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Biyani's Think Tank Concept based notes

Maths [Class-X]

Mrs. Poonam Fatehpuria Assistant Professor Department of Science Biyani Girls College, Jaipur

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2

Published by :

Think Tanks Biyani Group of Colleges

Concept & Copyright :

Biyani Shikshan Samiti Sector-3, Vidhyadhar Nagar, Jaipur-302 023 (Rajasthan) Ph : 0141-2338371, 2338591-95 Fax : 0141-2338007 E-mail : [email protected] [email protected] Website :www.gurukpo.com; :www.gurukpo.com; www.biyanicolleges.org www.biyanicolleges.org

First Edition : 2011 Sign up to vote on this title

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While every effort is taken to avoid errors or omissions in this Publication, an mistake or omission that may have crept in is not intentional. It may be taken note

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Maths

Preface

I

am glad to present this book, especially designed to serve the needs of th students. The book has been written keeping in mind the general weakness

understanding the fundamental concept of the topic. The book is self-explanato

and adopts the “Teach Yourself” style. It is based on question -answer pattern. Th

language of book is quite easy and understandable based on scientific approach.

Any further improvement in the contents of the book by making correction

omission and inclusion is keen to be achieved based on suggestions from the reade for which the author shall be obliged.

I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay Biyan

Director (Acad.) Biyani Group of Colleges, who is the backbone and main concep

provider and also have been constant source of motivation throughout th

endeavour. We also extend our thanks to Biyani Shikshan Samiti, Jaipur, who playe

an active role in co-ordinating the various stages of this endeavour and spearheade the publishing work.

Sign up to vote on this title of variou I look forward to receiving valuable suggestions from professors

Useful useful the Notfor educational institutions, other faculty members and students improvement o

the quality of the book. The reader may feel free to send in their comments an suggestions to the under mentioned address.

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4

Chapter 1

Real Numbers

(Euclid’s Division Lemma) : Given positive integers a and b, there exi unique integers q and r satisfying satisfying a = bq + r, 0≤ r < b.

Every composite number can be expressed( factorised) as a product of prime and this factorisatio factorisation n is unique, apart from from the order order in which the prime facto facto occur.

Let p be a prime number. If p divides a 2 , then p divides a, where a is a positiv integer. Q.1

Use Euclid‟s division algorithm to find the HCF of:

Ans.

196 and 38220 Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain 38220 = 196 × 195 + 0

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Since the remainder is zero, the process p rocess stops.

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Maths

Hence, product of two numbers = HCF × LCM Q.3 Prove that Ans. Let

is irrational.

is rational.

Therefore, we can find two integers a, b (b (b ≠ 0) such that

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Maths

Chapter 2

Polynomials p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of polynomiall p(x) polynomia p(x ) A polynomial of degree 1 is called a linear polynomial. A polynomial of degree 2 is called a quadratic polynomial. A polynomial of degree 3 is called a cubic polynomial. Sum of its zeroes :

Product of its zeroes:

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Thus

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8

Ans.

The value of Therefore, the zeroes of

is zero when x when x − 4 = 0 or x or x + 2 = 0, i.e., when x when x = = 4 or x or x = −2 are 4 and −2.

Sum of zeroes =

Product of zeroes

Q.2 Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

Ans Sign up to vote on this title

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Since the remainder is 0, Hence,

is a factor of

Q.3 Find a cubic polynomial with the sum, sum of the product of its zeroes taken two a a time, and the product of its zeroes as 2, − 7, − 14 respectively. Ans. Let the polynomial be

and the zeroes be

.

It is given that Sign up to vote on this title

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10 Q.4 It two two zeroes of the the poly olynom nomial zeroes. Ans. Given that 2 +

Therefore,

and 2

are

, fin find ot

are zeroes of the given polynomial. 2

= x + 4 − 4 x − 3 = x

2

= x − 4 x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient b dividing

2

by x − 4 x + 1. by x

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1

Therefore, the value of the polynomial is also zero when

or

Or x = 7 or −5 Hence, 7 and −5 are also zeroes of this polynomial.

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12

Chapter 3

Pair of Linear Equations in two Variables An equation which can be put in the form ax +by +c = 0, where a, b and c are real number numbers, s, and a and b are not not both zero, is called called a linear linear equation equation in in two variables x and y.

A pair of linear equations in two variables can be represented, and solved, by the: (i) graphical method (ii) algebraic method Graphical Method : (i) The graph of a pair of linear equations in two variables is represented by two lines. If the lines intersect at a point, then that point gives the unique solution of the two equations. In this case, the pair of equations is consistent . (ii) If the lines coincide, then there are infinitely many solutions — solutions — each each point on the line being a solution. In this case, the pair of equations is dependent (consistent). Sign up to vote on this title (iii)If the lines are parallel, then the pair of equations no useful solution. In this Useful has Not case, the pair of equations is inconsistent .

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Maths

1

solutions graphically. 10 students of Public School took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Ans (i) Let the number of girls be x be x and the number of boys be y be y.. According to the question, the algebraic representation is + y = y = 10 − y = y = 4 For x For x + + y y = = 10, = 10 − y − y x x 5

4

6

y y 5

6

4

For x For x − y = y = 4, = 4 + y + y Sign up to vote on this title

x x 5

4

3

y

0

−1

1

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14

From the figure, it can be observed that these lines intersect each other at point (7, 3). Therefore, Therefore, the the number of girls girls and and boys in the the class are are 7 and 3 respectively respectively.. Q2 Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Ans. Let the width of the garden be x be x and and length be y be y.. According to the question, − x = 4 + x = 36

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Useful……………(1) Not useful

…………….(2)

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1

+ x = x = 36 x

0

36

16

y

36

0

20

Hence, the graphic representation is as follows.

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From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m

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16 Ans.

From equation (1), we obtain

Substituting this value in equation (2), we obtain

Substituting this value in equation (3), we obtain =0 x = x = 0, y 0, y = = 0

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1

From equation (1), we obtain

Substituting this value in equation (2), we obtain

Substituting this value in equation (3), we obtain Sign up to vote on this title

Hence, = 2, = 3

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18

Ans By elimin ation method method

Subtracting equation (2) from equation (1), we obtain

Substituting this value in equation (1), we obtain

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Hence, x Hence, x = = 2, y 2, y = −3

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1

5 y = −15

Substituting the value in equation (5), we obtain

x = ∴ x =

2, y 2, y = −3

Q.6 Sushil tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you ill be.”Solve by the method of substitution. Ans. Let s and t be the ages (in years) of Sushil and his daughter, respectively. Then, the pair of linear equations that represent the situation is S – – 7 = 7 ( t – t – 7) 7) s – 7 t + 42 = 0 (1)) (1 and s+ 3 = 3 ( t + 3) s – 3 t= 6 (2) Using Equation (2), we gets = 3t + 6. Putting this value in Equation (1), we get Sign up to vote on this title (3t + 6) – 6) – 7t 7t + 42 = 0, Useful Not useful i.e., 4t = 48, which gives t = 12. Putting this value oft in Equation (2), we get

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20

(ii)

Ans

Therefore, the given sets of lines are parallel to each other. Therefore, they will no intersect each other and thus, there will not be any solution for these equations.

