Aptitude - Chain Rule
11.
10 pipes of the same diameter can fill a tank in 24 minutes. If 2 pipes go out of order, how long will the remaining pipes take to fill the tank ?
Answer
Let the required time be x minutes.
Less pipes, more time (Indirect)
8 : 10 :: 24 : x ⇒ 8x = (10 x 24)
⇒ x = (10 x 24)8 = 30 min
Answer :
Option CExplanation :
Let the required time be x minutes.
Less pipes, more time (Indirect)
8 : 10 :: 24 : x ⇒ 8x = (10 x 24)
⇒ x = (10 x 24)8 = 30 min
12.
3 men or 5 women can finish a piece of work in 12 days. How long will 6 men and 5 women take to finish the work ?
Answer
3 men ≡ 5 women ⇒ 6 men ≡ 10 women
⇒ (6 men + 5 women) ≡ 15 women
Let the required number of days be x.
More women, less days (Indirect)
15 : 5 :: 12 : x ⇒ 15x = (5 x 12)
⇒ x = 5 x 1215 = 4 days
Answer :
Option AExplanation :
3 men ≡ 5 women ⇒ 6 men ≡ 10 women
⇒ (6 men + 5 women) ≡ 15 women
Let the required number of days be x.
More women, less days (Indirect)
15 : 5 :: 12 : x ⇒ 15x = (5 x 12)
⇒ x = 5 x 1215 = 4 days
13.
8 women can complete a work in 10 days and 10 children take 16 days to complete the same work. How many days will 10 women and 12 children take to complete the work ?
Answer
1 women can complete the work in (10 x 8) days = 80 days
1 child can complete the work in (16 x 10) days = 160 days
1 woman's 1 day's work = 180
1 child's 1 day's work = 1160
(10 women + 12 children)'s 1 day's work =
10 x 180 + 12 x 1160 
= 18 + 340 = 840 = 15
∴ 10 women and 12 children will finish the work in 5 days.
Answer :
Option DExplanation :
1 women can complete the work in (10 x 8) days = 80 days
1 child can complete the work in (16 x 10) days = 160 days
1 woman's 1 day's work = 180
1 child's 1 day's work = 1160
(10 women + 12 children)'s 1 day's work =
10 x 180 + 12 x 1160 =
18 + 340 = 840 = 15∴ 10 women and 12 children will finish the work in 5 days.
14.
A garrison had provisions for a certain number of days. After 10 days, 1 5th of men desert and it is found that the provisions will now last just as long as before. How long was that ?
Answer
Initially, let x men had provisions for y days. Then, x men had provisions for (y − 10) days.
x − x5 men had provisions for y days.
∴ x(y − 10) = 4x5 x y ⇒ 5xy − 50x = 4xy
xy − 50x = 0 ⇒ x(y − 50) = 0 ⇒ y = 50
Answer :
Option DExplanation :
Initially, let x men had provisions for y days. Then, x men had provisions for (y − 10) days.
x − x5 men had provisions for y days.∴ x(y − 10) = 4x5 x y ⇒ 5xy − 50x = 4xy
xy − 50x = 0 ⇒ x(y − 50) = 0 ⇒ y = 50
15.
A contractor undertakes to do a piece of work in 40 days. He engages 100 men at the begining and 100 more after 35 days and completes the work in stipulated time. If he had not engaged the additional men, how many days behind the schedule would it be finished ?
Answer
[(100 x 35) + (200 x 5)] men can finish the work in 1 day.
4500 men can finish it in 1 day.
∴ 100 men can finish it in 4500100 days = 45 days.
This is 5 days behind schedule.
Answer :
Option BExplanation :
[(100 x 35) + (200 x 5)] men can finish the work in 1 day.
4500 men can finish it in 1 day.
∴ 100 men can finish it in 4500100 days = 45 days.
This is 5 days behind schedule.
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