SSC CGL 2025 Tier-I · 12 September 2025, Shift 1 · Quantitative Aptitude · Question 18 of 100

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 11 September 2026.

Where this paper and its answers come from

SSC does not publish CGL question papers or a public answer key. From 2025 the candidate response sheet prints every question and option as a picture, so the text here is transcribed from those pictures for this shift and checked against them independently; where the layout carries meaning, the picture is shown as well. Each answer is the key SSC colour-marked on that sheet. SSC released a tentative key for challenge and later its final one, both only to candidates behind a login, and the sheet does not say which of the two it carries. The solutions are written by UnlockIAS.

Three pipes, A, B, and C are capable of filling a tank in 6, 8, and 12 hours, respectively. When all three pipes are opened together, they operate for 2 hours before pipe C is closed. How much additional time will it take to completely fill the tank after that?

(a)6⁄7 hours
(b)2⁄6 hours
(c)3⁄5 hours
(d)9⁄2 hours

Answer

Why

Correct — A. Take the tank as the LCM of 6, 8 and 12 = 24 units.

Rates: A = 24 ÷ 6 = 4, B = 24 ÷ 8 = 3, C = 24 ÷ 12 = 2 units an hour
All three together: 4 + 3 + 2 = 9 units an hour
In 2 hours: 9 × 2 = 18 units filled
Left: 24 − 18 = 6 units

C closes, so A + B work on: 4 + 3 = 7 units an hour
Additional time = 6 ÷ 7 = 6⁄7 hours → option (a)

Why the others are wrong

(b)
2⁄6 hours — 2⁄6 hour is 20 minutes. At A + B's 7 units an hour that fills 7 × 1⁄3 ≈ 2.3 units, well short of the 6 units still empty.
(c)
3⁄5 hours — 3⁄5 hour at 7 units an hour fills 7 × 3⁄5 = 4.2 units. The tank has 6 units left, so 1.8 units would still be empty.
(d)
9⁄2 hours — 9⁄2 hours at 7 units an hour pours in 31.5 units — more than the whole 24-unit tank, let alone the 6 units left after the first 2 hours.

Concept

Pipe problems are work-rate problems. Give the tank a convenient size — the LCM of the filling times — so that every rate becomes a whole number of units an hour.

Rates add while pipes run together. Work done = rate × time, and whatever is left is finished at the rate of the pipes still open.

When a pipe is closed partway, split the job into phases: all pipes first, then the remaining pipes on the remaining work.

The question asks for the additional time after C closes, not the total. The total from the start is 2 + 6⁄7 = 2 6⁄7 hours, which is not among the options.

Key facts

Taking the tank as the LCM of the filling times turns every rate into whole units an hour.
Rates of pipes working together add: 1⁄6 + 1⁄8 + 1⁄12 = 9⁄24 = 3⁄8 of the tank an hour.
After 2 hours of all three pipes, 18 of the 24 units — three-quarters of the tank — are full.
A and B alone fill 7⁄24 of the tank an hour, so they would fill an empty tank in 24⁄7 ≈ 3.43 hours.

Study next

Pipe problems with an outlet pipe, whose rate is subtractedPipe problems where one pipe is closed a set time before the tank is fullWork-and-time problems where a worker joins or leaves partway

Common traps

Giving the total time, 2 6⁄7 hours, when the question asks for the additional time after C closes.
Keeping C's 2 units an hour in the second phase after it has been closed, which gives 6 ÷ 9 = 2⁄3 hour.

The same two-phase structure, with the same 6-hour and 8-hour pipes, is on 18 Sep 2024, 12:30, Quant Q.13: A closes after 3 hours, B alone needs another hour, and the keyed 4 hours is the total time.

Work done, then work left, is also asked on 12 Sep 2025, 09:00, Quant Q.16, where A and B work together for 4 days and 8⁄15 of the job remains.

Related PYQs

No directly related past PYQ was found.

Topic: Time & Work — all 39 SSC CGL questions on it, every solved shift.