° and tan 5° = cot 85°.
7. 13 . 4 Hint: sin 30° = cos 45° =
1 3 = cos 60°, sin 60° = , 2 2
... (i)
Now, sin6 θ + cos6 θ = sin2 θ + cos2 θ –3
θ
θ
(sin2
)
3
=
θ + cos2 θ) 2
I O N
)
(
)
sec θ + tan 2 θ + 1 + 2 sec θ tan θ tan 2 θ + tan 2 θ + 2 sec θ tan θ sec 2 θ + sec 2 θ + 2 sec θ tan θ
tan θ = tan θ cos θ sec θ
=
sin θ . cos θ = sin θ = RHS. cos θ Hence proved. OR
T O
sin 2 A – sin 2 B + cos2 A – cos2 B (cos A + cos B )(sin A + sin B)
(sin
2
A + cos 2 A) − (sin 2 B + cos 2 B
(cos A + cos B)(sin A + sin B)
)
1–1
(cos A + cos B )(sin A + sin B )
= 0 which is an integer.
a2 – 1 = 13 – 3 (1) 2
I N T R O D U C T
2
θ − 1 + tan 2 θ + 2 sec θ tan θ
=
=
a2 – 1 2
cos2
2
(sin A – sin B )(sin A + sin B ) + (cos A – cos B )( cos A + cos B ) = (cos A + cos B )(sin A + sin B )
sin2 θ + cos2 θ + 2 sin θ cos θ = a2 ⇒ 2 sin θ cos θ = a2 – 1
(
(sec
2 tan θ (tan θ + sec θ) 2 sec θ (sec θ + tan θ)
Square both sides.
sin2
sec 2 θ + tan 2 θ + 2 sec θ tan θ – 1 sec 2 θ + tan 2 θ + 2 sec θ tan θ + 1
sin A – sin B cos A – cos B + cos A + cos B sin A + sin B
1 = sin 45°, sin 90° = 1. 2
sin θ cos θ =
(sec θ + tan θ )2 – 1 (sec θ + tan θ )2 + 1
=
=
8. sin θ + cos θ = a
⇒
)
❑❑ T R
I G O N O M E T R Y
87
Chapter
6
STATISTICS WORKSHEET–62
1. (B) Hint: 3 Median = Mode + 2 Mean. 2. (C) Since the mode is 7 ∴ 2k – 1 = 7 ⇒ k = 4. 3. In such case, mean will increase by 3. ∴ New mean = 18 + 3 = 21.
Σf i x i 4. Hint: Mean = . Σf i 5. Class Cumulative Frequency ( f ) interval Frequency (cf ) 0-8 8-16 16-24 24-32 32-40 40-48
8 10 16 24 15 7 N = 80
8 18 34 58 73 80
N 80 = 40 = 2 2 In the cumulative frequency column, 58 is just greater than 40.
For median class,
So, 24-32 is the median class. N = 40, h = 8 Here, l = 24, cf = 34, f = 24, 2 Using formula: N – cf Median = l + 2 f
× h
40 – 34 = 24 + × 8 = 26 24 Hence, median of the given distribution is 26. 6. 36.25 Hint: Here maximum class frequency is 32. So, the modal class is 30-40.
88
Now, l = 30, f1 = 32, f0 = 12, f2 = 20, h = 10 Use the formula:
f1 – f0 Mode = l + × h. 2 f1 – f0 – f2 7. Hint: Production No. of yield farms 50-55 55-60 60-65 65-70 70-75 75-80 Total
2 8 12 24 38 16 100
Production yield more more more more more more more
than than than than than than than
50 55 60 65 70 75 80
c.f. 100 98 90 78 54 16 0
∴ For more than ogive plot following points. (50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16), (80, 0).
WORKSHEET–63 1. (B) 2. (B) median. 3. The given distribution can be represented as: Marks obtained
No. of students
0-10 10-20 20-30 30-40 40-50 More than 50
5 3 4 3 6 42
Clearly, the frequency of the class 30-40 is 3. 4. Let us rewrite the given table with cumulative frequencies. Class interval
f
cf
0-5 5-10 10-15 15-20 20-25
10 15 12 20 9 N = 66
10 25 37 57 66
M A T H E M A T
I C S – X
N = 66
∵
N = 2 ∴ Median class = Mode class = Required sum = ∴
33 10-15 15-20 10 + 15 = 25.
5. In the given distribution, maximum class frequency is 20, so the modal class is 40-50. Here, lower limit of modal class: l = 40 Frequency of the modal class: f1 = 20 Frequency of the class preceding the modal class: f0 = 12 Frequency of the class succeeding the modal class: f2 = 11 Size of class: h = 10 Using the formula:
f1 – f0 Mode = l + × h 2 f1 – f0 – f2 20 – 12 × 10 = 40 + 2 × 20 – 12 – 11 = 40 + 4.70 = 44.70. Hence, mode of the given data is 45 cars. 6. Let us rewritten the table with class intervals. Class interval
f
cf
36-38 38-40 40-42 42-44 44-46 46-48 48-50 50-52
0 3 2 4 5 14 4 3 N = 35
0 3 5 9 14 28 32 35
We mark the upper class limits on x-axis and with a suitable scale cumulative frequencies on y-axis with a suitable scale. We plot the points (38, 0); (40, 3); (42, 5); (44, 9); (46, 14); (48, 28); (50, 32) and (52, 35). These points are joined by a free hand smooth curve to obtain a less than type ogive as shown in the figure. S T A T
I
S T LI C S
To obtain median from the graph: We first locate the point corresponding to
N 35 = = 17.5 students on the y-axis. From 2 2 this point, draw a line parallel to the x-axis to cut the curve at P. From the point P, draw a perpendicular PQ on the x-axis to meet it at Q. The x-coordinate of Q is 46.5. Hence, the median is 46.5 kg. Let us verify this median using the formula. N − cf 2 Median = l + ×h f 17.5 − 14 ×2 = 46 + 14
= 46 +
7 = 46 + 0.5 14
= 46.5 kg. Thus, the median is the same in both methods. 7. In the given data, the class intervals are formed by exclusive method. But we can
89
convert the series into inclusive form as:
Let us assumed mean be a = 52 and h = 20 Using the formula: Mean = a +
Σfi ui ×h Σfi
⇒ 50 = 52 + Let the assumed mean be a = 57 and h = 3 Now, using the formula: Mean = a +
Σfi ui ×h Σfi
= 57 +
25 ×3 400
⇒
1.1 f1 – 0.9 f2 = 9.2 68 + f1 + f2 = 120
⇒
f1 + f2 = 52
5 + 8+ 3+ 2 N = = 9. 2 2
2. (D) We have Mode = 3 Median – 2 Mean ⇒ ⇒
45 = 3 Median – 2 × 27 Median = 33.
3. (C) mid-points of the classes. 4. Required number of athletes = 2 + 4 + 5 + 71 = 82. 5.
90
...(ii)
Solving (i) and (ii), we obtain f1 = 28 and
WORKSHEET–64
Hint:
...(i)
But
= 57.19.
1. (A)
– 2.8 – 1.1 f1 + 0.9 f 2 × 20 120
f2 = 24. 6. Let us convert the given data into less than type distribution. Class interval
f
Lifetimes (in hrs.)
0-20
10
less than 20
10
20-40
35
less than 40
45
40-60
52
less than 60
97
60-80
61
less than 80
158
80-100
38
less than 100
196
100-120
29
less than 120
225
cf
We mark the upper class, limits along the x-axis with a suitable scale and the cumulative frequencies along the y-axis with a suitable scale. For this, we plot the points A(20, 10), B(40, 45), C(60, 97), D(80, 158), E(100, 196) and F(120, 225) on a graph paper. These points are joined by a free hand smooth curve to obtain a less than type ogive as shown in the following figure. M A T H E M A T
I C S – X
∴
Mean =
7260 = 145.20. 50
Hence, the mean is Rs. 145.20 Median: N − cf ×h Median = l + 2 f
∵ N = 50,
N = 25, f = 14, cf = 12, 2
l = 120 and h = 20 25 − 12 × 20 Median = 120 + 14
∴
= 120 + 18.57 = 138.57 Hence, the median is Rs. 138.57. Mode:
f1 − f0 Mode = l + ×h 2 f 1 − f0 − f2 ∵ l = 120, f1 = 14, f0 = 12 7. The given distribution can be again represented with the cumulative frequencies as given below:
f2 = 8 and h = 20 14 − 12 × 20 Mode = 120 + 2 × 14 − 12 − 8
∴
Class interval
fi
xi
cf
100-120
12
110
12
1320
120-140
14
130
26
1820
140-160
8
150
34
1200
160-180
6
170
40
1020
180-200
10
190
50
1900
Further
7260
⇒
x + y + 45 = 60
⇒
x + y = 15
50
fi x i
Mean: Mean =
Σ fi = 50 and Σfixi = 7260
∵ S T A T
Σfi xi Σfi
I
S T LI C S
= 120 +
40 = 125 8
Hence, the mode is Rs. 125. 8. Given that median is 28.5. It lies in the class interval 20-30, so 20-30 is the median class. N = 60
N – cf Median = l + 2 f
.... (i) ×h
91
30 – ( 5 + x ) 28.5 = 20 + × 10 20
⇒ ⇒
x=8
...(ii)
From equations (i) and (ii), we have x = 8 and y = 7.
4. C.I.
fi
xi
fixi
1-3 3-5 5-7 7-10
9 22 27 17
2 4 6 8.5
18 88 162 144.5
WORKSHEET–65
Σfi = 75
1. (B) Hint:
Mean =
Class interval (C.I.) 0-10 10-20 20-30 30-40 40-50 50-60 60-70
Frequency (f)
Cumulative Frequency
4 4 8 10 12 8 4
4 8 16 26 38 46 50
2. (A) Hint: Draw a line parallel to the x-axis at the point 40 y= = 20. This line cuts the curve at a 2 point. From this point, draw a perpendicular to the x-axis. The point of intersection of this perpendicular with the x-axis determines the median of the data. 3. The given distribution can also be represented as follows:
Σfi xi 412.5 = = 5.5 . 75 Σf i
5. In the given distribution, the classes are in the inclusive form. Let us convert it into exclusive form by subtracting
163 − 162 , i.e., 2
0.5 from lower limit and adding the same to upper limit of each class. Class interval
f
159.5-162.5
15
162.5-165.5
118
165.5-168.5
142
168.5-171.5
127
171.5-174.5
18
Here, the maximum frequency is 142. ∴ l = 165.5, fl = 142, f0 = 118, f2 = 127, h = 3 Now,
f1 − f0 ×h mode = l + 2 f1 − f0 − f2
Class interval
Frequency
0-10
3
10-20
9
142 − 118 ×3 = 165.5 + 284 − 118 − 127
20-30
15
30-40
30
= 165.5 + 1.85 = 167.35
40-50
18
50-60
5
As the maximum frequency is 30, the modal class is 30-40.
92
Σfixi = 412.5
Hence, the modal height of the students is 167.35 cm. 6. The given data may be re-tabulated by the following manner with corresponding cumulative frequencies. M A T H E M A T
I C S – X
Heights (cm.) C.I.
No. of Girls ( f) (cf )
Cumulative frequency
Below 140 140-145 145-150 150-155 155-160 160-165
4 7 18 11 6 5
4 11 29 40 46 51
Σfi ui Mean = A + h × Σfi 14 = 60 + 20 × 50
N = 51
N = 25.5. 2 This observation lies in the class 145-150. Then l = 145, cf = 11, f = 18, h = 5
Now, N = 51. So,
Now,
N – cf median = l + 2 f
Let assumed mean be A = 60 h = 20
× h
= 60 + 5.6 = 65.6 Hence, the required arithmetic mean is 65.6. 8. 69.5. Hint: Change the given distribution into less than type and more than type distributions. For drawing the ‘less than type’ ogive, take upper class limits and corresponding cumulative frequencies; and drawing the ‘more than type’ ogive take lower class limits and corresponding cumulative frequencies.
25.5 – 11 = 145 + × 5 18 = 149.03. Hence, the median height of the girls is 149.03 cm. 7.
Class interval
MidFrevalues quency (xi) (fi)
ui = xi − A h
f iu i
ASSESSMENT SHEET– 11 1. (A) Here, a = 25, h = 10. ∴
Σfi ui x =a+h Σfi 20 = 25 + 10 = 27. 100
10-30
20
5
−
40 = − 2 – 10 20
30-50
40
8
−
20 = −1 – 8 20
Sum of first 6 numbers = 6 × 32 = 192
50-70
60 = A
12
0 =0 20
0
∴ 6th number = 192 + 222 – 385 = 29.
