Time & Work, Pipes & Cisterns: previous-year questions

Answer: (B) 21 hours

The pipe fills $\frac{1}{3}$ of the tank per hour, while pipe and leak together fill $\frac{2}{7}$ per hour. The leak therefore empties $\frac{1}{3} - \frac{2}{7} = \frac{1}{21}$ per hour, i.e. the full tank in 21 hours. 12 or $10\frac{1}{2}$ hours would make the leak too strong to allow filling in $3\frac{1}{2}$ hours.

Answer: (A) 30 days

Together Ramesh and Rahman do $\frac{1}{20} + \frac{1}{25} = \frac{9}{100}$ of the work a day, so in 10 days they finish $\frac{90}{100}$. Suresh does the remaining $\frac{1}{10}$ in 3 days, so he alone needs 30 days. The other options do not match $\frac{1}{10}$ in 3 days.

Answer: (D) 8 days

A's daily work = $\frac{1}{6} - \frac{1}{12} = \frac{1}{12}$ and C's = $\frac{1}{6} - \frac{1}{8} = \frac{1}{24}$. Together A and C do $\frac{3}{24} = \frac{1}{8}$ a day, so they need 8 days. 6 days is the time for all three together.

Answer: (C) ₹ 200

Payment is shared in proportion to the work done, and C did $1 - \frac{7}{11} = \frac{4}{11}$ of it. C's share is $\frac{4}{11} \times 550$ = ₹ 200. ₹ 350 is the combined share of A and B.

Answer: (D) 14 min 40 sec

In 4 minutes both pipes fill $4 \times (\frac{1}{15} + \frac{1}{20}) = \frac{7}{15}$ of the tank. B alone fills the remaining $\frac{8}{15}$ in $\frac{8}{15} \times 20 = 10\frac{2}{3}$ minutes, i.e. 10 min 40 sec. Total time = 4 min + 10 min 40 sec = 14 min 40 sec.

Answer: (B) 6 days

Together they do $\frac{1}{12} + \frac{1}{15} = \frac{3}{20}$ a day, so in 4 days they finish $\frac{3}{5}$ of the work. B alone does the remaining $\frac{2}{5}$ in $\frac{2}{5} \times 15 = 6$ days. 10 days would be B's time for $\frac{2}{3}$ of the work.

Answer: (D) 3 hours 45 minutes

One tap fills the first half in 3 hours. Four taps together fill $\frac{4}{6}$ of the tank per hour, so the second half takes $\frac{1}{2} \div \frac{2}{3} = \frac{3}{4}$ hour = 45 minutes. Total = 3 hours 45 minutes.

Answer: (B) 10 days

Adding the three pairs counts each person twice: $2(A + B + C) = \frac{1}{12} + \frac{1}{15} + \frac{1}{20} = \frac{1}{5}$. So A + B + C = $\frac{1}{10}$ per day, i.e. 10 days. 5 days forgets to halve the sum.

Answer: (C) 15 days

B does 1·4 times A's daily work, so B needs 21 ÷ 1·4 = 15 days. Cutting 40% off A's time gives about 12·6 days, which is the usual trap.

Answer: (B) 40

The total work gives 8(4m + 6w) = 10(3m + 7w), so 32m + 48w = 30m + 70w and one man equals 11 women. Then 4 men + 6 women = 50 women, who take 8 days, so the work is 400 woman-days. 10 women need 400 ÷ 10 = 40 days.

Answer: (A) 30 days

A's rate = $\frac{1}{2} \div \frac{3}{4} = \frac{2}{3}$ of B's rate. Together they work at $\frac{5}{3}$ of B's rate, which equals $\frac{1}{18}$ per day, so B alone does $\frac{1}{30}$ per day. B needs 30 days; 45 days is A's time.

Answer: (D) 24 minutes

In 30 minutes A fills $\frac{30}{60} = \frac{1}{2}$ of the cistern, so B must fill the other half. B fills half in $\frac{48}{2} = 24$ minutes, so B is turned off after 24 minutes.

Answer: (C) 900 liters

For capacity C, A fills $\frac{C}{15}$ litres a minute and together they fill $\frac{C}{45}$ a minute. So $\frac{C}{15} - 40 = \frac{C}{45}$, giving $\frac{2C}{45} = 40$ and C = 900 litres. Check: A gives 60 litres a minute, net 20, and 900 ÷ 20 = 45 minutes.

Answer: (C) 24 days

Together Bulu and Tathagata do $\frac{1}{20} + \frac{1}{30} = \frac{1}{12}$ a day, so in 7 days they finish $\frac{7}{12}$. Anita does the remaining $\frac{5}{12}$ in 10 days, so the whole work takes her $10 \times \frac{12}{5} = 24$ days.

Answer: (B) 15 days

In each 3-day cycle P does $\frac{3}{20}$ and Q and R together add $\frac{1}{30} + \frac{1}{60} = \frac{1}{20}$ on the third day, a total of $\frac{1}{5}$. Five such cycles finish the job, which is 15 days. 16 or 17 days overshoot the exact finish.

Answer: (D) $8\frac{1}{4}$ minutes

In 3 minutes both pipes fill $3 \times (\frac{1}{12} + \frac{1}{15}) = \frac{9}{20}$ of the cistern. Q alone fills the remaining $\frac{11}{20}$ in $\frac{11}{20} \times 15 = 8\frac{1}{4}$ minutes. The other fractions come from slips in this arithmetic.

Answer: (C) 200 days

After 30 days the food left would last 4000 men for $190 - 30 = 160$ days, i.e. $4000 \times 160$ man-days. With $4000 - 800 = 3200$ men it will last $\frac{4000 \times 160}{3200} = 200$ days. Rule: men and days are inversely proportional, so $M_1 D_1 = M_2 D_2$. Option D (170) is a subtraction trap.

Answer: (B) $3\frac{3}{4}$ hrs.

Mohan does $\frac{1}{6}$ of the job per hour and Laltu $\frac{1}{10}$. Together: $\frac{1}{6}+\frac{1}{10}=\frac{5+3}{30}=\frac{4}{15}$ of the job per hour, so the time taken is $\frac{15}{4}=3\frac{3}{4}$ hours (3 hours 45 minutes). Shortcut for two workers: $\frac{ab}{a+b}=\frac{6\times10}{16}=3.75$. The answer must be less than the faster worker's time (6 h) and more than half of it (3 h), which rules out $1\frac{1}{2}$ and $2\frac{1}{3}$.

Answer: (D) Rs. 400

Manoj types $\frac{100}{10}=10$ pages an hour and Bobby $\frac{100}{15}=\frac{20}{3}$ pages an hour. In 5 hours together they type $5\times\left(10+\frac{20}{3}\right)=83\frac{1}{3}$ pages, i.e., $\frac{5}{6}$ of the work. Lucy does the remaining $\frac{1}{6}$ (about $16\frac{2}{3}$ pages). Payment is shared in proportion to the work done, so Lucy gets $\frac{1}{6}\times2400=$ Rs. 400; Manoj gets Rs. 1,200 and Bobby Rs. 800. The '2 hours' is extra data - wages depend on work done, not on time taken.

Answer: (A) 8 minutes

The second pipe works throughout, and after the first is closed it needs 10 more minutes, in which it fills $\tfrac{10}{30} = \tfrac{1}{3}$ of the cistern. So the two pipes together must fill $\tfrac{2}{3}$ before that; their joint rate is $\tfrac{1}{20}+\tfrac{1}{30} = \tfrac{1}{12}$ per minute, which takes $\tfrac{2}{3}\times12 = 8$ minutes. So the first pipe is turned off after 8 minutes (total time 18 minutes).