# Time and Work Questions

Q:

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

Explanation:

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

B + C =50%

$\inline&space;\fn_jvn&space;\Rightarrow$ B= 20%     A= 50%        and   C=30%

Hence A is most efficient

Report Error

8 1518
Q:

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?

Answer

150 men in 25 days do = $\inline \frac{1}{4}$ work

1 man in 1 day does = $\inline \frac{1}{4}\times \frac{1}{25}\times \frac{1}{150}$ work

$\inline \therefore$ 100 men in 60 days do = $\inline \frac{1}{4}\times \frac{1}{25}\times \frac{1}{150}\times 100\times 60=\frac{2}{5}$ work

Total work done =$\inline \frac{1}{4}+\frac{2}{5}=\frac{5+8}{20}=\frac{13}{20}$

$\inline \therefore$ Remaining work =$\inline 1-\frac{13}{20}=\frac{7}{20}$

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

$\inline \therefore\: \frac{1}{4}$work is done in 25 days  by 150 men

$\inline \therefore\: \frac{7}{20}$ work is done in 35 days by $\inline \frac{150\times 4\times 25\times 7}{35\times 20}=150$ men

Report Error

1493
Q:

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

Explanation:

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

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

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

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

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

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

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

Report Error

3 1484
Q:

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

Explanation:

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

Report Error

7 1389
Q:

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

Explanation:

Rate of leakage = 8.33% per hour

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

Time required = 100/25 = 4 hours

Report Error

5 1327