Practice MCQ Questions and Answer on Time and Work

351.

If A and B together can complete a piece of work in 15 days and B alone in 20 days, in how many days can A alone complete the work?

(A) 60
(B) 45
(C) 40
(D) 30

Solution: 1st method: A and B complete a work in = 15 days One day's work of (A + B) = $\frac{1}{15}$ B complete the work in = 20 days; One day's work of B = $\frac{1}{20}$ Then, A's one day's work $\begin{aligned} = \frac{1}{15} - \frac{1}{20} \\ = \frac{4 - 3}{6} \\ = \frac{1}{60} \end{aligned}$ Thus, A can complete the work in = 60 days. 2nd method: (A + B)'s one day's % work = $\frac{100}{15}$ = 6.66% B's one day's % work = $\frac{100}{20}$ = 5% A's one day's % work = 6.66 - 5 = 1.66% Thus, A need = $\frac{100}{1.66}$ = 60 days to complete the work.

352.

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)

353.

A takes twice as much time as B and thrice as much time as C to finalise a task. Working together, they can complete the task in 8 days. The time (in days) taken by A, B and C, respectively, to complete the task is:

(A) 42, 21, 14
(B) 60, 30, 20
(C) 54, 27, 18
(D) 48, 24, 16

Solution: Efficiency of A : B : C = 1 : 2 : 3 Total work = 8 × (A + B + C) = 48 Time taken by A → $\frac{48}{1}$ = 48 days Time taken by B → $\frac{48}{2}$ = 24 days Time taken by C → $\frac{48}{3}$ = 16 days

354.

If 450 men can finish construction of an apartment in 20 days, then how many men are needed to complete the same work in 30 days?

(A) 150
(B) 300
(C) 400
(D) 250

355.

The time taken by 4 men to complete a job is double the time taken by 5 children to complete the same job. Each man is twice as fast as a woman. How long will 12 men, 10 children and 8 women take to complete a job, given that a child would finish the job in 20 days ?

(A) 4 days
(B) 2 days
(C) 1 day
(D) 10 days

Solution: $\begin{aligned} By using formula \\ M_{1} D_{1} H_{1} = M_{2} D_{2} H_{2} \\ According to the question, \\ 5C = 4M \times 2 \\ \frac{M}{C} = \frac{5}{8} (Efficiency ratio) \\ Again, \\ \frac{M}{W} = \frac{2}{1} (Efficiency ratio) \\ M : W : C \\ 2_{\times 5} : 1_{\times 5} \\ 5_{\times 2} : 8_{2} \\ \overset{―}{10 : 5 : 16} \\ Now, \\ 20C = (12M + 10C + 8W) \times D \end{aligned}$ $20 \times 16 =$ $(12 \times 10 + 10 \times 16 + 8 \times 5)$ $\times D$ $\begin{aligned} 320 = 320 D \\ D = \frac{320}{320} = 1 \\ D = 1 day \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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