Aptitude SimplificationPage 3
17. | | What is the value of (P + Q)/(P - Q), if P/Q = 7. | | (a)4/3 | | (b)13/17 | | (c)7/2 | | (d)1/17 |
|
Answer is: ALet, P + Q/P - Q
= [Q(P/Q) + 1]/[Q(P/Q) - 1]
= [(P/Q) + 1]/[(P/Q) - 1]
= [7 + 1]/[7 - 1]
= 8/6
= 4/3
18. | | If X/Y = 4/5 , then the value of [4/7 + (2Y - X)÷(2Y + X)] is ? | | (a)1 | | (b)10 | | (c)2 | | (d)20 |
|
Answer is: A⇒ [4/7 – (2Y - X)÷(2Y + X)] = [4/7 – (2 - X/Y)÷(2 + X/Y)]
⇒ 4/7 + (2 - 4/5)÷(2 + 4/5) = 4/7 + (6/5)÷(14/5)
⇒4/7 + ( 6/5 x 5/14 )
⇒ 4/7 + 3/7 = 7/7 = 1
19. | | If a/b = 4/3 , then the value of (6a + 4b)÷(6a - 5b) is ? | | (a)2 | | (b)4 | | (c)6 | | (d)8 |
|
Answer is: B⇒ a/b = 4/3
⇒ (6a + 4b)÷(6a - 5b) = [6(a/b)+ 4]÷[6 (a/b) - 5]
⇒(6 x 4/3 + 4)÷(6 x 4/3 - 5) = (8 + 4)÷(8 - 5 ) = 12/3 = 4
20. | | If a/3 = b/4 = c/7 , then the value of (a + b + c)/c is: | | (a)2 | | (b)4 | | (c)6 | | (d)8 |
|
Answer is: ALet, a/3 = b/4 = c/7 = k (say).
Then, a = 3k, b = 4k and c = 7k.
∴ (a + b + c )/c = (3k + 4k + 7k )/7k = 14k/7k = 2
21. | | The value of 1/(2 + 1/(2 + 1/(2 - 1/2))) is: | | (a)7 / 19 | | (b)8 / 19 | | (c)9 / 19 | | (d)11 / 19 |
|
Answer is: BGiven exp. = 1/(2+1/(2+1/(3/2))) = 1/(2+1/(2 + 2/3))
⇒ 1/(2 + 1/(8/3)) = 1/(2 + 3/8)
⇒ 1/(( 19/8 )) = 8/19
22. | | Simplified : 3/4(1 + 1/3) (1 + 2/3) (1 – 2/5) (1 + 6/7) (1 – 12/13) = ? | | (a)1/7 | | (b)1/8 | | (c)1/9 | | (d)1/10 |
|
Answer is: A⇒ 3/4(1 + 1/3) (1 + 2/3) (1 – 2/5) (1 + 6/7) (1 – 12/13) = ( 3/4 x 4/3 x 5/3 x 3/5 x 13/7 x 1/13 ) = 1/7
23. | Simplified the product
(1 – 1/2) (1 – 1/3) (1 – 1/4)……….(1 – 1/n) gives:
| | (a)n | | (b)n + 1 | | (c)n - 1 | | (d)1/n |
|
Answer is: DLet, (1 – 1/2) (1 – 1/3) (1 – 1/4)……….(1 – 1/n)
⇒ 1/2 x 2/3 x 3/4 x 4/5 x……..x (n – 1)/n = 1/n
24. | Find the sum :
1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/43 + 1/56 + 1/72 + 1/90 + 1/110 +1/132
| | (a)9/10 | | (b)10/11 | | (c)11/12 | | (d)12/13 |
|
Answer is: CGiven = 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/43 + 1/56 + 1/72 + 1/90 + 1/110 +1/132
⇒ (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1/4) + ………+ (1/11 – 1/12)
⇒ (1 – 1/12) = 11/12
Comments