Two pipes S1 and S2 alone can fill an empty tank in 15 hours and 20 hours respectively. Pipe S3 alone can empty that completely filled tank in 40 hours. Firstly both pipes S1 and S2 are opened and after 2 hour pipe S3 is also opened. In how much time tank will be completely filled after S3 is opened?

Question

Question

This question was previously asked in

SSC CGL 2023 Tier-I Official Paper (Held On: 21 Jul 2023 Shift 2)

Answer 1

Given:

wo pipes S1 and S2 alone can fill an empty tank in 15 hours and 20 hours respectively.

Pipe S3 alone can empty that completely filled tank in 40 hours.

Firstly both pipes S1 and S2 are opened and after 2 hour pipe S3 is also opened.

Concept used:

Efficiency = (Total work / Total time taken)

Efficiency = work done in a single day

Calculation:

Let total work is 120 units ( LCM of 15 , 20 and 40)

The efficiency of pipe s1 is 120 /15 = 8 units

The efficiency of pipe s2 is 120 / 20 = 6 units

The efficiency of pipe S1 + S2 is 8 + 6= 14 units

In 2 hrs they will fill 14 × 2 = 28 units

Remaining amount 120 - 28 = 92 units

The efficiency of pipe s3 is 120 / 40 = - 3 units

Now combined efficiency of all three pipes is 14 - 3 = 11 units

Time to fill remaining 92 units is

⇒ 92 / 11 hr

∴ The correct option is 4

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