Practice MCQ Questions and Answer on Time and Work

181.

Dinesh and Rakesh are working on an Assignment, Dinesh takes 6 hours to type 32 pages on a computer, while Rakesh takes 5 hours to type 40 pages. How much time will they take working together on two different computers to type an assignment of 110 page ?

(A) 7 hours, 30 minutes
(B) 8 hours
(C) 8 hours, 15 minutes
(D) 8 hours, 25 minutes

Solution: $\begin{aligned} Dinesh's one hour work \\ = \frac{32}{6} \\ = \frac{16}{3} pages/hour \\ Rakesh's one hour work \\ = \frac{40}{5} \\ = 8 pages/hour \\ Dinesh's and Rakesh's one hour work \\ = \frac{16}{3} + 8 \\ = \frac{40}{3} pages/hour \\ They will finish the work together \\ \frac{Total work}{Efficiency} \\ = \frac{110}{\frac{40}{3}} \\ = 8 \frac{1}{4} \\ = 8 hours, 15 minutes \end{aligned}$

182.

Two persons having different productivity of labour, working together can reap a field in 2 days. If one-third of the field was reaped by the first man and rest by the other one working alternatively took 4 days. How long did it take for the faster person to reap the whole field working alone?

(A) 3
(B) 6
(C) 8
(D) 12

Solution: Total efficiency of two persons = 50% [As they complete work in 2 days. ] First Person completes work = $\frac{1}{3}$ = 33.33% [In 2 days] Rest work will be completed by Second man = $\frac{2}{3}$ = 66.66% [In 2 days] So, efficiency of second person is greater. Efficiency of second person = $\frac{66.66}{2}$ = 33.33% per day Then, Second person will complete whole work in, = $\frac{100}{33.33}$ = 3 days.

183.

A can do three times the work done by B in one day. They together finish $\frac{2}{5}$ of the work in 9 days. The number of days by which B can do the work alone are ?

(A) 120 days
(B) 100 days
(C) 30 days
(D) 90 days

Solution: Efficiency of A = 3 efficiency of B $(A = 3, B = 1)$ $\frac{2}{5}$ th of the work done by (A + B) in 9 days Total work (A + B) = $\frac{45}{2}$ days will be completed $\begin{aligned} Total work \\ = days \times efficiency (A + B) \\ = \frac{45}{2} \times 4 \\ = 90 \\ Number of days for B \\ = \frac{Total work}{Efficiency} \\ = \frac{90}{1} \\ = 90 days \end{aligned}$

184.

A and B can together finish a piece of work in 30 days. They worked on it 20 days and then B left. The remaining work was done by A alone in 20 days. A alone can finish the work in = ?

(A) 60 days
(B) 54 days
(C) 48 days
(D) 50 days

Solution: (T.W.)60 2 units/day ↓ 30 days(A + B) $\begin{aligned} (A + B) worked for 20 days \\ So, they completed \\ 20 \times 2 = 40 units \\ Work left = 60 - 40 = 20 units \end{aligned}$ (T.W.) 2 0 1 units/day ↓ 20 days (A) A's efficiency 1 units/day A will complete the whole work in $\begin{aligned} = \frac{Total work}{Efficiency} \\ = \frac{60}{1} \\ = 60 days \\ A l t e r n a t e : \\ (A + B) \times 10 = A \times 20 \\ \frac{A + B}{A} = \frac{2}{1} \\ Total work \\ = 2 \times 30 \\ = 60 units \\ Time taken by A alone \\ = \frac{60}{1} \\ = 60 days \end{aligned}$

185.

If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do the work in ?

(A) $5 \frac{6}{13} months$
(B) $6 months$
(C) $5 \frac{7}{13} months$
(D) $11 \frac{1}{13} months$

Solution: 40 men = 60 women = 80 children 2 men = 3 women = 4 children 2 men = 3 women 1 women = $\frac{2}{3}$ men → 10 women $\to \frac{2}{3} \times 10 = \frac{20}{3} men$ Similarly 2 men = 4 children 1 children = $\frac{1}{2}$ men → 10 children $= \frac{10}{2} = 5 men$ 10 men = 10 women = 10 children $\begin{aligned} 10 men + \frac{20}{3} + 5 \\ \Rightarrow \frac{30 + 20 + 15}{3} \end{aligned}$ 10 men + 10 women + 10 children = $\frac{65}{3}$ men 40 men can do a piece of work in 6 months 1 men can do a piece of work in 6 × 40 $\frac{65}{3}$ men can do a piece of work in $\begin{aligned} = \frac{6 \times 40}{\frac{65}{3}} \\ = 11 \frac{1}{13} months \end{aligned}$ Alternate : 40 men = 60 women = 80 children 2 men = 3 women = 4 children men : women : children = 6 : 4 : 3 (efficiency) ∴ Total work $\begin{aligned} = 40 \times 6 \times 6 \\ = 1440 units \end{aligned}$ Total time taken by (40 men + 60 women + 80 children) $\begin{aligned} = \frac{Total work}{Efficiency} \\ = \frac{1440}{130} \\ = 11 \frac{1}{13} months \end{aligned}$