70-90
80
20
20 =1 20
20 6 6
90-110
100
3
40 =2 20
110-130
120
2
60 =3 20
Σfi = 50 S T A T
I
S T LI C S
Σfiui = 14
2. (B) Sum of 11 numbers = 11 × 35 = 385 Sum of last 6 numbers = 6 × 37 = 222 3. The modal class is 30-40. h = 10, f1 = 32, f0 = 12, f2 = 20, l = 30. f1 − f 0 Mode = l + ×h f 2 1 − f0 − f 2 32 − 12 = 30 + × 10 64 − 12 − 20
= 30 + 6.25 = 36.25.
93
4. False, the reason is given below: N = 5 + 15 + 30 + 8 + 2 = 60 N = 30 ∴ 2 Since 30 lies in the interval 30-60, so the median will lie in 30-60. 5.
xi
fi
f ix i
3 6 7 4 p+3 8
5 2 3 2 4 6 Σfi = 22
15 12 21 8 4p + 12 48 Σfixi = 4p + 116
Mean = ⇒ ⇒ 6.
Point
0-10 10-20 20-30 30-40 40-50 50-60
5 13 23 32 38 45
(10, 5) (20, 13) (30, 23) (40, 32) (50, 38) (60, 45)
5 8 10 9 6 7
10 20 30 40 50 60
On plotting these points on a graph paper and joining them by a free hand smooth curve, we get a curve called less than ogive.
4 p + 116 ⇒ 132 = 4p + 116 22 4p = 16 ⇒ p = 4.
6=
Class interval
Frequency (f)
0-20 20-40 40-60 60-80 80-100
4 6 18 8 14
fi − f0 ×h Mode = l + 2 f1 − f0 − f2 Here, l = 40, f1 = 18, f0 = 6, f2 = 8 and h = 20 18 − 6 Mode = 40 + × 20 2 × 18 − 6 − 8
= 40 +
12 × 20 = 50.91. 22
7. We notice classes are continuous. We form cumulative frequency table by less than method.
94
cf
Σfi xi Σfi
The class corresponding to the maximum frequency is 40-60. So, 40-60 is the modal class.
∴
Marks Number Marks (C.I.) of stuless dents than
8. f1 + f2 = 25
... (i)
as median is 32 which lies in 30-40 So median class is 30-40. ∴ l = 30; h = 10; f = 30; N = 100 c.f. = 10 + f1 + 25 = f1 + 35. ∴ Median = l + ⇒ f1 = 9
(N/2 – cf ) ×h f
∴ from (i) ⇒ f2 = 16.
ASSESSMENT SHEET–12 1. (D) x1 + x2 + ................ + xn = n × x x n x x ⇒ 1 + 2 + .......... + n = x k k k k (Dividing through by k) M A T H E M A T
I C S – X
x1 x2 xn + + ......... + k k = x ⇒ k k n (Dividing throught by n) x ⇒ Required mean = . k 2. (B) The first ten prime numbers are: 2, 3, 4, 7, 11, 13, 17, 19, 23, 29.
Mean =
⇒
16 − f 6 = 10 20 − f
⇒ ⇒
Σf i xi Σf i
5 × 6 + 10 × k + 15 × 6 + 20 × 10 + 25 × 5 15 = 6 + k + 6 + 10 + 5
⇒ ⇒
16 − f 36 = 30 + × 10 32 − f − 12
445 + 10k = 15 27 + k
⇒
k = 8.
4. False, because the values of these three measures depends upon the type of data, so it can be the same. 5. Let us use the assumed mean method to find the mean of the given data. Marks No. of Class di = (C.I.) students mark xi – 35 (xi) ( fi ) 0-10 10-20 20-30 30-40 40-50 50-60
4 6 8 10 12 30 Σfi = 70
5 15 25 35 45 55
–30 –20 –10 0 10 20
fi d i –120 –120 –80 0 120 600 Σfidi = 400
Here, assumed mean, a = 35 Σf d Now, required mean = a + i i Σf i 400 = 35 + = 35 + 5.71 = 40.71. 70 6. Since mode = 36, which lies in the class interval 30-40, so the modal class is 30-40. ∴ f1 = 16, f0 = f, f2 = 12, l = 30 and h = 10.
Now, S T A T
f1 − f0 ×h mode = l + 2 f 1 − f0 − f 2
I
S T LI C S
120 – 6f = 160 – 10 f 4f = 40 ⇒
f = 10.
7. The cumulative frequency table for the given data is given below:
11 + 13 24 Median = = = 12. 2 2 3.
⇒
Marks (C.I.)
No. of students (f)
Cumulative frequency (cf )
0-10 10-20 20-30 30-40 40-50 50-60
10 f1 25 30 f2 10
10 10 + f1 35 + f1 65 + f1 65 + f1 + f2 72 + f1 + f2
N = 75 + f1 + f2 Clearly, N = 75 + f1 + f2 But N = 100 ∴ f1 + f2 = 25
.... (i)
N = 50. 2 The median is 32 which lies in the class 30-40. So, l = 30, f = 30, cf = 35 + f1, h = 10. Using the formula: ∴
N − cf Median = l + 2 f ⇒
×h
50 − 35 − f1 32 = 30 + × 10 30
2 15 − f1 = ⇒ 75 – 5f1 = 30 10 30 75 − 30 ⇒ f1 = ⇒ f1 = 9 5 Substitute f1 = 9 in equation (i) we get ⇒
9 + f2 = 25 ⇒ f2 = 16 Hence, f1 = 9 and f2 = 16.
95
8. We prepare the cumulative frequency table by less than method as given below: Scores
Fre- Score Cumuquency less lative than frequency (f) (f)
200-250 250-300 300-350 350-400 400-450 450-500 500-550 550-600
30 15 45 20 25 40 10 15
250 300 350 400 450 500 550 600
30 45 90 110 135 175 185 200
Point
(250, 30) (300, 45) (350, 90) (400, 110) (450, 135) (500, 175) (550, 185) (600, 200)
We plot the points given in above table on a graph paper and then joint them by free hand smooth curve to draw the cumulative frequency curve by less than method. Similarly can be drawn the cumulative frequency curve by more than method. We prepare the corresponding frequency table. Scores
200-250 250-300 300-350 350-400 400-450 450-500 500-550 550-600
Fre- Score Cumulaquency more tive (f) than frequency (cf) 30 15 45 20 25 40 10 15
200 250 300 350 400 450 500 550
200 170 155 110 90 65 25 15
Point
CHAPTER TEST (200, 200) (250, 170) (300, 155) (350, 110) (400, 90) (450, 65) (500, 25) (550, 15)
We plot the points given in this last table on the same graph and join them by free hand smooth curve to draw the cumulative frequency curve by more than method. Median: The two curves intersect each other a point. From this point, we draw a perpendicular on the x-axis. The foot of this perpendicular is P(375, 0). The abscissa of the point P, i.e., 375 is the required median. Hence, the median is 375.
96
1. (C) Let us rewrite the given distribution in the other manner. Marks
No. of students
0-10 10-20 20-30 30-40 40-50 50-60
3 9 15 30 18 5
Clearly, the modal class is 30-40. 2. (A) Let
Σfi = N Σ(fixi – x ) = Σfi xi – N x
Σfi xi − x =N N = N ( x – x ) = 0. M A T H E M A T
I C S – X
3. Hint: First, find the cumulative frequency table and N = 13 + 10 + 15 + 8 + 11 = 57
A = 850
N = 28.5. 2
∴ 4.
Let assumed mean be h = 20
Monthly income (in Rs.)
No. of families
10000-13000 13000-16000 16000-19000 19000-22000 22000-25000
15 16 19 17 18
Σui fi ×h Mean = A + Σfi 5 = 850 + × 5 64 = 850 + 0.391 = 850.391. Hence, the required mean is 850.391.
Hence, required number of families is 19. 5. No, because an ogive is a graphical representation of a cumulative frequency distribution. 6. Yes; as we know mode = 3 median – 2 mean ⇒ 3 median = mode + 2 mean 2 1 mode + mean ⇒ Median = 3 3 2 2 = mode – mode + mean 3 3 2 = mode + (Mean – Mode). 3 7.
C.I.
xi
fi
ui =
45 − 30 Mode = 30 + × 10 90 − 30 − 12
= 30 + 3.125 = 33.125 marks. 9. 31.5 marks. Hint: Cumulative frequency
0-10 10-20 20-30 30-40 40-50 50-60
5 8 6 10 6 6
5 13 19 29 35 41
−
40 = −2 20
– 14
820-840 830
14
−
20 = −1 20
– 14
840-860 850
19
0 =0 20
860-880 870
15
20 =1 20
15
880-900 890
9
40 =2 20
18
S T LI C S
∴
No. of students
7
I
f0 = 30, f2 = 12, h = 10
Classes
800-820 810
S T A T
Here, l = 30, f1 = 45,
f iu i
xi − A h
Σfi = 64
f1 − f0 ×h Mode = l + 2 f1 − f0 − f2
8.
Draw the ogive by plotting the points: 0
Σfiui = 5
(10, 15), (20, 13), (30, 19), (40, 29), (50, 35) N = 20.5. Locate the 2 point on the ogive whose ordinate is 20.5. The x-coordinate of this point will be the median.
and (60, 41). Here
❑❑
97
Solutions to P RACTICE PAPERS (SUMMATIVE ASSESSMENTS) [FIRST TERM]
PRACTICE PAPERS Practice Paper-1
⇒
SECTION-A
14587 14587 8 116696 × = 1. (D) ... = 10000 1250 8 1250 = 11.6696. 2. (C) ... LCM (x, y) = =
x×y HCF( x , y) 1800 = 150. 12
3. (B) Let the zeroes of ax3 + bx2 + cx + d be α, β and γ. We are given γ = 0. c ∴ αβ + βγ + γα = a c ⇒ αβ + 0 + 0 = a c αβ = a . c1 a b 10 6 –3 = 3, 1 = = 3, = 4. (D) 1 = c a2 b2 2 9 2 –1 ⇒ i.e.,
c b a1 = 1 ≠ c1 2 b2 b2
∠D = ∠Q and ∠E = ∠R
⇒ ∆DEF ~ ∆QRP (AA rule of similarity) ⇒ 6. (A)
) (3 – cot 30° ) = ( 3 + 1) ( 3 – 3 ) = 3 ( 3 + 1)( 3 – 1)
3 +1
= 2 3.
7. (C) 9α < 90° ⇒ α < 10° ⇒ α is an acute angle. cos 9α = sin α
P R A C T
I
C E
∴ 8. (B)
9. (B) Consider,
⇒ ⇒ ∴
3 sin θ = cos θ sin θ 1 1 = ⇒ tan θ = cos θ 3 3 θ = 30° cos θ = cos 30° =
3 . 2
10. (B) Given expression =
sin 2 35° + sin 2 (90° – 35°) + sin2 63° cos 2 35°+ cos 2 (90° – 35°) + cos 63°sin (90° – 63°)
(sin 2 35°+ cos 2 35°) + (sin2 63° + cos2 63°) (cos 2 35°+ sin 2 35°)
1 + 1 (∵ sin2 θ + cos2 θ = 1) 1 = 2. =
SECTION-B
DE EF ≠ . PQ RP
(
⇒
=
⇒ The given lines are parallel. 5. (B)
⇒
π cos 9α = cos – α 2 π –α 9α = 2 π α = 20 π tan 5α = tan = 1. 4 Median.
P A P E R S
11. True, because out of any two consecutive positive integers, one is even and the other one is odd; and the product of an even and an odd is even. 12. No, if two zeroes are α and β of polynomial x2 + kx + k, then α + β = – k and α . β = k ⇒ 2α = – k and α2 = k (when α = β) –k ⇒ α= and α2 = k. 2
99
k2 =k (Comparing both) 4 ⇒ k2 = 4k ⇒ k2 – 4k = 0 k(– k – 4) = 0 ∴ k = 4, 0.
⇒
13. For infinitely many solutions,
k k 13 = = k+4 6 39 ⇒
⇒ ⇒
Hence, distance between tops of the poles is 13 metres. 16. False, because the range of sin θ is [–1, 1] 1 but a + ≥ 2. a OR See worksheet-50, sol. 6. 17.
13 k = and 39 6
xi 3 5 7 9 11 13
k 13 = k + 4 39 k = 2 and 3k = k + 4 k = 2 and k = 2, i.e., k = 2.
14. Yes, because converse of Basic proportionality theorem is applicable here as 2 PB PA = = . 3 AQ BR
fi 6 8 15 p 8 4 Σfi = p + 41
Mean = ⇒ ⇒ ∴
f i xi 18 40 105 9p 88 52 Σfixi = 9p + 303
∑ fi xi 9p + 303 ⇒ 7.5 = ∑ fi xi p + 41
7.5p + 307.5 = 9p + 303 1.5p = 4.5 p = 3.
