Practice MCQ Questions and Answer on Time and Work

111.

A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in :

(A) 4 days
(B) 6 days
(C) 8 days
(D) 18 days

Solution: $\begin{aligned} Ration of rates of working of A and B \\ = 2 : 1 \\ So, ratio of times taken = 1 : 2 \\ B's 1 day's work = \frac{1}{12} \\ ∴ A's 1 day's work \\ = \frac{1}{3}; (2 times of B's work) \\ (A + B)'s 1 day's work \\ = \frac{1}{6} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \\ So, A and B together can finish the \\ work in 4 days . \end{aligned}$

112.

A can build up a wall in 8 days while B can break it in 3 days. A has worked for 4 days and then B joined to work with A for another 2 days only. In how many days will A alone build up the remaining part of the wall ?

(A) $6 \frac{1}{3} days$
(B) 7 days
(C) $7 \frac{1}{3} days$
(D) $13 \frac{1}{3} days$

Solution: Part of wall built by A in 1 day = $\frac{1}{8}$ Part of wall broken by B in 1 day = $\frac{1}{3}$ Part of wall built by A in 4 days $\begin{aligned} = (\frac{1}{8} \times 4) \\ = \frac{1}{2} \end{aligned}$ Part of wall broken by B and built by A in 2 days $\begin{aligned} = 2 (\frac{1}{3} - \frac{1}{8}) \\ = \frac{5}{12} \end{aligned}$ $\begin{aligned} Part of wall built in 6 days \\ = (\frac{1}{2} - \frac{5}{12}) \\ = \frac{1}{12} \\ Remaining part to be built \\ = (1 - \frac{1}{12}) \\ = \frac{11}{12} \end{aligned}$ Now, $\frac{1}{8}$ part of wall built by A in 1 day $\begin{aligned} ∴ \frac{11}{12} part of wall built by A in \\ = (8 \times \frac{11}{12}) \\ = \frac{22}{3} \\ = 7 \frac{1}{3} day \end{aligned}$

113.

A contractor decided to complete a work in 80 days and employed 60 men at the beginning and 20 men additionally after 20 days and got the work completed as per schedule. If he had not employed, the additional men, how many extra days would he have needed to complete the work (round off to the nearest integer)?

(A) 32 days
(B) 20 days
(C) 30 days
(D) 26 days

114.

A tyre has 3 punctures. The first puncture alone would have made the tyre flat in 9 minutes, the second alone would have done it in 18 minutes, the third alone would have done it in 6 minutes. If the air leaks out at a constant rate, then how long (in minutes) does it take for all the punctures together to make it flat?

(A) 2
(B) 6
(C) 4
(D) 3

Solution: $∴ \frac{18}{2 + 1 + 3} = 3 minutes$

115.

Two pipes A and B can fill a tank in 36 min. and 45 min. respectively. Another pipe C can empty the tank in 30 min. First A and B are opened. After 7 minutes, C is also opened. The tank filled up in:

(A) 39 min.
(B) 46 min
(C) 40 min.
(D) 45 min.

Solution: Pipe A can fill empty tank in 36 min. Pipe A can fill the tank = $\frac{100}{36}$ = 2.77% per minute Pipe B can fill empty tank in 45 min. Pipe B can fill the tank = $\frac{100}{45}$ = 2.22% per min. A and B can together fill the tank = (2.77 + 2.22) ≈ 5% per minute So, A and B can fill the tank in 7 min. = 7 × 5 = 35% of the tank Rest tank to be filled = 100 - 35 = 65% C can empty the full tank in 30 min. C can empty the tank = $\frac{100}{30}$ = 3.33% per min. C is doing negative work i.e. emptying the tank A, B and C can together fill the tank, = 2.77% + 2.22% - 3.33% = 1.67% tank per minute So, A, B and C will take time to fill 65% empty tank, = $\frac{65}{1.67}$ = 39 min. (Approx)

116.

Two spinning machines A and B can together produce 300000 metres of cloth in 10 hours. If machine B alone can produce the same amount of cloth in 15 hours. Then how much cloth can machine A produce alone in 10 hours ?

