SSC CGL 2024 Tier-I · 12 September 2024, Shift 1 · Quantitative Aptitude · Question 7 of 99

Text as printed on a candidate's response sheet for this shift; answer per SSC's tentative key as marked on that sheet. Solution last reviewed 6 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. The questions and options here are reproduced as printed on a candidate's own response sheet for this shift — the Commission's text, unchanged. Each answer is SSC's tentative key as marked on that sheet; SSC allowed challenges to it and never publishes a final key. The solutions are written by UnlockIAS.

Two pipes, A and B, can fill a tank in 10 minutes and 20 minutes, respectively. The pipe C can empty the tank in 30 minutes. All the three pipes are opened at a time in the beginning. However, pipe C is closed 2 minutes before the tank is filled. In what time, will the tank be full (in minutes)?

(a)10
(b)8
(c)12
(d)6

Answer

Why

Correct — B. Work in units. Take the tank as LCM(10, 20, 30) = 60 units.

A fills 60 ÷ 10 = 6 units a minute
B fills 60 ÷ 20 = 3 units a minute
C drains 60 ÷ 30 = 2 units a minute

Let T be the total time. A and B run for T; C runs for T − 2.

9T − 2(T − 2) = 60
7T + 4 = 60 → 7T = 56 → T = 8 minutes

Check: 6 minutes at the net 7 units gives 42, then A and B alone add 18 in 2 minutes — 60 exactly. So option (b).

Why the others are wrong

(a)
10 — Put T = 10 into 7T + 4 and you get 74 units poured into a 60-unit tank. 10 minutes is A's solo time, and B more than covers what C removes.
(c)
12 — T = 12 gives 7(12) + 4 = 88 units, further past the 60 the tank holds. The tank overflowed long before minute 12.
(d)
6 — T = 6 gives 7(6) + 4 = 46 units, short of 60. C would still have been running at minute 4, and the tank is not yet full.

Concept

Fill-and-drain problems are cleanest in units of work: set the tank to the LCM of the stated times so every rate is a whole number.

Here the tank is 60 units, A does 6 a minute, B does 3, and C removes 2. With all three open the net rate is 7 units a minute.

The complication is that C stops early, and the stopping point is defined backwards from the end. Let T be the total time, give C a run of T − 2, and solve one equation — do not guess how the time splits.

Because C closes 2 minutes before the end rather than after a stated number of minutes, the two-phase split (6 minutes, then 2) is an output of the equation, not an input to it.

Key facts

LCM of 10, 20 and 30 is 60, so one unit is 1/60 of the tank.
A and B together fill 9 units a minute, and the net rate with C open is 7 units a minute.
9T − 2(T − 2) = 60 gives 7T = 56, so T = 8 minutes.
With all three open throughout, the tank would instead take 60 ÷ 7, about 8.57 minutes.

Study next

Leak problems where the drain rate is inferred from the delay it causesPipes opened in alternate minutes or in rotationTwo-phase filling where a pipe is opened late rather than closed early

Common traps

Reading the 2 minutes as measured from the start instead of from the end
Using the net rate for the whole time and answering 60 ÷ 7
Dropping the minus sign on C and adding its 2 units instead of subtracting them

Two fill pipes against one drain is the frame. The plain version — all three open throughout — is asked at 24 Sep 2024, 16:00, Quant Q.7 (fill in 4 h and 5 h, empty in 3 h) and 17 Sep 2024, 16:00, Quant Q.9 (fill in 12 h and 18 h, empty in 8 h).

This item adds the pipe that closes 2 minutes before the end, and that single change is the whole difficulty.

Related PYQs

No directly related past PYQ was found.