SSLC Maths Question Bank & Solution All Chapters

SSLC Ma Mathematics

SSLC

(Plus Two Maths & Science Teacher)

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SSLC Ma Mathematics

Arithmetic Sequences Equations:  First term: f  Common difference: d  Nth term:

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 1     (a-common difference; a+b first term) Common difference          Sum of sequence:   2 2    1    2      2       2   2            etc.    ;        

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Common difference =d 8th term = 35 11th term = 47 8th term + 3d = 11th term 3d=11th term - 8th term = 47- 35= 12 14th term = 11th term + 3d =47 + 12 = 59 5th term = 8 th term – 3d =35-12 = 23 17th term = 8th term + 9d 17th term = 8th term + 3 x12 = 8 th term + 36

Common difference d = 26 – 22 = 4 10 th term is  = 46

 22 7  14

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If any term of this sequence is divided by the common difference 4, the remainder is 2. 50 divided by 4 gives remainder 2. So, 50 is a term of this sequence The difference between any two terms of an arithmetic sequence will be the multiple of its common difference. Here, 50 is not a multiple of 4. So, for this sequence, 50 cannot be the difference of two terms.

To get the first term, first find the remainder when 100 is divided by 7, remainder is 2. Then subtract 2 from 100 and add 7 So, 1st term 100-2+7 105 To get the last term, find the remainder when 500 is divided by 7, remainder is 3. Subtract 3 from 500 So, last term 500-3 497 b) Common difference,

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  49771057 n  (497-98)/7 57 So, number of term 57

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First term a 6 Sum of first two terms 10 a+a+d 10 d 10-12 -2 Third term a 3 a+2d 2

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First term a = 7 Common difference d = 29 – 7 = 22 nth term = 685 a + (n-1)d = 685 7 + 22n - 22 = 685 n = 31.82 Thus 685 is not a term of the arithmetic sequence 7, 29, 51 … Because n is not a whole number

First term, f =205 and Common difference d = 6 Let the nth term of the given AP be the first negative term. Then, x n < 0 f + (n-1) d < 0 205+ (n-1) x 6 < 0 205 + 6n - 6 < 0 9048332443


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6n > 199 n > 33.17 i.e., n ≥ 34 Thus, the 34 th term of the given sequence is the first negative term.

According to the given information, x 5 = f + 4d = 34 and x 1515 = f + 14d = 9 Solving the two equations, we get f = 44 d = -2.5 Therefore, rth term term = f + (r-1)d = 44 -(r-1)2.5 = 46.5 - 2.5 r

Natural numbers starting from 1 are 1, 2, 3... Multiplying with 5; 5, 10, 15... Adding 2;

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7, 12, 17... This is an Arithmetic sequence with common difference 5. If any term of this sequence is divided by the common difference 5, the remainder is 2. 100 divided by 5 gives remainder 0, not 2. So, 100 is not a term of this sequence.

7, 5, 3... ) x n = f + (n-1) d = 7+ (n-1) × -2 = -2n +9 Take -30 as the nth term. If we get n as a natural number, it will be a term of the sequence. -2n + 9 = -30 n = -39 / -2 n = 19.5 Here, n is not a natural number. So, -30 is not a term of this sequence.

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a) Common difference,

               188  (147-67)/(18-8)  80/10 8

b) 1st term + 25th term= 8th term + 18th term =67+147 = 214 c) 13th term = (12th term + 14th term)/2 = (1st term + 25th term)/2 = 214 / 2 = 107 d) Sum of the terms = Number of terms x Middle term Sum of first 25 terms = 25 x 13th term = 25 x 107 = 2675 For what value of n, the nth terms of the arithmetic sequence 63, 65, 67… and 3, 10, 17… are equal?

Here, f=63, d=2 Therefore x n=63 + (n-1)2 Here, f=3, d=7 Therefore, a n= 3 + (n-1)7 Therfore, 63 + (n-1)2 = 3 + (n-1)7 60 = 5(n-1)

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n -1 = 12 n = 12+1 = 13

Here, f = 5 and d = 10 x 3131= f + 30d = 5 + 30 x 10 = 305 Let the required term be the nth term. Then, x n = 130 + x 3131 f + (n-1)d = 130+305 5 + (n-1)10 = 435 (n-1)10 = 430 n-1 = 43 n = 44 Hence, the 44th term of the given AP is 130 more than its 31st term.

f = -4, x n = 29, Sn = 150 Given that the last term is 29. 29 -4 + (n - 1)d 33 (n - 1)d … (1) Also, given that that the sum of all all its term is 150. 150 

   24   1 300  n [-8 + 33] [Using (1)]

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 

300 n (25) n 12 So, from (1), we get: d = 3

x n= 3+2n Now, put n=1,2,3 x 1= 3+2(1) = 5 x 2= 3+2(2) = 7 x 3= 3+2(3) = 9 Thus, the terms of the AP are 5,7,9 Here, f = 5 and d = 2

   2 25 2412  12[10+46]  12 56  672

-5, -8, -11, … -230 forms an AP with f=-5, d=-8-(-5) =-3 Let -230 = x n = f+(n-1)d = -5+(n-1)(-3) -230 + 5 = (n-1)(-3) n -1 = 75 n = 76

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  [(-5) + (-230)]  38 (-235)  - 8930 Find the sum of all two-digit odd positive numbers.

Two digit odd positive numbers are 11, 13, 15, ...., 99. First term, f = 11 Common difference, d = 13 - 11 = 2 Last term, x n = l =99 =99 Now, x n f + (n - 1)d 99 11 + (n - 1)2 99 11 + 2n - 2 2n 90 n 45

       2  45   2 1199 2475

Thus, the sum of all two-digit odd positive numbers is 2475.

