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A leaves five days before the end

A can do a work in 20 days and B in 30 days. They begin together, but A leaves 5 days before the work is finished. In how many days is the work completed?

Show the answer and the worked solution

Correct answer: B.

15 days

Worked solution

  1. Let the total time be T days. Then (T − 5)/20 + T/30 = 1.
  2. Multiply by 60: 3(T − 5) + 2T = 60 → 3T − 15 + 2T = 60 → 5T = 75.
  3. T = 15 days.

Answer: 15 days.

The idea

Set up one equation in the total time, remembering that the person who leaves works fewer days. The one who stays always works the full duration.

Shortcut. Pretend A stays to the end: he would add 5/20 = 1/4 of a job, so the pair (12 days a job, 20 × 30 ÷ 50) would finish 1¼ jobs in the time: 1.25 × 12 = 15 days.

Common trap. 12 days — the time if both worked throughout, ignoring that A leaves early.

A leaves five days before the end — Aptitude Question

A leaves five days before the end: A can do a work in 20 days and B in 30 days. They begin together, but A leaves 5… Answer and worked solution on CodeKairo.

Question

A can do a work in 20 days and B in 30 days. They begin together, but A leaves 5 days before the work is finished. In how many days is the work completed?

Options

  1. 12 days
  2. 15 days
  3. 18 days
  4. 20 days
Show the answer and the worked solution

Correct answer: B. 15 days

Worked solution

  1. Let the total time be T days. Then (T − 5)/20 + T/30 = 1.
  2. Multiply by 60: 3(T − 5) + 2T = 60 → 3T − 15 + 2T = 60 → 5T = 75.
  3. T = 15 days.

Answer: 15 days.

The idea

Set up one equation in the total time, remembering that the person who leaves works fewer days. The one who stays always works the full duration.

Shortcut. Pretend A stays to the end: he would add 5/20 = 1/4 of a job, so the pair (12 days a job, 20 × 30 ÷ 50) would finish 1¼ jobs in the time: 1.25 × 12 = 15 days.

Common trap. 12 days — the time if both worked throughout, ignoring that A leaves early.

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