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X-STD MATHEMATICS IMPORTANT 5 MARKS PROBLEMS

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1 X-STD MATHEMATICS IMPORTANT 5 MARKS PROBLEMS
PREPARED BY: R.RAJENDRAN. M.A., M. Sc., M. Ed., K.C.SANKARALINGA NADAR HR. SEC. SCHOOL, CHENNAI-21

2 Find the sum of all numbers between 300 and 500 divisible by 11
2 7 2 2 8 0 7 7 3 4 5 4 4 6 0 5 5 5 n = 18 First number = (11 – 3 ) = 308 Last number = 500 – 5 = 495 The AP is ……+ 495 a = 308, l = 495, d = 11 9 = 9  803 = 7227

3 The 10th and 18th terms of an A. P. are 41 and 73 respectively
The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find the 27th term. tn = a + (n – 1)d 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 = (4) = = 109 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)  d = 32

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

5 Find the three consecutive terms in an A. P
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 d2 = 140 2d2 = 140 – 108 2d2 = 32 d2 = 32/2 = 16 d = 4 The given numbers are 2, 6, 10 (or) 10, 6, 2

6 If the 4th and 7th terms of a G. P
If the 4th and 7th terms of a G.P. are 2/3 and 16/81 respectively, find the G.P. 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) Sub r = 2/3 in eqn (1) a = 9/4, d = 2/3 The GP is

7 If the 4th and 7th terms of a G. P
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) 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,……. r3 = 27 r = 3

8 The sum three numbers in GP is 39/10. Their product is 1
The sum three numbers in GP is 39/10. Their product is 1. Find the numbers. 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/ (or) 5/2, 1, 2/5 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 r = 39/10 Multiplying by r 1 + r + r2 = 39r/10 Multiplying by 10

9 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. 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 a2 + a2r2 + a2 r = 39r a2(1 + r2 + r) = 39r……(B) 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 = (xy + yz + zx) 169 – 91 = 2 (xy + yz + zx) xy + yz + zx = 78/2 xy + yz + zx = 39…….(2)

10 a2(1 + r + r2) = 39r……(B) a (1 + r + r2) = 13r……(A)
the GP is 1, 3, 9, ….. (or) 9, 3, 1, ……… 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

11 Find the sum of 122 + 132 + 142 + ……+ 352 122 + 132 + 142 + ………+ 352
= ( …..+352) – ( …+112) 6 2 = 35  6  71 – 11  2  23 = 210  71 – 22  23 = – 506 = 14404

12 Find the sum of 12 + 32 + 52 + ……+ 512 12 + 32 + 52 + ………+ 512
= ( …..+512) – ( …+502) = ( …..+512) – 22( …+252) 17 26 2 17 3 3 = 17  26  103 – 2  25  26  17 = 442  103 – 50  442 = – = 23426

13 Find the sum of 113 + 123 + 133 +………+ 283 113 + 123 + 133 +…….+ 283
= ( …….+ 283) – ( ……+ 103) 14 5 = (14  29)2 – (5  11)2 = 4062 – 552 = – 3025 =

14 Find the Square Root of 9x4 – 6x3 + 7x2 – 2x + 1
(-) 6x2 – x – 6x x2 – 6x x2 (+) (-) 6x2 – 2x + 1 6x2 – 2x 6x2 – 2x (-) (+) (-) Square root = |3x2 – x + 1|

15 Find the Square Root of x4 – 4x3 + 10x2 – 12x + 9
+ 3 x2 x4 – 4x x2 – 12x + 9 x2 (–) 2x2 – 2x – 4x x2 – 4x x2 (+) (–) 2x2 – 4x + 3 6x2 – 12x 6x2 – 12x (–) (+) (–) Square root = |x2 – 2x + 3|

16 Find the Square Root of x4 – 6x3 + 19x2 – 30x + 25
+ 5 x2 x4 – 6x x2 – 30x x2 (–) 2x2 – 3x – 6x x2 – 6x x2 (+) (–) 2x2 – 6x + 5 +10x2 – 30x 10x2 – 30x (–) (+) (–) Square root = |x2 – 3x + 5|

