X-STD MATHEMATICSIMPORTANT

5 MARKS PROBLEMS

PREPARED BY:R.RAJENDRAN. M.A., M. Sc., M. Ed.,K.C.SANKARALINGA NADAR HR. SEC. SCHOOL,CHENNAI-21

Find the sum of all numbers between 300 and 500 divisible by 11

First number = 300 + (11 – 3 ) = 308

Last number = 500 – 5 = 495

The AP is

308 + 319 +……+ 495

a = 308, l = 495, d = 11

1d

a-ln

111

308495n

117111

187

n = 18

)al(2

nsn

)803(2

18

)308495(2

18s18

= 9 803

= 7227

9

2 7

11 3 0 0

2 2

8 0

7 7

3

4 5

11 5 0 0

4 4

6 0

5 5

5

t10 = 41 t18 = 73

a + (10 – 1)d = 41

a + 9d = 41……(1)

a + (18 – 1)d = 73

a + 17d = 73……(2)

(2) a + 17d = 73

(1) a + 9d = 41

(2)-(1) 8d = 32

The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term.

d = 32/8

d = 4

substitute d = 4 in eqn (1)

a + 9 (4) = 41

a + 36 = 41

a = 41 – 36

a = 5

t27 = a + (27 – 1)d

= 5 + 26(4)

= 5 + 104 = 109

tn = a + (n – 1)d

The sum of three terms in an AP is 6 and their product is – 120. Find the numbers.

Let a – d, a, a + d be the

three terms of the AP.

sum = 6.

(a – d) + a + (a + d) = 6.

3a = 6.

a = 6/3.

a = 2.

Product = – 120.

(a – d) a (a + d) = – 120.

a(a2 – d2) = – 120.

2(22 – d2) = – 120.

4 – d2 = –60.

– d2 = –60 – 4.

– d2 = – 64.

d2 = 64.

d = 8.

The three numbers are.

–6 , 2 , 10 (or) 10 , 2 , –6 .

Find the three consecutive terms in an A. P. whose sum is 18 and the sum of their squares is 140.

Let a – d, a, a + d be three numbers in an AP

Sum = 18

a – d + a + a + d = 18

3a = 18

a = 18/3

a = 6

Sum of squares = 140

(a – d)2 + a2 + (a + d)2 = 140

a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 140

3a2 + 2d2 = 140

3(6)2 + 2d2 = 140

3(36) + 2d2 = 140

108 + 2d2 = 140

2d2 = 140 – 108

2d2 = 32

d2 = 32/2 = 16

d = 4The given numbers

are

2, 6, 10 (or) 10, 6, 2

If the 4th and 7th terms of a G.P. are 2/3 and 16/81 respectively, find the G.P.

a = 9/4, d = 2/3

The GP is

In a GP, tn = ar(n-1)

t4 = 2/3

ar(4 – 1) = 2/3

ar3 = 2/3………(1)

t7 = 16/81

ar(7 – 1) = 16/81

ar6 = 16/81…….(2)

32

8116

ar

ar

)1(

)2(3

6

27

8

2

3

81

16r3

3

2r

Sub r = 2/3 in eqn (1)

3

2

3

2a

3

3

2

27

8a

4

9

8

27

3

2a

,.....1,2

3,

4

9

If the 4th and 7th terms of a G.P. are 54 and 1458 respectively, find the G.P.tn = arn–1

Given t4 = 54

a r 4–1 = 54

a r 3 = 54……(1)

t7 = 1458

a r 7–1 = 1458

a r 6 = 1458……(2)

54

1458

ar

ar

)1(

)2(3

6

r3 = 27

r = 3

Substituting r = 3 in eqn (1)

a(3)3 = 54

a(27) = 54

a = 54/27 = 2

First term a = 2

Common ratio r = 3

The GP is 2, 6, 18,…….

The sum three numbers in GP is 39/10. Their product is 1. Find the numbers.