1) 2)

Find the 8th term of an arithmetic sequence whose 15th term is 47 and the common difference is 4. [19] Find the 31st term of an arithmetic sequence whose 11th term is 38 and the 16th term is 73. [178] 9048332443


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3)

4) 5)

6) 7) 8)

9) 10) 11) 12)

The 10th term of an arithmetic sequence is 52 and 16th 16 th term is 82. (a) Find the 32nd term (b) Obtain the general term. [x 3232 = 162, x n = 5n + 2] Which term of the arithmetic sequence 3, 8, 13, 18 ... is 248? [50] The 7th term of an arithmetic sequence is – 4 and its 13th term is –16. (a) What is common difference? difference? (b) Find the arithmetic sequence. [-2; 8, 6, 4, ..... ] The 17th term of an arithmetic sequence exceeds its 10th term by 7. Find the common difference. [1] If the 10th term of an arithmetic sequence is 52 and 17th term is 20 more than the 13th term, find the arithmetic sequence. [7, 12, 17, 22.......] The 9th term of an arithmetic sequence is equal to 7 times the 2nd term and 12th term exceeds 5 times the 3rd term by 2. (a) Find the first term (b) Find the common difference. [a = 1, d = 6] Find the middle term of an arithmetic sequence with 17 terms whose 5th term is 23 and the common difference is –2. [15] Find 20th and 25th terms of an arithmetic sequence 2, 5,8,11… [59, 74] For what value of n the nth term of the arithmetic sequence 23, 25, 27, 29 ….and -17,-10,-3, 4 ….are equal? [9] The sixth term of an AP is -10 and tenth term is -26.Find its 15th term. [-46] 9048332443


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13) 14) 15) 16) 17) 18) 19) 20) 21)

22)

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The angles of a triangle are in arithmetic sequence. If the greatest angle equals the sum of the other two, find the angles. [30°, 60°, 90°] Three numbers are in arithmetic sequence. If the sum of these numbers is 27 and the product is 648, find the numbers. [6, 9, 12] Find sum of 1 + 3 + 5 + ....... to 50 terms. [2500] Find the sum of first 30 even natural numbers. [930] The 10th term of an arithmetic sequence is 29 and and sum of the first 20 terms is 610. Find the sum of the first 30 terms. [1365] How many terms of the arithmetic sequence 3, 5, 7, 9 … must be added to get the sum 120? [10] The 3rd term of arithmetic sequence is –40 and 13th term is 0. Find the common difference. [4] The sum of three numbers of an AP is 15. Find its first term. [5] The seats in a theatre are arranged in 20 rows. Each row contains 8seats more than the number of seats in the previous row. If the 1st row contains 70seats find the total number of seats in the theatre. [2920] In a flower decoration, flowers are arranged in 10 concentric circles, such that their numbers from an AP. There are 40 flowers in 3rd circle and 60 flowers in 5th circle. Find the total number of flowers. [650] If (x – 3), (x + 6), (2x – 3) are 1st three consecutive terms of an AP, find x. [18] 9048332443


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Circles Points:  The ends of a diameter of a circle are joined to some point and we get a right angle at that point. At points within the circle we get an angle larger than a right angle. For points outside the circle, we get an angle smaller than aright angle.  Any diameter of the circle divides it into two equal arcs; and we get a pair of right angles by joining points on each to the ends of the diameter.  Two points on a circle divide it into a pair of arcs. The angle got by joining these two points to a point on one of these arcs is equal to half the central angle of the alternate arc.  The angle made by an arc at any point on the alternate arc is equal to half the angle made at the centre.  Every chord of a circle divides it into two pruts. Such a part is called a segment of a circle.  Angles in the same segment of a circle are equal.  A chord divides a circle into a pair of segments.  Angles in alternate segments are supplementary.  If all vertices of a quadrilateral are on a circle, then its opposite angles are supplementary.  Suppose a circle is drawn through three vertices of a quadrilateral. If the fourth vertex is outside this circle, then the sum of the angles at this vertex and the opposite vertex is less than 180°; if the fourth vertex is inside the circle, then this sum is greater than 180°.  A quadrilateral for which a circle can be draVvn through all the four vertices is called a cyclic quadrilateral.  All rectangles are cyclic quadrilaterals. Isosceles trapeziums are also cyclic quadrilaterals.  If the opposite angles of a quadrilateral are supplementary, then we can draw a circle through all four of its vertices.  The exterior angle at a vertex of a cyclic quadrilateral is equal to the interior opposite angle.

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Join OP Since OA = OP ACO = OAP = 35o Similarly, OB = OP and OPB = OBP = 40o APB = 35o + 40o = 75o

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∠AOB2x75 150 o

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Join OP. In ∆OQP, OQ = OP = r OQP = OPQ = 35o

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In ∆OPR, OR = OP = r ORP = OPR = 45o QPR = 35o + 45o = 80o

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∠QOR = 2∠QPR = 2

80o

QOR = 160o

= [180 - (2 x 42)]/2 = 48

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∠R=90 ∠Q=90-30=60

The ratio of sides of triangle is 1:√3:2 Thus PR=6√3

Join OP. In ∆OAP, OA = OP = radius

∠OAP = ∠OPA = 32

o

In ∆OPR, OB = OP = radius

∠OBP = ∠OPB = 47 ∠ APB = 32 + 47 = 79 ∠ AQB = 180 -79 =10 o

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Take a point E on the circle, join AE and CE. AEC=100/2=50o AEC + ABC = 180o (Opposite Angles of a cyclic quadrilaterals) ABC = 130o ABC + CBD = 180o (linear pair) 130o + CBD = 180o CBD = 50o

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(a) 180o (b) EAD

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24)

ABCD is a cyclic quadrilateral. If BCD = 100,  ABD = 70, find  ADB. [30]

25)

ABCD is a cyclic quadrilateral with DBC = 80, BAC = 40, find BCD.