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x = ∴ x =

2

2, y 2, y = = 1

Q. 8 Solve the following pairs of equations by reducing them to a pair of linea equations:

(a)

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(b )

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22 (a)

Putting

and

in the given equations, we obtain

Multiplying equation (1) by 3, we obtain 6 p + p + 9q 9q = 6 (3) Adding equation (2) and (3), we obtain

Putting in equation (1), we obtain

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2

Hence, (b)

Putting

in these equations, we obtain Sign up to vote on this title

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24

Substituting in (2), we obtain

Adding equations (3) and (4), we obtain

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Substituting in (3), we obtain

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2

Ans The difference between the the ages of Gita Gita and Sita is 3 years. years. Either Gita Gita is 3 yea older than Sita or Sita is 3 years years older than Gita. However, However, it is obvious that in cases, Sita’s Sita’s father’s age will be 30 years more than that of Baby’s Baby’s age. Let the age of Sita and Gita be x be x and and y y years years respectively. Therefore, age of Sita’s Sita’s f ather, ather, Ramesh = 2 × x × x = = 2 x years x years

And age of Gita’s Gita’s sister Baby

years

By using the information given in the question, Case (I) When Sita is older than Gita by 3 years, x − y =

3

(i)

4 x − y = 60

(ii) ii)

Subtracting (i (i) from (ii (ii), ), we obtain 3 x = 60 − 3 = 57 Sign up to vote on this title

Therefore, age of Sita = 19 years

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26 4 x − y = 60 (ii (ii)) Adding (i (i) and (ii (ii), ), we obtain 3 x = x = 63 x = x = 21 Therefore, age of Sita = 21 years And age of Gita = 21 + 3 = 24 years

Q.10 In a ΔABC, C = 3 B = 2 ( A + B). Find the three angles.

Ans Given that, ∠C

= 3∠B = 2(∠A + ∠B)

3∠B = 2(∠A + ∠B) 3∠B = 2∠A + 2∠B ∠B

= 2∠A

2 ∠A − ∠B = 0 … (i (i)

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We know that the sum of the measures of all angles of a triangle is 180°. Therefore,

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2

Adding equations (ii (ii)) and (iii (iii), ), we obtain 9 ∠A = 180° ∠A

= 20°

From equation (ii (ii), ), we obtain 20° + 4 ∠B = 180° 4 ∠B = 160° ∠B

= 40°

∠C

= 3 ∠B

= 3 × 40° = 120° Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

Q.11 ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral .

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28 4 y + 20 − 4 x = x = 180 − 4 x + x + 4 y = y = 160 x − y = − 40 (i (i) Also, ∠B + ∠D = 180 3 y − 5 − 7 x + x + 5 = 180 − 7 x + x + 3 y = y = 180 (ii (ii)) Multiplying equation (i (i) by 3, we obtain 3 x − 3 y = − 120 (iii (iii)) Adding equations (ii (ii)) and (iii (iii), ), we obtain − 7 x + x + 3 x = 180 − 120 − 4 x = x = 60 x = −15 By using equation (i (i), we obtain x − y = − 40 −15 − y − y = − 40 = −15 + 40 = 25

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30

Chapter 4

Quadratic Equations

+c = 0 , where a, b, c ar A quadratic equation in the variable x is of the form ax 2 +bx +c real numbers and a ≠ 0. 2 If we can factorise ax +bx +c = 0 ,a ≠ 0, into a product of two linear factors, the the roots of the quadratic equation ax 2 +bx +c = 0 can be found by equating each factor t zero. A quadratic equation can also be solved b y the method of completing the square. 2 Quadratic formula: The roots of a quadratic equation ax +bx +c = 0 are given by

4ac ≥ 0. provided b2 - 4ac 2 A quadratic equation a x +b x +c = 0 has Sign up to vote on this title 2 4ac > 0, (i) two distinct real roots, if b - 4ac Useful Not useful 4ac = 0, and (ii) two equal roots (i.e., coincident roots), if b2 - 4ac 2 4ac < 0. (iii) no real roots, if b - 4ac

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Ans.(a)

It is not of the form

.

Hence, the given equation is not a quadratic equation. (b)

It is of the form

.

Hence, the given equation is a quadratic equation. Q.2 Represent the following situations in the form of quadratic equations.

(a) The product of two consecutive positive integers is 306. We need to find the integers. (b) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. W need to find the speed of the train. Ans. (a) Sign up to vote on this title

Let the consecutive integers be x be x and and x x + + 1. It is given that their product is 306.

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32 In second condition, let the speed of train =

km/h

It is also given that the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km =

hrs

Speed × Time = Distance

Q.3 Find the roots of the following quadratic equations by factorisation:

(a)

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(b)

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Roots of this equation are the values for which ∴

= 0 or

i.e., x i.e., x =

=0

=0

or x = x =

(b)

Roots of this equation are the values for which

=0

Therefore,

i.e., Sign up to vote on this title

Q. 4 Find two numbers whose sum is 27 and product is Useful 182.

be x and the second number is 27 − x − x.. Ans. Let the first number be x

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34

Either

= 0 or x or x − 14 = 0

i.e., x i.e., x = = 13 or x or x = = 14 If first number = 13, then Other number = 27 − 13 = 14 If first number = 14, then Other number = 27 − 14 = 13 Therefore, the numbers are 13 and 14.

Q.5 Find the roots of the following quadratic equation, by the method of completin the square and by applying the quadratic formula : Sign up to vote on this title

Ans. By the method of completing the square:

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By applying the quadratic formula :

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36 2

(A) If b − 4ac 4ac > 0 → two distinct real roots 2

(B) If b − 4ac 4ac = 0 → two equal real roots 2

(C) If b − 4ac 4ac < 0 → no real roots 2

(I) 2 x −3 x + 5 = 0 2

Comparing this equation with ax + bx + c = 0, we obtain a = 2, b = −3, c = 5 Discriminate

2

2

= b − 4ac 4ac = (− 3) − 4 (2) (5) = 9 − 40 = −31

As

2

b − 4ac 4ac < < 0,

Therefore, no real root is possible for the given equation. (II) 2

Comparing this equation with ax + bx + c = 0, we obtain

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Discriminate

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Maths

Therefore, the roots are

3

and and

.

Q. 7 Find the roots of the following equation:

Ans.

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38

Q.8 A train travels 360 km at a uniform speed. If the speed had had been 5 km/h km/h more, i would have taken 1 hour less for the same journey. Find the speed of the train. Ans Let the speed of the train be x be x km/hr. km/hr.

Time taken to cover 360 km

hr

According to the given question,

Speed cannot be negative. Therefore, the speed of train is 40 km/h

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Q. 9 Sum of the areas of two squares is 468 m . If the difference of their perimeters is 2 m, find the sides of the two squares

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3 2

2

Areas will be x be x and y and y respectively. It is given that 4 x − 4 y = y = 24 x − y = y = 6 x = y + y + 6

However, side of a square cannot be negative. Hence, the sides of the squares are 12 m and (12 + 6) Sign m =up 18tomvote on this title

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Q.10 Find the values of k for for kx (x − 2) + 6 = 0 , so that they have two equal roots.

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40 2

= 4k 4k − 24k 24k 2

For equal roots,

b − 4ac 4ac = = 0 2

4k − 24k 24k = = 0 4k (k (k − − 6) = 0 Either

4k = 4k = 0 or k = = 6 = 0 k = = 0 or k = = 6 2

However, if k = = 0, then the equation will not have the terms ‘ x ’ and ‘ x’. x’. Therefore, if this equation has two equal roots, k should should be 6 only.

Q.11. The sum of the ages of two friends is 20 years. Four years ago, the product o their ages in years was 48. Find the present age. Ans

Let the age of one friend

= x years. x years.

Age of the other friend

=(20 − x − x)) years.

st

4 years ago, age of 1 friend nd

And, age of 2 friend

=( x x − 4) years = (20 − x − x − 4)

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= (16 − x − x)) years Given that

( − 4) (16 − ) = 48

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4 a = 1, b = −20, c = 112 2

2

Discriminate = b − 4ac 4ac = (− 20) − 4 (1) (112) = 400 − 448 = −48 As

2

b − 4ac 4ac < < 0,

Therefore, no real root is possible for this equation and hence, this situation is no possible.