Let a/r, a, ar be three terms of a GP

Product = 1

(a/r)(a)(ar) = 1

a3 = 1 a = 1

Sum = 39/10

a/r + a + ar = 39/10

1/r + 1 + r = 39/10

Multiplying by r

1 + r + r2 = 39r/10

Multiplying by 10

10r2 + 10r + 10 – 39r = 0

10r2 – 29r + 10 = 0

10r2 – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(2r – 5)(5r – 2) = 0

r = 5/2, 2/5

The given numbers are

2/5, 1, 5/2 (or) 5/2, 1, 2/5

The sum of the first three terms of a GP is 13 and the sum of their squares is 91. Find the first five terms of the GP. Let x = a/r, y = a, z = ar be

three terms of a GP

Sum = 13

x + y + z = 13……(1)

Sum of square = 91

x2 + y2 + z2 = 91

(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx

132 = 91 + 2(xy + yz + zx)

169 – 91 = 2 (xy + yz + zx)

xy + yz + zx = 78/2

xy + yz + zx = 39…….(2)

From equation (1)

a/r + a + ar = 13

Multiplying by r

a + ar + ar2 = 13r

a (1 + r + r2) = 13r……(A)

From equation (2)

(a/r)(a) + (a)(ar) + (ar)(a/r) = 39

a2/r + a2r + a2 = 39

Multiplying by r

a2 + a2r2 + a2 r = 39r

a2(1 + r2 + r) = 39r……(B)

a2(1 + r + r2) = 39r……(B)

a (1 + r + r2) = 13r……(A)

r

r

13

39

)r r (1 a

)r r (1 a

(A)

(B)2

22

a = 3

Sub a = 3 in eqn (A)

3(1 + r + r2) = 13r

3r2 + 3r + 3 – 13r = 0

3r2 – 10r + 3 = 0

3r2 – 9r – r + 3 = 0

3r(r – 3) – 1 (r – 3) = 0

(r – 3)(3r – 1) = 0

r = 3, 1/3

the GP is

1, 3, 9, ….. (or)

9, 3, 1, ………

Find the sum of 122 + 132 + 142 + ……+ 352

122 + 132 + 142 + ………+ 352

= (12 + 22 + 32+…..+352) – (12 + 22 + 32 +…+112)

6

1)1)(2nn(nn2

6

)1112)(111(11

6

)1352)(135(35

6

231211

6

713635

= 35 6 71 – 11 2 23

= 210 71 – 22 23

= 14910 – 506 = 14404

6 2

Find the sum of 12 + 32 + 52 + ……+ 512

12 + 32 + 52 + ………+ 512

= (12 + 22 + 32+…..+512) – (22 + 42 + 62 +…+502) = (12 + 22 + 32+…..+512) – 22(12 + 22 + 32 +…+252)

6

1)1)(2nn(nn2

6

)1252)(125(252

6

)1512)(151(51 2

6

5126254

6

1035251

= 17 26 103 – 2 25 26 17

= 442 103 – 50 442

= 45526 – 22100 = 23426

3

2617

3

2 17

Find the sum of 113 + 123 + 133 +………+ 283

113 + 123 + 133 +…….+ 283

= (13 + 23 + 33 +…….+ 283) – (13 + 23 + 33 +……+ 103)

2

3

2

1nnn

22

2

11010

2

12828

22

2

1110

2

2928

= (14 29)2 – (5 11)2

= 4062 – 552

= 164836 – 3025 = 161811

14 5

Find the Square Root of 9x4 – 6x3 + 7x2 – 2x + 1

9x4 – 6x3 + 7x2 – 2x + 13x2

3x2

9x4(-)

6x2 – 6x3 + 7x2– x

– x

– 6x3 + x2

(+) (-)

6x2 – 2x + 16x2 – 2x + 1

+ 1

6x2 – 2x + 1(-) (+) (-)

0Square root = |3x2 – x + 1|

Find the Square Root of x4 – 4x3 + 10x2 – 12x + 9

x4 – 4x3 + 10x2 – 12x + 9x2

x2

x2(–)

2x2 – 4x3 + 10x2– 2x

– 2x

– 4x3 + 4x2

(+) (–)

6x2 – 12x + 92x2 – 4x + 3

+ 3

6x2 – 12x + 9(–) (+) (–)

0Square root = |x2 – 2x + 3|

Find the Square Root of x4 – 6x3 + 19x2 – 30x + 25

x4 – 6x3 + 19x2 – 30x + 25x2

x2

x2(–)

2x2 – 6x3 + 19x2– 3x

– 3x

– 6x3 + 9x2

(+) (–)

+10x2 – 30x + 252x2 – 6x + 5

+ 5

10x2 – 30x + 25(–) (+) (–)