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26) 27) 28) 29) 30)

In cyclic quadrilateral ABCD, AD BC. B = 70, find remaining angles. [110, 70, 110]

Draw a rectangle of sides 5cm and 3cm and draw a square of the same area. 2 chords AB and CD intersect at a point O. If AO = 3.5cm, CO = 5cm, DO = 7cm, find OB.

[10cm]

2 chords AB and CD intersect at O. If AO = 8cm, CO = 6cm, OD = 4cm, find OB.

[3cm]

31)

Two circles intersect at two points A and B. AD and AC respectively are diameters to the two circles. Prove that B lies on the line segment DC. How do we draw a 22½ o angle

32)

In the picture below, O is centre of the circle. Find  AOB

33)

Compute specified angle in each

[1000]

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34)

What is the central central angle of arc ABC?

35)

Find all angles of quadrilaterals.

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Second Degree Equations Points:  A quadratic equation in the variable x is of the form ax2 + bx + c = 0, where a, b, c are real numbers and a ≠ 0.  Different methods to findout the solutions of second degree equations  Completing the square method  Quadratic Formula (Shreedharacharya’s rule)method

 4    √  2 is the solutions of second degree equation ax +bx+c=0,a≠0 2

b2 4ac is the discriminant of ax2+bx+c=0  If b2 4ac=0, then the equation has only one solution and the solution is –b/2a.  If b2 4ac <0 (a negative number),then the equation has no solution  If b2 4ac >0 (a positive number) then the equation has two different solutions 

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We have

2  7   3  0   72   32  0   2    74   32

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  7 7 3 7   2    4  4   2   4  49 3 7   4  16  2  49 3 7   4  16  2  49 24 25 7   4  16  16  16   74   54 1   3    2

Let the side of the squares be x and y meters. According to the condition, x 2 + y 2 = 500 ….(1)

4x 4x - 4y = 40 (x – y) = 10 y = x - 10 Substituting the value of y in (1), we get, x 2 + (x - 10)2 = 500 2x 2 - 20x - 400 = 0 x 2 - 10x - 200 = 0 x = 20 or x = -10 As the side cannot be negative, x = 20 Hence, side of the first square, x = 20m

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Side of the second square, y = (20 - 10) = 10 m

Let the three consecutive positive integers be x, x + 1, x + 2. x 2 + (x + 1) (x + 2) = 232 x 2 + (x 2 + 3x + 2) = 232 2x 2 + 3x - 230 = 0

 4 √  2 3  1849 1 849 √  4 x = 10 or -11.5

But, x is a positive integer, so, x = 10. Thus, the numbers are 10, 11, and 12.

Let the consecutive numbers numbers be x, x  2. x2  x  22  164 x2  x2  4x  4  164 2x2  4x ‐ 160  0 x2  2x ‐ 80  0 x  8 or x  ‐10

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Neglecting the negative negative value, we get, x  8. The numbers are 8 and 10.

Let the number of children be x. It is given that Rs 250 is divided amongst x children. So, money received by each child  250/x If there were 25 children more, then Money received by each child 250/x25 From the given information, 250  250  50  25 100  25125000  = -125 or 100 Since, the number of children cannot be negative, so, x  100. Hence, the number of children is 100. 1 00.

Let speed of the bus be x km/hr Time t = 450/x If speed is x + 10, then time T = 450/(x + 10)

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By question, 450/x - 450/(x + 10) =3/2 450 (x + 10) -450 x = (3/2) (x 2 + 10x) 4500 2 = 3 x 2 + 30x x 2 + 10x - 3000 = 0 x = 50 or x = -60 x = 50

Let the length and breadth of garden be x m and y m respectively. Area of the garden garden = 100 sq m xy = 100 m2 or y =100/x Suppose the person builds his house along the breadth of the garden. Then, we have: 2x + y = 30 2x + (100/x) =30 2x 2 - 30x + 100 = 0 x = 10, x = 5 When x = 10 m, we we have: y = 10 m When x = 5 m, we have: y = 20 m Thus, the dimensions of the garden are 10 m × 10 m or 5 m × 20 m. 8 2 e g a P

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x 2 + y 2 = 202 x 2 + y 2 = 400 Also x - y = 4 x = 4 + y

(4 + y)2 + y 2 = 400 2 y 2 + 8 y – 384 = 0 y = 12 y = – 16 Sides are 12cm and 16cm

36) 37) 38) 39) 40)

41) 42)

Perimeter of a rectangle is 40cm. If area is 96cm 2, find the sides. [12, 8] If from a number, twice its reciprocal is subtracted we get 1. What is the number? [2 or -1] The sum of the squares of two consecutive odd positive integers is 290. Find them. [11, 13] Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164. [10, 6] The speed of a boat in still water is 8 km/hr. It can go 15 km upstream 22 km downstream in 5 hours. Find the speed of the stream. [3 km/hr] Find two consecutive numbers whose squares have the sum 85. [6, 7] Sum of 2 numbers is 12. If the sum of their squares is 90, find the numbers. [9, 3] 9048332443


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43) 44) 45)

46) 47) 48) 49) 50) 51) 52) 53) 54)

If the sum of the squares of 2 consecutive even natural numbers is 244. Find the numbers. [10, 12] Square of a number is 60 more than 7times the number. Find the number. [12 or -5] The sum of squares of 2 consecutive odd numbers is 74. Which is the smaller of the numbers? [5 or 7] Anu is 4 years years older older than Vinu. If 4 is added to the product of their their ages, the result is 169. What are their ages? To the square of a natural number, four times the next natural number is added and the result is 36. What is number? The difference of two numbers is 6 and their product is 16. What are the numbers If from the squre of a number, six time the number is subtracted, we get 40. What is the number? How many terms of the arithmetic progression 3, 7, 11… must be added to get 253? If the product of a number with 6 more than the number is 160. What is the number? number? If the product of a number with 8 less than the number is 65, what is the number? How many terms of the arithmetic progression 4, 10, 16… starting from the first, are to be added to get 252? The width of a rectangle is 7metre more than its height and its area is 60 squre metres. Find the dimensions of rectangle. 0 3 e g a P

