A tap can fill a cistern in 10 minutes and another tap can empty it in 12 minutes. If both the taps are open, the time (in hours) taken to fill the tank will be:
- (a)2 hours
- (b)2.5 hours
- (c)1 hour
- (d)1.5 hours
Answer
Why
Correct — C. Turn each tap into a rate per minute and add them with signs — the emptying tap is negative.
Filling rate = 1⁄10 cistern per minute
Emptying rate = −1⁄12 cistern per minute
Net = 1⁄10 − 1⁄12 = (6 − 5)⁄60 = 1⁄60 per minute
So the cistern fills in 60 minutes = 1 hour → option (c).
The unit method agrees. Take the cistern as LCM(10, 12) = 60 units: the taps move +6 and −5 units a minute, a net of 1 unit, so 60 units take 60 minutes.
Why the others are wrong
- (a)2 hours — 2 hours is the difference 12 − 10 read off as an answer. It is the rates that subtract, not the times — 120 minutes would need a net of 1⁄120 per minute, half the actual 1⁄60.
- (b)2.5 hours — 2.5 hours is 150 minutes, which needs a net rate of 1⁄150 per minute. The actual net, 1⁄10 − 1⁄12, has denominator LCM(10, 12) = 60, so the fill time is 60 minutes.
- (d)1.5 hours — 1.5 hours is 90 minutes, needing a net of 1⁄90. The outlet is the weaker tap, so it slows the fill from 10 minutes to 60, and the arithmetic lands on 60 exactly, not near it.
Concept
Pipe questions are rate questions. A tap that fills in t minutes contributes 1⁄t of the cistern per minute, and an emptying tap contributes the same quantity with a minus sign.
Open several together and the rates add: net = 1⁄10 − 1⁄12 = 1⁄60. Invert the net rate and you have the time, 60 minutes.
The LCM route avoids fractions entirely. Call the cistern 60 units, and the taps become +6 and −5 units a minute, so one unit of progress is made per minute.
The sign matters more than the arithmetic. If the outlet were faster than the inlet the net rate would be negative and the cistern would never fill, whatever the options say.
Note also the unit switch: the data is in minutes and the answer is wanted in hours, so 60 must become 1.
Key facts
- A pipe that fills a tank in t units of time works at 1⁄t of the tank per unit of time.
- An emptying pipe carries a negative rate, so rates are added with signs rather than subtracted case by case.
- 1⁄10 − 1⁄12 = 1⁄60, so the pair fills one sixtieth of the cistern each minute.
- For one inlet and one outlet the time is the product over the difference: (10 × 12) ⁄ (12 − 10) = 60 minutes.
Study next
Common traps
- Answering 60 when the options are in hours.
- Adding the two rates instead of subtracting, which gives 60⁄11 minutes.
- Using product over sum, (10 × 12) ⁄ 22, instead of product over difference.
SSC keeps the numbers small and moves the difficulty into the unit — minutes given, hours wanted, or the reverse. Inlet-and-outlet items also run at 17 Sep 2024, 16:00, Quant Q.9, at 24 Sep 2024, 16:00, Quant Q.7 and at 12 Sep 2024, 09:00, Quant Q.7.
Related PYQs
No directly related past PYQ was found.