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Hardpipes

~105s

Closing one tap early

Two pipes fill a tank in 20 minutes and 30 minutes respectively. Both are opened together, but the first is closed after some time and the tank fills in 18 minutes in all. After how long was the first pipe closed?

Show the answer and the worked solution

Correct answer: B.

8 min

Worked solution

  1. The second pipe contributes 18/30 = 3/5 of the tank.
  2. The first must supply the remaining 2/5: t/20 = 2/5 → t = 8.
  3. The first pipe was closed after 8 minutes.

Answer: 8 min.

The idea

The pipe that runs throughout is computed first; the other fills whatever is left. One equation in the unknown running time.

Shortcut. Had the first pipe run all 18 minutes, 18/20 + 18/30 = 1.5 tanks would flow; the extra half tank is 10 minutes of the first pipe, so it shut after 18 − 10 = 8 minutes.

Common trap. 12 min — giving the first pipe the 3/5 that the second pipe supplies (3/5 × 20 = 12) instead of the remaining 2/5.

Closing one tap early — Time & Work Aptitude Question

Closing one tap early: Two pipes fill a tank in 20 minutes and 30 minutes respectively. Answer and worked solution on CodeKairo.

Question

Two pipes fill a tank in 20 minutes and 30 minutes respectively. Both are opened together, but the first is closed after some time and the tank fills in 18 minutes in all. After how long was the first pipe closed?

Options

  1. 6 min
  2. 8 min
  3. 10 min
  4. 12 min
Show the answer and the worked solution

Correct answer: B. 8 min

Worked solution

  1. The second pipe contributes 18/30 = 3/5 of the tank.
  2. The first must supply the remaining 2/5: t/20 = 2/5 → t = 8.
  3. The first pipe was closed after 8 minutes.

Answer: 8 min.

The idea

The pipe that runs throughout is computed first; the other fills whatever is left. One equation in the unknown running time.

Shortcut. Had the first pipe run all 18 minutes, 18/20 + 18/30 = 1.5 tanks would flow; the extra half tank is 10 minutes of the first pipe, so it shut after 18 − 10 = 8 minutes.

Common trap. 12 min — giving the first pipe the 3/5 that the second pipe supplies (3/5 × 20 = 12) instead of the remaining 2/5.

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