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Trigonometry Points: of a triangle triangle are 600,600,600; then its sides will  If the angles of be in the ratio 1:1:1 of a triangle triangle are 450,450,900; then its sides will  If the angles of be in the ratio 1:1:√2 of a triangle triangle are 300,600,900; then its sides will  If the angles of be in the ratio 1:√3:2  In triangle ABC Sin A=BC/AC Cos A= AB/AC Tan A=BC/AB  The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above a bove the horizontal level  The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level

Triangle side ratio is ratio 1:√3:2

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Altitude = hypotenuse × √3/2 = 4 × √3/2 = 3.46 cm

From figure, sin60 = 4/BD 0.8660 = BC/BD BD = 4.62 cm Therefore radius = 2.31 cm

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CD = AC sin60 = 6×0.8660 = 5.2 AD = AC cos60 = 6×0.5 = 3 BD = BA + AD = 3+7=10 In triangle BDC BC2 = BD2+CD2 = 102+5.22 = 100+27.04 = 127.04 BC = 11.3 Third side is 11.3cm ◦

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(a) A=300 (b) sin 60 = AB/AC = 10/AC AC = 10/0.8660=11.55 cm tan 60 = AB/BC BC = 10/1.73 =5.78 cm ◦

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(a) tan x= AD/DC=6/9=2/3 (b) x (c) tan x = BD/AD=BD/6 2/3=BD/6 BD=(2/3)× 6=4 cm ◦

◦

cos 60 = OP/AO=(r-3)/r 0.5 r = r - 3 r - 0.5 r = 3 r = 3/0.5 = 6cm 4 3 e g a P

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In ∆ ABD, AB/BD = tan 30 h /(45+x)=1/√3 /(45+x)=1/√3 x = (√3 (√3hh - 45) …..(1) In ∆ABC, AB/BC = tan 60 h / x = √3 x = h/√3 h/√3 …….(2) From equation (1) and (2), we get (√3h - 45) = h/√3 h = 38.97m

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In the figure AB is tower. tan 50 = AB/BC AB = CB tan 55 AB = 24 × 1.4281 = 34.2744 Height of the tower = 34.3 m

In figure PR = 28.5 In ∆PAR, PR/AR = tan 30 28.5/AR = 0.5773 AR = 49.3634

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In ∆PRB, PR/BR = tan 60 28.5/BR = √3 BR = BR = 16.4545 ST = AR – BR = 49.3634 – 16. 3634 = 32.9089 m

In ∆ABC, AB/BC=tan 60° BC=50/√3 …..(1) In ∆DCB, DC/BC = tan 30 h /BC = 1/√3 h = BC/√3 h = 16.67m

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55) 56) 57) 58)

59)

60)

61)

In ∆ABC, B = 90, C = 70, BC = 16cm, find AC. [Sin70 = 0.94; Cos 70 = 0.34; tan70 = 2.75] [47.05cm]

One angle of a triangle is 110 and the side opposite to it is 4cm long. What is its circumradius?

[2.13cm]

The angle between the radius and slant height of a cone is 600. Find the radius of the cone, if its slant height is 14 cm

[7cm]

A man standing on the deck of a ship, which is 10 m above water level. He observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. (a) Calculate the distance of the hill from the ship (b) Find the height of the hill.

[10√3m, 40m]

An observer in a lighthouse 100 m above the sea-level is watching the ship sailing towards the lighthouse. The angle of depression of the ship from the observer is 30°. How far is the ship from the lighthouse?

[100√3m]

A ladder is placed along a wall such that its upper end is touching the top of the wall. The foot of the ladder is 2m away from the wall and the ladder is making an angle of 60° with the level ground. Find the height of the wall.

[3.46m]

The top of a tower is seen at an angle of elevation of 40 0 from a point 30m away from the base of the tower. What is the height of the tower? [Sin 40 = 0.64; Cos 40 = 0.77; tan 40 = 0.84]

[25.20m]

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62)

63)

64)

65)

66)

The angle of elevation of a tower at a point is 45°. After going 40 m towards the foot of the tower, its angle of elevation becomes 60°. Find the height of the tower. [94.64m]

Two men are on the opposite sides of a tower. They measure the angles of elevation of the top of the towers as 30° and 45°. If the height of the tower is 60 m, find the distance between them.

[163.92 m]

From the top of a building 60m high the angles of depression of the top and bottom are observed to be 30º and 60º.Find the height of the tower.

[40m]

The horizontal distance between 2 towers is 70m.The angle of depression of the top of first tower when seen from the top of second tower is 30º. If the height of the second tower is 120m, find the height of the first tower.

[79.6m]

An aero plane when 3000m high passes vertically above another aero plane at an instant when the angle of elevation of the two aero planes from the same point on the ground are 60  and 45 respectively. Find the vertical distance between the aero planes.

[1268m]

67) 68) 69) 70)

A long pole leans against a short short wall, making a 400 angle with the ground. The foot of the pole is 2metres away from the bottom of the wall. What is the height of the wall? A man 1.7metres tall standing standing 10 metres away from from a tree sees the top of the tree at an angle of elevation 50 0. What is the height of the tree? When the sun is at an angle of elevation 480, the shadow of a tree is 18metres long. What is the height of the tree? The difference in the lengths of the shadow of a tower when the sun is at angles of elevations 300 and 600 is 45 metres. Compute the height of the tower. 9048332443


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Solids Points: Square Pyramid:  Lateral surface area = 4×½(Base edge × Slant height) L.S.A  Total Surface Area = Base Area × Lateral surface area  T.S.A  Relations connecting base edge a, lateral edge e, slant height l , height h and base diagonal d :

 2    2 

1  41      4       14   Volume = Base area × height  Volume          

 

Cones  The radius of the sector becomes the slant height of the cone; the arc length of the sector becomes the base circumference of the cone.  Suppose that a cone of base radius r and slant height l , radius of the sector l and and the central angle x ,then

   

  360  L.S.A of Cone  T.S.A of Cone       Volume     

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Spheres  T.S.A    

 4  Volume =    Hemispheres L.S.A  2  T.S.A  3   Volume =   

2=2×16×10=320 cm    2 = 16×16 + 320=576 cm      =100 – 64   6 Volume =   = 256 × 6=1536 cm

L.S.A = T.S.A=

2

2

3

44.