186.

The ratio of the amount of work done by (x - 1) labours in (x + 1) days and (x + 1) labours in (x + 2) days is 5 : 6. Then the value of x is ?

(A) 16
(B) 15
(C) 17
(D) 14

Solution: $\begin{aligned} From M_{1} D_{1} = M_{2} D_{2} \\ \Rightarrow \frac{M_{1} D_{1}}{M_{2} D_{2}} = \frac{5}{6} \\ \Rightarrow \frac{(x - 1) (x + 1)}{(x + 1) (x + 2)} = \frac{5}{6} \\ \Rightarrow \frac{(x - 1)}{(x + 2)} = \frac{5}{6} \\ \Rightarrow 6 x - 6 = 5 x + 10 \\ \Rightarrow x = 16 \end{aligned}$

187.

Working together, Asha and Sudha can complete an assigned task in 20 days. However, if Asha worked alone and completed half the work and then Sudha takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Asha take to complete the if she worked alone ? Assume that Sudha is more efficient than Asha.

(A) 25 days
(B) 30 days
(C) 60 days
(D) 65 days

Solution: Suppose, Asha takes x days to complete the task alone while Sudha takes y days to complete it alone Since Sudha is more efficient than Asha, we have x > y $\begin{aligned} Asha's 1 day's work = \frac{1}{x} \\ Sudha's 1 day's work = \frac{1}{y} \\ (Asha + Sudha)'s 1 day's work \\ = \frac{1}{x} + \frac{1}{y} \\ = \frac{x + y}{x y} \end{aligned}$ If Asha and Sudha each does half of the work alone, time taken $\begin{aligned} = (\frac{x}{2} + \frac{y}{2}) days \\ = (\frac{x + y}{2}) days . \\ ∴ \frac{x + y}{2} = 45 \\ \Rightarrow x + y = 90 \\ From (i) and (ii), \\ We have : \frac{x y}{20} = 90 or x y = 1800 \\ Now, x y = 1800 and x + y = 90 \\ \Rightarrow x = 60, y = 30 [∵ x > y] \end{aligned}$ Hence, Asha alone will take 60 days to complete the task.

188.

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

189.

A can do a piece of work in 20 days and B in 15 days. With help of C, they finish the work in 5 days. In how many days C alone can do the same work ?

(A) 5 days
(B) 6 days
(C) 10 days
(D) 12 days

Solution: L.C.M. of Total Work =60 One day work of A = $\frac{60}{20}$ = 3 unit/day One day work of B = $\frac{60}{15}$ = 4 unit/day One day work of A + B + C = $\frac{60}{5}$ = 12 unit/day $\begin{aligned} C's efficiency \\ = 12 - 3 - 4 \\ = 5 \\ C will complete total work in \\ = \frac{60}{5} \\ = 12 days \end{aligned}$

190.

A's 2 days work is equal to B's 3 days work. If A can complete the work in 8 days, then to complete the work B will take ?

(A) 15 days
(B) 12 days
(C) 16 days
(D) 14 days

Solution: $\begin{aligned} According to the question, \\ \Rightarrow 2 A = 3B \\ \Rightarrow \frac{A}{B} = \frac{3}{2} \\ \Rightarrow Then efficiency ratio \\ A : B = 3 : 2 \end{aligned}$ ⇒ We know that time is inversely proportional to efficiency ⇒ Then time taken by them in ratio $\begin{aligned} A : B = \underset{8 days}{\underset{4 \times ↓}{2}} : \underset{12 days}{\underset{↓ \times 4}{3}} \\ ∵ A can do the work in 8 days \\ \Rightarrow i .e, 2 units \to 8 \\ 1 unit \to 4 \\ \Rightarrow Time taken by B \to 3 units \\ = 3 \times 4 \\ = 12 days \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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