2

Q.12 A rectangular park of perimeter 80 and area 400 m ? Find its length an breadth. Ans.

Let the length and breadth of the park be l and b. Perimeter

= 2 (l + + b) = 80 l + + b = 40

Or,

b = 40 − l − l 2

Area

= l × × b = l (40 l (40 − l ) = 40l 40l − l 2

40l 40l − l = 400 2

l − 40l 40l + + 400 = 0 2

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42 Therefore, this equation has equal real roots. And hence, this situation is possible. Root of this equation,

Therefore, length of park, l = = 20 m And breadth of park, b = 40 − l = l = 40 − 20 = 20 m

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4

Chapter 5 Arithmetic Progression

An arithmetic progression (AP) is a list of numbers in which each term i obtained by adding a fixed numbered to the preceding term, except the first term. The fixe numbered is called the the common difference. difference. The form form of an AP is a, a +d, a + 2d, 2 d, a + 3d, . a1 , a 2 , a 3 , . . is an AP, if the difference A given list of numbers a 2 -a1 , a 3 -a 2 , a 4 - a 3 , . . ., give the same value, i.e., if a k +1 - a k

is the same fo

different values of k. In an AP with first term a and common difference, then th term (or the gener term) is given by a n = a + (n – (n – 1)d 1)d The sum of the first n terms of an AP is given by

If l is the last term of the finite AP, say then th term, term, then the sum of all terms Sign up to vote on this title the AP is given by Useful Not useful

s

n

(a

l )

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44 Let the series be a1, a2, a3, a4 … a1 = a = −1.25 a2 = a1 + d = d = − 1.25 − 0.25 = −1.50 a3 = a2 + d = d = − 1.50 − 0.25 = −1.75 a4 = a3 + d = d = − 1.75 − 0.25 = −2.00 Clearly, the series will be will be 1.25, −1.50, −1.75, −2.00 …….. First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00. (b) a = 4, d 4, d = = −3 Let the series be a1, a2, a3, a4 … a1 = a = 4 a2 = a1 + d = d = 4 − 3 = 1 a3 = a2 + d = d = 1 − 3 = −2 a4 = a3 + d = d = − 2 − 3 = −5 Therefore, the series will be 4, 1, −2 −5 …

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Useful First four terms of this A.P. will be 4, 1, −2 and −5.

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Q.2 For the following A.P.s, write the first term and the common difference.

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Here, first term, a = −5 Common difference, d = d = Second term − First term = (−1) − (−5) = − 1 + 5 = 4

(b)

Here, first term, Common difference, d = d = Second term − First term

Q.3 Which of the following are APs? If they form an A.P. find the common difference and write three more terms.

(a) − 1.2, − 3.2, − 5.2, − 7.2 … (b) 0.2, 0.22, 0.222, 0.2222 …. Ans. (a) −1.2, −3.2, −5.2, −7.2 …

Here a2 − a1 = (−3.2) − (−1.2) = −2

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46 a5 = − 7.2 − 2 = −9.2 a6 = − 9.2 − 2 = −11.2 a7 = − 11.2 − 2 = −13.2 (b) 0.2, 0.22, 0.222, 0.2222 …. It can be observed that a2 − a1 = 0.22 − 0.2 = 0.02 a3 − a2 = 0.222 − 0.22 = 0.002 a4 − a3 = 0.2222 − 0.222 = 0.0002 = 0.0002 i.e., ak +1 is not the same every time. +1 − ak is Therefore, the given numbers are not in A.P. th

Q. 4 Find the 30 term of the A.P: 10, 7, 4, ……… Ans.

Given that A.P.

10, 7, 4, …

First term,

a = 10

Common difference, d = = a2 − a1 = 7 – 7 – 10 10 = −3 We know that,

+ ( − 1) d

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Q.5 In the following AP AP find the missing term in the boxes

Ans.

For this A.P.,

We know that,

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48 Q.6 Which Which term of the A.P. 3, 8, 13, 18, … is 78? Ans. 3, 8, 13, 18, …

For this A.P., a = 3 d = = a2 − a1 = 8 − 3 = 5 th

Let n term of this A.P. be 78. an = a + (n (n − 1) d 78 = 3 + (n (n − 1) 5 75 = (n (n − 1) 5 (n − 1) = 15 n = 16 th

Hence, 16 term of this A.P. is 78. th

th

Q.7 If 17 term of an A.P. exceeds its 10 term by 7. Find the common difference. Ans. We know that,

For an A.P., an = a + (n (n − 1) d + (17 − 1) d

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(a + 16d 16d ) − (a (a + 9d 9d ) = 7 7d = = 7 d = = 1 Therefore, the common difference is 1.

th

Q.8 Which term of the A.P. 3, 15, 27, 39, … will be 132 more than its 54 term? Ans. Given A.P. is 3, 15, 27, 39, …

a=3 d = = a2 − a1 = 15 − 3 = 12 a54 = a + (54 − 1) d = 3 + (53) (12) = 3 + 636 = 639 132 + 639 = 771 We have to find the term of o f this A.P. which is 771. th

Let n term be 771. an = a + (n (n − 1) d

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50

Q. 9 Hari saved Rs 5 in the first week of a year and then increased her weekly saving by th Rs 1.75. If in the n week, his week, his weekly savings become Rs 20.75, find n .

. Given that, A n s

a = 5, d = 1.75, an = 20.75 , n = ?

an = a + (n (n − 1) d

n−1=9 n = 10 Hence, n is 10. Q.10 Find the sum of the AP 0.6, 1.7, 2.8 ,…….., to 100 terms Ans.

0.6, 1.7, 2.8 ,…, to 100 terms Sign up to vote on this title

Here

a = 0.6 d = = a2 − a1 = 1.7 − 0.6 = 1.1

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5

Q.11 In an AP a = = 5, d = = 3, a = 50, find n and and S . n n n n Ans. Ans. Given that,

As ∴

a = 5, d = = 3, an = 50

an = a + (n (n − 1)d 1)d , 50 = 5 + (n − 1)3 45 = (n (n − 1)3 15 = n − 1 n = 16

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52 As

an = a + (n (n − 1)d 1)d , a3 = a + (3 − 1)d 1)d 15 = a + 2d 2d

(i)

(ii) On multiplying equation (1) by 2, we obtain 30 = 2a 2a + 4d 4d

(iii)

On subtracting equation (iii) from (ii), we obtain −5 = 5d 5d d = d = −1 From equation (i), 15 = a + 2(−1) 15 = a − 2

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Q. 13 Given a = = 2, d = = 8, S = 90, find n and and a . n n n n Ans. Given that,

a = 2, d = = 8, S n = 90

As

,

90 = n [2 + (n (n − 1)4] 90 = n [2 + 4n 4n − 4] 2

90 = n (4n (4n − 2) = 4n 4n − 2n 2n 2

4n − 2n 2n − 90 = 0 2

4n − 20 20n + 18n 18n − 90 = 0 4n (n − 5) + 18 (n (n − 5) = 0 (n − 5) (4n (4n + 18) = 0 Either

n − 5 = 0 or 4n 4n + 18 = 0 Sign up to vote on this title

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54 = 2 + (4) (8) = 2 + 32 = 34

Q.14 In an A.P a = = 3, n = = 8, S = = 192, find d . Ans. Given that, a = 3, n = 8, S = = 192

192 = 4 [6 + 7d 7d ] 48 = 6 + 7d 7d 42 = 7d 7d d=6

Q.15 Given l = = 28, S = 144 and there are total 9 terms. Find a = . Sign up to vote on this title

Ans. Given that, l = = 28, S = = 144 and there are total of 9 terms. Useful

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Q.16 The first term of an AP is 5, the last term is 45 and the sum is 400. Find th number of terms and the common difference . Ans. Given that, a = 5, l = = 45, S n = 400

n = 16 l = a + (n − 1) d 45 = 5 + (16 − 1) d 40 = 15d 15d

Q. 17 The first term of an AP is 5, the last term is Sign 45 and 400. Find the up tothe vote sum on thisistitle number of terms and the common difference. Useful Not useful Ans. Given that,a that,a = 5, l = = 45, S n = 400

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56 l = a + (n − 1) d 45 = 5 + (16 − 1) d 40 = 15d 15d

Q.18 The sum of the third and the seventh terms of an A.P is 6 and their product is Find the sum of first sixteen terms of the A.P. Ans.

