Page 119 - AFCAT Solved Papers 2011 - 2019 SSBCrack eBook

50. (c) A. S. Kiran Kumar is the current chairman of Indian 57. (d) Let salary of B is 100 `
Space Research Organisation (ISRO). So salary of A is 150 `
51. (d) Ricky Kej won the Grammy at the 57th Annual Grammy The percentage B’s salary is less than to A’s
awards for his album Winds of Samsara, a collaboration
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1
with South African flautish Wouter Kellerman. salary = 150 100 ´ 100 = 50 ´ 100 = 33 %
52. (c) Let length of the train is x. 150 150 3
Distance = 12 km = 12000 m 58. (a) Let the principle amount be ‘P’
Time = 10 min = 10 × 60 = 600 sec P.r.t
S.I. =
12000 100
Speed of train = = 20 m / sec
600 According to question –
when train passes a telegraph post it covers distance P.4.Q
equal to its length. P – 340 =
using distance = time × speed 100
x = 6 × 20 = 120 m QP
53. (d) Let the weight of the boy who left the class is x. P – 340 =
Total weight = 50 × 45 = 2250 kg 25
when one boy left the class total weight is (2250 – x) QP 17P
and no. of student left is 49. P – 25 = 340 Þ 25 = 340
According to question – P = 500 `
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2250 x æ 100 ö 59. (b) Let A, B and C can work x, y, z units work daily
= ç 45 - ÷ kg
49 è 1000 ø According to question –
-
2250 x 2x = y + z
2z = x + y
49 = 44.9 total work = No. of days taken by ‘c’ × unit work of ‘c’
2250 – x = 2200.1 = 40z
x = 2250 – 2200.1 time taken when A, B and C working together
x = 49.9 kg
54. (c) Relative speed between man and train = 68 – 8 total work 40z
= =
+
5 50 totalunit work daily x + y z
= 60 km /h = 60× m / sec = m/s as x + y = 2z so,
10 3
150 40z 40z 40 1
time taken to cross the man = = 9 sec time taken = = = 3 = 13 days.
+
3
50 / 3 2z z 3z
55. (a) Let the price of item be `100. 60. (a) In 35 litre mixture
price after 25% decrement = 75 ` 1
Now % change to achieve 100 ` again water is = 35 ´ = 7 litre
5
-
(100 75) milk is = 35 – 7 = 28 litre
= ´ 100
75 after adding 7 litre of more water, water is 7+7 = 14 litre.
25 1 28 2
= ´ 100 = 33 % so ratio of milk and water in new mixture is = = = 2 : 1
75 3 14 1
56. (c) Let the sum be ‘P’ and rate be ‘r’ simple interest (S.I.) 1000
P.r.t 61. (b) Cost price of item where Anmol gained 10% is =
= 1.1
100 10000
= `
P.r.3 3Pr 11
S.I. = = ...(1)
100 100 1000
again. Cost price of item where Anmol lost 10% is = 0.9
+
P.(r 4).3 = 10000
(S.I. + 600) = 9 `
100 Amount Anmol received is = 1000 + 1000 = 2000 `
putting S.I. value from eq. ...(1) 10000
3Pr + 3P(r + 4) Amount Anmol paid is = 10000 + 9
11
100 600 = 100 20
3Pr + 3Pr + 12P = 10000´ 99 `
100 600 = 100 100
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12P æ ç 2000 10000´ 20 ö ÷
600 = profit or loss % = ç 99 ´ 100
100 20 ÷
P = 5000 ` ç ç 10000´ ÷ ÷
è 99 ø