18. The maximum class frequency is 20. ⇒ The modal class is 15-20 15. Let the given pole be AB and CD with their feet B and D respectively (see figure).
Class
f
cf
0-5 5-10 10-15 15-20 20-25
10 15 12 20 9 N = 66
10 25 37 57 66
N 66 = = 33 2 2 cf just more than 33 is 37. PD = AD = 6 m ∴ CP = 11 m – 6 m = 5 m And AP = BD = 12 m Now, in right-angled triangle ACP, AC2 = CP2 + AP2 ⇒
100
AC =
2
5 + 12
2
=
169 = 13
⇒ Median class is 10-15 Now, required sum = 15 + 10 = 25. SECTION-C 19. On the contrary let us assume that 2 3 – 3 2 is a rational number. Then, we can take coprime a and b such that
M A T H E M A T
I C S – X
a = 2 3 – 3 2 b
⇒
⇒ ⇒
a2 = 12 + 18 – 12 6 b2 (Squaring both the sides)
21. Since x – 5 is a factor of f (x) = x3 – 3 5 x2 + 13x – 3 5 , so as f (x) may be rewritten. f(x) = x3 – 3 5 x2 + 13x – 3 5 = x3 – 5 x2 – 2 5 x2 + 10x
2
12 6 = 30 – a 2 b
30 b 2 – a 2 . 6 = 12 b 2
incorrect assumption that 2 3 – 3 2 is a rational number.
I
C E
P A P E R S
= x– 5
) (x
) + 3 (x – 5 ) 5x + 3)
2
–2
To find zeroes of f(x), put f(x) = 0.
(x – 5 ) (x
2
)
– 2 5x + 3 = 0
x – 5 = 0 or x2 – 2 5 + 3 = 0
⇒ ⇒
x=
So, we conclude that 2 3 – 3 2 is an
P R A C T
)
(
x=
irrational number. 20. Let a be any odd positive integer. Then it is of the form 6m + 1, 6m + 3 or 6m + 5, where m is an integer. Here, 3 cases arise. Case I: When a = 6m + 1, a2 = (6m + 1)2 = 36m2 + 12m + 1 = 12m (3m + 1) + 1 = 6q + 1, where q = 2m (3m + 1). Case II: When a = 6m + 3, a2 = (6m + 3)2 = 36m2 + 36m + 9 = 36m2 + 36m + 6 + 3 = 6(6m2 + 6m + 1) + 3 = 6q + 3, where q = 6m2 + 6m + 1. Case III: When a = 6m + 5, a2 = (6m + 5)2 = 36m2 + 60m + 25 = 36m2 + 60m + 24 + 1 = 12(3m2 + 5m + 2) + 1 = 6q + 1, where q = 2(3m2 + 5m + 2). Hence, a2 is of the form 6q + 1 or 6q + 3. OR See worksheet-3, sol. 9.
(
(
= x2 x – 5 – 2 5 x x – 5
Since, a and b are integers, therefore, RHS of this last equation is rational and so LHS must be rational. But this contradicts the fact that 6 is irrational. This contradiction has arisen due to
+ 3x – 3 5
5 or
2 5 ± 20 – 4 × 1× 3 2
⇒
x=
5 or x =
⇒
x=
5 or
x=
5 +
2 5 ±2 2 2
2 or
5 –
Hence all the zeroes of f (x) are 5 +
2 and
5 –
2 5,
2.
22. The given pair of equations may be rewritten as
x+y 10 2 2x – y = ; = – xy 2 xy 3 3 i.e,
2 1 1 1 10 4 + = ;– + = – y y x x 3 3
Adding this last pair, we get 3 = –2 y ⇒
y= –
3 2
101
3 in the first equation 2 of last pair, we get
Substituting y = –
1 1 4 + = x –3 3 2
⇒
24.
2 1 4 – = x 3 3
1 1 =2 ∴ x = x 2
⇒
Hence, x = solution.
3 1 , y = – is the required 2 2
AD BD = CD AD (Corresponding parts) ⇒ BD . CD = AD2. OR See worksheet-36, sol. 8. ⇒
23. Let the length of each side of the given equilateral triangle be a, then ∴
AB = BC = CA = a a BD = 3
∠BAD = 90° – ∠CAD (∵ ∠BAC = 90°) = 90° – (90° – ∠ACD) (∵ ∠ADC = 90°) ⇒ ∠BAD = ∠ACD ...(i) ∠BDA = ∠ADC = 90° ...(ii) Using equations (i) and (ii) in ∆ABD and ∆CAD, we have ∆ABD ~ ∆CAD (AA rule of similarity)
...(i) ...(ii)
cos θ – sin θ 1– = cos θ + sin θ 1+
25.
Draw AP⊥BC to meet BC at P. P will be the mid-point of BC, that is a BP = ...(iii) 2 a a a ∴ DP = BP – BD = – = ...(iv) 2 3 6 [Using (ii) and (iii)] Now, in right-angled triangle APB, AP2 = AB2 – BP2 ⇒
AP2 = a2 –
a 4
AP2 =
AD2 ⇒
=
AD2 =
AP2 3a 4
2
+ +
DP2 27 a2 + a2 a2 = 36 36 [From (iv) and (v)]
⇒ 36AD2 = 28a2 ⇒ 9AD2 = 7a2 ⇒
102
9AD2
=
7AB2.
⇒
1– 1 – tan θ = 1 + tan θ 1+
⇒
tan θ =
Hence proved.
3 3
3 ∴ θ = 60°.
26. In ∆ABC,
∴ AC =
2
3
cos θ sin θ – cos θ cos θ 1– 3 = sin θ cos θ 1+ 3 + cos θ cos θ
tan A =
2
3a ...(v) 4 Also, in right-angled triangle APD,
⇒
⇒
3
= sin A =
BC 1 = AB 2 2 (1)2 + (2 2)2
1+ 8 = 3
1 2 2 , cos A = , 3 3
1 2 2 , cos C = 3 3 Now, sin A. cos C + cos A. sin C sin C =
=
1 8 1 1 2 2 2 2 × + × = + = 1. 9 9 3 3 3 3
M A T H E M A T
I C S – X
27. We will use the step-deviation method. ui = Marks No. of Class- d = i students mark d (C.I.) xi – 45 i ( fi ) xi 10 0-10 5 5 – 40 – 4 10-20 4 15 – 30 – 3 20-30 8 25 – 20 – 2 30-40 12 35 – 10 – 1 40-50 16 45 0 0 50-60 15 55 10 1 60-70 10 65 20 2 70-80 8 75 30 3 80-90 5 85 40 4 90-100 2 95 50 5 Σfi = 85 Let
fiui – 20 – 12 – 16 – 12 0 15 20 24 20 10
Σfiui = 29
SECTION-D 29. Let f (x) = 3 2 x2 + 13x + 6 2 = 3 2 x2 + 9x + 4x + 6 2 (Spliting middle term) = 3x( 2 x + 3) + 2 2 ( 2 x + 3) = (3x + 2 2 ) ( 2 x + 3) To find the zeroes of f (x), we have 3x + 2 2 = 0 or
–2 2 –3 or 3 2 ∴ Zeroes of the given polynomial are
⇒
x=
–3 –2 2 and . 3 2 –3 –2 2 + 2 3 – 13 = 3 2
a = 45. Here h = 10
Now, sum of zeroes =
∑ fi ui =a+ ×h ∑ fi
Mean
29 × 10 85 = 45 + 3.41 = 45 +
=– Product of zeroes
= 48.41. 28. The given data is 10 15
30
80
40
20
5
=
From the table, maximum occuring frequency is 80. So, modal class is 15-20. ∵
f1 − f 0 Mode = l + ×h 2 f1 − f 0 − f 2
80 − 30 Mode = 15 + ×5 160 − 30 − 40
= 15 +
250 = 17.78 90
P R A C T
I
C E
P A P E R S
6 2 3 2
Constant term . Coefficient of x 2 Hence proved.
Case I: According to question, we get ⇒
300 – 60 60 + =4 y x
⇒
240 60 + =4 y x
⇒
See worksheet-62, sol. 5.
–3 –2 2 × 2 3
Distance = Time Speed
Hence, modal size is 17.78 hectares. OR
Coefficient of x Coefficient of x 2
30. Let speed of the train be x km/hr and that of the bus be y km/hr.
Here, l = 15, f1 = 80, f0 = 30, f2 = 40, h = 5 ∴
= =
C.I. 0-5 5-10 10-15 15-20 20-25 25-30 30-35 f
2x + 3 = 0
4 1 60 15 1 + =1 ⇒ + = y x 15 y x
...(i)
103
Case II: According to the given conditions, we get
100 300 – 100 10 =4+ + 60 x y 100 25 200 + = x y 6 4 1 8 ⇒ + = x y 6 Dividing by 4, we get
⇒
1 1 2 + = y x 24 Use equation (i) – equation (ii),
... (ii)
1 1 2 4 – = – y y 24 15 8–5 3 2 = = y 120 120 120 ×2 ∴ y= = 80 km/hr 3 Put y = 80 in equation (i), we get 4 1 1 + = x 80 15 1 4–3 1 1 1 ⇒ = – = = x 60 60 15 20 ∴ x = 60 km/hr. Hence, speed of the train = 60 km/hr and speed of the bus = 80 km/hr.
Now,area of a triangle
1 × base × height 2 1 ∴ ar(∆APQ) = × AP × MQ ...(i) 2 1 Also ar(∆APQ) = × AQ × NP ...(ii) 2 Comparing equations (i) and (ii), we get 1 1 × AP × MQ = × AQ × NP 2 2 =
AP NP = AQ MQ
⇒
1 × PB × MQ ...(iv) 2 1 And ar(∆CQP) = × QC × NP ...(v) 2 But triangles BPQ and CQP are on the same base PQ and between the same parallels PQ and BC, so their areas must be equal.
Further, ar(∆BPQ) =
i.e., ⇒
ar(∆BPQ) = ar(∆CQP) 1 1 × PB × MQ = × QC × NP 2 2 [Using equations (iv) and (v)]
PB NP = ...(vi) MQ QC From, equations (iii) and (vi), we get
⇒
OR See worksheet-22, sol. 9. 31. We are given a triangle ABC in which a line PQ parallel to BC is drawn to intersect the sides AB and AC at P and Q respectively. We need to prove AQ AP = QC PB Join PC and QB. Draw QM ⊥ AB and PN ⊥ AC.
AP PB = AQ QC AQ AP = . Hence proved. QC PB
⇒
32. sin θ + cos θ =
(Given) 3 ⇒ + + 2 sin θ cos θ = 3 (Squaring both the sides) ⇒ 1 + 2 sin θ cos θ = 3 ⇒ sin θ cos θ = 1 ...(i) sin θ cos θ Now, tan θ + cot θ = + cos θ sin θ sin2 θ
cos2 θ
1 sin 2 θ + cos2 θ = 1 sin θ. cos θ
[Using (i)]
= 1.
Hence proved.
M A T H E M A T
I C S – X
=
104
...(iii)
33. Let us take LHS of the given identity.
tan 2 θ tan 2 θ – 1
=
=
= =
+
cos ec 2 θ sec 2 θ – cos ec 2 θ
sin 2 θ cos 2 θ sin 2 θ –1 cos 2 θ
1 sin 2 θ
+
1 1 – cos2 θ sin 2 θ
sin 2 θ cos 2 θ
+
sin 2 θ – cos 2 θ cos 2 θ
sin 2 θ sin 2 θ – cos 2 θ
+
1 sin 2 θ sin 2 θ – cos 2 θ sin 2 θ.cos 2 θ
cos2 θ sin 2 θ – cos2 θ
sin 2 θ + cos 2 θ
sin 2 θ – cos2 θ 1 = [... sin2 θ + cos2 θ = 1] 2 sin θ – cos 2 θ = RHS.
Hence proved. OR See worksheet-55, sol. 10. 34. Let us prepare the cumulative frequency table by more than method as given below: Production fi yield (in kg/ha) 50-55 55-60 60-65 65-70 70-75 75-80 Total
2 6 8 14 15 5 50
Production yield (in kg/ha) cf more than or equal to 50 55 60 65 70 75
50 48 42 34 20 5
I
C E
Practice Paper–2 SECTION-A 1. (C) ...
343 3
3
2 × 5 ×7
3
Point
(50, 50) (55, 48) (60, 42) (65, 34) (70, 20) (75, 5)
We plot the points mentioning in the table such that lower class limits are on the x-axis and the cumulative frequencies are on the y-axis. By joining these points by free hand smooth curve, We obtain more than type ogive as shown in the adjoining graph.
