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Time And Work - Aptitude MCQ Questions and Solutions with Explanations

Time And Work MCQ Questions and answers with easy and logical explanations.Arithmetic Ability provides you all type of quantitative and competitive aptitude mcq questions on Time And Work with easy and logical explanations. Time And Work MCQ is important for exams like Banking exams,IBPS,SCC,CAT,XAT,MAT etc. Page-8 section-2

Each section contains 100 MCQs on Time and Work .
71
Kamal can do a work in 15 days. Bimal is 50% more efficient then Kamal. The number of days, Bimal will take to do the same piece of work, is = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Ratio of times taken by Kamal and Bimal
$$\eqalign{ & = 150:100 \cr & = 3:2 \cr} $$
Suppose Bimal takes x days to do the work
$$\eqalign{ & \Rightarrow 3:2::15:x \cr & \Rightarrow x = \left( {\frac{{2 \times 15}}{3}} \right) \cr & \Rightarrow x = 10{\text{ days}}{\text{.}} \cr} $$
72
A does 20% less work than B. If A can complete a piece of work in $${\text{7}}\frac{1}{2}$$ hours, then B can do it in ?
Discuss
Answer & Solution
Answer: Option C
Solution:
Ratio of times taken by A and B
$$\eqalign{ & = 100:80 \cr & = 5:4 \cr} $$
Suppose B takes x hours to do the work
$$\eqalign{ & \Rightarrow 5:4::\frac{{15}}{2}:x \cr & \Rightarrow x = \left( {\frac{{4 \times 15}}{{2 \times 5}}} \right) \cr & \Rightarrow x = 6{\text{ hours}}{\text{}} \cr} $$
73
A, B and C can separately do a work in 12, 15 and 20 days respectively. They started to work together but C left after 2 days. The remaining work will be finished in = ?
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{12}} + \frac{1}{{15}} + \frac{1}{{20}}} \right) \cr & = \frac{{12}}{{60}} \cr & = \frac{1}{5} \cr & \left( {{\text{A}} + {\text{B}} + {\text{C}}} \right){\text{'s 2 day's work}} \cr & = \left( {\frac{1}{5} \times 2} \right) \cr & = \frac{2}{5} \cr & {\text{Remaining work }} \cr & = \left( {1 - \frac{2}{3}} \right) \cr & = \frac{3}{5}{\text{ }} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{12}} + \frac{1}{{15}}} \right) \cr & = \frac{9}{{60}} = \frac{3}{{20}} \cr & {\text{Now, }}\frac{3}{{20}}{\text{ work is done by A and B in 1 day}} \cr & \therefore \frac{3}{5}\,{\text{work is done by A and B in }} \cr & = \left( {\frac{{20}}{3} \times \frac{3}{5}} \right) \cr & = 4{\text{ days}}{\text{.}} \cr} $$
74
A can do a piece of work in 15 days, which B can do it in 10 days. B worked at it for 8 days. A can finish the remaining work in?
Discuss
Answer & Solution
Answer: Option B
Solution:
$$\eqalign{ & {\text{B's 8 day's work}} \cr & = \left( {\frac{1}{{10}} \times 8} \right) \cr & = \frac{4}{5} \cr & {\text{Remaining work}} \cr & = \left( {1 - \frac{4}{5}} \right) \cr & = \frac{1}{5} \cr & {\text{Now,}} \cr & \frac{1}{{15}}{\text{work is done by A in 1 day}}{\text{.}} \cr & \therefore \frac{1}{5}{\text{ work is done by A in}} \cr & = \left( {15 \times \frac{1}{5}} \right) \cr & = 3{\text{ days}}{\text{.}} \cr} $$
75
A does half as much work as B in one-sixth of the time. If together they take 10 days to complete a work. How much time shall B alone take to do it ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Suppose B takes x days to do the work.