(n − 1) d an = a + (n a3 = a + (3 − 1) d a3 = a + 2d 2d Similarly,

a7 = a + 6d 6d

Given that, a3 + a7 = 6 (a + 2d 2d ) + (a (a + 6d 6d ) = 6 2a + 8d 8d = = 6 a + 4d 4d = = 3

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5

From equation (i (i),

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58

Q.19 A ladder has rungs 25 cm apart. (See figure). The rungs decrease uniformly i length from 45 cm at the bottom to 25 cm at the top. If the top and bottom rungs ar m apart, what is the length of the wood required forupthe rungs? Sign to vote on this title

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Ans.

It is given given that that the the rungs rungs are 25 cm cm apart apart and and the the top top and and bottom bottom rungs rungs are

Total ∴ To

m apart apart

number of rungs

Sign up to vote this title Now, as the lengths of the rungs decrease uniformly, they will be inonan A.P.

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The length of the wood required for the rungs equals the sum of all the terms of this A.P

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60

Therefore, the length of the wood required for the rungs is 385 cm.

Q.20 Which term of the A.P. 121, 117, 113 … is its first negative term? Ans. Given A.P. is 121, 117, 113 …

a = 121 d = 117 − 121 = −4 an = a + (n (n − 1) d = 121 + (n (n − 1) (−4) = 121 − 4n 4n + 4 = 125 − 4n 4n We have to find the first negative term of this A.P .

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Therefore, 12, 16, 20, 24, …

All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 1 and common difference as 4.

When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisib by 4. The series is as follows. 12, 16, 20, 24, …, 248 th

Let 248 be the n term of this A.P.

Therefore, there are 60 multiples of 4 between 10 andSign 250.up to vote on this title

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62

Chapter 6

Triangle

Two figures having the same shape but not necessarily the same size are calle similar figures. All the congruent figures are similar but the converse converse is not true. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (i.e., proportion).

If a line is drawn parallel to one side of a triangle to intersect the other two side in distinct points, then the other two sides are divided in the same ratio.

If a line divides any two sides of a triangle in the same ratio, then the line i parallel to the third side.

If in two triangles, corresponding angles are equal, then their corresponding side are in the same ratio and hence the two triangles are similar (AAA similarit criterion).

the tw If in two triangles, two angles of one triangle are respectively equal to Sign up to vote on this title angles of the other triangle, then the two triangles are similar (AA similarit Useful Not useful criterion).

If in two triangles, corresponding sides are in the same ratio, then thei

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If a perpendicular is drawn from the vertex of the right angle of a right triangl to the hypotenuse, then the triangles on both sides of the perpendicular ar similar to the whole triangle and also to each other.

In a right triangle: triangle: the square of the the hypotenuse is equal equal to the sum of th squares of the other two sides (Pythagoras Theorem).

If in a triangle: square of one side is equal to to the sum of the squares of the oth two sides, then the angle angle opposite the first first side side is a right angle angle

Q.1 E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of th following cases, state whether EF || QR.

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm (ii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm Ans. (i)

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64

(ii)

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

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In ΔDOC and ΔBOA, ∠CDO

= ∠ABO [Alternate interior angles as AB || CD]

∠DCO

= ∠BAO [Alternate interior angles as AB || CD]

∠DOC

= ∠BOA [Vertically opposite angles]

∴

ΔDOC ∼ ΔBOA [AAA similarity criterion]

Q.3 If the areas of two similar triangles are equal, prove that they are congruent. Ans. Let us assume two similar triangles as ΔABC ∼ Sign up to vote on this title

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66

Q.4 ABC and BDE are two equilateral triangles such that D is the mid-point of BC Ratio of the area of triangles ABC and BDE is

Ans

We know that equilateral triangles have all its angles as 60º and all its sides of the sam length. Therefore, all equilateral triangles are similar to each other. Hence, the rati between the areas of these triangles will be equal to the square of the ratio between sides of these triangles. Let side of ΔABC = x = x

Therefore, side of

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6

Ans.

We have, DN || CB, DM || AB, and ∠B = 90° ∴ DMBN ∴ DN

is a rectangle.

= MB and DM = NB

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68 ∠1

+ ∠2 + ∠DMC = 180°

⇒ ∠1

+ ∠2 = 90°

… (2)

In ΔDMB, ∠3

+ ∠DMB + ∠4 = 180°

⇒ ∠3

+ ∠4 = 90°

… (3)

From equation (1) and (2), we obtain ∠1

= ∠3

From equation (1) and (3), we obtain ∠2

= ∠4

In ΔDCM and ΔBDM, ∠1

= ∠3 (Proved above)

∠2

= ∠4 (Proved above)

∴

ΔDCM ∼ ΔBDM (AA similarity criterion) Sign up to vote on this title

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Maths ∠6

6

+ ∠8 = 90°

… (5

D is the foot of the perpendicular drawn from B to AC. ∴ ∠ADB ⇒ ∠5

= 90°

+ ∠6 = 90° … (6)

From equation (4) and (6), we obtain ∠6

= ∠7

From equation (5) and (6), we obtain ∠8

= ∠5

In ΔDNA and ΔBND, ∠6

= ∠7

∠8

= ∠5

∴

ΔDNA ∼ ΔBND (AA similarity criterion)

Sign up to vote on this title 2 ⇒ DN =

AN × NB

2 ⇒ DN =

AN × DM (As NB = DM)

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70

Chapter 7

Coordinate Geometry The distance between P (x1 ,y1 ) and Q(x 2 ,y2 ) is

The distance of a point P(x,y) P(x ,y) from the origin is

The coordinates of the point P(x,y) which divides the line segment joining the po p o i n t s A(x1 ,y1 ) and B (x 2 ,y2 ) internally in the ratio m1 : m2 are

The mid-point of the line segment joining the points P( x1 ,y1 ) and Q( x 2 ,y 2 ) is Sign up to vote on this title

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(i) (2, 3), (4, 1) (ii) (−5, 7), (−1, 3) (iii) (a ( a, b), (− a, − b) Ans. i) Distance between the two points is given by

(ii) Distance between

(iii) Distance between

is given by

is given by

Sign upoftoan vote on this title Q.2 Check whether (5, − 2), (6, 4) and (7, − 2) are the vertices isosceles triangle.

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Ans. Let the points (5, −2), (6, 4), and (7, −2) are representing the vertices A, B, and C of th given triangle respectively.

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72 As two sides are equal in length, therefore, A BCis an isosceles triangle. Q.3 In the following find the value of ‘k ‘k ’, ’, for which the points are collinear.

(i) (7, − 2), (5, 1), (3, − k ) (ii) (8, 1), (k , − 4), (2, − 5) Ans. (i) For collinear points, area of triangle formed b y them is zero.