P R A C T
To obtain median from graph: Draw a line parallel to x-axis passing 50 through y = = 25. This line meets the 2 ogive at (68.2, 25). ∴ Median = 68.2.
P A P E R S
2. (B) ...
=
343 3
2 × 53 × 343 1 = = 0.001. 1000
p(x) = 3x2 – (2x + 1) x + 3 = x2 – x + 3 –1 3 – 1 1 = 1 – 3 = – 2.
∴ (α + β) – αβ = –
ar(∆ABC) BC 2 = ar(∆DEF) EF 2 (Result on areas of similar triangles)
3. (A) As
D A
B
54 32 ⇒ = 2 ar(∆DEF) 4
C
E
F
⇒ ar(∆DEF) = 96 cm2.
105
10. (B) As sin (45° + θ) – cos (45° – θ)
4. (C) As cosec2 θ – cot2 θ = 1 ⇒ 16 – 3k2 = 1 15 ⇒ k2 = 3 ⇒
k= cos (90o – 70o )
5. (C)
sin 70o
o
sin 70o
⇒
3=
[∵ sin (90° – α) = cos α] = sin (45° + θ) – sin (45°+ θ) = 0.
5.
k 2 cos θ + = 2 cos θ
sin 70
⇒
= sin (45° + θ) – sin {90° – (45° – θ)
+ 2=
k 2
k ⇒ k = 6. 2
1 – tan 2 45° 2
1 + tan 45°
=
1–1 =0 1+1
7. (B) Let us prepare the cumulative frequency table from the given data. Frequency Cumulative frequency
10-15 15-20 20-25 25-30 30-35
12. No, since the discriminant is zero for
4 7 20 8 1
4 11 31 39 40
N = 40 The cumulative frequency just greater N = 20 is 31 and its corresponding than 2 class is 20-25. Hence 20-25 is the median class.
a1 b c is a = 1 = 1 . 2 c2 b2 ⇒
2b – 1 3 – ( a + 1) = = 1 – 2a 3b 5
⇒
– ( a + 1) 2b – 1 3 3 = and = 1 – 2 a 3b 5 5
⇒ 3 – 6a = – 5a – 5 and 9b = 10b – 5 ⇒ a = 8 and b = 5. 14. Yes, because ∆PBC and ∆PDE are similar by SAS rule. BP PC 1 = = and DP PE 2 ∠BPC = ∠DPE.
as
15. In the given ∆ABC, DE || AB So by the Thales Theorem,
8. (D) 0. 9. (D) The condition for the line parallel is:
106
∵
a1 b c = 1 =≠ 1 a2 c2 b2
⇒
3 2c 2 = ≠ 2 5 1
⇒
c=
15 . 4
1 . 2
13. The condition for infinitely many solutions
= tan 0°.
Class interval
11. False, because 3 or 2 3 is an irrational and sum or difference of a rational number and an irrational number is an irrational number.
k=±
6. (D) ∵ tan 45° = 1 ∴
SECTION-B
⇒
CE CD = BE AD
⇒
x x+3 = 3x + 4 3x + 19
⇒ 3x2 + 19x = 3x2 + 4x + 9x + 12 ⇒
6x = 12
⇒
x = 2.
M A T H E M A T
I C S – X
16. We have the identity: sin2 θ + cos2 θ = 1 a2
⇒
b
2
+ cos2 θ = 1
1–
⇒ cos θ =
a2
=
b2
(∵ sin θ =
a ) b
b 2 – a2 . b
17. The empirical relationship among the three measures of central tendency is: Mode = 3 Median – 2 Mean = 3 × 55 – 2 × 50 = 165 – 100 = 65. OR See Assessment sheet-11, sol. 4. 18. Let us convert the given less than type distribution to normal distribution. Marks
No. of students
0-10 10-20 20-30 30-40 40-50 50-60
3 12 – 3 = 9 27 – 12 =15 57 – 27 =30 75 – 57 =18 80 – 75 = 5
From the table, the maximum frequency is 30 and its corresponding class is 30-40. Therefore, the modal class is 30-40. SECTION-C 19. To prove that the pair of numbers (847, 2160) is coprime by using Euclid’s algorithm, we have to prove that the highest common factor of the pair is 1. Since 2160 > 847 ∴ 2160 = 847 × 2 + 466 Since the remainder 466 ≠ 0. ∴ 847 = 466 × 1 + 381 Since the new remainder 381 ≠ 0. ∴ 466 = 381 × 1 + 85 Since the new remainder 85 ≠ 0. ∴ 381 = 85 × 4 + 41 Since the new remainder 41 ≠ 0. ∴ 85 = 41 × 2 + 3 Since the new remainder 3 ≠ 0. ∴ 41 = 3 × 13 + 2 Since the new remainder 2 ≠ 0.
P R A C T
I
C E
P A P E R S
∴ 3 =2 × 1 + 1 Since the new remainder 1 ≠ 0. ∴ 2 =1 × 2 + 0 Since, the remainder has now become zero, the divisor at this stage is 1, the HCF of 847 and 2160 is 1. OR See worksheet-1, sol. 9.
distance speed Time taken by Abhay to cover one complete round
20. We have time =
360 = 30 hours 12 Time taken by Ravi to cover one complete round =
360 = 24 hours 15 Abhay and Ravi reach the starting point respectively after 30 hours and 24 hours, and their respective multiples. Therefore, they will meet again at the starting point after the time given by least common multiple of 30 hours and 24 hours Let us determine the LCM of 30 hours and 24 hours. 30 = 2 × 3 × 5, 24 = 23 × 3 ⇒ LCM = 23 × 3 × 5 = 120 Hence, the required time is 120 hours.
=
21. Let α and β be the zeroes of 6x2 + x + k. ∴ α2 + β2 =(α + β)2 – 2αβ 2
1 k = – – 2 6 6 =
1 k – 36 3
But it is given that α2 + β2 = ∴
25 1 k = – 36 36 3
⇒
⇒
25 36
k – 24 = 3 36 k = – 2.
OR See worksheet-13, sol. 9.
107
22. To solve a system of equations graphically we need atleast two solutions of each equation. Two solutions of the equation 2x – y = 2 are given in the following table: x
0
3
y
–2
4
24. Draw RM ⊥ PQ. In ∆PRS, ∠PSR > 90°
⇒ PR2 = PS2 + RS2 + 2PS.SM ...(i) [Using result on obtuse-angled triangle]
Two solutions of the equation 4x – y = 8 are given in the following table: x
2
1
y
0
–4
In ∆QRS, ∠QSR < 90° ⇒ QR2 = RS2 + SQ2 – 2SQ.SM
...(ii)
[Using result on acute-angled triangle]
Let us draw the graph of the two given equations.
Add equations (i) and (ii) to get PR2 + QR2 = 4PS2 – 2SM(PS – SQ)
PQ = 4 2 ⇒
2
(∵ PS = SQ)
PR2 + QR2 = PQ2.
Hence proved.
OR See worksheet-41, sol. 7. sin θ =
25. ∵ ∴
From the graph, it is clear that the two lines intersect each other at the point (3, 4). Hence, the solution is x = 3, y = 4. 23. In right-angled triangle PQS, PS2 = PQ2 + QS2 ⇒ QS2 = PS2 – PQ2 2
QR = PS2 – PQ2 4 ⇒ QR2 = 4PS2 – 4PQ2 ... (i) In right-angled triangle PQR, PR2 = PQ2 + QR2 = PQ2 + 4PS2 – 4PQ2 [Using (i)] ⇒ PR2 = 4PS2 – 3PQ2. Hence proved.
⇒
108
3 4
... (i) Given
sin2 θ + cos2 θ = 1 cos2 θ = 1 – sin2 θ
⇒
3 cos2 θ = 1 – 4
⇒
cos2 θ =
7 16
⇒
sec2 θ =
16 7
2
[Using (i)]
... (ii)
Let us take LHS of the given equation. LHS =
cosec 2 θ – cot 2 θ 2
sec θ – 1
=
1 2
sec θ – 1
[... cosec2 θ – cot2 θ = 1] =
1 7 7 = = 16 9 3 –1 7
= RHS. M A T H E M A T
[Using (ii)]
Hence proved. I C S – X
3 1 , cos 60° = , 2 2
26. We have sin 60° =
2
sec 30° = ∴
f1 – f0 Now, mode = l + ×h 2 – – f f f 0 2 i Here, l = 40, fi = 20, f0 = 7, f2 = 15, h = 20
, cosec 30° = 2
3
3 + sin 2 60° + cosec 2 30°
20 – 7 ∴ Mode = 40 + × 20 40 – 7 – 15
5 + cos2 60° + sec 2 30° 2
=
3 2 3 + + ( 2 ) 2 2
1 2 5+ + 2 3
2
3 +4 4 = 1 4 5+ + 4 3
= 40 +
3+
= 54.44 marks. SECTION-D 29. Let f (x) = x4 + x3 – 9x2 – 3x + 18
Multiply Num. and Deno. by 12, 93 36 + 9 + 48 = = . 79 60 + 3 + 16
It is given that – 3 and zeroes of f (x).
27. Let us prepare the cumulative frequency table from the given data. C.I.
fi
xi
di = x i – A
fi d i
0-10 10-20 20-30 30-40 40-50
7 10 15 8 10
5 15 25 35 45
– 20 – 10 0 10 20
– 140 – 100 0 80 200 Σfi di = 40
Σfi = 50 Mean = A +
∑ fi di 40 = 25 + 50 ∑ fi
= 25 + 0.8 = 25.8. 28. Let us convert the more than type distribution to the normal distribution. Marks
No. of students
0-20 20-40 40-60 60-80 80-100
3 7 20 15 5
We observe from the table that the value 20 is the maximum frequency. So, the modal class is 40-60.
P R A C T
I
C E
260 = 40 + 14.44 18
P A P E R S
3 are two of
⇒ x + 3 and x – 3 are the factors of f(x) ⇒ (x + 3 ) (x – 3 ) is a factor of f(x) ⇒ x2 – 3 is a factor of f (x)
)
(
x2 – 3 x4 + x3 – 9x2 – 3x + 18 x2 + x – 6 – 3x2 x4 – + x3 – 6x2 – 3x + 18 x3 – 3x – + – 6x2 + 18 – 6x2 + 18 + – 0 ∴ p(x) = (x2 – 3) (x2 + x – 6) To find other zeroes of p(x) Let x2 + x – 6 = 0 ⇒ (x+ 3) (x – 2) = 0 ⇒ x = – 3 and 2 So all zeroes are ± 3 , 2, – 3. Hence, the zeroes of f(x) are ± 3 , 2, – 3. 30. Let the speed of rowing in still water and the speed of the current be u km/hr and v km/hr respectively. The speed of rowing in downstream = (u + v) km/hr. The speed of rowing upstream = (u – v) km/hr.
109
Using the formula: Distance = Time Speed
According to first condition of the question, 12 18 + =3 ... (i) u+v u – v According to second condition of the question, 40 36 + =8 u–v u+v Let us put,
... (ii)
1 1 = x and = y such that u+v u–v equations (i) and (ii) reduce to
18x + 12y = 3 ... (iii) And 36x + 40y =8 ... (iv) Equations (iii) and (iv) form a pair of linear equations. Multiply equation (iii) by 2 and subtract the result from equation (iv) to get 1 16y = 2 ⇒ y = 8 1 Substitute y = in equation (iv) to get 8 1 36x = 8 – 5 ⇒ x = 12 1 ∵x= ∴ u + v = 12 u+v ∵y=
1 u–v
CA AB BC = = RP PQ QR
...(i)
We have to prove ar ( ∆PQR )
110
=
AB2 PQ 2
=
BC2 QR 2
=
BC AM AB AM = ⇒ = ...(ii) [Using (i)] PN PN QR PQ
Since area of a 1 triangle = × base × height 2
1 × BC × AM ∴ = 2 1 ar ( ∆PQR ) × QR × PN 2 ar ( ∆ABC )
BC AM × QR PN Using equation (ii), we get =
ar ( ∆ABC ) ar ( ∆PQR )
CA2 RP2
=
BC2
... (iii)
QR 2
From equations (i) and (iii), we obtain ar ( ∆PQR )
31. We are given two triangles ABC and PQR such that ∆ABC ~ ∆PQR.
ar ( ∆ABC )
∴
ar ( ∆ABC )
∴ u–v=8
This last system gives u = 10 and v = 2 Hence, the speed of the rowing in still water = 10 km/hr and the speed of the current = 2 km/hr.
∴
Draw AM ⊥ BC and PN ⊥ QR In ∆ABM and ∆PQN, ∠B = ∠Q (∵ ∆ABC ~ ∆PQR) ∠M = ∠N (Each 90°) So, ∆ABM ~ ∆PQN (AA similarity criterion)
=
AB 2 PQ 2
=
BC2 QR 2
=
CA2 RP2
.
