Practice MCQ Questions and Answer on Time and Work

321.

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

322.

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

323.

If 28 men complete $\frac{7}{8}$ of a piece of work in a week, then the number of men, who must be engaged to get the remaining work completed in another week, is = ?

(A) 5
(B) 6
(C) 4
(D) 3

Solution: $\begin{aligned} \frac{28 M \times 1 Week}{\frac{7}{8}} = \frac{x \times 1 Week}{\frac{1}{8}} \\ \Leftrightarrow x = 4 men \end{aligned}$

324.

If one pipe A can fill a tank in 20 minutes, then 5 pipes, each of 20% efficiency of A, can fill the tank in:

(A) 80 min
(B) 100 min
(C) 20 min
(D) 25 min

Solution: Efficiency of pipe A, $\frac{100}{20}$ = 5% 20% of efficiency of A = 1% Then, efficiency of 5 such pipes = 5%. Then,Time taken to fill the tank = $\frac{100}{5}$ = 20 min.

325.

4 mat-weavers can weave 4 mats in 4 days. At the same rate how many mats would be woven by 8 mat-weavers in 8 days ?

(A) 4
(B) 8
(C) 12
(D) 16

Solution: $\begin{aligned} \frac{4_{mat - wevers} \times 4_{days}}{4_{mats}} = \frac{8_{mat - wevers} \times 8_{days}}{N_{mats}} \\ \Leftrightarrow N = 16 mats \end{aligned}$

326.

X can do a piece of work in 40 days. He works at it for 8 days and then Y finished it in 16 days. How long will they together take to complete the work?

(A) $13 \frac{1}{3}$ days
(B) 15 days
(C) 20 days
(D) 26 days

Solution: Work done by X in 8 days = $\frac{1}{40} \times 8$ = $\frac{1}{5}$ Remaining work = $1 - \frac{1}{5}$ = $\frac{4}{5}$ Now, $\frac{4}{5}$ work is done by Y in 16 days Whole work will be done by Y in = $16 \times \frac{5}{4}$ = 20 days ∴ X's 1 day's work = $\frac{1}{40}$ ∴ Y's 1 day's work = $\frac{1}{20}$ (X + Y)'s 1 day's work $\begin{aligned} = \frac{1}{40} + \frac{1}{20} \\ = \frac{3}{40} \end{aligned}$ Hence, X and Y will together complete the work in $\begin{aligned} = \frac{40}{3} \\ = 13 \frac{1}{3} days \end{aligned}$

327.

Two pipes can fill an empty tank separately in 24 minutes and 40 minutes respectively and a third pipe can empty 30 gallons of water per minute. If all three pipes are open, empty tanks become full in one hour. The capacity of the tank (in gallons) is:

(A) 800 gallons
(B) 600 gallons
(C) 500 gallons
(D) 400 gallons

Solution: Let capacity of the tank = x gallons; Part of the tank filled in 1 minute $\begin{aligned} = \frac{x}{24} + \frac{x}{40} - 30 \\ Or, \frac{x}{24} + \frac{x}{40} - 30 = \frac{x}{60} \\ Or, \frac{x}{24} + \frac{x}{40} - \frac{x}{60} = 30 \\ Or, \frac{10 x + 6 x - 4 x}{240} = 30 \\ Or, 12 x = 30 \times 240 \\ Or, x = 600 gallons \end{aligned}$

328.

A can do a piece of work in 12 days while B alone can do it in 15 days. With the help of C they can finish it in 5 days. If they are paid Rs. 960 for the whole work. How much money A gets ?

(A) Rs. 480
(B) Rs. 240
(C) Rs. 320
(D) Rs. 400

Solution: (A + B)'s 1 day work $\begin{aligned} = \frac{1}{12} + \frac{1}{15} \\ = \frac{5 + 4}{60} \\ = \frac{9}{60} = \frac{3}{20} \end{aligned}$ (A + B + C)'s 1 day work = $\frac{1}{5}$ ∴ C's 1 day work $\begin{aligned} = \frac{1}{5} - \frac{3}{20} \\ = \frac{4 - 3}{20} \\ = \frac{1}{20} \end{aligned}$ Ratio of their work $\begin{aligned} = \frac{1}{12} : \frac{1}{15} : \frac{1}{20} \\ = 5 : 4 : 3 \end{aligned}$ $\begin{aligned} A's share = \frac{5}{12} \times 960 \\ = Rs . 400 \end{aligned}$

329.

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.

330.

If 4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days, then 10 women complete it in ?

(A) 35 days
(B) 50 days
(C) 45 days
(D) 40 days

Solution: $\begin{aligned} According to the question, \\ \Rightarrow 4m + 6w = 8 days . . . . . (i) \\ or \\ \Rightarrow 32m + 48 w = 1 days \\ \Rightarrow 3m + 7 w = 10 days . . . . . (ii) \\ or \\ \Rightarrow 30m + 70 w = 1 day \\ ∴ 32m + 48w = 30 m + 70 w \\ \Rightarrow 2m = 22 w \\ \Rightarrow m = 11 w \\ From (i) \\ \Rightarrow 4m = 44 w \\ ∴ (44w + 6w) \times 8 = 10 w \times x \\ \Rightarrow 50w \times 8 = 10 w \times x \\ \Rightarrow x = 40 days \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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