Therefore, for points (7, −2) (5, 1), and (3, k ), ), area = 0

(ii) For collinear points, area of triangle formed by them is zero. Therefore, for points (8, 1), (k (k , −4), and (2, −5), area = 0

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Useful ofNot usefulwhose centr diameter Q.4 Find the coordinates of a point A, where AB is the circle is (2, − 3) and B is (1, 4). 4).

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Q.5 Median of a triangle divides it into two triangles of equal areas. Verify this resul for ΔABC whose vertices are A (4, − 6), B (3, − 2) and C (5, 2) Ans.

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2). Sign up to vote on this title

Let D be the mid- point point of side BC of ΔABC. Therefore, AD is the in ΔABC. Not useful Useful median

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74 However, area cannot be negative. Therefore, area of ΔABD is 3 square u nits.

However, area cannot be negative. Therefore, area of ΔADC is 3 square u nits. Clearly, median AD has divided ΔABC in two triangles of equal areas. Q.7 Find a relation between x and and y if if the points (x , y ), ), (1, 2) and (7, 0) are collinear Ans. If the given points are collinear, then the area of triangle formed by these points will be 0.

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This is the required relation between x between x and y and y..

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On adding equation (1) and (2), we obtain 10 y = −20 y = −2 From equation (1), we obtain 3 −2=7

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76

Ans.

Let ABCD be ABCD be a square having (−1, 2) and (3, 2) as vertices v ertices A and C respectively. Let ( y), y), ( x x1, y1) be the coordinate of vertex B and D respectively. We know that the sides of a square are equal to each other. ∴ AB

= BC

We know that in a square, all interior angles are of 90°. In ΔABC,

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2

2

AB + BC = AC

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Maths

7

y ( y y ⇒ y (

− 4) = 0

y = ⇒ y =

0 or 4

We know that in a square, the diagonals are of equal length and bisect each other at 90 Let O be the mid-point of AC. Therefore, it will also be the mid-point of BD.

y + y y1 = ⇒ y +

4

If y If y = = 0, y1 = 4 If y If y = = 4,

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y1 = 0

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78

Chapter 8

Introduction to Trigonometry In a right triangle ABC, right-angled at B,

Some Formulas: ,

,

,

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Q.1 In ΔABC right angled at B, AB = 24 cm, cm, BC = 7 m. Determine (i) sin A, cos A (ii) sin C, cos C Ans. Applying Pythagoras theorem for ΔABC, we obtain 2

2

2

AC = AB + BC

2

2

= (24 cm) + (7 cm) 2

= (576 + 49) cm 2

= 625 cm ∴

AC =

cm = 25 cm

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(i) sin A =

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80

sin C =

cos C =

Q.2 Find sin A and sec A if 15cotA=8. Ans. Consider a right-angled triangle, right-angled at B.

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Here cot A =

. AB=8k & BC = 15k Applying Pythagoras theorem in ΔABC, we obtain 2

2

2

AC = AB + BC 2

2

= (8k (8k ) + (15k (15k ) 2

2

= 64k 64k + 225k 225k 2

= 289k 289k AC = 17k 17k

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82

Ans It is given that 3cot A = 4

Or,

cot A =

Consider a right triangle ABC, right-angled at point B.

If AB is 4k 4k , then BC will be 3k 3k , where k is is a positive integer. Sign up to vote on this title

In ΔABC, ΔABC,

2

2

2

(AC) = (AB) + (BC)

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8

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2

cos A − sin A =

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84

(b) (c ) Ans (a) sin60° cos30° + sin30° cos 60°

2

2

2

(b) 2tan 45° + cos 30° − sin 60°

(c)

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(c )

Q.5 If tan 2A = cot (A− 18°), where 2A is an acute angle, find the value of A. Ans Given that, tan 2A = cot (A− 18°)

cot (90° − 2A) = cot (A −18°) − 18°) 90° − 2A = A− 18°

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86 = cos 23° + sin 15°

Q.7 Evaluate 2 2 (a) 9 sec A − 9 tan A (b) (1 + tan θ + sec θ) (1 + cot θ − cosec θ) (c ) (secA + tanA) (1 − sinA) (d) 2

2

Ans. (i) 9 sec A − 9 tan A 2

2

= 9 (sec A − tan A) 2

2

= 9 (1) [As sec A − tan A = 1] =9 (ii) (1 + tan θ + sec θ) (1 + cot θ − cosec θ)

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= cosA (iv)

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88

Chapter 9

Application of Trigonometry

The line of sight is the line drawn from the eye of an observer to the point in the object viewe by the observer. The angle of elevation of the point viewed is the angle formed by the line o sight with the horizontal when the point being viewed is above the horizontal level.

The angle of depression of a point on the object being viewed is the angle formed by the line o sight with the horizontal when the point is below the horizontal level

Q. 1 A tree breaks due to storm and the broken part bends so that the top of the tre touches the ground making an angle 30 ° with it. The distance between the foot o the tree to the point where the top touches the ground is 8 m. Find the height of th tree. Ans. Sign up to vote on this title

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8

Let AC was the original tree. Due to storm, it was broken into two parts. The broken pa is making 30° with the ground. In

,

Height of tree

=

+ BC

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Hence, the height of the tree is

.

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90

Let K be the kite and the string is tied to point P on the ground. In ΔKLP,

Hence, the length of the string is

.

Q.3 A contractor plans to install two slides for the children to play in a park. For th children below the age of 5 years, she prefers to have a slide whose top is at a heigh of 1.5 m, and is inclined at an angle of 30 ° to the ground, where as for the elde Sign voteinclined on this title children she wants to have a steep side at a height of 3 up m,toand at an angle o Usefulin each useful 60 ° to the ground. What should be the length of the case? slide Not

Ans. It can be observed that AC and PR PR are the slides slides for younger and elder childre

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In ΔABC,

In ΔPQR,

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Therefore, the lengths of these slides are 3 m and

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92

Ans

Let AC was the original tree. Due to storm, it was broken into two parts. The broken pa is making 30° with the ground. In

,

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Hence, the height of the tree is

9

.

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94

Chapter 10

Circles A tangent to a circle is a line that intersects the circle at only one point.

The tangent at any point of a circle is perpendicular to the radius through the poin of contact.

The length of the segment of the tangent from the external point P and the point o contact with the circle is called the length of the tangent from the point P to the circle.

The lengths of tangents drawn from an external point to a circle are equal.

Q.1 A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centr O at a point Q so that OQ = 12 cm. Length PQ is :

Ans. We know that the line drawn from the centre of the circle to the tangent is perpendicul Sign up to vote on this title to the tangent.

OP

PQ

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2

2

2

OP + PQ = OQ 2

2

2

5 + PQ =12 2

PQ =144 − 25 PQ =

cm.

Q.2 Draw a circle and two lines parallel to a given line such that one is a tangent an the other, a secant to the circle.

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96

Ans.

