PATHANTUPadho, Samjho, Badho
Menu

PATHANTU

Time and work: rates, remaining work and leaks

Solve joint work and delayed joining with consistent work units. Assume constant rates unless specified otherwise.

By Pathantu · Published 6 October 2026 · Original practice, not PYQ

Method

  1. One job in d days means a rate of 1/d job per day. Add rates, not completion times.

  2. Take the LCM of individual times as total work units. Time = remaining units ÷ combined daily rate.

  3. Subtract a leak’s rate. If it equals or exceeds filling, an initially empty tank cannot fill under those constant conditions.

Six questions: try first, then open the solution

1. A takes 10 days and B 15. Find their time together.

Show worked solution

Take 30 units: daily rates 3 and 2. Time=30/(3+2)=6 days.

2. A takes 12 days, B 18. A works alone for 3 days, then B joins. Find total time.

Show worked solution

Take 36 units. A completes 9; 27 remain. Joint rate=5. Total=3+27/5=8.4 days.

3. Together A and B need 8 days; A alone needs 24. Find B’s time.

Show worked solution

B’s rate=1/8−1/24=1/12. B needs 12 days.

4. A pipe fills in 6 hours; a leak empties in 9. How long with both open?

Show worked solution

Net rate=1/6−1/9=1/18 tank/hour. Filling takes 18 hours.

5. A is twice as efficient as B. Together they need 6 days. Find individual times.

Show worked solution

Rates 2:1, total work 18 units. A needs 18/2=9 days; B needs 18/1=18.

6. After 4 days on a 10-day job, what fraction remains?

Show worked solution

Completed=4/10=2/5. Remaining=1−2/5=3/5.

Check your reasoning

Convert hours and days to one unit. Twice the efficiency means half the time for the same job.

For each wrong answer, write the incorrect step and solve the question again without the solution. These open lessons need no login; timed quizzes require a student account.

Practise the free SSC subject quiz: 20 questions, 20 minutes, +2 / −0.50

Continue learning

Report a correction