Practice MCQ Questions and Answer on Time and Work

161.

There is provision of food in fort for 1200 soldiers for 60 days. After 15 days, 200 soldiers leave the fort. Remaining food will last for how many days?

(A) 56 days
(B) 50 days
(C) 54 days
(D) 48 days

Solution: Work equivalence method: 1200 × 45 = 1000 × x Hence, x = 54 days Variation Method: After 15 days 200 soldiers leaved. Soldiers Food for days 1200 ↓ 45 1000 ↑ x (let) Arrows show the opposite variation to each other $\frac{1200}{1000} = \frac{x}{45}$ Or, x = $\frac{1200 \times 45}{1000}$ = 54 days.

162.

A is thrice as good a workman as B and so takes 60 days less than B for doing a job. The time in which they can do the job together is = ?

(A) $22 \frac{1}{2} days$
(B) $30 days$
(C) $45 days$
(D) $60 days$

Solution: Ratio of times taken by A and B = 1 : 3 If difference of time is 2 days, B takes 3 days If difference of time is 60 days, B takes $\begin{aligned} = (\frac{3}{2} \times 60) \\ = 90 days \end{aligned}$ So, A takes 30 days to do the work $\begin{aligned} A's 1 day's work = \frac{1}{30} \\ B's 1 day's work = \frac{1}{90} \\ (A + B)'s 1 day's work \\ = (\frac{1}{30} + \frac{1}{90}) \\ = \frac{4}{90} \\ = \frac{2}{45} \end{aligned}$ ∴ A and B together can do the job in $\begin{aligned} = \frac{45}{2} \\ = 22 \frac{1}{2} days \end{aligned}$

163.

A and B undertake a contract of a task for Rs. 10,800. A and B can complete the task in 45 days and 60 days, respectively. However, to finish the work early, they take C's help and complete the entire work in 20 days. What is the value (in Rs.) of (a - b + 2c), where a, b and c are the shares of A, B and C, respectively, for their contribution to complete the task?

(A) 3000
(B) 6000
(C) 4000
(D) 8000

164.

Two typist of varying skills can do a job in 6 minutes if they work together. If the first typist typed alone for 4 minutes and then the second typist typed alone for 6 minutes, they would be left with $\frac{1}{5}$ of the whole work. How many minutes would it take the slower typist to complete the typing job working alone ?

(A) 10 minutes
(B) 15 minutes
(C) 12 minutes
(D) 20 minutes

Solution: Working efficiency of both typist together, = $\frac{100}{6}$ = 16.66% per minute Now, let work efficiency of first typist be x and then second typist will be (16.66 - x) First typist typed alone for 4 minutes and second typed alone for 6 minutes and they left with $\frac{1}{5}$ (i.e 20%) of job, means they have completed 80% job Now, First Typist typed in 4 minute + Second typed in 6 minutes = 80% 4 × x + 6 × (16.66 - x) = 80% 4x + 100% - 6x = 80% x = 10% First Typist typed 10% per minutes. Then second typed (16.66 - 10) = 6.66% per minute Then, Second typist complete the whole job in $\frac{100}{6.66}$ = 15.01 = 15 minutes.

165.

X can do a piece of work in 24 days. When he had worked for 4 days, Y joined him. If complete work was finished in 16 days, Y can alone finish that work in:

(A) 27 days
(B) 36 days
(C) 42 days
(D) 18 days

Solution: According to the question, X → 24 days ⇒ Work done by X in 4 days alone = 4 × 1 = 4 units ⇒ Remaining work = 24 - 4 = 20 units ⇒ 20 units done by both together in (16 - 4 days) = 12 days ⇒ Then efficiencies of (X + Y) $= \frac{work}{days} = \frac{20}{12} = \frac{5}{3} = 1 + \frac{2}{3}$ ⇒ Efficiency of Y = $\frac{2}{3}$ ⇒ Time taken by Y alone to complete the total work $= \frac{24}{\frac{2}{3}} = 36 days$ Alternate solution: $\begin{aligned} X \times 20 = (X + Y) \times 12 \\ \frac{X}{X + Y} = \frac{12}{20} \\ = \frac{3 \to Efficiency of X}{5 \to Efficiency of (X + Y)} \end{aligned}$ Efficiency of Y = 5 - 3 = 2 units/day Total work = 24 × 3 = 72 units Total time taken by Y = $\frac{72}{2}$ = 36 days

166.

5 men can do a piece of work in 6 days while 10 women can do it in 5 days. In how many days can 5 women and 3 men do it ?

(A) 4 days
(B) 5 days
(C) 6 days
(D) 8 days

Solution: $\begin{aligned} 5M \times 6 days = 10W \times 5 days \\ 3M = 5W \\ \frac{M}{W} = \frac{5}{3} \\ 1 M (work) = 5 units/day \\ 1W (work) = 3 units/day \\ Hence, \\ Total work \\ = (5M \times 6) \\ = 5 \times 5 \times 6 \\ = 150 units \\ Required time for (5W + 3M) \\ = \frac{Total work}{Work done/day} \\ = \frac{150}{(5 \times 3 + 3 \times 5)} \\ = \frac{150}{30} \\ = 5 days \end{aligned}$

167.

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) $\begin{aligned} = \frac{Total work}{Efficiency} \\ = \frac{1440}{130} \\ = 11 \frac{1}{13} months \end{aligned}$

168.

20 man can finish a work in 30 days. They started working, but 4 men left the work after 10 days. In how many days would the work be completed?

(A) 30
(B) 25
(C) 35
(D) 28

Solution: Total work = 30 × 20 = 600 After 10 days = 20 × 10 = 200 Remaining work = 600 - 200 = 400 4 men left, remaining = 16 men 16 × D = 400 D = $\frac{400}{16}$ D = 25 days Total days = 25 + 10 = 35 days

169.

A and B can do a piece of work in 18 days. B and C together can do it in 30 days. If A is twice as good a workman as C, find the how many days B alone can do the work?

(A) 90 days
(B) 100 days
(C) 80 days
(D) 75 days

Solution: A : C = 2 : 1 A = 4 unit C = 2 unit B = 1 unit B = $\frac{90}{1}$ = 90 days

170.

If 12 carpenters working 6 hours a day can make 460 chairs in 240 days, then number of chairs made by 18 carpenters in 360 days each working 8 hours a day ?

(A) 1320
(B) 1380
(C) 1260
(D) 920

Solution: $According to the question,$ $\Rightarrow \frac{12 \times 6 \times 240}{460}$ = $\frac{18 \times 360 \times 8}{x}$ $\begin{aligned} \Rightarrow x = \frac{18 \times 360 \times 8 \times 460}{12 \times 6 \times 240} \\ \Rightarrow x = 1380 \end{aligned}$

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Practice MCQ Questions and Answer on Time and Work

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