Presentation is loading. Please wait.

Presentation is loading. Please wait.

X-STD MATHEMATICS IMPORTANT 5 MARKS PROBLEMS Find the sum of all numbers between 300 and 500 divisible by 11 First number = 300 + (11 – 3 ) = 308 Last.

Similar presentations


Presentation on theme: "X-STD MATHEMATICS IMPORTANT 5 MARKS PROBLEMS Find the sum of all numbers between 300 and 500 divisible by 11 First number = 300 + (11 – 3 ) = 308 Last."— Presentation transcript:

1

2 X-STD MATHEMATICS IMPORTANT 5 MARKS PROBLEMS

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

4 t 10 = 41t 18 = 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 10 th and 18 th terms of an A.P. are 41 and 73 respectively. Find the 27 th term. d = 32/8 d = 4 substitute d = 4 in eqn (1) a + 9  (4) = 41 a + 36 = 41 a = 41 – 36 a = 5 t 27 = a + (27 – 1)d = (4) = = 109 t n = a + (n – 1)d

5 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(a 2 – d 2 ) = – (2 2 – d 2 ) = – – d 2 = –60. – d 2 = –60 – 4. – d 2 = – 64. d 2 = 64. d =  8.  The three numbers are. –6, 2, 10 (or) 10, 2, –6.

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 + a 2 + (a + d) 2 = 140 a 2 – 2ad + d 2 + a 2 + a 2 + 2ad + d 2 = 140 3a 2 + 2d 2 = 140 3(6) 2 + 2d 2 = 140 3(36) + 2d 2 = d 2 = 140 2d 2 = 140 – 108 2d 2 = 32 d 2 = 32/2 = 16 d =  4 The given numbers are 2, 6, 10 (or) 10, 6, 2

7 If the 4 th and 7 th 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, t n = ar (n-1) t 4 = 2/3 ar (4 – 1) = 2/3 ar 3 = 2/3………(1) t 7 = 16/81 ar (7 – 1) = 16/81 ar 6 = 16/81…….(2) Sub r = 2/3 in eqn (1)

8 If the 4 th and 7 th terms of a G.P. are 54 and 1458 respectively, find the G.P. t n = ar n–1 Given t 4 = 54 a r 4–1 = 54 a r 3 = 54……(1) t 7 = 1458 a r 7–1 = 1458 a r 6 = 1458……(2) r 3 = 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,…….

9 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 a 3 = 1  a = 1 Sum = 39/10 a/r + a + ar = 39/10 1/r r = 39/10 Multiplying by r 1 + r + r 2 = 39r/10 Multiplying by 10 10r r + 10 – 39r = 0 10r 2 – 29r + 10 = 0 10r 2 – 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

10 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 x 2 + y 2 + z 2 = 91 (x + y + z) 2 = x 2 + y 2 + z 2 + 2xy + 2yz + 2zx 13 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 + ar 2 = 13r a (1 + r + r 2 ) = 13r……(A) From equation (2) (a/r)(a) + (a)(ar) + (ar)(a/r) = 39 a 2 /r + a 2 r + a 2 = 39 Multiplying by r a 2 + a 2 r 2 + a 2 r = 39r a 2 (1 + r 2 + r) = 39r……(B)

11 a 2 (1 + r + r 2 ) = 39r……(B) a (1 + r + r 2 ) = 13r……(A) a = 3 Sub a = 3 in eqn (A) 3(1 + r + r 2 ) = 13r 3r 2 + 3r + 3 – 13r = 0 3r 2 – 10r + 3 = 0 3r 2 – 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, ………

12 Find the sum of …… ……… = ( … ) – ( …+11 2 ) = 35  6  71 – 11  2  23 = 210  71 – 22  23 = – 506 =

13 Find the sum of …… ……… = ( … ) – ( …+50 2 ) = ( … ) – 2 2 ( …+25 2 ) = 17  26  103 – 2  25  26  17 = 442  103 – 50  442 = – =

14 Find the sum of ……… …… = ( …… ) – ( …… ) = (14  29) 2 – (5  11) 2 = – 55 2 = – 3025 =

15 Find the Square Root of 9x 4 – 6x 3 + 7x 2 – 2x + 1 9x 4 – 6x 3 + 7x 2 – 2x + 13x 2 9x 4 (-) 6x 2 – 6x 3 + 7x 2 – x – 6x 3 + x 2 (+)(-) 6x 2 – 2x + 1 6x 2 – 2x+ 1 6x 2 – 2x + 1 (-)(+)(-) 0 Square root = |3x 2 – x + 1|

16 Find the Square Root of x 4 – 4x x 2 – 12x + 9 x 4 – 4x x 2 – 12x + 9x2x2 x2x2 x2x2 (–) 2x 2 – 4x x 2 – 2x – 4x 3 + 4x 2 (+)(–) 6x 2 – 12x + 9 2x 2 – 4x+ 3 6x 2 – 12x + 9 (–)(+)(–) 0 Square root = |x 2 – 2x + 3|

17 Find the Square Root of x 4 – 6x x 2 – 30x + 25 x 4 – 6x x 2 – 30x + 25x2x2 x2x2 x2x2 (–) 2x 2 – 6x x 2 – 3x – 6x 3 + 9x 2 (+)(–) +10x 2 – 30x x 2 – 6x+ 5 10x 2 – 30x + 25 (–)(+)(–) 0 Square root = |x 2 – 3x + 5|