0Square root = |x2 – 3x + 5|

Find the Square Root of x4 – 10x3 + 37x2 – 60x + 36

x4 – 10x3 + 37x2 – 60x + 36x2

x2

x2(–)

2x2 – 10x3 + 37x2– 5x

– 5x

– 10x3 + 25x2

(+) (–)

+12x2 – 60x + 362x2 – 10x + 6

+ 6

12x2 – 60x + 36(–) (+) (–)

0Square root = |x2 – 5x + 6|

Find the Square Root of 4x4 + 8x3 + 8x2 + 4x + 1

4x4 + 8x3 + 8x2 + 4x + 12x2

2x2

4x4(–)

4x2 + 8x3 + 8x2+ 2x

+ 2x

+ 8x3 + 4x2

(–) (–)

+4x2 + 4x + 14x2 + 4x + 1

+ 1

+4x2 + 4x + 1(-) (+) (-)

0Square root = |2x2 + 2x + 1|

If 4x4 – 12x3 + 37x2 + ax + b is a perfect square, find the values of a, b

4x4 – 12x3 + 37x2 + ax + b2x2

2x2

4x4(–)

4x2 – 12x3 + 37x2– 3x

– 3x

– 12x3 + 9x2

(+) (–)

+28x2 + ax + b4x2 – 6x + 7

+ 7

+28x2 –42x + 49(–) (+) (–)

0 a = –42, b = 49

If x4 – 4x3 + 10x2 – ax + b is a perfect square, find the values of a, b

x4 – 4x3 + 10x2 – ax + bx2

x2

x4(–)

2x2 – 4x3 + 10x2– 2x

– 2x

– 4x3 + 4x2

(+) (–)

+6x2 – ax + b2x2 – 4x + 3

+ 3

+6x2 – 12x + 9(–) (+) (–)

0 a = 12, b = 9

If ax4 – bx3 + 40x2 + 24x + 36 is a perfect square, find the values of a, b

36 + 24x + 40x2 – bx3 + ax4 66

36(–)

12 +24x + 40x2+ 2x

+ 2x

+24x + 4x2

(+) (–)

+36x2 – bx3 + ax412 + 4x + 3x2

+ 3x2

+36x2 + 2x3 + 9x4

(–) (-) (–)

0 a = – 2 , b = 9

If ax4 + bx3 + 109x2 – 60x + 36 is a perfect square, find the values of a, b

36 – 60x + 109x2 + bx3 + ax4 66

36(–)

12 –60x + 109x2– 5 x

– 5 x

–60x + 25x2

(+) (–)

+84x2 + bx3 + ax412 – 10x + 7x2

+ 7x2

+84x2 – 70x3 + 49x4

(–) (+) (–)

0 a = 49, b = –70

Find the Square Root of 4 + 25x2 – 12x – 24x3 + 16x4

16x4 – 24x3 + 25x2 – 12x + 44x2

4x2

16x4(-)

8x2 – 24x3 + 25x2– 3x

– 3x

– 24x3 + 9x2

(+) (-)

+ 16x2 – 12x + 48x2 – 6x + 2

+ 2

+16x2 – 12x + 4(+) (-) (-)

0Square root = |4x2 – 3x + 2|

Simplify:189

127

124

84

2

822

2

2

2

xx

xx

xx

x

x

xx

x2 – 2x – 8 = (x – 4)(x + 2); 4x – 8 = 4(x – 2)

x2 – 4x – 12 = (x – 6)(x + 2); x2 – 7x + 12 = (x – 4)(x – 3)

x2 – 9x + 18 = (x – 6)(x – 3)

189

127

124

84

2

822

2

2

2

xx

xx

xx

x

x

xx

)3)(6(

)3)(4(

)2)(6(

)2(4

2

)2)(4(

xx

xx

xx

x

x

xx

)3)(4(

)3)(6(

)2)(6(

)2(4

2

)2)(4(

xx

xx

xx

x

x

xx

= 4

Simplify:34

2

65

1

23

1222

xxxxxx

x2 + 3x + 2 = (x + 1)(x + 2); x2 + 5x + 6 = (x + 2)(x + 3)

x2 + 4x + 3 = (x + 1)(x + 3)