Let O be the centre of the circle. Given that, OQ = 25cm and PQ = 24 cm As the radius is perpendicular to the tangent at the point of contact, Therefore, OP ⊥ PQ Applying Pythagoras theorem in ΔOPQ, we obtain 2

2

2

OP + PQ = OQ 2

2

2

OP + 24 = 25 2

OP = 625 − 576 2

OP = 49 OP = 7

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9

Ans

Let us consider a circle centered at point O. AB is a tangent drawn on this circle from point A. Given that, OA = 5cm and AB = 4 cm In ΔABO, OB ⊥ AB (Radius ⊥ tangent at the point of contact) Applying Pythagoras theorem in ΔABO, we obtain 2

2

2

AB + BO = OA 2

2

2

4 + BO = 5 2

16 + BO = 25 2

BO = 9

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98

Ans

Let us consider a circle centered at point O. Let P be an external point from which tw tangents PA and PB are drawn to the circle which are touching the circle at point A and respectively and AB is the line segment, joining point of contacts A and B together suc that it subtends ∠AOB at center O of the circle. It can be observed that OA (radius) ⊥ PA (tangent) Therefore, ∠OAP = 90° Similarly, OB (radius) ∠OBP

⊥ PB

(tangent)

= 90°

In quadrilateral OAPB, Sum of all interior angles = 360º ∠OAP

+∠APB+∠PBO +∠BOA = 360º

90º + ∠APB + 90º + ∠BOA = 360º

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9

Ans

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches th circle at point P, Q, R, S. Let us join the vertices of the quadrilateral ABCD to the cente of the circle. Consider ΔOAP and ΔOAS, AP = AS (Tangents from the same point) OP = OS (Radii of the same circle) OA = OA (Common side) ΔOAP ≅ ΔOAS (SSS congruence criterion) Therefore, A ↔ A, P ↔ S, O ↔ O And thus, ∠POA = ∠AOS ∠1

= ∠8

Similarly,

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100 2(∠1 + ∠2) + 2(∠5 + ∠6) = 360º (∠1 + ∠2) + (∠5 + ∠6) = 180º ∠AOB

+ ∠COD = 180º

Similarly, we can prove that ∠BOC + ∠DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementar angles at the centre of the circle.

Q.7 A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that th segments BD and DC into which BC is divided by the point of contact D are o lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.

Ans

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10

Let the given circle touch the sides AB and AC of the triangle at point E and respectively and the length of the line segment AF be x. be x. In

ABC,

CF = CD = 6cm (Tangents on o n the circle from point C) BE = BD = 8cm (Tangents (Tangen ts on the circle from point B) AE = AF = x = x (Tangents (Tangents on the circle from point A) AB = AE + EB = x = x + + 8 BC = BD + DC = 8 + 6 = 14 CA = CF + FA = 6 + x + x 2 s = AB + BC + CA = x + x + 8 + 14 + 6 + x + x = 28 + 2 x s = s = 14 + x + x

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102

Area of ΔOAB = Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

Either x+ Either x+14 14 = 0 or x or x − 7 =0 Therefore , x = −14and 7 However, x However, x = −14 is not possible as the length o f the sides si desup will be negative. Sign to vote on this title Therefore, x Therefore, x = = 7

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Ans Let us join point O to C.

In ΔOPA and ΔOCA, OP = OC (Radii of the same circle) AP = AC (Tangents from point A) Sign up to vote on this title

AO = AO (Common side) ΔOPA

ΔOCA (SSS congruence criterion)

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104 Therefore, ∠POA + ∠COA + ∠COB + ∠QOB = 180º From equations (i (i) and (ii (ii), ), it can be observed that 2∠COA + 2 ∠COB = 180º ∠COA

+ ∠COB = 90º

∠AOB

= 90°

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Chapter 11

Construction

Q.1 Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts. Give the justification of the construction. Ans. A line segment of length 7.6 cm can be divided in the ratio of 5:8 as follows.

Step 1 Draw line segment AB of 7.6 cm and draw a ray AX making an acute angle wit line segment AB. Step 2 Locate 13 (= 5 + 8) points, A 1, A2, A3, A4 …….. A13, on AX such that AA A1A2 = A2A3 and so on. Step 3 Join BA13.

Step 4 Through the point A5, draw a line parallel to BA13 (by making an angle equal t ∠AA13B) at A5 intersecting AB at point C.

C is the point dividing line segment AB of o f 7.6 cm in the required ratio of 5:8. Sign up to vote on this title

The lengths of AC and CB can be measured. It comes out to 2.9 cm and 4.7 c Useful Not useful respectively.

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106

Justification

The construction can be justified by proving that

By construction, we have A5C || A13B. By applying Basic proportionality theorem for th triangle AA13B, we obtain

… (1)

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From the figure, it can be observed that AA5 and A5A13 contain 5 and 8 equal divisions o line segments respectively.

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This justifies the construction.

Q.2 Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose side are triangle.

times the corresponding sides of the isoscele

Give the justification of the construction.

Ans Let us assume that ΔABC is an isosceles triangle having CA and CB of equal length base AB of 8 cm, and AD is the altitude of 4 cm.

A ΔAB'C' whose sides are

times of ΔABC can be drawn as follows.

Step 1

Draw a line segment AB of 8 cm. Draw arcs of same radius on both sides of the lin segment while taking point A and B as its centre. Let these arcs intersect each other at O and O'. Join OO'. Let OO' intersect AB at D. Step 2

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Taking D as centre, draw an arc of 4 cm radius whichcuts the extended line segment OO at point C. An isosceles ΔABC is formed, having CD (altitude) as 4 cm and AB (base) a 8 cm. Useful

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108 Locate 3 points (as 3 is greater between 3 and 2) A 1, A 2, and A3 on AX such that AA A1A2 = A2A3. Step 5

Join BA2 and draw a line through A3 parallel to BA2 to intersect extended line segmen AB at point B'. Step 6

Draw a line through B' parallel to BC intersecting the extended line segment AC at C ΔAB'C' is the required triangle.

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Justification

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Maths ∠BAC ∴

10 = ∠B'AC' (Common)

ΔABC ∼ ΔAB'C' (AA similarity criterion)

… (1) In ΔAA2B and ΔAA3B', ∠A2AB

= ∠A3AB' (Common)

∠AA2B

= ∠AA3B' (Corresponding angles)

∴

ΔAA2B ∼ ΔAA3B' (AA similarity criterion)

On comparing equations (1) and (2), we obtain

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This justifies the construction.

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110 Step 1

Draw a circle of 4 cm radius with centre as O on the given plane. Step 2

Draw a circle of 6 cm radius taking O as its centre. Locate a point P on this circle an join OP. Step 3

Bisect OP. Let M be the mid-point of PO. Step 4

Taking M as its centre and MO as its radius, draw a circle. Let it intersect the given circl at the points Q and R. Step 5

Join PQ and PR. PQ and PR are the required tangents.

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PO = 6 cm QO = 4 cm Applying Pythagoras theorem in ΔPQO, we obtain 2

2

2

PQ + QO = PQ 2

2

PQ + (4) = (6)

2

2

PQ + 16 = 36 2

PQ = 36 − 16 2

PQ = 20 PQ PQ = 4.47 cm Justification

The construction can be justified by proving that PQ and PR are the tangents to the circl (whose centre is O and radius is 4 cm). For this, let us join OQ and OR. Sign up to vote on this title

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112 ⇒ OQ ⊥ PQ

Since OQ is the radius of the circle, PQ has to be a tangent of the circle. Similarly, PR a tangent of the circle.

Q.4 Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radiu 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangent to each circle from the centre of the other circle. Give the justification of th construction . Ans. The tangents can be constructed on the given circles as follows. Step 1

Draw a line segment AB of 8 cm. Taking A and B as centre, draw two circles of 4 cm an 3 cm radius. Step 2

Bisect the line AB. Let the mid-point of AB be C. Taking C as centre, draw a circle o AC radius which will intersect the circles at points P, Q, R, and S. Join BP, BQ, AS, an AR. These are the required tangents.

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∠ASB

11

is an angle in the semi-circle. We know that an angle in a semi-circle is a rig

angle. ∴ ∠ASB

= 90°

⇒ BS ⊥ AS

Since BS is the radius of the circle, AS has to be a tangent of the circle. Similarly, AR BP, and BQ are the tangents.