(A) 50000 metres
(B) 100000 metres
(C) 150000 metres
(D) 200000 metres

Solution: Length of cloth produced by A and B in 10 hours $= 300000 metres$ Length of cloth produced by B in 10 hours $\begin{aligned} = (\frac{300000}{15} \times 10) \\ = 200000 metres \end{aligned}$ ∴ Length of cloth produced by A in 10 hours $\begin{aligned} = (300000 - 200000) \\ = 100000 metres \end{aligned}$

117.

Working together, Asha and Sudha can complete an assigned task in 20 days. However, if Asha worked alone and completed half the work and then Sudha takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Asha take to complete the if she worked alone ? Assume that Sudha is more efficient than Asha.

(A) 25 days
(B) 30 days
(C) 60 days
(D) 65 days

Solution: Suppose, Asha takes x days to complete the task alone while Sudha takes y days to complete it alone Since Sudha is more efficient than Asha, we have x > y $\begin{aligned} Asha's 1 day's work = \frac{1}{x} \\ Sudha's 1 day's work = \frac{1}{y} \\ (Asha + Sudha)'s 1 day's work \\ = \frac{1}{x} + \frac{1}{y} \\ = \frac{x + y}{x y} \end{aligned}$ If Asha and Sudha each does half of the work alone, time taken $\begin{aligned} = (\frac{x}{2} + \frac{y}{2}) days \\ = (\frac{x + y}{2}) days . \\ ∴ \frac{x + y}{2} = 45 \\ \Rightarrow x + y = 90 \\ From (i) and (ii), \\ We have : \frac{x y}{20} = 90 or x y = 1800 \\ Now, x y = 1800 and x + y = 90 \\ \Rightarrow x = 60, y = 30 [∵ x > y] \end{aligned}$ Hence, Asha alone will take 60 days to complete the task.

118.

A can do a piece of work in 4 hours, B and C together in 3 hours, and A and C together in 2 hours. How long will B alone take to do it ?

(A) 8 hours
(B) 10 hours
(C) 12 hours
(D) 24 hours

Solution: $\begin{aligned} A's 1 hour's work = \frac{1}{4} \\ (B + C)'s 1 hour's work = \frac{1}{3} \\ (A + C)'s 1 hour's work = \frac{1}{2} \\ (A + B + C)'s 1 hour's work \\ = \frac{1}{4} + \frac{1}{3} \\ = \frac{7}{12} \\ ∴ B's 1 hour's work \end{aligned}$ = (A + B + C)'s 1 hour's work - (A + C)'s 1 hour's work $\begin{aligned} = \frac{7}{12} - \frac{1}{2} \\ = \frac{1}{12} \end{aligned}$ So, B alone can complete the work in 12 hours.

119.

A and B together can do a job in 2 days; B and C can do it in 4 days; A and C in $2 \frac{2}{5}$ days. The number of days required for A to do the job alone is = ?

(A) 1
(B) 3
(C) 6
(D) 12

Solution: $\begin{aligned} (A + B)'s 1 day's work = \frac{1}{2} \\ (B + C)'s 1 day's work = \frac{1}{4} \\ (A + C)'s 1 day's work = \frac{5}{12} \\ Adding, we get: \\ 2 (A + B + C)'s 1 day's work \\ = (\frac{1}{2} + \frac{1}{4} + \frac{1}{12}) \\ = \frac{14}{12} \\ = \frac{7}{6} \\ \Rightarrow (A + B + C)'s 1 day's work = \frac{7}{12} \\ So, A's 1 day's work \\ = (\frac{7}{12} - \frac{1}{4}) \\ = \frac{4}{12} \\ = \frac{1}{3} \\ ∴ A alone can do the work in 3 days . \end{aligned}$

120.

Shyam can complete a task in 12 days by working 10 hours a day. How many hours a day should he work to complete the task in 8 days?

(A) 14
(B) 16
(C) 12
(D) 15

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Practice MCQ Questions and Answer on Time and Work

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