Practice MCQ Questions and Answer on Time and Work

281.

A and B separately can build a wall in 12 and 16 days, respectively. If they work for each day alternatively, starting with A, in how many days will the wall be built?

(A) $7 \frac{2}{3} days$
(B) $12 \frac{2}{3} days$
(C) $13 \frac{2}{3} days$
(D) $6 \frac{3}{4} days$

Solution: 2 days → 4 + 3 = 7 units 12 days → 42 units 13 days → 42 + 4 = 46 units Total time taken = $13 \frac{2}{3} days$

282.

The labourers A, B, C were given a contract of Rs. 750 for doing a certain piece of work. All the three together can finish the work in 8 days, A and C together can do it 12 days, while A and B together can do it in $13 \frac{1}{3}$ days. The money will be divided in the ratio ?

(A) 4 : 5 : 6
(B) 4 : 7 : 5
(C) 5 : 7 : 4
(D) 5 : 6 : 8

Solution: L.C.M. of Total Work = 120 One day work of A + B + C = $\frac{120}{8}$ = 15 unit/day One day work of A + B = $\frac{120}{\frac{40}{3}}$ = 9 unit/day One day work of A + C = $\frac{120}{12}$ = 10 unit/day $\begin{aligned} C's efficiency = 15 - 9 = 6 \\ B's efficiency = 15 - 10 = 5 \\ A's efficiency = 15 - (6 + 5) \\ = 15 - 11 \\ = 4 \\ So, A : B : C = 4 : 5 : 6 \end{aligned}$

283.

If 72 men can build a wall of 280 m length in 21 days, how many men could take 18 days to build a similar type of wall of length 100 m ?

(A) 30
(B) 10
(C) 18
(D) 28

Solution: $\begin{aligned} Here work is 280 m length of wall \\ And 100 m length of wall \\ Let, M men will finish 100 m wall . \\ \frac{72 \times 21}{280} = \frac{M \times 18}{100} \\ \Leftrightarrow M = 30 \end{aligned}$

284.

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

285.

A completes $\frac{7}{10}$ of the work 15 days. Then he completes the remaining work the help of B in 4 days. The time required for A and B together to complete the entire work is = ?

(A) $8 \frac{1}{4} days$
(B) $10 \frac{1}{2} days$
(C) $12 \frac{2}{3} days$
(D) $13 \frac{1}{3} days$

Solution: $\begin{aligned} (A + B)'s 4 day's work \\ = (1 - \frac{7}{10}) \\ = \frac{3}{10} \\ (A + B)'s 1 day's work \\ = (\frac{3}{10} \times \frac{1}{4}) \\ = \frac{3}{40} \\ Remaining work \\ = (1 - \frac{3}{10}) \\ = \frac{7}{10} \\ (B + C)'s 1 day's work \\ = (\frac{1}{15} + \frac{1}{20}) \\ = \frac{7}{60} \end{aligned}$ Hence, A an B together take $= \frac{40}{3} = 13 \frac{1}{2}$ days to complete the entire work.

286.

A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :

(A) 4 days
(B) 6 days
(C) 8 days
(D) 12 days

Solution: $\begin{aligned} (A + B + C)'s 1 day's work = \frac{1}{6} \\ (A + B)'s 1 day's work = \frac{1}{8} \\ (B + C)'s 1 day's work = \frac{1}{12} \\ ∴ (A + C)'s 1 day's work \\ = (2 \times \frac{1}{6}) - (\frac{1}{8} + \frac{1}{12}) \\ = \frac{1}{3} - \frac{5}{24} \\ = \frac{3}{24} \\ = \frac{1}{8} \end{aligned}$ So, A and C together will do the work in 8 days

287.

If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:

(A) 4 days
(B) 5 days
(C) 6 days
(D) 7 days

Solution: $\begin{aligned} Let 1 man's 1 day's work = x and \\ 1 boy's 1day's work = y \\ Then, 6 x + 8 y = \frac{1}{10} and \\ 26 x + 48 y = \frac{1}{2} \\ Solving these two equations, we get \\ x = \frac{1}{100} a n d y = \frac{1}{200} \\ (15 men + 20 boy)'s 1 day's work \\ = \frac{15}{100} + \frac{20}{200} = \frac{1}{4} \\ ∴ 15 men and 20 boys can do the work in 4 days \end{aligned}$

288.

Twenty women together can complete a piece of work in 16 days, 16 men together can complete the same work in 15 days. The ratio of the working capacity of a man to that of a women is = ?

(A) 3 : 4
(B) 4 : 3
(C) 5 : 3
(D) 4 : 5

Solution: $\begin{aligned} 20 w \times 16 = 16 m \times 15 \\ 20 w = 15 m \\ 4 w = 3 m \\ \frac{m}{w} = \frac{4}{3} \\ ∴ Man : Women \\ 4 : 3 \end{aligned}$

289.

12 men can do a piece of work in 24 days. How many days are needed to complete the work, if 8 men do this work ?

(A) 28 days
(B) 36 days
(C) 48 days
(D) 52 days

Solution: 12 men can do a piece of work in 24 days $\Rightarrow M_{1} = 12 and D_{1} = 24$ 8 men can do this work in $D_{2}$ days $\begin{aligned} \Rightarrow M_{2} = 8 \\ M_{1} D_{1} = M_{2} D_{2} \\ \Rightarrow 12 \times 24 = 8 \times D_{2} \\ \Rightarrow D_{2} = \frac{12 \times 24}{8} \\ \Rightarrow D_{2} = 36 days \end{aligned}$

290.

A and B working separately can do a piece of work in 9 and 15 days respectively. If they work for a day alternatively, with A beginning, then the work will be completed in ?

(A) 10 days
(B) 11 days
(C) 9 days
(D) 12 days

Solution: L.C.M. of Total Work = 45 One day work of A = $\frac{45}{9}$ = 5 unit/day One day work of B = $\frac{45}{15}$ = 3 unit/day $\begin{aligned} (A + B)'s 2 days work \\ = 5 + 3 \\ = 8 units \\ They will do in \\ = \frac{40}{8} \times 2 \\ = (5 \times 2) \\ = 10 days \\ ∴ Work left \\ = 45 - 40 \\ = 5 units \\ Now, \\ A's turn and he will complete in \\ = \frac{5}{5} \\ = 1 days \\ Then total work completed in \\ = 10 + 1 \\ = 11 days \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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