17 Find the Square Root of x4 – 10x3 + 37x2 – 60x + 36
+ 6 x2 x4 – 10x x2 – 60x x2 (–) 2x2 – 5x – 10x x2 – 10x x2 (+) (–) 2x2 – 10x + 6 +12x2 – 60x 12x2 – 60x (–) (+) (–) Square root = |x2 – 5x + 6|

18 Find the Square Root of 4x4 + 8x3 + 8x2 + 4x + 1
(–) 4x2 + 2x + 8x x2 + 8x x2 (–) (–) 4x2 + 4x + 1 +4x2 + 4x +4x2 + 4x (-) (+) (-) Square root = |2x2 + 2x + 1|

19 If 4x4 – 12x3 + 37x2 + ax + b is a perfect square, find the values of a, b
+ 7 2x2 4x4 – 12x x2 + ax + b 4x4 (–) 4x2 – 3x – 12x x2 – 12x x2 (+) (–) 4x2 – 6x + 7 +28x2 + ax b +28x2 –42x (–) (+) (–) a = –42, b = 49

20 If x4 – 4x3 + 10x2 – ax + b is a perfect square, find the values of a, b
+ 3 x2 x4 – 4x x2 – ax + b x4 (–) 2x2 – 2x – 4x x2 – 4x x2 (+) (–) 2x2 – 4x + 3 +6x2 – ax b +6x2 – 12x (–) (+) (–) a = 12, b = 9

21 If ax4 – bx3 + 40x2 + 24x + 36 is a perfect square, find the values of a, b
x + 40x2 – bx3 + ax4 36 (–) 12 + 2x +24x x2 +24x x2 (+) (–) 12 + 4x + 3x2 +36x2 – bx ax4 +36x x x4 (–) (-) (–) a = – 2 , b = 9

22 If ax4 + bx3 + 109x2 – 60x + 36 is a perfect square, find the values of a, b
36 – 60x + 109x2 + bx3 + ax4 36 (–) 12 – 5 x –60x x2 –60x x2 (+) (–) 12 – 10x + 7x2 +84x2 + bx ax4 +84x2 – 70x x4 (–) (+) (–) a = 49, b = –70

23 Find the Square Root of 4 + 25x2 – 12x – 24x3 + 16x4
(-) 8x2 – 3x – 24x x2 – 24x x2 (+) (-) 8x2 – 6x + 2 + 16x2 – 12x +16x2 – 12x (+) (-) (-) Square root = |4x2 – 3x + 2|

24 Simplify: 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) = 4

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

26 Factorize 3x3 – 10x2 + 11x – 4 3 – –4 1 3 –7 4 3 –7 4  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)

27 Factorize x3 – 7x + 6 1 0 – 7 6 1 1 1 –6 1 1 –6  x – 1 is a factor
1 1 1 –6 1 1 –6  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)

28 Factorize x3 + 13x2 + 32x + 20 –1 – 1 –12 –20 1 12 20  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)

29 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 Cylinder Diameter = 5mm Radius = 2.5mm Height = 14 – 2  2.5 = 9mm Curved surface area = 2rh = 2  3.14  2.5  9 = 141.3sq.mm Hemispherical part Radius = 2.5cm Curved surface area = 2r2 = 2  3.14  2.5  2.5 = 39.25sq.mm Surface area of the capsule = (39.25) = = 219.8sq.mm

30 A solid wooden toy is in the form of a cone surmounted on a hemisphere
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 = 3.5cm Conical part: Radius = 3.5cm Height = 17.5 – 3.5 = 14cm Volume of wood = volume of hemisphere + volume of cone 2 2

31 A cup is in the form of a hemisphere surmounted by a cylinder
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 = 2rh 2 2 = 22  3.5 = 77sq.cm Total surface area of the cup = = 253sq.cm = 22  8 = 176sq.cm Hemispherical part Radius = 3.5cm Surface area = 2r2

32 A solid is in the shape of a cylinder surmounted on a hemisphere
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 0.5 1.5 1.5 = 2  22  0.5  10.5  10.5 = 22  = cu.cm Total volume of the cup = = 7626=3cu.cm = 22  1.5  10.5  15 = 33  = cu.cm Hemispherical part Radius = 12.5cm Volume = ⅔ r3

33 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(  8) = ⅓   63( ) =   21  409 = 22  3  409 = 66  409 = 26994cu.cm = litres 3


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