Practice MCQ Questions and Answer on Time and Work

91.

A man undertakes to do a certain work in 150 days. He employs 200 men. He finds that only a quarter of the work is done in 50 days. The number of additional men that should be appointed so that the whole work will be finished in time is = ?

(A) 75
(B) 100
(C) 125
(D) 50

Solution: Let 'n' number of men can required $\begin{aligned} \frac{(200_{men} \times 50_{days})}{\frac{1}{4}} = \frac{(200 + n) \times 100_{days}}{\frac{3}{4}} \\ \Rightarrow 3 \times 100 = (200 + n) \\ \Rightarrow n = 100 \end{aligned}$

92.

A can complete work in 25 days and B can complete the same work in 20 days. They started the work together but B left after 4 days and A continued to work. In how many days will the entire work be completed?

(A) 25
(B) 20
(C) 28
(D) 22

Solution: Total work of (A + B) in 4 days = 4(4 + 5) = 36 Remaining work = 100 - 36 = 64 A = $\frac{64}{4}$ = 16 days Total days = 4 + 16 = 20 days

93.

If 12 carpenters working 6 hours a day can make 460 chairs in 240 days, then number of chairs made by 18 carpenters in 360 days each working 8 hours a day ?

(A) 1320
(B) 1380
(C) 1260
(D) 920

Solution: $According to the question,$ $\Rightarrow \frac{12 \times 6 \times 240}{460}$ = $\frac{18 \times 360 \times 8}{x}$ $\begin{aligned} \Rightarrow x = \frac{18 \times 360 \times 8 \times 460}{12 \times 6 \times 240} \\ \Rightarrow x = 1380 \end{aligned}$

94.

Raju can finish a piece of work in 20 days. He worked at it for 5 days and then Jakob alone finished the remaining work in 15 days. In how many days can both finish it together?

(A) 10 days
(B) 12 days
(C) 16 days
(D) 20 days

Solution: Let total work = 60 unit Work by Raju in one day = $\frac{60}{20}$ = 3 unit 5 days → 5 × 3 = 15 unit Remaining work = 60 - 15 = 45 unit 15 days → 45 unit 1 day → 3 unit (Jakob) Work done by Raja and Jakob in one day = 3 + 3 = 6 Time = $\frac{60}{6}$ = 10 days

95.

A can do a piece of work in 18 days. He worked at it for 12 days and B finished the remaining work in 8 days. B alone can do the whole work in ?

(A) 16 days
(B) 24 days
(C) 35 days
(D) 28 days

Solution: $\begin{aligned} 18A = 12A + 8B \\ 6A = 8B \\ \frac{A}{B} = \frac{4}{3} \\ So, \\ Efficiency of A and B are 4, 3. \\ Total work \\ = 18 \times 4 \\ = 72 units \\ So, B will do \\ = \frac{72}{3} \\ = 24 days \end{aligned}$

96.

A can do a piece of work in 6 days. B is 25% more efficient than A. How long would B alone take to finish this work ?

(A) $4 \frac{4}{5} days$
(B) $3 \frac{1}{3} days$
(C) $5 \frac{1}{4} days$
(D) $2 \frac{2}{3} days$

Solution: A : B Efficiency → 100% : 125% 4 : 5 Time 5 : 4 ×1.2↓ ↓×1.2 Actual time 6 days $4 \frac{4}{5}$ days

97.

Two men and women are entrusted with a task. The second man needs three hours more to cope up with the job than the second man and the woman would need working together. The first man, working alone, would need as much time as second man and the woman working together. The first man working alone, would spend eight hours less than the double period of the time second man would spend working alone. How much time would the two men and the women need to complete the task if they all asked together?

(A) 1 hour
(B) 2 hours
(C) 3 hours
(D) 4 hours

98.

There was a leakage in the container of the refined oil. If 11 kg oil is leaked out per day then it would have lasted for 50 days, if the leakage was 15 kg per day, then it would have lasted for only 45 days. For how many days would the oil have lasted, if there was no leakage ant it was completely used for eating purpose?

(A) 80 days
(B) 72 days
(C) 100 days
(D) 120 days

Solution: Let x kg of oil be used for the eating purpose, daily, then (x + 11) × 50 = (x +15) × 45 → x = 25 Thus, Total quantity of oil, = (25 +11) × 50 = 1800 Hence, required number of days, = $\frac{1800}{25}$ = 72 days

99.

Two pipes A and B can fill a cistern in 12 min and 16 min respectively. Both the pipes are opened together for a certain time but due to some obstruction the flow of water was restricted to $\frac{7}{8}$ of full flow in pipe A and $\frac{5}{6}$ of full in pipe B. This obstruction is removed after some time and tank is now filled in 3 min from that moment. How long was it before the full flow.

(A) 8 min
(B) 3 min
(C) 5.6 min
(D) 4.5 min

Solution: Let the obstruction remain for x min. Hence, Part of cistern filled in X min + part of cistern filled in 3 min = full cistern $[\frac{7 x}{8 \times 12} + \frac{5 x}{6 \times 16}]$ $+$ $[\frac{3}{12} + \frac{3}{16}]$ = 1 $\frac{12 x}{96} + \frac{7}{16} = 1$ Thus, X = 4.5 min.

100.

If A and B together can complete a work in 18 days, A and C together in 12 days, and B and C together in 9 days, then B alone can do the work in:

(A) 18 days
(B) 24 days
(C) 30 days
(D) 40 days

Solution: $\begin{aligned} One day's work of \\ (A + B) = \frac{1}{18} . . . . . . . (1) \\ One day's work of \\ (A + C) = \frac{1}{12} . . . . . . . (2) \\ One day's work of \\ (B + C) = \frac{1}{9} . . . . . . . (3) \\ Adding (1), (2) and (3) \\ 2 \times (A + B + C) = \frac{1}{18} + \frac{1}{12} + \frac{1}{9} \\ 2 (A + B + C) = \frac{1}{4} \\ One day's work of \\ A + B + C = \frac{1}{8} \\ B = \frac{1}{8} - (A + C) \\ B = \frac{1}{8} - \frac{1}{12} \\ One day's work of \\ B = \frac{3 - 2}{24} = \frac{1}{24} \\ B need 24 days \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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