        41  40 =81 9 ∴ Diameter = 18cm

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Volume

       9  40 = 3394.3 cm3

Diameter of a football is 30cm. Surface area of a football

 4  4  15 900 The least area of leather required to make 50 footballs 50 90045000 Volume of air inside 50 such footballs   = 50    50    15   = 225000  cm 3

   2           3.5 12   3.5 641.67 cm Total Surface Area of solid 24 23.51243.5

Volume of solid

3

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= 418 cm2

Diameter and height of the cylindrical part of the tent are 126m and 5m. Total height of the tent is 21m.

      16  63  2563969 = 4225  65m The total surface area 2 26356365  4725 The total cost of the tent  4725  12 56700  178200 178200   



Volume of cylinder with height 2cm = n Volume of iron spheres

       4    6  2    3   1.5 72  4.5  72  16   4.5

3 4

Number of spheres are 16.

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Volume of one gulab jamun = volume of cylindrical part + 2 × (volume of hemispherical part)

   2  23        1.42.2  1.4 Volume of such 45 gulab jamun  7.97  Volume of Syrup  30% Volume of such 45 gulab jamun 338 cm 3

71)

A solid is hemispherical at the bottom and conical above. If the curved surface area of the two parts are equal, then from the ratio of the radius and height of the conical part.

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72)

A toy is in form of a cone mounted mounted on a hemi-sphere of radius 3.5 cm. The total height of the toy is 15.5 cm. (a) Find slant height. (b) Find its total surface area.

73)

A tent is of the shape of a right circular cylinder upto a height of 3 metres and conical above it. The total height of the tent is 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs. 2 per square metre, if the radius of the base is 14 metres.

74)

A solid sphere of radius 3cm is melted and then cast into small spherical balls each of diameter 0.6cm.Find the number of small balls thus obtained.

75)

A cylinder of radius 12 cm contains water to a depth of 20cm, a spherical iron ball is dropped into the cylinder and thus the level of water is raised to 6.75cm.Find the radius of the ball.

76)

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.

77)

A glass cylinder with diameter 20 cm has water to the height of 9 cm. A metal cube of 8 cm edge is immersed in it completely. Find the height by which the water will rise in the cylinder.

78)

The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm, find the height of the cylinder. 9048332443


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79)

If the number of square centimetres on the surface of a sphere is equal to the number of cubic centimetres in its volume, what is the diameter of the sphere?

80)

The diameter of a metallic solid sphere is 12 cm. It is melted and drawn into a wire having diameter of the cross-section 0.2 cm. Find the length of the wire.

81)

A semi-circular thin sheet of metal of diameter 28 cm, is bent to make an open conical cap. Find the capacity of the cap.

82)

An ice-cream cone has a hemispherical top. If the height of the cone is 9 cm and base radius is 2.5 cm, find the volume of ice cream cone.

83)

The volume of a square pyramid is 720 cubic centimetres and its base edge is12 centimetres. Find its height. Two square pyramids are of equal volume and the base edge of the first is double that of the the second. What fraction of the height of the second is the height of the first? Two square pyramids are of equal volume and the height of the first is double that of the second. What fraction of the base edge of the second is the base edge of the first? Compute the curved surface area of a cone of base radius 12 cm and slant height 25cm. What is the surface area of a cone with the diameter of the base 30cm and height 40cm? If the volumes of two spheres are 27 cubic centimetres and 64 cubic centimetres, what is the ratio of their radii? The ratio of the surface areas of two spheres is 3: 5. What is the ratio of their volumes? Compute the surface area of the largest sphere that can be cut out of a cube of side 15 centimetres.

84) 85) 86) 87) 88) 89) 90)

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91)

The base edges of two square pyramids are in the ratio 1: 2 and their heights are are in the ratio 1: 3. The volume of the first pyramid is 180 cubic centimetres. Compute the volume of the second. 92) A cylindrical vessel of base radius 10 centimetres and height 85 centimetres is completely filled with water. Spheres of radius 10 centimetres, as many as can be completely immersed in water, are put into the vessel. Find the volume of water remaining in the vessel. 93) A cylindrical rod of length 4 centimetres and diameter 4 centimetres is melted and recast into spheres of radius 2 centimetres. How many such spheres can be made? made? 94) A metal metal sphere of diameter diameter 24 centimetres centimetres is melted melted and and recast recast into cones of base radius and height 6 centimetres. How many such cones are made? 95) A metal metal sphere of diameter diameter 24 centimetres centimetres is melted melted and recast recast into cones of base radius and height 6 centimetres. How many such cones are made? 96) The cost of painting a hemispherical paper weight was 80 rupees. What will be the cost to paint a hemisphere of triple the radius at the same rate? 97) If the surface area of a solid hemisphere is 432π square centimetres, what is its radius? 98) If the inner radius of a hemispherical bowl is 60 centimetres, how many litres of water can it contain? 99) A petrol tank is in the shape of a cylinder with hemispheres of the same radius attached to both ends. If the total length of the tank is 6 metres and the radius is 1 metre, what is the capacity of the tank in litres? 100) A rocket is in the shape of a cylinder with a cone attached to one end and a hemisphere attached to the other. All of them of the same radius of 1.5 metre. The total length of the rocket is 7 metres and the height of the cone is 2 metres. Compute the volume of the rocket. 9048332443


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Coordinates Points  The coordinates of a point are the distances of the point from x-axis and y-axis.  The coordinates of a point on the x-axis are of the form ( x, 0) and of a oint oint on the -axis -axis are of the form form 0 .

