Practice MCQ Questions and Answer on Time and Work

241.

12 men can do a piece of work in 15 days and 20 women can do the same work in 12 days. In how many days can 5 men and 5 women complete the same work ?

(A) $20 \frac{4}{7} days$
(B) $2 \frac{4}{7} days$
(C) $20 \frac{3}{7} days$
(D) 18 days

Solution: 12 man × 15 = 20 women × 12 = Total work 3 man = 4 women $\begin{aligned} \frac{Men}{Women} = \frac{4}{3} \to Efficiency \\ Total work = 12 \times 4 \times 15 = 720 \\ (5 man + 5 women) \times D = 720 \\ D (5 \times 4 + 5 \times 3) = 720 \\ D = \frac{720}{35} = 20 \frac{4}{7} days \end{aligned}$ Alternate: According to the question, 12 men can do the 1 day work in $\frac{1}{15}$ days So, 1 men can do the 1 day work in $\frac{1}{15 \times 12}$ = $\frac{1}{180}$ days 5 men can do the 1 day work in $\frac{1}{180} \times 5$ = $\frac{5}{180}$ days Similarly 5 women do the 1 dys work $\frac{5}{240}$ days ∴ 5 men & 5 women together work in 1 day $\begin{aligned} = \frac{5}{180} + \frac{5}{240} \\ = \frac{20 + 15}{720} \\ = \frac{35}{720} \end{aligned}$ Hence, they complete the work $\frac{720}{35}$ = $20 \frac{4}{35}$ days

242.

Raj can do a piece of work in 20 days. He started the work and left after some days, when 25% work was done. After it Abhijit joined and completed it working for 10 days. In how many days Raj and Abhijit can do the complete work, working together?

(A) 6
(B) 8
(C) 10
(D) 12

Solution: Efficiency of Raj = $\frac{100}{20}$ = 5% Work completed by Raj = 25% Rest work = 75% Efficiency of Abhijit = $\frac{75}{10}$ = 7.5% Combined efficiency = 5 + 7.5 = 12.5% They will complete the whole work by working together in,= $\frac{100}{12.5}$ = 8 days

243.

A conveyor belt delivers baggage at the rate of 3 tons in 5 minutes and second conveyor belt delivers baggage at the rate of 1 ton in 2 minutes. How much time will it take to get 33 tons of baggage delivered using both the conveyor belts together ?

(A) 25 minutes 30 seconds
(B) 30 minutes
(C) 35 minutes
(D) 45 minutes

Solution: Baggage delivered by first belt in 1 minute $= (\frac{3}{5}) tons$ Baggage delivered by second belt in 1 minute $= (\frac{1}{2}) tons$ Baggage delivered by both belt in 1 minute $\begin{aligned} = (\frac{3}{5} + \frac{1}{2}) tons \\ = \frac{11}{10} tons \\ ∴ Required time \\ = (33 \div \frac{11}{10}) minutes \\ = (33 \times \frac{10}{11}) minutes \\ = 30 minutes \end{aligned}$

244.

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 $\begin{aligned} = \frac{96}{13} \\ = 7 \frac{5}{13} days \end{aligned}$

245.

A can do a piece of work in 20 days and B in 40 days. If they work together for 5 days, then the fraction of the work that is left is ?

(A) $\frac{5}{8}$
(B) $\frac{8}{15}$
(C) $\frac{7}{15}$
(D) $\frac{1}{10}$

Solution: L.C.M of total work = 40 One day work of A = $\frac{40}{20}$ = 2 unit/day One day work of B = $\frac{40}{40}$ = 1 unit/day (A + B)'s one day work is (2 + 1) units (A + B)'s 5 day work is 3 × 5 = 15 units Work left = 40 - 15 = 25 ∴ Fraction of work left $\begin{aligned} = \frac{Work left}{Total work} \\ = \frac{25}{40} \\ = \frac{5}{8} \end{aligned}$

246.

A, B and C completed a work costing Rs. 1800. A worked for 6 days, B for 4 days and C for 9 days. If their daily wages are in the ratio of 5 : 6 : 4, how much amount will be received by A ?

(A) Rs. 600
(B) Rs. 750
(C) Rs. 800
(D) Rs. 900

Solution: Let the daily wages of A, B and C be Rs. 5x, Rs. 6x and Rs. 4x respectively. Then, ratio of their amounts $\begin{aligned} = (5 x \times 6) : (6 x \times 4) : (4 x : 9) \\ = 30 : 24 : 36 \\ = 5 : 4 : 6 \\ ∴ A's amount \\ = Rs . (1800 \times \frac{5}{15}) \\ = Rs .600 \end{aligned}$

247.

60 men could complete a piece of work in 250 days. They worked together for 200 days. After that work had to be stopped for 10 days due to bad weather. How many more men should be engaged to complete the work in time ?

(A) 10
(B) 15
(C) 18
(D) 20

248.

A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it 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: $\begin{aligned} A's 1 hour's work = \frac{1}{4}; \\ (B + C)'s 1 hour's work = \frac{1}{3}; \\ (A + C)'s 1 hour's work = \frac{1}{2} . \\ (A + B + C)'s 1 hour's work \\ = \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \\ B's 1 hour's work \\ = \frac{7}{12} - \frac{1}{2} = \frac{1}{12} \\ ∴ B along will take 12 hours to do the work \end{aligned}$

249.

20 men can do a piece of work in 18 days. They worked together for 3 days, then 5 men joined. In how many days is the remaining work completed ?

(A) 12 days
(B) 14 days
(C) 13 days
(D) 15 days

Solution: 20 men → 18 days ⇒ Work done by 20 men working Together = 1 work ⇒ Work done by them in 3 days working Together = 1 × 3 = 3 work ⇒ Remaining work = 18 - 3 = 15 work ⇒ 15 work is to be done by (20 + 5) = 25 men $\begin{aligned} ∴ Efficiency of 1 man = \frac{1}{20} \\ \Rightarrow Efficincy of 5 men \\ = \frac{5}{20} \\ = \frac{1}{4} \\ \Rightarrow So, efficiency of (20 + 5) \\ \Rightarrow 25 men = 1 + \frac{1}{4} \\ = \frac{5}{4} working days \\ Required time \\ = \frac{Work}{Efficiency} \\ = \frac{15}{\frac{5}{4}} \\ = 12 days \end{aligned}$ Therefore, 12 more days will be taken to finish the remaining work Alternate : 20 men can do 18 days So, total work = 18 × 20 = 360 20 men 3 days work = 20 × 3 = 60 Remaining work = 360 - 60 = 300 After joining 5 men total men = 20 + 5 = 25 So, finish the remaining work in = $\frac{300}{25} = 12 days$

250.

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 = $\frac{51}{5}$

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Practice MCQ Questions and Answer on Time and Work

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