Hence proved. 32. Consider, left hand side of the given equation,
1 tan θ tan θ cot θ tan θ + = + 1 1 – tan θ 1 – cot θ 1 – tan θ 1– tan θ =
tan 2 θ 1 + tan θ – 1 tan θ (1 – tan θ )
=
tan 3 – 1 tan θ ( tan θ – 1)
M A T H E M A T
I C S – X
( tan θ – 1) ( tan 2 θ + 1 + tan θ ) = tan θ ( tan θ – 1) [∵ a3 – b3 = (a – b) (a2 + b2 + ab] = =
tan θ +
1 sin θ cos θ +1 = + +1 tan θ cos θ sin θ
sin 2 θ + cos2 θ + 1 = sec θ. cosec θ + 1 sin θ cos θ
= RHS.
Hence proved.
Adding equations (v) and (vi), we get (m2n)2/3 + (n2m)2/3 = cos2 θ + sin2 θ i.e., (m2n)2/3 + (n2m)2/3 = 1. Hence proved. 34. Let us prepare the cumulative frequency distribution from the given distribution. Class interval (C.I.) 0-10 10-20 20-30 30-40 40-50 50-60
OR See Assessment sheet-9, sol. 8. 33.
m = cosec θ – sin θ
(Given) 2
⇒ m=
1 1 – sin θ – sin θ = sin θ sin θ
⇒ m=
cos2 θ sin θ
... (i) (∵ 1 – sin2 θ = cos2 θ)
⇒ m2 =
cos 4 θ
... (ii) (Squaring) sin 2 θ Further, n = sec θ – cos θ (Given)
1 – cos2 θ cos θ
⇒ n=
sin 2 θ cos θ
⇒ n2 =
... (iii)
sin 4 θ
... (iv) cos2 θ Multiplying equations (ii) and (iii), we get m2n
cos 4 θ
sin 2 θ × = = cos3 θ sin 2 θ cos θ
... (v) ∴ (m2n)2/3 = cos2 θ Multiplying equations (i) and (iv), we get
cos2 θ = sin3 θ sin θ cos2 θ ∴ (n2m)2/3 = sin2 θ n2m =
sin 4 θ
P R A C T
I
×
C E
... (vi)
P A P E R S
( f)
Cumulative frequency (c f )
5 5 x 5+x 20 25 + x 14 39 + x y 39 + x + y 8 47 + x + y N = 47 + x + y
Since 27 is the median, therefore, 20-30 is the median class. ∴ l = 20, cf = 5 + x, f = 20, h = 10 Given: N = 68 Using the formula:
N – cf 2 Median = l + ×h f
1 – cos θ ⇒ n= cos θ
⇒ n=
Frequency
⇒
68 –5– x 2 27 = 20 + ×10 20
⇒
27 = 20 +
⇒
29 – x 2
29 – x = (27 – 20) × 2
⇒ x = 15
From the table, N = 47 + x + y But N = 68 (Given) ∴ x + y + 47 = 68 ⇒ 15 + y + 47 = 68 (Substituting x = 15) ⇒ y=6 Thus, x = 15, y = 6. OR See worksheet-65, sol. 7.
111
Practice Paper –3
B + C 180° – A Now, tan = tan 2 2
SECTION-A 1. (D) There are infinitely many real numbers of both types rational and irrational between 3 and 5 . 2. (D) Decimal representation of an irrational number is always non-terminating, nonrepeating.
6 =4 a 6 3 ⇒ a= = . 4 2 4. (B) For infinite number of solutions: 5 3 = , i.e., k = 6. 10 k 5. (A) ∆ABC ~ ∆DEF
A A = tan 90° – = cot . 2 2 10. (B) Given expression = cos (40° + θ) – cos {90°– (50°– θ)} +
3. (B) Product of zeroes =
∠B = ∠E =180° – (40° + 65°) = 75°. 6. (C) AC2 = BC2 – AB2 ⇒ AC2 = 2 – 1 = 1 AC = 1 BC 2 = AB 1 = 2.
+
cos2 40° + sin 2 40° 2
2
sin 40° + cos 40°
=0+
1 = 1. 1
SECTION-B 11. 8n can be rewritten as 23n. Clearly, the prime factor of 8n is only 2. To end with the digit 0, one of the prime factors of 8n must be 5. Hence, 8n cannot end with the digit zero for any n∈N. 12. True, because we find the remainder zero when 3x4 + 5x3 – 7x2 + 2x + 2 is divided by x2 + 3x + 1. 13. Infinite number of solutions because the system obeys the following condition:
14. Yes, because converse of Basic Proportionality Theorem holds as
sin θ (1 – cos θ ) 2
sin θ
=
1 – cos θ . sin θ
8. (D) Using empirical formula, 2 Mean = 3 Median – Mode ⇒ 2 Mean = 3 × 500 – 400 1100 ⇒ Mean = = 550. 2 9. (A) A + B + C = 180° (Angle sum property of a triangle)
112
= cos (40°+ θ) – cos (40°+ θ)
sin θ sin θ 1 – cos θ = × 1 + cos θ 1 + cos θ 1 – cos θ =
sin 2 40° + sin 2 (90° – 40°)
–3 –3 1 a1 b c = 1 = 1 , i.e., = = . – 9 –9 3 b2 a2 c2
cosec C =
7. (C)
cos2 40° + cos2 (90° – 40°)
PM PN 3 = = . MQ NR 2 15. In ∆AOD and ∆COB,
1 AO DO = = 2 OC OB and∠AOD = ∠COB ⇒ ∆AOD ~ ∆COB ∴ ⇒
AD 1 4 1 = ⇒ = BC BC 2 2
BC = 8 cm.
M A T H E M A T
I C S – X
16. No. 1 and cot θ = 3 3 10 1 tan2 θ + cot2 θ = + 3 = ≠ 2. 3 3
For θ = 30°, tan θ =
17. Since the maximum frequency is 8, so modal class is 4-8.
From (i) and (iv), we get 4c2 = 2b2 ⇒ b2 = 2c2 ⇒ b2 is divisible by 2 ⇒ b is divisible by 2
∴ l = 4, f1 = 8, f0 = 4, f2 = 5, h = 4 f1 – f0 Now, mode = l + ×h 2 f1 – f0 – f2 8–4 =4 + ×4 16 – 4 – 5
= 4 + 2.29 = 6.29. OR See worksheet-65, sol. 4. 18. Let us prepare cumulative frequency table: C.I.
f
cf
60-70 70-80 80-90 90-100 100-110 110-120 120-130
2 5 12 31 39 10 4
2 7 19 50 89 99 103
2 is 2 can be written
a rational number. Then a as 2 = , where a and b are coprime b and b ≠ 0.
P R A C T
I
C E
(Squaring) ... (i)
P A P E R S
a = 3q
∴
a3 = (3q)3 = 27q3 = 9 (3q3) = 9m,
where
m = 3q3.
Case II:
19. Let us assume on the contrary that
⇒
that 2 is an irrational number.
Case I:
SECTION-B
a b2 a2 = 2b2
From results (ii) and (v), we have a and b both are divisible by 2. But this contradict the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that 2 is a rational number. Thus, we conclude
20. Let a be a positive integer. Then it is of the form 3q, 3q + 1 or 3q + 2. Now, three cases arise.
103 N = = 51.5 2 2 Cumulative frequency just greater than 51.5 is 89. So, median class is 100-110.
2=
.... (v)
OR See worksheet-4, sol. 9.
N = 103
2
⇒ a2 is divisible by 2 ...(ii) ⇒ a is divisible by 2 ...(iii) [If a prime (here 2) divides d2, then the same prime divides d, where d is a positive integer.] ⇒ a = 2c ⇒ a2 = 4c2 ... (iv)
∴
a = 3q + 1 a3 = (3q + 1)3 = 27q3 + 1 + 27q2 + 9q = 9 (3q3 + 3q2 + q) + 1 = 9m + 1,
where Case III: ∴
m = 3q3 + 3q2 + q. a = 3q + 2 a3 = (3q + 2)3 = 27q3 + 8 + 54q2 + 36q = 9 (3q3 + 6q2 + 4) + 8
= 9m + 8, where m = 3q3 + 6q2 + 4. Hence, a3 is of the form 9m, 9m + 1 or 9m + 8.
113
21. p(t) = t2 – 15 To obtain zeroes of p(t), put p(t) = 0 i.e., t2 – 15 = 0 t2 – ⇒
(t +
15
(
)(t –
⇒
Adiagonal =
2
15
Sum of zeroes = – 15 + 15 = 0 =
–0 – Coefficient of t = 1 Coefficient of t2
Product of zeroes = – 15 × 15 = – 15 =
–15 1
Constant term . Coefficient of t Hence verified. 22. The given system of equations can be rewritten as 4x + 3y – 48 = 0 40x – 6y – 192 = 0 Applying the method of cross multiplication to solve the system. –y 1 x ⇒ = = – 768 + 1920 – 24 – 120 – 576 – 288 =
⇒ ⇒ ⇒
y 1 x = = – 1152 – 144 – 864 1152 864 and y = 144 144 x = 6 and y = 8.
x=
23. We are given a square of side length a. Then length of its diagonal will be a 2 . We know that area of an equilateral 3 2 x . 4 ∴ Area of the equilateral triangle described of the side of the square.
triangle of side length x is
Aside =
4 3 2 A a ... (i) ⇒ a2 = 3 side 4
And area of the equilateral triangle described on the diagonal of the square
114
[Using (i)]
1 × Adiagonal . Hence proved. 2 24. In the figure drawn, AB || DC and ∆AED ~ ∆BEC. ∆ADC and ∆BDC both are on the same base DC and lie between same parallels AB and DC. So, ⇒
t = – 15 , 15
So, zeroes of p(t) are – 15 and
3 4 Aside × 2 3
⇒ Adiagonal =
) =0 15 ) = 0
15
3 3 2 a (a 2 )2 = 4 2
Aside =
ar(∆ADC) = ar(∆BDC) ⇒ ar(∆AED) + ar(∆DEC) = ar(∆BEC) + ar(∆DEC) ⇒ ar(∆AED) = ar(∆BEC) (i) Now,
ar(∆AED) (AD)2 = ar(∆BEC) (BC)2 (∵ ∆AED ~ ∆BEC) ⇒
1=
⇒
( AD )2 (BC )2
[From (i)]
AD = BC.
Hence proved.
25. Given expression
1 4 1 4 = 2 + – 2 2
{(
3
)
2
2
+ (1)
}
2 + 3 3
2
1 4 1 = 2 + – ( 3 + 1) + 3 × 16 16 3
=
1 1 –4+4= . 4 4
26. Consider left hand side of the given equation. LHS = (cosec A – sin A) (sec A – cos A) 1 – sin A = sin A
=
1 – cos A cos A
1 – sin 2 A 1 – cos2 A . sin A cos A
M A T H E M A T
I C S – X
N = 40 2 Cumulative frequency just more than 40 is 58. So 24-32 is the median class. ∴ l = 24, cf = 34, f = 24, h = 8
cos2 A sin 2 A . = sin A cos A sin A cos A Also, taking right hand side,
Here, N = 80
=
1 1 = sin A cos A tan A + cot A + cos A sin A sin A cos A = = sin A cos A sin 2 A + cos2 A
RHS =
Hence, LHS = RHS. OR See worksheet-51, solution-9. 27. Let us use step-deviation method to obtain the mean. C.I.
fi
xi
0-20 20-40 40-60 60-80 80-100
17 p 32 24 19
10 30 50 70 90
di = xi – a u i = – 40 – 20 0 20 40
–2 –1 0 1 2
di u i – 34 –p 0 24 38
Σfi ui = 28 – p
Σfi = 92+p Here
di h
a = 50, h = 20 ∑ fi ui Mean = a + ×h ∑ fi 28 – p 50 = 50 + 92 + p
⇒ ⇒
28 – p = 0
∴
N – cf ∴ Median = l + 2 ×h f 48 40 – 34 = 24 + = 26. × 8 = 24 + 24 24
OR See Assessment sheet-12, sol. 6. SECTION-D 29. Let f (x) = 3x4 + 6x3 – 2x2 – 10x – 5 x=
5 5 and x = – are the zeroes of f (x) 3 3
5 5 ⇒ x– and x + are factors of f (x) 3 3 5 5 5 ⇒ x– x+ = x2 – is a factor of f (x) 3 3 3
Let us divide f (x) by x2 –
5 3
∴ p = 28.