The coordinates of rectangle are (0,0),(5,0),(5,4) and (0,4)

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(a) (b)

(6,0) and (-6,0) (0,6) and (0,-6)

Parallel to the x-axis A (4, 3) and C (−6, 3) Parallel to the y-axis B (3, 5) and D (3,−2),

Distance =|-3-4|=7

(a) (b) (c)

5 (5,0) Distance = |2-0|=2 9 4 e g a P

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Let the x coordinate of other end be x. Given, distance of A(-3,2) from B(x,10) is 10 units, i.e., i .e., AB =10. AB2 = (-3-x)2 + (2-10)2 102 =9 + x 2 +6x + 64 or 100 = x 2 + 6x + 73 x 2 + 6x - 27 = 0 x = -9 or 3 Thus, the x coordinate of the other end is -9 or 3. 101) Find a relation between x and y such that the point (x, y) is equidistant from the points A (7,1) and B(3,5). [x - y = 2] 102) Find the point on y – axis which is equidistant from (-5, -2) and (3, 2). [-2, 0] 103) Find the point on y axis which is equidistant from (-5,2) and (3,2) [(0,-2)] 104) Draw x and y axis then mark the following points. (4,3),(-4,7),(-4,-6),(5,9),(6,-4) 105) Draw x and y axis and mark the points P (-1, 6) and Q (6, 6) then join PQ. Test whether whether the following points are on the line PQ (3, 4), (-6, 6), (4, 6), (5, 6) [(-6, 6), (4, 6), (5, 6)] 106) Find the distance from x axis (4, 4), (4, 3), (5, 7),(4,-3) [4, 4, 5, 4] 107) Draw a circle passing through the points (0,-3),(0,-2),(2,3),(0,1) 108) Mark the following points without drawing axis [(5, 7),(3,4)] 9048332443


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109) Draw a triangle joining the points with coordinates (0,0),(4,0) and (2,5) 110) Draw a quadrilaterals joining the points with coordinates (0,0),(3,2),(4,6),(1,4) 111) Draw a quadrilaterals joining the points with coordinates (0,0),(1,4),(5,4) and (6,0)

Mathematics of Chance Points  An outcome of a random experiment is called an elementary event. outcomes/ Number  The probability of an event = Number of outcomes/ Number of all possible outcomes  The probability of an impossible event is 0, and that of sure

Total possibilities = {HH, HT, TH, TT} Number of possible outcomes = 4 The probability of getting at least one tail =3/4 57.

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Red balls=6 Green balls=8 White balls=8 Total number of balls= 6 + 8 + 8 = 22 The probability of getting a white or green ball =16/20 The probability of getting neither green ball nor a red ball =8/20

Number of red balls = 4 Number of blue balls = x Total Number of balls = 4 + x Probability of drawing a red ball from the bag = 5/ (4 + x) Probability of drawing a blue ball from the bag = x/ (4 + x) By condition

 3 4 4   4  8480   4  12 12

Hence the number of blue balls in the bag = 12

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Total number of outcomes = 20 (i) Prime numbers from 1 to 17 are 2,3,5,7,11,13,17,19. Number of outcomes = 8 The probability that the the card drawn is prime number = 8/20 (ii) Numbers are divisible by 3 are a re 3,6,9,12,15,18. Number of outcomes = 6 The probability that the card drawn is divisible by 3 =6/20 (iii) Numbers are divisible by 5 are a re 5,10,15,20. Number of outcomes = 4 The probability that the card drawn is divisible by 3 = 4/20

Possible outcomes = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Number of total outcomes = 8 (a) 1/8 (b) 1/8 (c) 3/8 (d) 4/8 = 1/2



Total probable pairs=6 6=36 (1,4),(2,3),(3,2) and (4,1) are the four pairs whose sum is 5 9048332443


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The probability of getting sum 5=4/36=1/9.

In a leap year, there are 366 days. We have, 366 days days = 52 weeks + 2 days. days. Thus, a leap year has always 52 sundays. probability = 2/7

(i) (ii) (iii) (iv)

1/2 14/100=7/50 The perfect squares are 4, 9, 16.....100 Probability=9/100 The prime numbers less than 25 are 2, 3, 5, 7, 11, 13, 17,19,23 Probability=9/100

112) A card is drawn from a well shuffled pack of 52 cards. Find the probability that the card is neither a red card nor a queen. [6/19] 113) A bag contains 3 red balls and 5 black balls. A ball is drawn at random from a bag. What is the probability that the ball drawn is (i) red (ii) not red 9048332443


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[3/8; 5/8] 114) A bag contains 5 red balls and some blue balls. If the probability of of drawing a blue ball from the bag is four times that of a red ball, find the number of blue balls in the bag. [20] 115) A bag contains 12 balls out of which x are white. (i) If one ball is drawn at random, what is the probability that it will be a white ball? (ii) If 6 more white balls are put in the bag, the probability of drawing a white ball will be double than that in (i). Find x. [x/12; x=3] 116) A bag contains 5 red, 8 white and 7 black balls. A ball is drawn at random from the bag. Find the probability that the ball drawn is ; (i ) red or white ( ii ) not black (iii ) neither white nor black [13/20, 13/20,1/4] 117) Find the probability that a number selected from the number 1 to 25 is not a prime number when each of the given numbers is equally likely to be selected. [16/25] 118) Find the probability that a number selected at a random from the numbers 1, 2, 3, ...., 35 is a (i) prime number (ii) multiple of 7 (iii) a multiple of 3 or 5 [11/35,1/7,16/35] 119) A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is : (i) white or blue (ii) red or black (iii) not white 9048332443


