SSC CGL 2024 Tier-I · 24 September 2024, Shift 3 · 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 10 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.

Pipe A can fill a cistern in 4 hours and Pipe B can fill the same cistern in 5 hours. Pipe C can empty a full cistern in 3 hours. If all three pipes are opened together, then the time (in minutes) taken to fill the tank is: (round to the nearest minute)

(a)625
(b)445
(c)514
(d)800

Answer

Why

Correct — C.

Take the cistern as 60 units (the LCM of 4, 5 and 3).

A fills 60 ⁄ 4 = 15 units per hour
B fills 60 ⁄ 5 = 12 units per hour
C drains 60 ⁄ 3 = 20 units per hour

Net rate with all three open = 15 + 12 − 20 = 7 units per hour

Time = 60 ⁄ 7 hours
= 8.571… hours × 60 = 514.28… minutes
To the nearest minute, 514 → option (c).

Why the others are wrong

(a)
625 — 625 minutes is 10 h 25 min, which needs a net rate of 0.096 of the cistern an hour. The three pipes together manage 7 ⁄ 60 ≈ 0.117, so the tank fills sooner.
(b)
445 — 7 ⁄ 60 of a cistern an hour cannot finish inside 8½ hours. 445 minutes is 7 h 25 min, while 60 ⁄ 7 hours is 8 h 34 min.
(d)
800 — 800 minutes is 13 h 20 min, the time you would need at a net rate of 3 ⁄ 40 an hour. The actual net rate, 7 ⁄ 60, is larger, so the fill is quicker.

Concept

Pipe questions are rate arithmetic. Turn each time into a rate, give the outlet a minus sign, and add.

As fractions of the cistern: 1 ⁄ 4 + 1 ⁄ 5 − 1 ⁄ 3 = 15 ⁄ 60 + 12 ⁄ 60 − 20 ⁄ 60 = 7 ⁄ 60 per hour, and the filling time is its reciprocal, 60 ⁄ 7 hours.

The LCM trick removes the fractions: call the cistern 60 units and the pipes move 15, 12 and −20 units an hour. The tank fills at all only because the inflow of 27 units beats the outflow of 20.

The times are given in hours and the options in minutes, so the ×60 conversion is part of the question rather than an afterthought.

Key facts

A pipe filling in t hours works at 1 ⁄ t of the cistern per hour, and an emptying pipe counts as negative.
With all three open the net rate is 1 ⁄ 4 + 1 ⁄ 5 − 1 ⁄ 3 = 7 ⁄ 60 of the cistern per hour.
60 ⁄ 7 hours is 8.571 hours, which is 514.28 minutes and rounds to 514.
The cistern fills only because 1 ⁄ 4 + 1 ⁄ 5 is greater than 1 ⁄ 3.

Study next

Inlet and outlet pipes handled as signed ratesThe LCM-of-times method for work and pipe problemsProblems where one pipe is closed part-way through the filling

Common traps

Adding all three rates and treating the outlet as an inlet, which gives 47 ⁄ 60 and a far shorter time.
Leaving the answer in hours when the options are in minutes.
Rounding 8.57 hours up to 9 before converting, which produces 540 minutes.

The three-pipe build with one drain recurs with different numbers.
17 Sep 2024, 16:00, Quant Q.9 uses 12 h, 18 h and an 8 h outlet and asks the same plain question; 17 Sep 2024, 09:00, Quant Q.17 closes pipe A after 5 hours; 12 Sep 2024, 09:00, Quant Q.7 works in minutes and shuts the outlet before the tank is full.

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