Time & Work Questions and Answers

Answer: A

The work is completed in 14 1/3 days. A does 1/12 and B does 1/18 of the work a day. Two days (A then B) complete 1/12 + 1/18 = 5/36. Seven such pairs, 14 days, complete 35/36. The remaining 1/36 is done by A on the 15th day in (1/36)/(1/12) = 1/3 day. So the total is 14 1/3 days. 15 1/3 would add a whole extra day and the other two values give wrong remainders. Because they work alternately and A is faster, the pair of days does not end in a whole number of days, and the fractional last day is done by A alone. Option 15 1/3 would add an extra day with B, which is not needed because only 1/36 of the work remains.

Answer: D

75 more men are needed. In the first 45 days 150 men did 2/5 of the work, leaving 3/5 to be done in the remaining 45 days. If 150 men do 2/5 in 45 days then the 3/5 needs 150 x (3/5)/(2/5) = 225 men for 45 days. So 225 - 150 = 75 more are required. 150 and 125 are too many, and 95 is not enough. The remaining work is 3/5 of the whole in 45 days. Since 150 men finished 2/5 in 45 days, the work done per man in 45 days is 2/5 divided by 150, so for 3/5 the men needed are 150 x 3/2 = 225, requiring 75 additional men.

Answer: B

X and Z can complete 35 % of the work in 10 days. X, Y and Z together do 1/20 of the work per day. X and Y together do 40 % in 10 days, so 1/25 per day, and Y and Z together do 1/40 per day.

\[ Z = \frac{1}{20} - \frac{1}{25} = \frac{1}{100} \]

\[ X = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \]

\[ X + Z = \frac{1}{40} + \frac{1}{100} = \frac{7}{200} \]

\[ \text{Time} = \frac{0.35}{7/200} = 10 \text{ days} \]

The options 9, 12 and 15 days do not equal 0.35 divided by 7/200, so the second option is the answer.

Answer: C

Shalini needs 50 days.

Rakesh is twice as efficient, so he takes half of Shalini's time. If Shalini takes S days, Rakesh takes S/2 days, and the difference is 25 days.

\[ S-\frac{S}{2}=25 \Rightarrow S=50 \]

35, 45 and 60 do not satisfy this condition.

The check is easy: if Shalini takes 50 days, Rakesh takes half of that, 25 days, and 50 − 25 = 25 days less, as stated. Twice the efficiency always means half the time for the same work.

Answer: A

The ratio is 10:7.

The work equals 30 × 24 = 720 woman-days and also 24 × 21 = 504 man-days. So 720 woman-days equal 504 man-days, which means one man does as much as 720/504 = 10/7 women.

\[ \text{man : woman}=\frac{10}{7} \]

7:10 is the inverse and 3:5 and 5:3 do not match the numbers.

You can check the answer by computing the work of one man and one woman in a day. One woman does 1/720 of the work per day and one man does 1/504. The ratio man to woman is (1/504) : (1/720), which equals 720 : 504, and dividing by 72 gives 10 : 7.

Answer: C

After 3 hours.

Take the book as 1 unit. Ananth reads \(1/6\) per hour and Bharath \(1/4\) per hour, so after \(t\) hours the parts still to be read are \(1 - t/6\) and \(1 - t/4\).

\[ \begin{aligned} 1 - \frac{t}{6} &= 2\left(1 - \frac{t}{4}\right) \\ \frac{t}{2} - \frac{t}{6} &= 1 \\ \frac{t}{3} = 1 &\Rightarrow t = 3 \end{aligned} \]

Check: after 3 hours Ananth has half the book left and Bharath a quarter, and a half is twice a quarter. After 1 or 2 hours Ananth's remainder is less than twice Bharath's (5/6 against 3/2, 2/3 against 1); at 4 hours Bharath has finished while Ananth still has a third to read.

Answer: B

It takes 7 minutes.

Seven machines make 7 toys in 7 minutes, so each machine makes one toy in 7 minutes. A hundred machines working at the same time each make one toy in those 7 minutes, giving 100 toys.

100 and 700 come from scaling the time with the number of toys and forgetting that the number of machines grows in the same proportion. 1 minute would need each machine to work seven times faster.

Answer: C

Working together 8 hours a day, they need about 3 days.

Convert each worker's time to hours: A needs \(7 \times 9 = 63\) hours and B needs \(6 \times 7 = 42\) hours.

\[ \begin{aligned} \text{per hour} &= \frac{1}{63} + \frac{1}{42} = \frac{5}{126} \\ \text{hours} &= \frac{126}{5} = 25.2 \\ \text{days} &= \frac{25.2}{8} = 3.15 \end{aligned} \]

The exact figure is 3.15 days of 8 hours, which the options round to 3 days. 2 days (16 hours) is too short, since together they need more than 25 hours. 5 and 6 days are too long: B alone needs 42 hours, about 5.25 days of 8 hours, and adding A can only shorten that.

Answer: B

Mohit needs 20/3 days more.

Praveen does 1/30 of the work a day and Mohit 1/25. Together in 10 days they finish:

\[ \begin{aligned} 10 \left( \tfrac{1}{30} + \tfrac{1}{25} \right) &= 10 \times \tfrac{11}{150} \\ &= \tfrac{11}{15} \end{aligned} \]

That leaves \(\tfrac{4}{15}\) of the work for Mohit alone:

\[ \tfrac{4}{15} \times 25 = \tfrac{20}{3} \text{ days} \]

That is 6 days and 8 hours. 16/3 and 14/3 days would leave work undone, and 22/3 days is more than the remaining work needs.

Answer: D

None of the statements is correct.

Statement 1: 5 workers × 10 days = 50 worker-days, so 10 workers need \(50 / 10 = 5\) days, not 6.

Statement 2: four pencils for ₹ 20 is ₹ 5 each, so 10 pencils cost ₹ 50, not ₹ 55.

Statement 3: two litres for ₹ 50 is ₹ 25 a litre, so 4 litres cost ₹ 100, not ₹ 120.

Each statement is a simple proportion with a wrong figure, so every option claiming one or more correct statements fails.