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(iv) neither white nor black [1/2,13/20,2/5,9/10] 120) A card is drawn at random from a well-shuffled deck of playing cards. Find the probability that the card drawn is : (i) a card of spade or an ace (ii) a red king (iii) neither a king nor a queen (iv) either a king or a queen [4/13,1/26,11/13,2/13] 121) Cards marked with numbers 3, 4, 5, ...., 50 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that number on the card drawn is : (i) divisible by 7 (ii) (ii ) a number which is a perfect square [7/48,1/8]

122) A box contains 100 bulbs out of which 10 are defective. What is the probability that if a bulb is drawn, it is (i) defective (ii) nondefective? 123) An integer is chosen from 1 to 15. Find the probability that the integer chosen is divisible by 4. 124) A bag contains 8 red, 6 white and 4 black balls. A ball is drawn at random from the bag. Find the probability that the drawn ball is: (i) red or white (ii) neither white nor black. 125) A bag contains 5 red balls and some black balls. If the probability of drawing a black ball is double that of a red ball, find the number of black balls in a bag. b ag. 126) A pair of dice di ce is thrown once. Find the probability of getting the sum of numbers on two dice as 11.

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Tangents Points:  Tangent to a circle at a point is perpendicular to the radius through the point of contact.  From a point, lying outside a circle, two and only two tangents can be drawn to it.  The tangent at a point on a circle is perpendicular to the radius through the point. a nd chord  Each of the two angles made by a tangent to a circle and through the point of contact is equal to an angle in the segment on the other side of the chord.  The lengths of the tangents from a point outside a circle are equal.  The bisectors of the three angles of a triangle intersect at a point.

Let PQ be a tangent to the circle from point P and OQ be the radius at the point of contact. OQP = 90° OP2 = OQ 2 + PQ 2 OQ 2 = OP2 – PQ 2 = 262 – 102 = (26 + 10) (26 – 10) = 36 × 16 OQ = 6 × 4 = 24

∠

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Radius of the circle

Since tangents from an exterior point to a circle are equal in length. BP = BQ ... (1) CP = CR ... (2) and, AQ = AR ... (3) from eqn. (3), we have AQ = AR AB + BQ = AC + CR AB + BP = AC + CP ... (4) Now, Perimeter of ∆ABC = AB + BC + AC = AB + (BP + PC) + AC = (AB + BP) + (AC + PC) Using (4) Perimeter of ∆ABC = 2 (AB + BP) = 2 AQ AQ =1/2 (Perimeter of ∆ABC).

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Join the points A, A, B and P to the the centre O of the the circle.

Look at the triangles ∆OAPand ∆OBP. Since the tangents PA and PB are perpendicular to the radii OA and OB these triangles are right right angled. angled. Both share share the the hypotenuse hypotenuse OP. Also OA = OB being radii of the circle. So by Pythagoras Theorem PA 2=OP2-OA 2=OP2-OA 2=PB2 From which we have PA = PB. Find all the angles of of the triangle in the figure below

From figure, ∠C=∠PAB=60 ∠B=∠QAC=40

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∠BAC+180-(60 +40 )=80 0

0

∠

0

∠

∠ ABC=90

o

Since AB being diameter is perpendicular to tangent BC at the point of contact. So ABP + PBC =90o (1) Also APB =90o (angle in the semi-circle) So BAP+ ABP = 90o (2) From (1) and (2), PBC = BAP

∠ ∠ ∠ ∠ ∠

∠

∠

127) A point P is 13 cm from from the centre of the the circle. The length of tangent drawn from P to the circle is 12 cm. Find the radius of the circle. [5cm] 128) The tangents at the points A and B on a circle centred at a point O meet at P. Prove that the line OP bisects APB 129) The tangents at the points A and B on a circle centred at the point O meet at P . Prove that the line OP bisects the line AB. 130) In the figure below the tangent to the circle at A and the chord BC extended meet meet at P . Prove that PB × PC = PA 2.

∠

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Hint:

∠PAB=∠PCA ∠PBA=∠CAB+∠PCA ∠PBA=∠CAB+∠PAB=∠PAC

∆PAB and ∆PAC are similar PB/PA=PA/PC PB × PC = PA 2 131) In the figure AB is a diameter of the circle and PQ is the tangent at A. Find the angles of ∆ABC

132) In the figure AB is a diameter of the circle and PQ P Q is the tangent at B. Prove that AP × AR = AQ × AB

133) In the figure PQ is the tangent to the circle at A. Find the angles of the quadrilateral ABCD

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134) Drawing the figures below according to the specifications.

135) Compute the length of the tangent from a point 29 cm away from the centre of a circle of radius 20 cm. [21cm] 136) Draw a circle of radius 3 centimetres and mark a point P which is 7 centimetres away from its centre. Draw the tangents tangents from P to the circle. 137) The tangents from a point P to a circle centred at 0 with radius r touch the circle at A and B. The line cuts 0P at Q. Prove that OP × OQ = r2 138) The length of the tangent from a point to a circle of radius 12 centimetres is 16 centimetres. How far away is this point from the centre of the circle? [20cm] 9048332443


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139) The length of the tangent from a point 61 centimetres away from the centre of a circle is 60 centimetres. What is the radius of the circle [11cm] 140) Draw triangle ABC with the specifications given below and draw the incircle of each (a) AB = 5 cm, BC = 6 cm, CA = 7 cm (b) AB = 7 cm A = 70 B = 50 (c) AB = BC = 6 cm B = 40 141) Draw an equilateral triangle with each side 6 centimetre and draw its circumcircle and incircle. 142) Draw the pictures given below according to the specifications.