28. We prepare cumulative frequency table from the given data. C.I.
Frequency ( f)
Cumulative Frequency (cf )
8 10 16 24 15 7
8 18 34 58 73 80
0-8 8-16 16-24 24-32 32-40 40-48
i.e.,
N = 80
P R A C T
I
C E
P A P E R S
To obtain other two zeroes, put the quotient = 0 3x2 + 6x + 3 = 0
115
⇒ ⇒ ⇒
3 (x2 + 2x + 1) = 0 3 (x + 1)2 = 0 x = – 1 or – 1
Hence, all the zeroes of f(x) are ± 5 , 3 – 1, –1.
x . y On adding 1 to both the numerator and
30. Let the original fraction be
x 4 the denominator of , it becomes y 5 i.e., i.e.,
x +1 4 = , i.e., 5x + 5 = 4y + 4 5 y +1 5x – 4y = –1 ...(i)
On subtracting 5 from both the numerator
1 x and the denominator of , it becomes y 2 i.e.,
x–5 1 = , y–5 2
i.e., 2x – 10 = y – 5
i.e.,
8x – 4y = 20
...(ii)
Subtracting equation (i) from equation (ii), we get 3x = 21 ⇒ x = 7 Substituting x = 7 in equation (i), we get 5 × 7 – 4y = –1 ⇒ y = 9 Hence, the required fraction is OR See worksheet-27, sol. 9 (OR).
7 . 9
31. Pythagoras Theorem: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Proof: We are given, a ∆ABC in which ∠A = 90° We need to prove BC2 = AB2 + AC2. Draw AD ⊥ BC (see figure) In ∆ABC and ∆DBA, ∠ABC = ∠DBA (Common) ∠BAC = ∠BDA (Each 90°)
116
So, ∆ABC ~ ∆DBA (A A criterion of similarity) Therefore,
AB BC = BD AB
AB2 = BD . BC
or
...(i)
Similarly, ∆ABC ~ ∆DAC
AC BC = AC DC AB2 = DC . BC
Therefore,
...(ii)
Adding equations (i) and (ii), we get BD . BC + DC . BC = AB2 + AC2 or (BD + DC) . BC = AB2 + AC2 or BC . BC = AB2 + AC2 (∵ BC = BD + DC) BC2
or
=
AB2
+ AC2. Hence proved.
OR See worksheet-38, sol. 9. 32. We have
x3 = sec A – cos A =
1 1 – cos2 A – cos A = cos A cos A 1
2 3 x = sin A cos A
∴
Similarly, y3 = cosec A – sin A =
1 – sin 2 A 1 – sin A = sin A sin A 1
2 3 y = cos A sin A
LHS = x2 y2 (x2 + y2) = x4y2 + x2y4 1 sin 2 A 3 = cos A
4
1 2 cos A 3 sin A
M A T H E M A T
2
I C S – X
1 sin 2 A 3 + cos A 4 sin 2 A 3
2
1 2 cos A 3 sin A
4
2 cos2 A 3
= cos A sin A 2
4
sin 2 A 3 cos2 A 3 + cos A sin A 1 cos4 A 3
sin 8 A × = 4 2 cos A sin A
1 cos8 A 3
sin 4 A × + 2 4 cos A sin A 1
1
(sin6 A )3 + (cos6 A )3
=
= sin2A + cos2 A = 1 = RHS. Hence proved. sec θ = x +
33. or
1 4x
... (i) (Given)
1 sec2 θ = x + 4x
2
(Squaring)
tan2 θ = x2 +
or
tan2
or
1 1 – 2 2 16x
1 θ = x – x 4
2
1 tan θ = ± x – 4x
... (ii)
Add equations (i) and (ii) to get sec θ + tan θ = x +
P R A C T
I
C E
x+
1 1 1 –x+ = 2x or . 4x 4x 2x Hence proved.
34. We prepare the cumulative frequency table by less than method as given below: Marks Frequ- Marks ency less than
1 1 +x– 4x 4x
P A P E R S
cf
Point
0-10
4
10
04
(10, 4)
10-20
10
20
14
(20, 14)
20-30
16
30
30
(30, 30)
30-40
22
40
52
(40, 52)
40-50
20
50
72
(50, 72)
50-60
18
60
90
(60, 90)
60-70
8
70
98
(70, 98)
70-80
2
80
100
(80, 100)
We take marks on the x-axis and cumulative frequency on the y-axis and then plot the points mentioned in the table. On joining these points by free hand smooth curve, we get less than ogive. Further, we prepare the cumulative frequency table by more than method as given below: Marks Frequ- Marks ency more than or equal to
1 1 + or 1 + tan2 θ = x2 + 16 x 2 2
or
or
cf
Point
0-10
4
0
100
(0, 100)
10-20
10
10
96
(10, 96)
20-30
16
20
86
(20, 86)
30-40
22
30
70
(30, 70)
40-50
20
40
48
(40, 48)
50-60
18
50
28
(50, 28)
60-70 70-80
8 2
60 70
10 2
(60, 10) (70, 2)
117
We will plot the points mentioned in this table on the same graph. On joining these points by free hand smooth curve, we get more than ogive.
Practice Paper–4 SECTION-A 1. (D)
2− 3 = 2+ 3
Y
=
105 100 95
(0, 100) (10, 96)
90 85
(20, 86)
(30, 70)
3. (B) p(x) = (x – 1)2 + 2x + 1 = x2 + 2
(50, 72)
Sum of zeroes =
60
2 = 2. 1 4. (C) 2 + k = 1 and p + 3 = 2, i.e., k = – 1 and p = – 1. So, p + k = – 2.
(40, 52) (40, 48)
45 40
5. (C)
35 (30, 30)
0 = 0; 1
product of zeroes =
55
30
2+3−2 6 2−3
n l ∴ HCF (x, y) is p2 p3 .
65
50
2− 3
2. (D) As p2 and p3 are common factors.
less than ogive
75 70
2− 3
As 2 6 − 5 is irrational, it has non-terminating, non-repeating decimal form.
(60, 90)
80
2+ 3
×
= 2 6 −5
(80, 100)
(70, 98)
2− 3
(50, 28)
⇒
25 20
AD AE 1 = = BD 3 CE
DE || BC
more than ogive
15 10 5 0
(20, 14) (10, 4)
(60, 10) (39, 0)
(70, 2)
10 20 30 40 50 60 70 80 90 100
X
Median: The abscissa of the point of intersection of both the ogives determines the median of the given distribution. To find such abscissa, we draw a perpendicular from the point of intersection of both the ogives to the x-axis, which meet the axis at (39, 0). Hence the required median is 39 marks.
118
⇒ ∆ADE ∼ ∆ABC ∴
DE AD = BC AB
⇒
x 3.5 = 7.5 14
⇒
x=
15 cm. 8
M A T H E M A T
I C S – X
6. (D) If x = 30°, 3 cos 30° – 4 cos3 30° =3×
3 3 3 –4× = 0. 2 8
7. (A) The value of cosec θ is minimum at an angle where sin θ is maximum as sin θ is maximum if θ = 90°. ∴ sin 90° = 1 So minimum value of cosec θ = 1. 8. (B) 52 is the median as the abscissa of the point of intersection of the two ogives determines the median of the data. 9. (A) We know that sin (90° – θ) = cos θ, cos (90° – θ) = sin θ, tan (90° – θ) = cot θ ∴ Given expression cot θ sin θ cos θ tan θ = + sin θ cos θ tan θ cot θ = 1 + 1 = 2. 10. (B) tan 2θ = cot (θ + 12°) ⇒ tan 2θ = tan {90° – (θ + 12°)} [∵ tan (90° – α) = cot α] ⇒ tan 2θ = tan (78° – θ) ⇒ 2θ = 78° – θ ⇒ θ = 26°. SECTION-B 11. The required highest number will be the HCF of 120, 224 and 256. 120 = 23 × 3 × 5; 224 = 25 × 7; 256 = 28 Therefore, HCF = 23 = 8. 12. Let the required polynomial be f(x). Then f (x) = k[x2 – (sum of zeroes) x + product of zeroes] = k[x2 – (–5 + 2)x + (– 5)(2)] = k(x2 + 3x – 10), k being a real number. f(x) is not a unique polynomial as k is any real number. 13. Condition for infinite number of solutions:
b a1 c = 1 = 1 c2 a2 b2
P R A C T
I
C E
2k + 1 4(k − 1) 2k + 3 = , = 4 3 3 5 k= . 2
i.e.,
P A P E R S
i.e., 14. Yes. In ∆ ABC,
1 AD AE = = EC 3 DB So, by converse of Thales Theorem, DE || BC. 15. No. In ∆AOB and ∆DOC, ∠AOB = ∠DOC (Vertically opposite angles)
AO OB ≠ DO OC 5 3 5 6 , i.e., ≠ ≠ 3 5 3 10 Therefore, ∆AOB is not similar to ∆DOC.
as
16. Yes. Let us take left hand side of the given equation. LHS =
1 + sin 2 θ cos 2 θ
=
1 cos 2 θ
+
sin 2 θ cos 2 θ
= sec2 θ + tan2 θ = (1 + tan θ)2 + tan2 θ = 1 + 2 tan2 θ = RHS. OR See worksheet-54, sol. 5. 17. Let us convert the given more than type distribution to the normal distribution. Marks
No. of students
0-20 20-40 40-60 60-80 80-100
3 7 20 15 5
119
The maximum valuable frequency is 20. So, modal class is 40-60.
Case I: When n = 3q, n + 1 = 3q + 1 and n + 2 = 3q + 2.
∴ l = 40, f1 = 20, f0 = 7, f2 = 15, h = 20
Here, only n is divisible by 3.
f1 − f0 Now, mode = l + ×h 2 f1 − f0 − f2
Here, only n + 2 is divisible by 3.
20 − 7 = 40 + × 20. 40 − 7 − 15
Case III: When n = 3q + 2,
= 40 + 14.44
and n + 2 = 3q + 5 = 3(q + 1) + 2 Hence only n + 1 is divisible by 3.
= 54.44 marks.
n + 1 = 3q + 3 = 3(q + 1)
OR See worksheet-4, sol. 11.
18. As N = 50 ∴
Case II: When n = 3q + 1, n + 1 = 3q + 2 and n + 2 = 3q + 3 = 3(q + 1)
N = 25 2
∴ Draw a line parallel to x-axis from y = 25 which meet to ogive at points P(40, 25). ∴ Abscissa of P is 40. ⇒ Median = 40. SECTION-C
21 By the division algorithm, Dividend = Divisor × Quotient + Remainder 3 2 ∴ 6x + 8x – 3x + 8 = g(x) × (3x + 4) + 6x + 20 ⇒
g(x) =
19. Let us assume to the contrary that 3 − 4 5 is rational. Then we can take integers a and b ≠ 0 such that a = 3– 4 5, b
i.e.,
4 5 = 3–
i.e.,
5=
a b
3b − a 4b
RHS of this last equation is rational as a and b are integers, but LHS of it is irrational. This is an incorrect statement due to our wrong assumption that 3 − 4 5 is rational.
2x 2 (3x + 4) − 3(3x + 4) 3x + 4
(3 x + 4)(2 x 2 − 3) 3x + 4
⇒
g(x) =
⇒
g(x) = 2x2 – 3.
22. To draw a line, we need atleast two solutions of its corresponding equation. x
0
2
x
1
0
y
–4
0
y
0
–1
Two solutions of
Two solutions of
2x – y = 4
x–y=1
So, we conclude that 3 − 4 5 is
From the graph, the two lines intersect each other at the point A(3, 2).
irrational.
∴ x = 3 and y = 2
20. Any positive integer is either of the form 3q, 3q + 1 or 3q + 2. There are 3 cases now:
120
=
6 x 3 + 8 x 2 − 9x − 12 3x + 4
Shaded region is ∆ABC. ∴ Height of ∆ABC = 3 units And its base = BC = 3 units. M A T H E M A T
I C S – X
24. Let median AD passes through the point O on PQ. To prove: PO = QO Proof: In ∆APO and ∆ABD, ∠PAO = ∠BAD
(Common angle)
∠APO = ∠ABD and ∠AOD = ∠ADB
Corresponding angles
∴ ∆APO ~ ∠ABD (By AAA similarity)
PO AO = BD AD Similarly, In ∆AQO and ∆ACD, ⇒
⇒ ∴
QO AO = CD AD
...(ii)
From equations (i) and (ii), we have
1 ×3×3 2 9 = square units. 2
ar(∆ABC) =
OR See worksheet-29, sol. 9. 23. Let us draw MN parallel to AB, which passes through P. So, AM = BN and DM = CN.