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114

Chapter 12

Areas Related to Circles Perimeter(circumference) and Area of a Circle: Perimeter(circumference) = 2 Area = r 2 Area of the sector of angle θ = θ =

r

2

360

r

Length of an arc of a sector of angle θ =

360

2

r

Q.1 The radii of two circles are 19 cm and 9 cm respectively. Find the radius of th circle which has circumference equal to the sum of the circumferences of the two circles. st

Ans Radius (r (r 1) of 1 circle = 19 cm nd

Radius (r (r 2) or 2 circle = 9 cm

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11 rd

st

nd

Circumference of 3 circle = Circumference of 1 circle + Circumference of 2 circle 2πr 2πr = = 38π + 18π = 56π

Therefore, the radius of the circle which has circumference equal to the sum of th circumference of the given two circles is 28 cm.

Q.2 If the perimeter and the area of a circle are numerically equal. Find the radius. Ans. Let the radius of the circle be r .

Circumference of circle = 2πr 2πr 2

Area of circle = πr πr

Given that, the circumference of the circle and the area of the circle are equal. 2

This implies 2πr 2πr = = πr πr 2 = r

Therefore, the radius of the circle is 2 units.

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Q.3 Find the area of a quadrant of a circle whose circumference is 22 cm.

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116 Circumference = 22 cm 2πr 2πr = = 22

Quadrant of circle will subtend 90° angle at the centre of the circle.

Area of such quadrant of the circle

Q.4 A car has two wipers which do not overlap. Each wiper has blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of th blades.

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Area swept b y 2 blades =

Q.5 Area of a sector of angle p (in degrees) of a circle with radius R is (in

Ans.

We know that area of sector of angle θ =

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Area of sector of angle P =

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118

Ans.

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Radius of inner circle = 7 cm

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Q.7 In the given figure, OACB is a quadrant of circle with centre O and radius radius 3.5 cm If OD = 2 cm, find the area of the (i) Quadrant OACB (ii) Shaded region

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120

(i) Since OACB is a quadrant, it will subtend 90° angle at O.

Area of quadrant OACB

(ii) Area of ΔOBD Sign up to vote on this title

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Q.8 AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see the given figure). If AOB = 30°, find the area of the shade region.

Ans

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122

Q.9 In the given figure, AB and CD are two diameters of a circle (with centre perpendicular to each other and OD is the diameter of the smaller circle. If OA = cm, find the area of the shaded region.

Ans

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Radius (r (r 1) of larger circle = 7 cm

Radius (r (r 2) of smaller circle Area of smaller circle

Area of semi-circle AECFB of larger circle

Area of

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Area of the shaded region

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124

Chapter 13

Surface Area & Volumes

Q. 1 A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Ans

It can be observed that radius (r (r ) of the cylindrical part and the hemispherical part is th same (i.e., 7 cm). Sign up to vote on this title

Height of hemispherical part

= Radius = 7 cm

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Q.2 From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area o 2 the remaining solid to the nearest cm .

Ans.

Height

= 2.4 cm

Diameter

= 1.4 cm

Therefore, radius (r (r ) = 0.7 cm Sign up to vote on this title

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126

2

2

The total surface area of the remaining solid to the nearest cm is 18 cm .

Q.3 Johny, an engineering student, was asked to make a model shaped like a cylinde with two cones attached at its two ends by using a thin aluminum sheet. Th diameter of the model is 3 cm and its length is 12 cm. if each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume th outer and inner dimensions of the model to be nearly the same.)

Ans.

From the figure, it can be observed that Height (h (h1) of each conical part = 2 cm

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Height (h (h2) of cylindrical cylindrical part = 12 − 2 × Height of conical part

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Q.4 A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the 3 mass of the pole, given that 1 cm of iron has approximately 8 g mass. [Use π = 3.14]

Ans

From the figure, it can be observed that Height (h (h1) of larger cylinder = 220 cm

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128

Mass of 1

iron = 8 g

Mass Mass of 11153 111532.8 2.8

iron iron = 1115 111532 32.8 .8 × 8 = 8922 892262 62.4 .4 g = 892. 892.26 262 2 kg

Q. 5 A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinde of radius 6 cm. Find the height of the cylinder. Ans Radius (r (r 1) of hemisphere = 4.2 cm

Radius (r (r 2) of cylinder = 6 cm Let the height of the c ylinder be h. The object formed by recasting the hemisphere will be the same in volume. Volume of sphere = Volume of cylinder c ylinder

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Q. 6 A drinking glass is in the shape of a frustum of a cone of height 14 cm. Th diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Ans.

Radius (r (r 1) of

Radius (r (r 2) of Capacity of glass = Volume of frustum of cone

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130

Therefore, the capacity of the glass is

.

Q.7 A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm water in it. Porous bricks are placed in the water until the cistern is full to the brim Each brick absorbs one-seventeenth of its own volume of water. How many brick can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm? Ans. Volume of cistern

= 150 × 120 × 110 3

= 1980000 cm

Volume to be filled in cistern = 1980000 − 129600 3

= 1850400 cm

Let n numbers of porous bricks were placed in the cistern. Volume of n bricks

= n × 22.5 × 7.5 × 6.5 = 1096.875n 1096.875n

As each brick absorbs one-seventeenth of its volume, therefore, volume absorbed b these bricks

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Q.8 A right triangle whose sides are 3 cm and 4 cm (other than hypotenuse) is made t revolve about its hypotenuse. Find the volume and surface area of the double con so formed. (Choose value of π as found appropriate.)

Ans

The double cone so formed by revolving this right-angled triangle ABC about i hypotenuse is shown in the figure. Hypotenuse = 5 cm

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132 Volume of double cone = Volume of cone 1 + Volume of cone 2

3

= 30.14 cm

Surface area of double cone = Surface area of cone 1 + Surface area of cone 2 = πrl πrl 1 + πrl πrl 2

2

= 52.75 cm

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13

Chapter 14

Statistics The mean for grouped data can be found by : (a) the direct method :

(b) the assumed mean method

© the step deviation method

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134

where l = lower limit limit of median median class, class, n = number number of observat observations, ions, cf = cumulative frequency of class preceding the median class, f = frequency frequency of median median class, class, h = class size size (assuming (assuming class class size to be equal) equal)

Q.1 A survey was conducted by a group of students as a part of their environmen awareness programme, in which they collected the following data regarding th number of plants in 20 houses in a locality. Find the mean number of plants pe house. Number of plants

0−2

2−4

4−6

6−8

8 − 10

10 − 12

12 − 14

Number of houses

1

2

1

5

6

2

3

Which method did you use for finding the mean, and why? Ans. To find the class mark ( xi) for each interval, the following relation is used. Sign up to vote on this title

Class mark ( x xi) = xi and f and f x i i can be calculated as follows.

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13

4−6

1

5

1×5=5

6−8

5

7

5 × 7 = 35

8 − 10

6

9

6 × 9 = 54

10 − 12

2

11

2 ×11 = 22

12 − 14

3

13

3 × 13 = 39

Total

20

162

From the table, it can be observed that

Mean,

Therefore, mean number of plants per house is 8.1. Sign up to vote on this title

Not f useful Here, direct method has been used as the values of class marks ( x x and Useful i) and f i are small.

Q.2 In a retail market, fruit vendors were selling oranges kept in packing boxes. Thes

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136

Find the mean number of mangoes kept in a packing box. Which method of findin the mean did you choose? Ans. Number of oranges

Number of boxes f i i

50 − 52

15

53 − 55 55

110

56 − 58 58

135

59 − 61 61

115

62 − 64

25

It can be observed that class intervals are not continuous. There is a gap of 1 between tw clas classs inte interv rval als. s. The There refo fore re,, has has to be be adde added d to the the uppe upperr clas classs limi limitt and and subtracted from the lower class limit of each interval.

has has to b

Class mark ( x xi) can be obtained by using the following relation.