∠

∠

∠

Polynomials Points:  The reminder on dividing the polynomial p(x) by the polynomial x-a is p(a)  If p(a)=0,then x-a is a factor of polynomial p(x).  If p(a)≠0,then x-a is not a factor of polynomial p(x)

3 6 e g a P

P (1) =3×1 -2×1 -3×1+2 = 0 3

2

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Therefore x -1 is a factor

P(x) =2x 2+5x+3 2x 2+5x+3=0 x =( -5 ± √1)/2= -1 or -3/2 Then the polynomial x + 2 , x + (3/2) are factors of 2x 2+5x+3 (x + 2) (x + 3/2) = x 2 + (5/2)x + 3 = ½(2x 2+5x+3) From this 2x 2+5x+3 = 2(x + 2) (x + 3/2)=(2x+3)(x+1) Find the value of k if remainder when 5x 3 + 4x 2 – 11x + k is divided by (x – 1) is 0. P(1) = 5x 3+4x 2-11x +k = 0 5×13+4×12-11×1 +k = 0 k=2

P(x) = x 3-2x 2+kx +7 = 11 P(4) = 43-2×42+k×4 +7 = 11 4k = -64+32-7+11=13-8=-28 k = -7

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143) When x 3 – 3x 2 – 5x + n is divided by (x + 3) remainder is 5. Find n. [-10] 144) Find the value of k if (x + 4) is a factor of 2x 3 + 11x 2 + kx – 36. [3] 145) If (x + 1) and (x ( x – 1) are factors of x 3 + 2x 2 + px + q, find p and q. [-1, -2] 146) When 3x 3 – 7x 2 + px + q is divided by (x – 2) and (x – 3), the remainders are 3 and 17 respectively. Find p and q. [-2, 5] 147) When P(x) = 3x 3 + kx 2 – 12x + 8 is divided by (x – 1), the remainder is 35. Find k and test whether (3x – 2) is a factor of P(x)? [36, no]

148) If (x + 2) and (x – 2) are factors of 2x 3 – 3x 2 – 8x + 12, find 3rd factor. 149) If (3x + 1) and (x + 3) are 2 factors of 3x 3 + 13x 2 + 13x +3, find 3rd factor. 150) Find the remainder when 2x 3 – 11x 2 + 3x + 5 is divided by (x – 3) 151) 3x 3 – 2x 2 – 7x + 5 is divided by (x – 4) 152) 5x 3 + 7x 2 + x – 1 is divided by (x + 1) 153) 2x 3 + 7x 2 – 11x – 8 is divided by (x – 2) 154) 4x 3 + 5x 2 + 2x – 10 is divided by (x + 2) 155) Is (x + 2) a factor of 3x 3 – x 2 – 20 x – 12? 156) Is (x – 2) a factor of 2x 3 – 11x 2 + 17x -6? 157) Is (x + 1) a factor of x 3 – 5x 2 + 2x + 3? 158) Prove that (x + 2) is a factor of x 3 + 3x 2 – 4x – 12.

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Geometry and Algebra 

Distance  Slope of a line  Equation of a line

Radius r = Distance between r2 = (3-2)2 + (3-6)2 r = √10

AB2= (6 -5)2 + (4 - 2)2 = 12 + 62 = 1+36 =37 BC2 = (7 + 6)2 + (-2 - 4)2 = 12 + (-6)2 = 1+ 36 = 37 AC2 = (7 + 5)2 + (-2 +2)2 = 22 + 02 = 4 Since, AB = BC = √37 We observe that A, B, C are the vertices of an isosceles triangle. triangle. Prove the points A(2,3),B(7,5),C(9,8) and D(4,6) are the vertices of parallelogram.

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Slope AB = 2/5 9048332443


SSLC Ma Mathematics

Slope BC = 3/2 Slope CD = 2/5 Slope AD = 3/2 AB = √29; BC = √29; CD = √13; AD = √13 √13 Since the length of opposite sides and their slopes are equal the points are the vertices of parallelogram.

Slope of the the line joining the points (1, 4) and (2, 6) = 2/1=2 If (x,y) be any point on this line, then (y - 4)/(x - 1) = 2 y – 4 = 2 (x - 1) y – 4 = 2x – 2 2x + y +2 =0 159) Find the value of y if the distance between the points (2, –3) and (10, y) be 10 units. [y = 3 or –9] 160) Find the point on x-axis which whic h is equidistant from the points (–4, 6) and (5, 9). [3, 0] 161) Find the area of rhombus ABCD, if A(2, 0), B(5, -5), C(8, 0), D(5, 5) [30] 162) Show that A (-5, 5), B (7, 10), C (10, ( 10, 6), D (-2, 1) are vertices of a parallelogram. 163) Show that the points (2,-2) ,(14,10) (11,13)and (1 1,13)and (-1,1) are vertices of a rectangle. 9048332443


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SSLC Ma Mathematics

Statistics Points:  Mean :  Median : is median,

    /2          

54 56 58 55 50 47 44 41

54 56 58 55 50 47 44 41

3 5 6 3 2 4 5 2

3 5 6 3 2 4 5 2

162 280 348 165 100 188 220 82

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Mean = 1545/30=51.5mm 9048332443


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SSLC Ma Mathematics

Below 30 Below 35 Below 40 Below 45 Below 50 Below 55

6 8 12 20 16 6

Below 30 Below 35 Below 40 Below 45 Below 50 Below 55

6 14 26 46 62 68

Median =? y = 68/2=34 Using Proportionality assumption, (x – 40)/(45-40)=(34-26)/(46-26) x = 42 Median = 42 164) If the mean of the following data is 5, find p. x 2 3 5 p 9 f 9 4 6 3 8

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165) Find the mean of the following distribution x 1 3 5 7 8 f 7 8 1 1 1 166) Find the median of the following frequency distribution x 0-10 10-20 20-30 30-40 40-50 50-60 f 5 3 10 6 4 2

[27]

167) Calculate the median income : Income No of employees

600700 40

700800 68

800900 86

[55]

9001000 120

1000- 1100- 1200 1100 1200 90 40 26

168) Calculate the median for the following data : x 1 2 3 4 5 6 f 6 1 2 4 6 7

[Rs. 934.17]

[35]

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Editing & Design Fassal Peringolam Peringolam

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