...(i)
PO QO = BD CD
⇒ PO = QO (... BD = CD) Hence, median AD also bisects PQ. Proved. OR See worksheet-44, sol. 5. 25.
cos θ − sin θ 1 − 3 = cos θ + sin θ 1 + 3 Using componendo and dividendo, we get
From right-angled triangles APM, BPN, CPN, DPM; we have respectively PA2 = PM2 + AM2 .... (i) PB2 = PN2 + BN2 .... (ii) PC2 = PN2 + CN2 .... (iii) PD2 = PM2 + DM2 .... (iv) From equations (i) and (ii), PA2 – PB2 = PM2 – PN2 .... (v) From equations (iii) and (iv), PC2 – PD2 = PN2 – PM2 .... (vi) Add equations (v) and (vi) to get PA2 + PC2 = PB2 + PD2. Hence proved.
P R A C T
I
C E
P A P E R S
⇒ ⇒ ⇒
cos θ − sin θ + cos θ + sin θ cos θ − sin θ − cos θ − sin θ 1− 3 + 1+ 3 = 1− 3 −1− 3 2 2 cos θ = −2 3 − 2 sin θ 1 cot θ = 3 cot θ = cot 60°
⇒
θ = 60°.
26. LHS =
2 2
cos θ
−
1 4
cos θ
−
2 2
sin θ
+
1 sin 4 θ
= 2 sec2 θ – sec4 θ – 2 cosec2 θ + cosec4 θ
121
= 2(1 + tan2 θ) – (1 + tan2 θ)2 – 2(1 + cot2 θ) + (1 + cot2 θ)2 = (1 + tan2 θ) × (2 – 1 – tan2 θ) – (1 + cot2 θ) × (2 – 1 – cot2 θ) = 1 – tan4 θ – (1 – cot4 θ) = cot4 θ – tan4 θ = RHS. Hence proved. 27. We convert the given data of less than type to the normal distribution. Marks
fi
xi
di = xi – 25
So,
cf = 200, f = 115, l = 3000.
Median
N − cf 2 = l+ f
250 − 200 = 3000 + × 500 115
SECTION-D
5
5
– 20
– 100
11
15
– 10
– 110
20-30
19
25
0
0
30-40
30
35
10
300
40-50
15
45
20
300 Σfidi = 390
Let us use the assumed mean method with assumed mean a = 25.
Σfi di 390 = 25 + 80 Σfi
= 3217.39 hours.
29. Let f (x) = 6x4 + 8x3 – 5x2 + ax + b f (x) is divisible by 2x2 – 5. ⇒ We obtain the remainder as zero when f (x) is divided by 2x2 – 5. Now we divide f (x) by 2x2 – 5. 3x 2 + 4 x + 5 2x 2 − 5 6 x 4 + 8 x 3 − 5x 2 + ax + b 6 x 4 − 15 x2 − + _____________________________ 8 x 3 + 10 x 2 + ax + b
= 25 + 4.875 = 29.88 marks. 28. We prepare the cumulative frequency table for the given data. Lifetimes (in hrs.)
Frequency (f) 24
24
2000-2500
86
110
2500-3000
90
200
3000-3500
115
315
3500-4000
95
410
4000-4500
72
482
4500-5000
18
500
Here,
122
h = 500
−
8x 3
− 20 x +
_________________________________ 2
10 x + ( a + 20) x + b
Cumulative frequency (cf )
1500-2000
N = 500
×h
= 3000 + 217.39
0-10
Now, mean = a +
N = 500, ∴
f i di
10-20
Σfi = 80
N = 250. 2
∵
10 x 2 − 25 − +
_______________________________
( a + 20) x + b + 25
_____________________
As remainder = 0, (a + 20)x + b + 25 = 0 i.e.,
(a + 20)x + (b + 25) = 0x + 0
i.e., a + 20 = 0 and b + 25 = 0 i.e.,
a = – 20 and b = – 25.
30. Let the present ages of father and his son are x years and y years respectively. According to the given conditions: x + y = 65 M A T H E M A T
I C S – X
After 5 years, the father’s age = (x + 5) years After 5 years, the son’s age = (y + 5) years Therefore, x + 5 = 2 (y + 5) i.e., x – 2y = 5 Thus, the required pair of linear equations is x + y = 65 .... (i) x – 2y = 5 .... (ii) Subtracting equation (ii) from equation (i), we get 3y = 60 ⇒ y = 20 Substituting y = 20 in equation (i), we get x + 20 = 65 ⇒ x = 45 Thus, present age of father = 45 years and present age of his son = 20 years. 31. We are given a ∆ABC in which a line PQ || BC intersects the sides AB at P and AC at Q. We need to prove
AP AQ . = PB QC
Draw QM ⊥ AB and PN ⊥ AC.
Join PC and BQ. Proof: Area of a triangle = ∴
P R A C T
1 × Base × Height 2
ar(∆APQ) =
1 × AP × QM 2
=
1 × AQ × PN 2
I
C E
P A P E R S
⇒
AP × QM = AQ × PN
⇒
QM AQ = AP PN
... (i)
Since ∆BPQ and ∆CQP are on the same base PQ and between the same parallels PQ and BC, therefore, their areas should be equal. i.e., ar(∆BPQ) = ar(∆CQP)
1 1 × PB × QM = × QC × PN 2 2
⇒
QC QM = PB PN
⇒
... (ii)
Form equations (i) and (ii), it is clear that QC AQ = , PB AP AP AQ = . PB QC
i.e.,
Hence proved. 32.
cot x 1 – tan x
+
tan x 1 – cot x
= 1 + sec x . cosec x
LHS cos x
=
cot x 1 – tan x
+
tan x 1 – cot x
=
sin x
sin x cos x + sin x cos x 1– 1– cos x sin x
=
cos x cos x sin x sin x + × × sin x cos x – sin x cos x sin x – cos x
=
cos 2 x sin 2 x + sin x (cos x – sin x) cos x (sin x – cos x)
=
cos3 x – sin 3 x sin x . cos x (cos x – sin x)
=
(cos x – sin x)(cos2 x + cos x . sin x + sin 2 x) sin x . cos x (cos x – sin x)
1 + cos x . sin x 1 = +1 sin x . cos x sin x . cos x = 1 + sec x . cosec x = RHS. Hence proved. =
123
33. It is given that sin θ + cos θ = a Squaring both the sides, we get sin2 θ + cos2 θ + 2 sin θ cos θ = a2 But sin2 θ + cos2 θ = 1 ∴ 1 + 2 sin θ cos θ = a2 ⇒ 2 sin θ cos θ = a2 – 1
34. We prepare a table for less than series with corresponding cumulative frequencies and points. .... (i)
a2 − 1 .... (ii) ⇒ sin θ cos θ = 2 Cubing both the sides of equation (i), we get sin6 θ + cos6 θ + 3 sin2 θ cos2 θ (sin2 θ + cos2 θ) = 1 6 ⇒ sin θ + cos6 θ + 3 (sin θ cos θ)2 = 1 [Using (i)]
Marks
Frequency
Marks Cumuless lative than frequency
Point
0-10
5
10
5
(10, 5)
10-20
8
20
13
(20, 13)
20-30
6
30
19
(30, 19)
30-40
10
40
29
(40, 29)
40-50
6
50
35
(50, 35)
50-60
5
60
40
(60, 40)
2
⇒
sin6
θ+
cos6
a 2 − 1 θ+3 =1 2
⇒ sin6 θ + cos6 θ = 1 – 3 × ⇒ sin6 θ + cos6 θ =
( a2 − 1)2 4
4 − 3( a 2 − 1)2 . 4 Hence proved.
OR Consider a sin3 α + b cos3 α = sin α . cos α ⇒ a sin α . sin2 α + b cos α . cos2 α = sin α . cos α 2 ⇒ b cos α . sin α + b cos α . cos2 α = sin α . cos α . . ( . a sin α = b cos α) ⇒ b sin2 α + b cos2 α = sin α ⇒ b (sin2 α + cos2 α) = sin α ∴ b = sin α ...(i) Again, a sin α = b cos α ⇒ a . b = b cos α [From (i)] ∴ a = cos α ...(ii) Now, squaring and adding equations (i) and (ii), we get b2 + a2 = sin2 α + cos2 α ∴ a2 + b2 = 1. Hence proved.
124
We take upper limits on the x-axis and cumulative frequencies on the y-axis. Then we plot the points on the graph paper. By joining these points by free hand smooth curve, we obtain less than ogive as shown in the above figure. OR See worksheet-62, sol. 7. M A T H E M A T
I C S – X
Practice Paper–5 SECTION-A
6. (B)
DE + EF + FD EF = AB + BC + CA BC
1. (D) 2 and 5. x + y is rational x + y = z; when z is
2. (B) Let us assume that number and let rational.
x + y + 2 xy = z2
⇒
2 xy = z2 – x – y
⇒
xy =
z2 – x – y 2
which given a contradiction as LHS is irrational but RHS is rational. ⇒
x+
y is an irrational number.
3. (C) Coincident lines is given by b c a1 = 1 = 1 b2 a2 c2
Here, a1 = p, b1 = q, c1 = – 4 a2 = 4, b2 = 3, c2 = – 5
p q –4 = = 4 3 –5 p q = = 0.8 4 3 p = 3.2, q = 2.4
Now, ⇒ ⇒
4. (A)
18
=
6 ≠ 24
p=
⇒
p = 3.
50 75
So, sin 30° – cos 2 × 30° = sin 30° – cos 60°
sin A =
4 CD = . AC 5
cos C =
4 CD = 5 AC
C E
16 4 4 . × = 25 5 5
SECTION-B 11. True, because product of an even number and an odd number is an even number.
1 1 – = 0. 2 2
I
1 or cos θ = q) q
AC = 32 + 42 = 5
sin A cos C =
∴ A = 30°
P R A C T
8. (C) Relation among mean, median and mode is given by the following impirical formula: Mode = 3 Median – 2 Mean. p sin θ 9. (C) tan θ = = . q cos θ
Now,
5. (B) Consider 3A = 90°
=
7. (C) cos A + cos2 A = 1 ⇒ cos A = 1 – cos2 A ⇒ cos A = sin2 A ⇒ cos2 A = sin4 A (Squaring) ⇒ 1 – sin2 A = sin4 A sin4 A + sin2 A = 1.
10. (B) In right triangle ADC,
6 18 3 24
⇒
4 × (3 + 2 + 2.5) 2
⇒ Perimeter of ∆DEF = 15 cm.
(... sin θ = p, sec θ =
Therefore, 3p + q = 12. 3p
⇒ Perimeter of ∆DEF =
P A P E R S
125
12. Let p(x) = ax2 + bx + c So, y =
ax2
⇒ ∆ABC ~ ∆DFE
+ bx + c should be satisfied by
But ∠B = 60°, so, ∠F = 60°.
(–1, 0), (0, –3) and (4, 0) Therefore,
⇒ ∠B = ∠F
0 = a – b + c; – 3 = c;
16. True,
0 = 16a + 4b + c ⇒
c = – 3, a =
9 3 ,b=– . 4 4
3 2 9 x – x –3 4 4 3 2 (x – 3x – 4) ⇒ p(x) = 4 This is the required expression.
Hence,
p(x) =
sin B =
AC AB
a1 c b = 1= 1 c2 a2 b2
sin Q =
PR PQ
a + b 18 2a = = , 5 2 9
sin B = sin Q
13. For infinite number of solutions,
i.e., i.e.,
a = 2 and a + b = 10
i.e.,
a = 2, b = 8.
14. In ∆ADP, and ∆EBQ, AD = EB
∠DAP = ∠BEQ (Corresponding angles) Therefore, ∆ADP ≅ ∆EBQ (ASA criterion) ∴ DP = BQ.
So, DPQB is a parallelogram. Therefore, PQ || DB. Hence, PQ || AB. 15.
126
BC AB CA 1 = = = EF DF DE 2
PR AC = PQ AB
⇒
AC AB = PR PQ
(Given)
∠ADP = ∠EBQ, (Corresponding angles as BQ || DP and AB is transversal)
Also, DP || BQ as DP || BC.
⇒
⇒
AB2 – AC2 AC 2 AB2 = = PR 2 PQ 2 PQ 2 – PR 2
⇒
AC 2 AB 2 BC 2 = = 2 2 PR PQ QR 2
⇒
AC AB BC = = PR PQ QR
∴
∆ACB ~ ∆PRQ
⇒
∠B = ∠Q.
p2
q2
–
OR = (a cot θ + b cosec θ)2 – (b cot θ + a cosec θ)2
= a2 cot2 θ + 2ab cot θ cosec θ + b2 cosec2 θ – b2 cot2 θ – 2ab cot θ cosec θ – a2 cosec2 θ = – a2 (cosec2 θ – cot2 θ) + b2 (cosec2 θ – cot2 θ) = – a2 (1) + b2 (1) = b 2 – a2 . M A T H E M A T
I C S – X
17. xi fi
15 17 19 2
3
20 + p
23
5p
6
4
∴ Σfi =15+5p
fixi 30 51 76 100p+5p2 138 Σfixi = 295 +100p+ 5p2 Mean =
∑ fi xi ∑ fi
295 + 100p + 5p2 15 + 5p 300 + 100p = 295 + 100p + 5p2 5p2 = 5 p = ± 1.
