A. 432
B. 288
C. 144
D. 72
E. None of these
Explanation:
Let the number of coins of each kind be x.
=> 5x + 2x + 1x = 1152
=> 8x = 1152 => x = 144
A. 432
B. 288
C. 144
D. 72
E. None of these
Explanation:
Let the number of coins of each kind be x.
=> 5x + 2x + 1x = 1152
=> 8x = 1152 => x = 144
A. 4
B. 5
C. 6
D. 8
E. None of these
Explanation:
If Bilal gives ‘x’ marbles to Arslan then Bilal and Arslan would have V – x and A + x marbles.
V – x = A + x — (1)
If Arslan gives 2x marbles to Bilal then Arslan and Bilal would have A – 2x and V + 2x marbles.
V + 2x – (A – 2x) = 30 => V – A + 4x = 30 — (2)
From (1) we have V – A = 2x
Substituting V – A = 2x in (2)
6x = 30 => x = 5.
A. Rs.25
B. Rs.22.50
C. Rs.20
D. Rs.17.50
E. None of these
Explanation:
Let the cost of each ice-cream cup be Rs.x
16(6) + 5(45) + 7(70) + 6(x) = 961
96 + 225 + 490 + 6x = 961
6x = 150 => x = 25.
A. 9, 11, 13
B. 11, 13, 15
C. 13, 15, 17
D. 7, 9, 11
E. None of these
Explanation:
Let the three consecutive odd integers be x, x + 2 and x + 4 respectively.
x + 4 + x + 2 = x + 13 => x = 7
Hence three consecutive odd integers are 7, 9 and 11.
A. 30
B. 46
C. 40
D. 44
E. None of these
Explanation:
Let the number of rabbits and peacocks be ‘r’ and ‘p’ respectively. As each animal has only one head, so r + p = 60 — (1)
Each rabbit has 4 legs and each peacock has 2 legs. Total number of legs of rabbits and peacocks, 4r + 2p = 192 — (2)
Multiplying equation (1) by 2 and subtracting it from equation (2), we get
=> 2r = 72 => r = 36.
A. 600
B. 620
C. 500
D. 520
E. None of these
Explanation:
Let the number of children in the school be x. Since each child gets 2 bananas, total number of bananas = 2x.
2x/(x – 360) = 2 + 2(extra)
=> 2x – 720 = x => x = 720.
A. 30 years
B. 40 years
C. 32 years
D. 48 years
E. None of these
Let the present ages of Zahid and Shahid be ‘Z’ and ‘S’ years respectively.
Z – 8 = 4/3 (S – 8) and Z + 8 = 6/5 (S + 8)
3/4(Z – 8) = S – 8 and 5/6(Z + 8) = S + 8
S = 3/4 (Z – 8) + 8 = 5/6 (Z + 8) – 8
=> 3/4 Z – 6 + 8 = 5/6 Z + 20/3 – 8
=> 10 – 20/3 = 10/12 Z – 9/12 Z
=> 10/3 = Z/12 => Z = 40.
A. Rs.400
B. Rs.500
C. Rs.300
D. Rs.600
E. None of these
Explanation:
Let the amounts to be received by P, Q and R be p, q and r.
p + q + r = 1200
p = 1/2 (q + r) => 2p = q + r
Adding ‘p’ both sides, 3p = p + q + r = 1200
=> p = Rs.400
q = 1/3 (p + r) => 3q = p + r
Adding ‘q’ both sides, 4q = p + q + r = 1200
=> q = Rs.300
r = 1200 – (p + q) => r = Rs.500.
A. Rs.3000
B. Rs.3600
C. Rs.2400
D. Rs.4000
E. None of these
Explanation:
Let the amount with R be Rs.r
r = 2/3 (total amount with P and Q)
r = 2/3(6000 – r) => 3r = 12000 – 2r
=> 5r = 12000 => r = 2400.
A. 60
B. 56
C. 64
D. 74
E. None of these
Explanation:
Let the number of one rupee coins in the bag be x.
Number of 50 paise coins in the bag is 93 – x.
Total value of coins
[100x + 50(93 – x)]paise = 5600 paise
=> x = 74