34

2

65

1

23

1222

xxxxxx

)3)(1(

2

)3)(2(

1

)2)(1(

1

xxxxxx

)3)(2)(1(

)2(2)1()3(

xxx

xxx)3)(2)(1(

4213

xxx

xxx

)3)(2)(1(

0

xxx0

Factorize 3x3 – 10x2 + 11x – 4

3 –10 11 –4

1

0

3

3

–7

–7

4

4

0 x – 1 is a factor

Quotient = 3x2 – 7x + 4

= 3x2 – 4x – 3x + 4

= x(3x – 4) – 1(3x – 4)

= (3x – 4)(x – 1)

3x3 – 10x2 + 11x – 4 = (x – 1)(3x – 4)(x – 1)

Factorize x3 – 7x + 6

1 0 – 7 6

1

0

1

1

1

1

–6

–6

0 x – 1 is a factor

Quotient = x2 + x – 6

= x2 + 3x – 2x – 6

= x(x + 3) – 2(x + 3)

= (x + 3)(x – 2)

x3 – 7x + 6 = (x – 1)(x + 3)(x – 2)

Factorize x3 + 13x2 + 32x + 20

1 13 32 20

–1

0

1

– 1

12

–12

20

–20

0 x + 1 is a factor

Quotient = x2 + 12x + 20

= x2 + 10x + 2x + 20

= x(x + 10) + 2(x + 10)

= (x + 10)(x + 2)

x3 + 13x2 + 32x + 20 = (x + 1)(x + 10 )(x + 12)

A capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. If the length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm, find its surface area

Surface area of the capsule

= 141.3 + 2(39.25)

= 141.3 + 78.5

= 219.8sq.mm

Cylinder Diameter = 5mmRadius = 2.5mmHeight = 14 – 2 2.5 = 9mmCurved surface area = 2rh

= 2 3.14 2.5 9= 141.3sq.mm

Hemispherical partRadius = 2.5cmCurved surface area = 2r2

= 2 3.14 2.5 2.5= 39.25sq.mm

A solid wooden toy is in the form of a cone surmounted on a hemisphere. If the radii of the hemisphere and the base of the cone are 3.5 cm each and the total height of the toy is 17.5cm, then find the volume of wood used in the toy.Hemispherical part: Radius = 3.5cmConical part: Radius = 3.5cmHeight = 17.5 – 3.5 = 14cmVolume of wood = volume of hemisphere + volume of cone

hrr 23 3

1

3

2

145.35.37

22

3

15.35.35.3

7

22

3

2

2

2

25.35.3223

15.35.322

3

1

)5395.269(3

1 cmcu.5.269)5.808(

3

1

A cup is in the form of a hemisphere surmounted by a cylinder. The height of the cylindrical portion is 8 cm and the total height of the cup is 11.5 cm. Find the total surface area of the cup.

Cylindrical part

Height = 8 cm

Radius = 11.5 – 8 = 3.5 cm

Surface area = 2rh

= 22 8= 176sq.cm

Hemispherical part

Radius = 3.5cm

Surface area = 2r2

85.37

222

25.35.3

7

222

2= 22 3.5

= 77sq.cm

Total surface area of the cup

= 176 + 77

= 253sq.cm

A solid is in the shape of a cylinder surmounted on a hemisphere. If the diameter and the total height of the solid are 21 cm, 25.5 cm respectively, then find its volume.

Cylindrical part

Diameter = 21cm

Radius = 10.5 cm

Height = 25.5 – 10.5 = 15 cm

Volume = r2h

= 2 22 0.5 10.5 10.5

= 22 110.25 = 2425.5cu.cm

Total volume of the cup

= 5197.5 + 2425.5

= 7626=3cu.cm

= 22 1.5 10.5 15

= 33 157.5 = 5197.5cu.cm

Hemispherical part

Radius = 12.5cm

Volume = ⅔ r3

155.105.107

22

1.5

5.105.105.107

22

3

2

1.5 0.5

The radii of two circular ends of a frustum shaped bucket are 15 cm and 8 cm. If its depth is 63 cm, find the capacity of the bucket in litres.

Radius of the top R = 15cm

Radius of the bottom r = 8cm

Height h = 63cm

Volume of the bucket

= ⅓ h(R2 + r2 +Rr)

= ⅓ 63(152 + 82 + 15 8)

= ⅓ 63(225 + 64 + 120)

= 21 409

409217

22

3

= 22 3 409= 66 409= 26994cu.cm

litres1000

26994

= 26.994litres