$$\eqalign{ & {\text{A takes}} = \left( {2 \times \frac{1}{6}x} \right) \cr & = \frac{1}{3}x{\text{ days to do it}}{\text{.}} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} = \frac{1}{{10}} \cr & \Rightarrow \frac{1}{x} + \frac{3}{x} = \frac{1}{{10}} \cr & \Rightarrow \frac{4}{x} = \frac{1}{{10}} \cr & \Rightarrow x = 40{\text{ days}}{\text{.}} \cr} $$
76
Some carpenters promised to do a job in 9 days but 5 of them were absent and remaining men did the job in 12 day. The original number of carpenters was ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let there were N carpenters in the beginning
According to the question,
$$\frac{{{{\text{N}}_{{\text{men}}}} \times {9_{{\text{days}}}}}}{{{{\text{1}}_{{\text{work}}}}}} = $$     $$\frac{{{{\left( {{\text{N}} - 5} \right)}_{{\text{men}}}} \times {{12}_{{\text{days}}}}}}{{{{\text{1}}_{{\text{work}}}}}}$$
$$\eqalign{ & {\text{3N}} = {\text{4N}} - 20 \cr & {\text{N}} = {\text{20 men}} \cr} $$
77
If 4 men or 6 women can do a piece of work in 12 days working 7 hours a day, how many days will it take to complete a work twice as large with 10 men and 3 women working together 8 hours a day ?
Discuss
Answer & Solution
Answer: Option B
Solution:
According to the question,
4 men = 6 women
2 men = 3 women
10 men + 3 women = 10 men + 2 men
10 men + 3 women = 12 men.....(i)
$$\frac{{{4_{{\text{men}}}} \times {{12}_{{\text{days}}}} \times {7_{{\text{hours}}}}}}{{{{\text{1}}_{{\text{work}}}}}} = $$      $$\frac{{{{12}_{{\text{men}}}} \times {{\text{D}}_{{\text{days}}}} \times {8_{{\text{hours}}}}}}{{{{\text{1}}_{{\text{work}}}}}}$$
$$\eqalign{ & {\text{After solving we get}} \cr & {\text{D}} = 7{\text{ days}} \cr} $$
78
A contractor undertook to finish a work in 92 days and employed 110 men. After 48 days, he found that he had already done $$\frac{3}{5}$$ part of the work, the number of men he can withdraw so that his work may still be finished in time is ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Let 'n' number of men can be withdrawn
$$\eqalign{ & \frac{{\left( {{{110}_{{\text{men}}}} \times {{48}_{{\text{days}}}}} \right)}}{{\frac{3}{5}{\text{work}}}} = \frac{{\left( {110 - {\text{n}}} \right) \times 44}}{{\frac{2}{5}{\text{work}}}} \cr & \Rightarrow 110 \times 16 = \left( {110 - {\text{n}}} \right) \times 22 \cr & \Rightarrow 160 = \left( {110 - {\text{n}}} \right) \times 2 \cr & \Rightarrow {\text{n}} = 30 \cr} $$
79
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 = ?
Discuss
Answer & Solution
Answer: Option B
Solution:
Let 'n' number of men can required
$$\eqalign{ & \frac{{\left( {{{200}_{{\text{men}}}} \times {{50}_{{\text{days}}}}} \right)}}{{\frac{1}{4}}} = \frac{{\left( {200 + {\text{n}}} \right) \times {{100}_{{\text{days}}}}}}{{\frac{3}{4}}} \cr & \Rightarrow 3 \times 100 = \left( {200 + {\text{n}}} \right) \cr & \Rightarrow {\text{n}} = 100 \cr} $$
80
A can complete a piece of work in 18 days, B in 20 days and C in 30 days. B and C together start the work and are forced to leave after 2 days. The time taken by A alone to complete the remaining work is ?
Discuss
Answer & Solution
Answer: Option C
Solution:
$$\eqalign{ & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{20}} + \frac{1}{{30}}} \right) \cr & = \frac{5}{{60}} \cr & = \frac{1}{{12}} \cr & \left( {{\text{B}} + {\text{C}}} \right){\text{'s 2 day's work}} \cr & = \left( {\frac{1}{{12}} \times 2} \right) \cr & = \frac{1}{6} \cr & {\text{Remaining work }} \cr & = \left( {1 - \frac{1}{6}} \right) \cr & = \frac{5}{6}{\text{ }} \cr & \left( {{\text{A}} + {\text{B}}} \right){\text{'s 1 day's work}} \cr & = \left( {\frac{1}{{12}} + \frac{1}{{15}}} \right) \cr & = \frac{9}{{60}} \cr & = \frac{3}{{20}} \cr} $$
Now, $$\frac{1}{{18}}$$ work is done by A in 1 day
$$\eqalign{ & \therefore \frac{5}{6}{\text{ work is done by A in }} \cr & = \left( {18 \times \frac{5}{6}} \right) \cr & = 15{\text{ days}} \cr} $$