18 Find the Square Root of x 4 – 10x x 2 – 60x + 36 x 4 – 10x x 2 – 60x + 36x2x2 x2x2 x2x2 (–) 2x 2 – 10x x 2 – 5x – 10x x 2 (+)(–) +12x 2 – 60x x 2 – 10x+ 6 12x 2 – 60x + 36 (–)(+)(–) 0 Square root = |x 2 – 5x + 6|

19 Find the Square Root of 4x 4 + 8x 3 + 8x 2 + 4x + 1 4x 4 + 8x 3 + 8x 2 + 4x + 12x 2 4x 4 (–) 4x 2 + 8x 3 + 8x 2 + 2x + 8x 3 + 4x 2 (–) +4x 2 + 4x + 1 4x 2 + 4x+ 1 +4x 2 + 4x + 1 (-)(+)(-) 0 Square root = |2x 2 + 2x + 1|

20 If 4x 4 – 12x x 2 + ax + b is a perfect square, find the values of a, b 4x 4 – 12x x 2 + ax + b 2x 2 4x 4 (–) 4x 2 – 12x x 2 – 3x – 12x 3 + 9x 2 (+)(–) +28x 2 + ax + b 4x 2 – 6x x 2 –42x + 49 (–) (+) (–) 0 a = –42, b = 49

21 If x 4 – 4x x 2 – ax + b is a perfect square, find the values of a, b x 4 – 4x x 2 – ax + b x2x2 x2x2 x4x4 (–) 2x 2 – 4x x 2 – 2x – 4x 3 + 4x 2 (+)(–) +6x 2 – ax + b 2x 2 – 4x+ 3 +6x 2 – 12x + 9 (–) (+) (–) 0 a = 12, b = 9

22 If ax 4 – bx x x + 36 is a perfect square, find the values of a, b x + 40x 2 – bx 3 + ax (–) x + 40x 2 + 2x +24x + 4x 2 (+)(–) +36x 2 – bx 3 + ax x+ 3x 2 +36x 2 + 2x 3 + 9x 4 (–) (-) (–) 0 a = – 2, b = 9

23 If ax 4 + bx x 2 – 60x + 36 is a perfect square, find the values of a, b 36 – 60x + 109x 2 + bx 3 + ax (–) 12 –60x + 109x 2 – 5 x –60x + 25x 2 (+)(–) +84x 2 + bx 3 + ax 4 12 – 10x+ 7x 2 +84x 2 – 70x x 4 (–) (+) (–) 0 a = 49,b = –70

24 Find the Square Root of x 2 – 12x – 24x x 4 16x 4 – 24x x 2 – 12x + 44x 2 16x 4 (-) 8x 2 – 24x x 2 – 3x – 24x 3 + 9x 2 (+)(-) + 16x 2 – 12x + 4 8x 2 – 6x x 2 – 12x + 4 (+)(-) 0 Square root = |4x 2 – 3x + 2|

25 Simplify: x 2 – 2x – 8 = (x – 4)(x + 2);4x – 8 = 4(x – 2) x 2 – 4x – 12 = (x – 6)(x + 2);x 2 – 7x + 12 = (x – 4)(x – 3) x 2 – 9x + 18 = (x – 6)(x – 3) = 4

26 Simplify: x 2 + 3x + 2 = (x + 1)(x + 2); x 2 + 5x + 6 = (x + 2)(x + 3) x 2 + 4x + 3 = (x + 1)(x + 3)

27 Factorize 3x 3 – 10x x – 4 3 –10 11 – –  x – 1 is a factor Quotient = 3x 2 – 7x + 4 = 3x 2 – 4x – 3x + 4 = x(3x – 4) – 1(3x – 4) = (3x – 4)(x – 1) 3x 3 – 10x x – 4 = (x – 1)(3x – 4)(x – 1)

28 Factorize x 3 – 7x – –6 0  x – 1 is a factor Quotient = x 2 + x – 6 = x 2 + 3x – 2x – 6 = x(x + 3) – 2(x + 3) = (x + 3)(x – 2) x 3 – 7x + 6 = (x – 1)(x + 3)(x – 2)

29 Factorize x x x – –12 20 –20 0  x + 1 is a factor Quotient = x x + 20 = x x + 2x + 20 = x(x + 10) + 2(x + 10) = (x + 10)(x + 2) x x x + 20 = (x + 1)(x + 10 )(x + 12)

30 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 = (39.25) = = 219.8sq.mm 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  r 2 = 2  3.14  2.5  2.5 = 39.25sq.mm

31 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

32 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 = 22  8 = 176sq.cm Hemispherical part Radius = 3.5cm Surface area = 2  r = 22  3.5 = 77sq.cm Total surface area of the cup = = 253sq.cm

33 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 =  r 2 h = 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  r

34 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(R 2 + r 2 +Rr) = ⅓   63(  8) = ⅓   63( ) =   21  = 22  3  409 = 66  409 = 26994cu.cm = litres


Download ppt "X-STD MATHEMATICS IMPORTANT 5 MARKS PROBLEMS Find the sum of all numbers between 300 and 500 divisible by 11 First number = 300 + (11 – 3 ) = 308 Last."

Similar presentations


Ads by Google