# Time and Work Questions

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?

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

1258
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

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

7 1238
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

Explanation:

Rate of leakage = 8.33% per hour

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

Time required = 100/25 = 4 hours

4 1214
Q:

Pipe A can fill the tank in 4 hours,while pipe B can fill it in 6 hours working separately.pipe C can empty whole the tank in 4 hours. He opened the pipe A and B simultaneously to fill the empty tank. He wanted to adjust his alarm so that he could open the pipe C when it was half-filled, but he mistakenly adjusted his alarm at a time when his tank would be 3/4th filled. what is the time difference between both the cases, to fill the tank fully:

 A) 48 min B) 54 min C) 30 min D) none of these

Explanation:

In ideal Case:

Time taken to fill the half tank by A and B = $\inline&space;\fn_jvn&space;\frac{50}{41.66}$ =$\inline&space;\fn_jvn&space;\frac{6}{5}$ hours

Time taken by A,B and C to fill rest half of the tank =$\inline&space;\fn_jvn&space;\frac{50}{16.66}$ = 3 hours

Total time = $\inline&space;\fn_jvn&space;\frac{6}{5}+3$ = 4 hours 12 min

In second case:

Time taken to fill $\inline&space;\fn_jvn&space;\frac{3}{4}$ tank by A and B =$\inline&space;\fn_jvn&space;\frac{75}{41.66}=\frac{9}{5}$ hours

Time taken by A,B and C to fill rest $\inline&space;\fn_jvn&space;\frac{1}{4}$ tank = $\inline&space;\fn_jvn&space;\frac{25}{16.66}=\frac{3}{2}$ hours

Total time =$\inline&space;\fn_jvn&space;\frac{9}{5}+\frac{3}{2}$ =3 hours 18 min

Therefore , difference in time = 54 minutes

0 1184
Q:

Four pipes A,B, C and D can fill a cistern  in 20,25, 40 and 50 hours respectively.The first pipe A was opened at 6:00 am, B at 8:00 am, C at 9:00 am and D at 10:00 am. when will the Cistern be full?

 A) 4:18 pm B) 3:09 pm C) 12:15 pm D) 11:09 am

Explanation:

Efficiency of P= 5%

Efficiency of Q= 4%

Efficiency of R= 2.5%

Efficiency of S= 2%

$\inline&space;\left.\begin{matrix}&space;Till&space;\;&space;10\:&space;am&space;\;&space;pipe\;&space;P&space;\;&space;filled\;&space;20\;&space;percent\\&space;Till&space;\;&space;10\:&space;am&space;\;&space;pipe\;&space;Q&space;\;&space;filled\;&space;8\;&space;percent\\&space;Till&space;\;&space;10\:&space;am&space;\;&space;pipe\;&space;R&space;\;&space;filled\;&space;2.5\;&space;percent&space;\end{matrix}\right\}30.5$ %

Thus, at 10 am pipe P,Q and R filled 30.5% of the cistern.

Now, the time taken by P,Q,R and S together to fill the remaining capacity of the cistern

=$\inline&space;\frac{69.5}{13.5}$ = $\inline&space;\frac{139}{27}$ =5 Hours and 9 minutes(approx)

Therefore, total time =4 hrs + 5hrs 9 mins

= 9 hrs and 9 mins

It means cistern will be filled up at 3:09 pm