Practice MCQ Questions and Answer on Time and Work

331.

The ratio of the efficiencies of A, B and C is 7 : 5 : 8. Working together, they can complete a piece of work in 42 days. B and C worked together for 21 days and the remaining work was completed by A alone. The whole work was completed in:

(A) 96 days
(B) 99 days
(C) 102 days
(D) 93 days

332.

A group of workers can complete a piece of work in 50 days, when they are working individually. On the first day one person works, on the second day another person joins him, on the third day one person joins them and this process continues till the work is completed. How many approximate days are needed to complete the work ?

(A) 8 days
(B) 9 days
(C) 10 days
(D) 11 days

Solution: Let a man complete 1 piece of work in a day. Then total work = 50 units Then by statement 1 st day = one man × 1 work/day = 1 Then by statement 2nd day = two men × 1 work/day = 2 Then by statement 3rd day = three men × 1 work/day = 3 Let the whole work will be completed in N days. Then total work 1 + 2 + 3 + ..... + N = 50 $\begin{aligned} \frac{N (N + 1)}{2} = 50 \\ N (N + 1) = 100 \\ Then, N = 10 days (approx) \end{aligned}$

333.

A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:

(A) 5 days
(B) 6 days
(C) 10 days
(D) $10 \frac{1}{2}$ days

Solution: $\begin{aligned} (B + C)'s 1 day's work \\ = \frac{1}{9} + \frac{1}{12} = \frac{7}{36} \\ Work done by B and C in 3 days \\ = \frac{7}{36} \times 3 = \frac{7}{12} \\ Remaining work \\ = 1 - \frac{7}{12} = \frac{5}{12} \\ Now, \frac{1}{24} work is done by A in 1 day \\ So, \frac{5}{12} work is done by A in \\ 24 \times \frac{5}{12} = 10 days \end{aligned}$

334.

X, Y and Z complete a work in 6 days. X or Y alone can do the same work in 16 days. In how many days Z alone can finish the same work ?

(A) 12
(B) 16
(C) 24
(D) 36

Solution: (X + Y)'s 1 day's work $\begin{aligned} = (\frac{1}{16} + \frac{1}{16}) \\ = \frac{2}{16} \\ = \frac{1}{8} \end{aligned}$ ∴ Z's 1 day's work (X + Y + Z)'s 1 day's work - (X + Y)'s 1 day's work $\begin{aligned} = \frac{1}{6} - \frac{1}{8} \\ = \frac{1}{24} \end{aligned}$ ∴ Z alone can finish the work in 24 days.

335.

If 2 men or 3 women or 4 boys can do a piece of work in 52 days, then the same piece of work will be done by 1 man, 1 woman and 1 boy in:

(A) 48 days
(B) 36 days
(C) 45 days
(D) 50 days

Solution: Work done by 2 men = 3 women = 4 boys 1 man = 2 boys 1 woman = $\frac{4}{3}$ boys Thus, Boys × days = 4 × 52 boys × days Again, 1 man + 1 women + 1 boys, $= 2 + \frac{4}{3} + 1$ $= \frac{13}{3}$ boys Using work equivalent method, boys × day = BOYS × DAYS $4 \times 52 = \frac{13}{3} \times x (let)$ x = 48 days

336.

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.

337.

A and B together can complete a work in 12 days. B and C together can complete the same work in 8 days and A and C together can complete it in 16 days. In total, how many days do A, B and C together take to complete the same work ?

(A) $3 \frac{5}{12}$
(B) $3 \frac{9}{13}$
(C) $7 \frac{5}{12}$
(D) $7 \frac{5}{13}$

Solution: $\begin{aligned} (A + B)'s 1 day's work = \frac{1}{12} \\ (B + C)'s 1 day's work = \frac{1}{8} \\ (A + C)'s 1 day's work = \frac{1}{16} \end{aligned}$ Adding, we get 2(A + B + C)'s 1 day's work $\begin{aligned} = (\frac{1}{12} + \frac{1}{8} + \frac{1}{16}) \\ = \frac{13}{96} \end{aligned}$ So, A, B and C together can complete the work in $\begin{aligned} = \frac{96}{13} \\ = 7 \frac{5}{13} days \end{aligned}$

338.

There are 3 taps A, B, and C in a tank. These can fill the tank in 10 hours, 20 hours and 25 hours, respectively. At first, all three taps are opened simultaneously. After 2 hours, tap C is closed and A and B keep running. After 4 hours from the beginning, tap B is also closed. The remaining tank is filled by tap A alone. Find the percentage of work done by tap A itself.

(A) 32%
(B) 75%
(C) 52%
(D) 72%

Solution: 2 hours total → 2 × 19 = 38 4 hours total = 38 + 2 × 15 = 68 Remaining work = 100 - 68 = 32 Total work done by A = 32 + 40 = 72 Required % = $\frac{72}{100}$ × 100 = 72%

339.

A and B together can complete a work in 3 days. They start together but after 2 days, B left the work. If the work is completed after two more days, B alone could do the work in

(A) 5 days
(B) 6 days
(C) 9 days
(D) 10 days

Solution: 1st Method: (A+B)'s one day's work = $\frac{1}{3}$ part (A+B) works 2 days together = $\frac{2}{3}$ part Remaining work = $1 - \frac{2}{3}$ = $\frac{1}{3}$ part $\frac{1}{3}$ part of work is completed by A in two days Hence, one day's work of A = $\frac{1}{6}$ Then, one day's work of B = $\frac{1}{3} - \frac{1}{6}$ = $\frac{1}{6}$ So, B alone can complete the whole work in 6 days. 2nd Method: (A+B)'s one day's % work = $\frac{100}{3}$ = 33.3% Work completed in 2 days = 66.6% Remaining work = 33.4% One day's % work of A = $\frac{33.4}{2}$ = 16.7% One day's work of B = 33.4 - 16.7 = 16.7% B alone can complete the work in, = $\frac{100}{16.7}$ = 5.98 days ≈ 6 days.

340.

While working 7 hour a day, A alone can complete a piece of work in 6 days and B alone in 8 days. In what time would they complete it together, 8 hour a day?

(A) 3 days
(B) 4 days
(C) 2.1 days
(D) 3.6 days

Solution: A can complete the work in 7 × 6 = 42 hours 1 hour's work of A = $\frac{1}{42}$ B can complete the work in 7 × 8 = 56 hours 1 hour's work of B = $\frac{1}{56}$ (A + B)'s 1 hour's work $\begin{aligned} = \frac{1}{42} + \frac{1}{56} \\ = \frac{4 + 3}{168} \\ = \frac{7}{168} \end{aligned}$ ∴ Time taken by (A + B) working 8 hours daily $\frac{168}{7 \times 8} = 3 days$

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Practice MCQ Questions and Answer on Time and Work

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