Time & Work Questions and Answers

Answer: B

All three together finish the work in 4 days.

Adding the three pair rates counts each person twice:

\[ \begin{aligned} 2(A + B + C) &= \frac{1}{6} + \frac{1}{12} + \frac{1}{4} = \frac{6}{12} \\ A + B + C &= \frac{1}{4} \end{aligned} \]

So they do a quarter of the work a day and need 4 days. 2 days is the trap of forgetting to halve the combined rate, and 8 or 10 days would be slower than C and A working as a pair. (The figures imply that B does no work at all, since A + B + C comes out equal to C + A, but the answer is still 4 days.)

Answer: C

The whole work was completed in 12 days.

In 4 days together they do

\[ \begin{aligned} 4\left(\frac{1}{12} + \frac{1}{18}\right) &= 4 \times \frac{5}{36} \\ &= \frac{5}{9} \end{aligned} \]

The remaining \(\frac{4}{9}\) is done by B alone at \(\frac{1}{18}\) per day, taking \(\frac{4}{9} \times 18 = 8\) days. The total is 4 + 8 = 12 days.

8 days is only B's share, the usual slip; 20 and 30 days are longer than B would take to do the whole job alone.

Answer: A

A does 20% less work than B, so A's rate is 0.8 of B's. If A takes 7.5 hours, B takes 7.5 x 0.8 = 6 hours, option A.

The other options come from using 1.25 or other wrong factors, such as \(5\frac{1}{2}\), \(6\frac{1}{2}\) and \(6\frac{3}{4}\) hours.

Work is rate times time, so with A's rate at 80 per cent of B's, B takes 80 per cent of A's time. Then 7.5 x 0.8 = 6 hours. The tempting answer of 9.375 hours comes from dividing by 0.8 instead of multiplying, and it is not offered. The other options come from similar slips.

Answer: B

4 men finish the work in 3 days, so the work is 12 man-days, and 3 women in 6 days make it 18 woman-days. Then 8 men and 6 women do 8/12 + 6/18 = 1 in a day, 12 men do it in a day, and 18 women do it in a day.

6 men and 8 women do 6/12 + 8/18 = 0.944 of the work in a day, which is less than the whole, so that statement is not true, option B.

Writing each statement in man-days makes the check easy: one man does 1/12 of the work a day and one woman 1/18. For 6 men and 8 women the sum is 6/12 + 8/18 = 17/18 of the work, so the statement that they finish in one day is not true, while the other three statements add up to exactly one day.

Answer: C

Let the daily work rates of a man, a woman and a boy be m, w and b. The three conditions are:

Twice the first minus the second gives:

The third gives:

The second gives:

Then:

so the time is 480/11 = 43 7/11 hours, option C.

Answer: D

The total work is 8 men x 9 hours x 20 days = 1,440 man-hours.

The new group works 12 men x 10 hours = 120 man-hours a day, so \[\text{days}=\tfrac{1440}{120}=12.\]

So the work takes 12 days, option D. The values 9, 10 and 11 days would give fewer than 1,440 man-hours.

The same method works for any problem of men, hours and days: men times hours times days stays constant for a fixed amount of work.

Answer: B

B alone would take 9 days.

Work in fractions of the job per day. Together they do \(\frac{1}{6}\) a day and A alone \(\frac{1}{18}\):

\[ \begin{aligned} \text{B's rate} &= \frac{1}{6} - \frac{1}{18} \\ &= \frac{3 - 1}{18} = \frac{1}{9} \end{aligned} \]

So B finishes the work in 9 days. Check: \(\frac{1}{18} + \frac{1}{9} = \frac{3}{18} = \frac{1}{6}\). 12 comes from subtracting days (\(18 - 6\)) instead of rates, a common mistake; 15 has no basis.

Answer: C

Working together they take \(3\frac{7}{11}\) days.

Add the work each person does in one day:

\[ \begin{aligned} \tfrac{1}{8} + \tfrac{1}{12} + \tfrac{1}{15} &= \tfrac{15 + 10 + 8}{120} \\ &= \tfrac{33}{120} = \tfrac{11}{40} \end{aligned} \]

So together they finish \(\frac{11}{40}\) of the work a day, and the whole work takes

\[ \frac{40}{11} = 3\tfrac{7}{11}\ \text{days} \]

The answer must be less than 8 days, the time of the fastest worker alone, and well under it with three people working. 5 and 6 days are too long, and \(4\frac{5}{11}\) comes from an arithmetic slip in the common denominator.

Answer: D

One worker alone takes 5 hrs. 20 min.

Two workers at the same speed, working on the task together, finish it in 2 hours 40 minutes, which is 160 minutes. One worker does half the work per minute, so takes twice as long: \(2 \times 160 = 320\) minutes, which is 5 hours 20 minutes.

1 hr. 20 min halves the time instead of doubling it. 4 hrs. 40 min and 5 hrs. come from adding instead of doubling, or from rounding.

Answer: A

The work is completed 1.6 hrs. after the second man joins.

The first man does \(\frac{1}{6}\) of the work per hour and the second \(\frac{1}{4}\). In his first 2 hours alone the first man does \(\frac{2}{6} = \frac{1}{3}\), leaving \(\frac{2}{3}\) for the two together:

\[ \begin{aligned} \text{joint rate} &= \frac{1}{6} + \frac{1}{4} = \frac{5}{12} \\ \text{time} &= \frac{2}{3} \div \frac{5}{12} = \frac{8}{5} = 1.6\ \text{hrs} \end{aligned} \]

The question asks for the time taken after the two start working together; counting the first man's 2 hours as well would give 3.6 hours, which is not offered. 2.4 hours is how long the two together would need for the whole job.