Page 17 - AFCAT Solved Papers 2011 - 2019 SSBCrack eBook

SOLVED PAPER: AFCAT 2019/I

54.(B) Let = , Then = = 6
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Amount = 2 × 60.(D) In 1 hour 314 weavers weave
where, = Principal = 6594 × 6 shawls
100× 6594 × 6
Rate = In 1 hour 1 weaver weaves = shawls
× 314
100 × = 126 shawls
= %
× 8 61.(D) Let Cost Price (CP) =
= 12.5% Then Selling Price (SP) if he sells at profit =
55.(C) Suppose they will meet after T hours. (1 + 8%) = 1.08
= × and SP, if he sells at loss = (1 − 8%)
Sum of distance traveled by them after T = 0.92%
hours Given, 1.08% − 0.92% = 12
6 + 4 = 20 0.16 = . 12
= 2 ℎ ⇒ = . 75
So, they will meet at 62.(A) Concentration of Sulphuric acid = 80%
7:00 + 2 ℎ = 9:00 So, in a 100cc, Sulphuric acid = 80cc and
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56.(D) + + + = water=20cc
200 400 600 800
25 4 Let be the amount of water added to
⇒ = make concentration of 50%
2400
⇒ = 384 ⇒ 80 = 50
∴ Average speed is 384 /ℎ 100 + 100
80 1
57.(B) Present average age of family=25year ⇒ =
100 + 2
3 year ago average of family =25-3=22 years
∴ = 60
58.(D) Let the total distance be y km.
63.(D) We know that speed is inversely
Then, proportional to time.
1 1 44
( ) × ( ) + ( ) × ( ) = Given that:
2 60 2 50 3
44 (Speed of A): (speed of B) = 2: 7
⇒ + = 1
120 100 3 ∴ (Time taken by A): (Time taken by B) = ∶
2
⇒ 5 + 6 = 8800 1 = 7 ∶ 2
⇒ = 800 7
64.(C) Let the total number of boys in the class
∴ Required distance = 800 km.
be
59.(B) Work done by 6 men = work done by 10
Then, according to the question,
women.
+ 45 = 95
Work done by 1 man = work done by 10/6
women = 5/3 women = 95 − 45 = 50
5 Hence, the required ratio of total number of
∴ 12 men + 5 women = 12 × ( ) + 5
3 boys to total number of girls
= 25 women = 50: 45 = 10: 9
∴ × = × 65.(A) Ram : Gopal = 7 : 17 = 49 : 119
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1
2
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W=women, D=days Gopal : Krishan = 7 : 17 = 119 : 289
10 × 15 = 25 × ∴ Ram: Gopal : Krishan = 49 : 119 : 289
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