Class size (h (h) of this data = 3

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Taking 57 as assumed mean (a (a), d i, ui, f iui are calculated as follows.

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58.5 − 61.5

115

60

3

1

115

61.5 − 64.5

25

63

6

2

50

Total

400

25

It can be observed that

Mean number of oranges kept in a packing packing box is 57.19.

Q.3 The following data gives the information on the observed lifetimes (in hours) of 225 electrical components: Sign up to vote on this title

Lifetimes (in hours)

0 − 20

20 − 40

40 − 60

Useful Not useful 60 − 80 80 − 100 100 − 120

Frequency

10

35

52

61

38

29

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138 Lower class limit (l (l ) = 60 Frequency ( f f 1) = 61 Frequency ( f f 0) = 52 Frequency ( f f 2) = 38 Class size (h (h) = 20

Therefore, modal lifetime of electrical components is 65.625 ho urs.

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Q.4 The given distribution shows the number of runs scored by some top batsmen o the world in one-day international cricket matches.

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6000 − 7000

7

7000 − 8000

6

8000 − 9000

3

9000 − 10000

1

10000 − 11000

1

Find the mode of the data.

Ans. From the given data, it can be observed that the maximum class frequency is 18 belonging to class interval 4000 − 5000.

Therefore, modal class = 4000 − 5000 Lower limit (l ) = 4000 Frequency ( f f 1) = 18 Frequency ( f f 0) = 4 Frequency ( f f 2) = 9 Class size (h (h) = 1000

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140

Q.5 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption (in units)

Number of consumers

65 − 85

4

85 − 105

5

105 − 125

13

125 − 145

20

145 − 165

14

165 − 185

8

185 − 205

4

Ans.

To find the class marks, the following relation is used.

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Taking 135 as assumed mean (a (a), d i, ui, f iui are calculated according to step deviatio Useful Not useful method as follows.

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class mark

135

4

75

− 60

−3

−1

85 − 105

5

95

− 40

−2

−1

105 − 125

13

115

− 20

−1

−1

125 − 145

20

135

0

0

0

145 − 165

14

155

20

1

14

165 − 185

8

175

40

2

16

185 − 205 205

4

195

60

3

12

Total

68

Monthly consumption (in units)

Number consumers (f i )

65 − 85

of

x i i

d i i = x i i −

7

From the table, we obtain

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142 Modal class = 125 − 145 Lower limit (l (l ) = 125 Class size (h (h) = 20 Frequency ( f f 1) = 20 Frequency ( f f 0) of class preceding modal class = 13 Frequency ( f f 2) of class succeeding the modal class = 14

To find the median of the given data, cumulative frequency is calculated as follows. Monthly consumption (in units)

Number of consumers

65 − 85

4

85 − 105

5

Cumulative frequency

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4+5=9

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14

From the table, we obtain n = 68

Cumulative frequency (cf (cf ) just greater than 125 − 145.

is 42, belonging to interv

Therefore, median class = 125 − 145 Lower limit (l (l ) of median class = 125 Class size (h (h) = 20 Frequency ( f f ) of median class = 20 Cumulative frequency (cf (cf ) of class preceding median class = 22

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Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectivel

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144 Number of letters

1−4

4−7

7 − 10

10 − 13

13 − 16

16 − 19

Number of surnames

6

30

40

6

4

4

Determine the median number of letters in the surnames. Find the mean number o letters in the surnames? Also, find the modal size of the surnames. Ans. The cumulative frequencies with their respective class intervals are a s follows. Number of letters

Frequency (f i i )

Cumulative frequency

1−4

6

6

4−7

30

30 + 6 = 36

7 − 10

40

36 + 40 = 76

10 − 13

16

76 + 16 = 92

13 − 16

4

92 + 4 = 96

16 − 19

4

96 + 4 = 100

Total (n (n)

100

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It can be observed that the cumulative frequency just greater than Not useful Useful 76, belonging to class interval 7 − 10.

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14

Median

= 8.05 To find the class marks of the given class intervals, the following relation is used.

Taking 11.5 as assumed mean (a (a), d i, ui, and f and f iui are calculated according to step deviatio method as follows. Number of letters

Number of surnames

x i i

d i i = x i i − 11.5

−Sign 9 up to vote on−this 3 title −18

f i i u i i

f i i

1−4

6

2.5

4−7

30

5.5

7 − 10

40

8.5

Useful

−6 −3

Not useful −2

− 60

−1

− 40

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146 From the table, we obtain ∑ f iui = −106 ∑ f i = 100

Mean,

= 11.5 − 3.18 = 8.32 The data in the given table can be written as Number of letters

Frequency (f i i )

1−4

6

4−7

30

7 − 10

40

10 − 13

16

13 − 16

4

16 − 19

4

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Class size (h (h) = 3 Frequency ( f f 1) of modal class = 40 Frequency ( f f 0) of class preceding the modal class = 30 Frequency ( f f 2) of class succeeding the modal class = 16

Therefore, median number and mean number of letters in surnames is 8.05 and 8.3 respectively while modal size of surnames is 7.88.

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148

Chapter 15

Probability Probability

The probability of a sure event (or certain event) is 1 and impossible event is 0. The probability of an event E is a number P(E) such that 0 ≤ P (E) ≤ 1 .

An event having only one outcome is called an elementary event. The sum of th probabilities probabilit ies of all the t he elementary elemen tary events even ts of an experimen experimentt is 1 For any event E, P (E) + P (E ) = 1 , where E stands for ‘not E’. E and E are called complementary events.

Q.1 Q.1 If P(E) = 0.05, what is the probability of „not E‟? Ans. We know that,

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(ii) an orange icecream ?

Ans (i) The bag contains orange icecreams only. It does not contain any mango icecream This implies that every time, she will take out only an orange icecream. Therefore, eve that Pinky will take out mango icecream is a impossible event.

Hence, P (a mango icecream) = 0

(ii)As the bag has an orange icecream, Pinky will take out only orange icecream Therefore, event that Pinky will take out an orange icecream is a sure event. P (an orange icecream) = 1

Q.3 Monty buys a fish from a shop for his aquarium. The shopkeeper takes out one fis at random from a tank containing 5 male fish and 8 female fish (see the given figure). What is the probability that the fish taken out is a male fish? Ans. Total number of fishes in a tank

= Number of male fishes + Number of female fishes = 5 + 8 = 13

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Q.4 A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Johny wins if all the tosses give the same result i.e., three heads or three tails

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150

P (Johny will win the game)

P (Johny will lose the game)

Q. 5 A box contains 12 pencils out of which x are black. If one pencil is drawn a random from the box, what is the probability that it will be a black pencils? Ans. If 6 more black pencils are put in the box, the probability of drawing a black pencil now double of what it was before. b efore. Find x Find x..

Ans. Total number of pencils = 12 Total number of black pencils = x = x

P (getting a black pencils) = If 6 more black pencils are put in the box, then Total number of pencils = 12 + 6 = 18 Total number of black pencils = x = x + + 6 Sign up to vote on this title

P (getting a black pencil now) According to the condition given in the question,

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Q.6 A box contains 24 pencils, some are green and others are blue. If a pencil is drawn a random random from from the the box , the proba probabilit bility y that it it is green green is pencils in the box

. Find the the number number of of blu

Ans Total number of pencils = 24

Let the total number of green pencils pen cils be x be x.. Then, total number of blue pencils = 24 − x − x

P (getting a given pencils) According to the condition given in the question,

Therefore, total number of green pencils in the box = 16 Hence, total number of blue pencils = 24 − x − x = 24 − 16 = 8

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