Time and Work Questions


When A, B and C are deployed for a task , A and B together do 70% of the work and B and C together do 50% of the work. who is most efficient?

A) A B) B
C) C D) can't be determined
Answer & Explanation Answer: A) A


A + B= 70%            \inline \fn_jvn \begin{bmatrix} \therefore (A+B)+(B+c)-(A+B+C)=B\\ 70+50-100=20 \end{bmatrix}

B + C =50% 


          \inline \fn_jvn \Rightarrow B= 20%     A= 50%        and   C=30%

Hence A is most efficient

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8 1518

A contractor undertakes to complete a work in 130 days. He employs 150 men for 25 days and they complete 1/4 of the work . He then reduces the number of men to 100, who work for 60 days, after which there are 10 days holidays.How many men must be employed for the remaining period to finish the work?


150 men in 25 days do =  work

   1 man in 1 day does =  work

 100 men in 60 days do =  work

Total work done = 

 Remaining work =

Remaining time = 130 - (25+60+10) = 35 days

work is done in 25 days  by 150 men

 work is done in 35 days by  men

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5 1493

A and  B can do  a piece of work in 30 days , while  B and C can do the same work in 24 days and C and A in 20 days . They all work together for 10 days when B and C leave. How many days more will A take to finish  the work?

A) 18 days B) 24 days
C) 30 days D) 36 days
Answer & Explanation Answer: A) 18 days


2(A+B+C)'s 1 day work = \inline {\color{Black}\left ( \frac{1}{30}+\frac{1}{24}+\frac{1}{20} \right )=\frac{15}{120}=\frac{1}{8} }

=>(A+B+C)'s  1 day's work=\inline {\color{Black}\frac{1}{16} }

work done by A,B and C in 10 days=\inline {\color{Black}\frac{10}{16}= \frac{5}{8}}

Remaining work=\inline {\color{Black}(1-\frac{5}{8})= \frac{3}{8}}

A's 1 day's work =\inline {\color{Black}(\frac{1}{16}-\frac{1}{24})=\frac{1}{48}}

Now, \inline {\color{Black}\frac{1}{48}} work is done by A in 1 day.

So, \inline {\color{Black}\frac{3}{8}} work  wil be done by A in \inline {\color{Black}(48\times \frac{3}{8})} = 18 days

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3 1484

A and B can do a work in 4 hours and 12 hours respectively. A starts the work at 6 AM and they work alternately for one hour each. When will the work be completed?

A) 4 days B) 5 days
C) 6 days D) 7 days
Answer & Explanation Answer: C) 6 days


Work donee by A and B in the first two hours, working alternatively = First hour A + Second hour B = (1/4) + (1/12) = 1/3.

Thus, the total time required to complete the work  = 2 (3) = 6 days 

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7 1389

A tank has an inlet and outlet pipe. The inlet pipe fills the tank completely in 2 hours when the outlet pipe is plugged. The outlet pipe empties the tank completely in 6 hours when the inlet pipe is pluggeed.

If there is a lekage also which is capable of draining out the liquid from the tank at half of the  rate of outet pipe,them what is the time taken to fill the emty tank when both the pipes are opened?

A) 3 hours B) 4 hours
C) 5 hours D) None of these
Answer & Explanation Answer: B) 4 hours


Rate of leakage = 8.33% per hour

Net efficiency = 50 - (16.66 + 8.33)= 25%

Time required = 100/25 = 4 hours

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