Practice MCQ Questions and Answer on Time and Work

51.

Ganga and Saraswati, working separately can mow field in 8 and 12 hours respectively. If they work in stretches of one hour alternately. Ganga is beginning at 9 a.m., when will the moving be completed?

(A) 6:20 PM
(B) 6:30 PM
(C) 6:36 PM
(D) 6:42 PM

Solution: According to question, Ganga begins at 9 am and she does 3 units/hours Saraswati begins at 10 am and she does 2 units/hours So by 11 am they complete 5 units Time $= \frac{T .W .}{3 + 2} = \frac{24}{5}$ (4 cycle of 2 hrs each + 4 units left) And now ganga will complete 3 unit out of 4 units in 1 hr Now, rest 1 unit work done by = $\frac{1}{2}$ hr Total time = 8 + 1 + $\frac{1}{2}$ = 9 $\frac{1}{2}$ hr Hence, Work finished at = 9 am + 9 $\frac{1}{2}$ hr = 6:30 PM Alternate Solution: Work done by Ganga in 1 hour = $\frac{1}{8}$ Work done by Saraswati in 1 hour = $\frac{1}{12}$ They are working alternatively with Ganga beginning the job. Work done in every two hours = $\frac{1}{8}$ + $\frac{1}{12}$ = $\frac{5}{24}$ Work done in 4 × 2 = 8 hours = $\frac{5 \times 4}{24}$ = $\frac{5}{6}$ Remaining work = 1 - $\frac{5}{6}$ = $\frac{1}{6}$ In 9th hour, Ganga starts the work and does $\frac{1}{8}$ of the work Work remaining = $\frac{1}{6}$ - $\frac{1}{8}$ = $\frac{1}{24}$ In 10th hour, Saraswati starts the work Time needed to finish the remaining work $\begin{aligned} = \frac{\frac{1}{24}}{\frac{1}{12}} \\ = \frac{1}{24} \times 12 \end{aligned}$ $=$ 0.5 hours $=$ 30 minutes i.e., work will be completed in 9 hour 30 minutes, after 9 AM i.e., at 6:30 PM

52.

639 persons can repair a road in 12 days working 5 hours a day. In how many days will 30 persons working 6 hours a day complete the work ?

(A) 210 days
(B) 213 days
(C) 214 days
(D) 215 days

Solution: $\begin{aligned} According to the question, \\ Let, D is number of days \\ \frac{639 \times 12 \times 5}{1 road} = \frac{30 \times 6 \times D}{1 road} \\ \Leftrightarrow D = 213 days \end{aligned}$

53.

42 women can do a piece of work in 18 days, How many women would be required do the same work in 21 days.

(A) 35
(B) 36
(C) 37
(D) 38

54.

A man undertakes to do a certain work in 150 days. He employs 200 men. He finds that only a quarter of the work is done in 50 days. The number of additional men that should be appointed so that the whole work will be finished in time is = ?

(A) 75
(B) 100
(C) 125
(D) 50

Solution: Let 'n' number of men can required $\begin{aligned} \frac{(200_{men} \times 50_{days})}{\frac{1}{4}} = \frac{(200 + n) \times 100_{days}}{\frac{3}{4}} \\ \Rightarrow 3 \times 100 = (200 + n) \\ \Rightarrow n = 100 \end{aligned}$

55.

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

56.

A and B can do a work in 12 days and 18 days, respectively. They worked for 4 days after which B was replaced by C and the remaining work was completed by A and C in the next 4 days. In how many days will C alone complete 50% of the same work?

(A) 24
(B) 18
(C) 21
(D) 36

Solution: 4 day work = 5 × 4 = 20 Remaining work = 36 - 20 = 16 (A + C) complete in 4 days A + C $\to \frac{16}{4} = 4$ 3 + C = 4 C = 1 50% of work = $\frac{36}{2}$ = 18 C do = $\frac{18}{1}$ = 18 days

57.

Two men undertake to do a piece of work for Rs. 1400. The first man alone can do this work in 7 days while the second man alone can do this work in 8 days. If they working together complete this work in 3 days with the help of a boy, how should the money be divided ?

(A) Rs. 600, Rs. 550, Rs. 250
(B) Rs. 600, Rs. 525, Rs. 275
(C) Rs. 600, Rs. 500, Rs. 300
(D) Rs. 500, Rs. 525, Rs. 375

Solution: $\begin{aligned} Boy's 1 day's work \\ = \frac{1}{3} - (\frac{1}{7} + \frac{1}{8}) \\ = (\frac{1}{3} - \frac{15}{56}) \\ = \frac{11}{168} \end{aligned}$ ∴ Ratio of wages of the first man, second man and boy $\begin{aligned} = \frac{1}{7} : \frac{1}{8} : \frac{11}{168} \\ = 24 : 21 : 11 \\ First man's share \\ = Rs . (\frac{24}{56} \times 1400) \\ = Rs .600 \\ Second man's share \\ = Rs . (\frac{21}{56} \times 1400) \\ = Rs .525 \\ Boy's man's share \\ = Rs . [1400 - (600 + 525)] \\ = Rs .275 \end{aligned}$

58.

A group of 12 men can do a piece of work in 14 days and other group of 12 women can do the same work in 21 days. They begin together but 3 days before the completion of work, man's group leaves off. The total number of days to complete the work is:

(A) $\frac{65}{4}$
(B) $\frac{93}{3}$
(C) $\frac{51}{5}$
(D) 60

Solution: Let x be the required number of days Given,12 men and 12 women can complete a work separately in 14 days and 21 days respectively Then, 12 men's 1 day work = $\frac{1}{14}$ And, 12 women's 1 day work = $\frac{1}{21}$ Then ,12 women's 3 days work = $\frac{3}{21}$ = $\frac{1}{7}$ The remaining work = $1 - \frac{1}{7}$ = $\frac{6}{7}$ Man's group leaves 3 days before the completion of work That is, they were working together for x - 3 days Thus, we have $\frac{1}{7}$ work left to be done in last 3 days by the women's group. This also means $\frac{6}{7}$ th of work has been done by both the groups (before men left) Now, (12 men + 12 women)'s 1 day work = $\frac{1}{14} + \frac{1}{21}$ = $\frac{5}{42}$ i.e., $\frac{5}{42}$ work is done by 2 groups in 1 day. So, $\frac{6}{7}$ of work is done by 2 groups together in $\frac{42}{5} \times \frac{6}{7}$ = $\frac{36}{5}$ days Total time take to complete the work will be = $\frac{36}{5}$ + 3 = $\frac{51}{5}$

59.

A manufacturer builds a machine which will address 500 envelopes in 8 minutes. He wishes to build another machine so that when both are operating together they will address 500 envelopes in 2 minutes. The equation used to find how many minutes x it would require the second machine to address 500 envelopes alone, is = ?

(A) $8 - x = 2$
(B) $\frac{1}{8} + \frac{1}{x} = \frac{1}{2}$
(C) $\frac{500}{8} + \frac{500}{x} = 500$
(D) $\frac{x}{2} + \frac{x}{8} = 1$

Solution: Number of envelopes addressed by first machine in 1 minute $= \frac{500}{8}$ Number of envelopes addressed by second machine in 1 minute $= \frac{500}{x}$ Number of envelopes addressed by both machine in 1 minute $\begin{aligned} = \frac{500}{2} \\ ∴ \frac{500}{8} + \frac{500}{x} = \frac{500}{2} \\ \Rightarrow \frac{1}{8} + \frac{1}{x} = \frac{1}{2} \end{aligned}$

60.

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)

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Practice MCQ Questions and Answer on Time and Work

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