⇒
20 =
⇒ ⇒ ⇒
But frequency cannot take negative value. So, p ≠ –1. Hence, p = 1. 18. Table for cumulative frequency: Class
Frequency
Cumulative frequency
0-10 10-30 30-60 60-80 80-90
5 15 30 8 2
5 20 50 58 60
N = 60 Modal class is 30-60. Class mark = Since
Required sum = 45 + 45 = 90. SECTION-C 19. Let a be any positive integer. We know that any positive integer is either of the form 2q or 2q + 1 for some integers q. ∴ a = 2q or 2q + 1 When a = 2q, a + 1 = 2q + 1.
I
20. Ram, Ravi and Nitin will meet next after the time given by the LCM of 5 days, 24 days and 9 days. Now, we find out the LCM of 5, 24 and 9 5 = 5; 24 = 23 × 3; 9 = 3 × 3 ∴ LCM = 23 × 3 × 3 × 5 = 360 They met last on Sunday. So, it will be Sunday after 7n days, where n is a natural number. So, it will be Sunday after 357 days. Therefore, it will be Wednesday after 360 days. Hence, they will meet on next Wednesday. OR See worksheet-2, Sol. 9. 21. Let zeroes are α, β, γ Let αβ = 8
... (i)
Also we know α + β + γ = 9
... (ii)
αβ + αγ + βγ = 26
C E
αβγ = 24 ∴
P A P E R S
8 (γ) = 24
⇒ (ii)
⇒ Class mark = 45
P R A C T
2q (2q +1) = 2r, where r = q (2q + 1) So, a (a + 1) is divisible by 2. Case II: When a = 2q + 1, a + 1 = 2q + 2 = 2 (q + 1) ∴ a (a + 1) = 2 (2q + 1) (q + 1) where r = (2q + 1) (q + 1) So, a (a + 1) is divisible by 2. Hence, multiplication of any two consecutive positive integers is divisible by 2.
30 + 60 = 45 2
N = 30 so, median class is 30-60. 2
Case I:
a (a + 1) = 2q × 2q + 1
(i) ∴
γ=3 ⇒
α+β=6 8 ⇒ β= α Use if in (iii) we get
... (iii)
8 =6 α α2 – 6α + 8 = 0 (α – 4) (α – 2) = 0 α+
⇒ ⇒
α=4
⇒
α = 2.
127
If α = 4 ⇒ β = 2 and if α = 2 ⇒
2 AB = 1 AD
β=4
∴ Zeroes are 2, 4 and 3.
Let
1 1 22. Put = u and = v in the given x–y x+y system of equation. we get 10u + 15u – i.e., 5u + 15u –
2v 5v v 5v
= =– = =–
4 2 2 2
... (i) ... (ii)
40u = 8 or u =
u= and
1 5
1 in equation (i) 5 v= 1
u=
BD =
(
2x – x =
(
BD = AB
)
2 –1 x
)
2 –1 x 2x
2 –1 × 2
=
2 2– 2 = . 2 2
24. In right-angled ∆ABC, AB2 + BC2 = AC2 ...(i) (By Pythagoras Theorem) In ∆ABN, AN2 = AB2 + BN2 2
BC = AB2 + 2 (... N is the mid-point of BC)
1 and v = 1 give x + y = 5 5 x – y= 1
On solving, we get
= AB2 +
x = 3 and y = 2 Hence, x = 3, y = 2 is the required solution. 23. In ∆ABC, DE || BC ∵∆ABC ~ ∆ADE ∴
2 x and AD = x, then from the figure,
Now,
Multiply equation (i) by 5 and add the result to (ii)
Substitute
AB =
1 BC2 4
A
M
ar ( ∆ABC ) AB2 = ar ( ∆ADE ) AD2
...(i) B
Again, DE || BC
N
C
⇒ 4 AN2 = 4AB2 + BC2 A
Similarly, in ∆CBM, 4 CM2 = AB2 + 4 BC2
D
...(ii) ...(iii)
Adding equations (ii) and (iii), we get
E
4AN2 + 4 CM2 B
C
= (4AB2 + AB2) + (BC2 + 4BC2)
∵ ar (∆ADE) = ar ( BCED) ∴ ar (∆ABC) = 2 ar (∆ADE) ⇒
ar ( ∆ABC ) =2 ar ( ∆ADE )
From equations (i) and (ii), we get
128
= 5 AB2 + 5BC2 ⇒ 4(AN2 + CM2) ... (ii)
= 5 (AB2 + BC2) = 5 AC2.
[From (i)] Hence proved.
M A T H E M A T
I C S – X
25.
LHS =
=
=
tan θ + sec θ – 1 tan θ – sec θ + 1
(
tan θ + sec θ – sec 2 θ – tan 2 θ
2 2 1 = 3 = = 4 4 2 3 Thus, LHS = RHS.
)
tan θ – sec θ + 1
(tan θ + sec θ) – (sec θ + tan θ) (sec θ – tan θ) tan θ – sec θ + 1 =
( tan θ + sec θ ) (1 – sec θ + tan θ ) tan θ – sec θ + 1 sin θ 1 + cos θ cos θ
= tan θ + sec θ = =
1 + sin θ cos θ
= RHS.
Hence proved.
cos ( 90° – 32° ) = 2 sin 32°
cos ( 90° – 52°) cosec 52° – 3 tan ( 90° – 75°) × 3 × tan 75° 1 sin 52° × sin 52° sin 32° = 2 – 2 cot 75° × 3 × 1 sin 32° cot 75° 3 ×
1 = 2 – 1 = 1. 3
OR LHS = cos 2θ = cos (2 × 30°) = cos 60° =
1 – tan 2 θ 1 – tan 2 30° RHS = = 2 1 + tan θ 1 + tan 2 30° 2
1 1– 3 = = 2 1 1+ 3
P R A C T
I
C E
27. First, we prepare the cumulative frequency table as given below: Class interval
Frequency ( f)
Cumulative frequency (cf )
85-100 100-115 115-130 130-145
11 9 8 5
11 20 28 33
N = 33
cos 58° cos 18° cosec 52° 26. 2 – 3 tan 15 tan 60 tan 75 ° ° ° sin 32 °
=2×1–
Hence proved.
1 2
1 1– 3 1 1+ 3
P A P E R S
N = 16.5 2 Cumulative frequency just greater than 16.5 is 20. So, median class is 100-115. ∵ N = 33
∴
∴ cf = 11, f = 9, l = 100, h = 15 N – cf Now, median = l + 2 ×h f 16.5 – 11 = 100 + × 15 = 100 + 9.17 9 = 109.17
Hence, the median speed is 109.17 km/hr. 28. Since, the maximum frequency is 41, so the modal class is 10000-15000. ∴ l = 10000, f1 = 41, f0 = 26, f2 = 16, h = 5000
f1 – f0 Now, mode = l + ×h 2 f1 – f0 – f2 41 – 26 = 10000 + × 5000 82 – 26 – 16
= 10000 +
75000 40
= 10000 + 1875 = 11875. Thus, the monthly modal income is Rs. 11875.
129
OR Let a = 50; h = 20 C.I.
fi
xi
0-20 20-40 40-60 60-80 80-100
4 10 28 36 50
10 30 50 70 90
We know that Time × Speed = Distance
ui =
xi – a h
–2 –1 0 1 2
–8 – 10 0 36 100
∑ fi ui x = a + h ∑ f = 50 + 20 × i
8x – 8y = 80 i.e.,
x – y = 10
80 km
1 a
(
)
2 – a2 + 1 1 a +1 Sum of zeroes = a + = =– a a a Coefficient of x = – Coefficient of x2 1 a Product of zeroes = a × = a a Constant term = . Coefficient of x 2 Hence proved. OR
32. We have q sin θ = p and p cos θ = q q p ⇒ sin θ = and cos θ = p q 6
6
(i)
q p + = sin6 θ + cos6 θ q p
= (sin2 θ)3 + (cos2 θ)3 = (sin2 θ + cos2 θ)3 – 3 sin2 θ . cos2 θ (sin2 θ + cos2 θ) . . 3 3 [ . a + b = (a + b)3 – 3ab (a + b)] = (1)3 – 3 .
2 p2 . q . 1 2 q2 p
= 1 – 3 = – 2.
See worksheet-12, Sol. 10. 30. Let the speeds of the cars be x km/hr and y km/hr.
B
1 × x + 1 × y= 80 or x + y = 80 ... (ii) To solve equations (i) and (ii), adding them and subtracting them respectively, we get 2x = 90 and 2y = 70 i.e., x = 45 and y = 35 Hence, the speed of the cars are 45 km/ hr and 35 km/hr. 31. See worksheet-34, Sol. 9 (1st part). OR See Assessment sheet-7, Sol. 8.
29. Given that p(x) = a (x2 + 1) – x (a2 + 1) i.e., p(x) = ax2 – (a2 + 1)x + a To find zeroes of p (x), put p (x) = 0. i.e., ax2 + a – a2x – x = 0 i.e., (ax2 – a2x) – (x – a) = 0 i.e., ax (x – a) – 1 (x – a) = 0 i.e., (x – a) (ax – 1) = 0
130
y km/hr
x km/hr A
1 are the zeroes of p (x). a
... (i)
When the cars run in the opposite directions:
118 128
x = a,
B
Difference of the distances covered by the two cars = 80 km
SECTION-D
Thus, a and
80 km
A
295 = 18.4 ≅ 68.4. 16
i.e.,
y km/hr
x km/hr
Σfi ui = 118
128
= 50 +
When the cars run in the same direction.
fi ui
(ii)
p q
6
6
+
q6 p6
Proved.
= –2
M A T H E M A T
I C S – X
⇒
p12 + q12 p6 q6
More than type cumulative frequency table:
= – 2 ⇒ p12 + q12 = – 2p6 q6
Marks
⇒ p12 + q12 + 2p6 q6 = 0 ⇒ (p6 + q6)2 = 0 ⇒ p6 + q6 = 0. Proved. 33. We have to prove sec θ + 1 sec θ – 1 + = 2 cosec θ sec θ – 1 sec θ + 1
LHS =
sec θ – 1 sec θ – 1 × + sec θ + 1 sec θ – 1 =
(sec θ – 1)2 sec 2 θ – 1
sec θ + 1 sec θ + 1 × sec θ – 1 sec θ + 1
+
=
( sec θ – 1) 2
tan θ
sec2 θ – 1
tan 2 θ
100
No. of Marks students less than 10 20 30 40 50 60
C E
100 (0, 100)
10-20
10
10
97 (10, 97)
20-30
23
20
83 (20, 83)
30-40
51
30
60 (30, 60)
40-50
6
40
9
(40, 9)
50-60
3
50
3
(50, 3)
cf
(60, 100)
(0, 100) (50, 97)
(10, 97) (40, 91)
Less than ogive
85
(20, 83)
80 75 70
34. To draw the less than type and more than type ogives, we prepare the cumulative frequency table by less than and more than methods as given below: Less than type cumulative frequency table:
I
0
90
= 2 cosec θ = RHS. Hence proved.
P R A C T
7
95
sec θ – 1 + sec θ + 1 tan θ 2 2 sec θ cos θ 2 = = = sin θ tan θ sin θ cos θ
7 10 23 51 6 3
0-10
Y
sec θ – 1 sec θ + 1 + tan θ tan θ
0-10 10-20 20-30 30-40 40-50 50-60
point
( sec θ + 1)
+
=
Marks
cf
2
2
=
( sec θ + 1)2
No. of Marks students more than or equal to
point
65 60
(30, 60)
55 50 45 (30, 40)
40 35 30 25
7 (10, 7) 17 (20, 17) 40 (30, 40) 91 (40, 91) 97 (50, 97) 100 (60, 100)
P A P E R S
20 15
(20, 17)
More than ogive (40, 9)
10 5 0
(10, 7) (50, 3)
(32, 0) 10
20
30
40
50
60
X
131
We plot the points as given in both of the tables on a graph, taking marks on the x-axis and the cumulative frequencies on the y-axis. On joining these points by free hand smooth curve, we obtain the less than and more than type ogives as shown in the figure. Median: The abscissa of the point of intersection of the two ogives determines the median of the given data. To obtain
132
the coordinates of this point of intersection, we draw a perpendicular from this point on the x-axis. The abscissa of the foot of this perpendicular is the required median. Here the coordinates of the foot of the perpendicular are (32,0), where 32 is the approximate value. Hence, the required median is nearly 32 marks.
M A T H E M A T
I C S – X
