# Time and Work Questions

Q:

A is twice efficient as B and together they do the same work in as much time as C and D together. If C and D can complete the work in 20 and 30 daysrespectively, working alone ,then in how many days A can complete the work individually:

 A) 12 days B) 18 days C) 24 days D) 30 days

Explanation:

A     +      B        =      C     +     D

|              |                 |             |

Ratio of efficiency         10x   +    5x               9x     +   6x

|________|                 |_________|

15x                           15x

Therefore , ratio of efficiency of A:C  =10:9

Therefore,  ratio of days taken by A:C = 9:10

Therefore, number of days taken by A = 18 days

3 1683
Q:

A contractor  undertook a project to complete it in 20 days which needed 5 workers to work continuously for all the days estimated. But before the start of the work the client wanted to complete it earlier than the scheduled time, so the contractor calculated that  he needed to increase 5 additional  men every 2 days to complete the work in the time the client wanted it:

If the work was further increased by 50% but the contractor continues to increase the 5 workers o every 2 days then how many more days are required over the initial time specified by the  client.

 A) 1 day B) 2 days C) 5 days D) None of these

Explanation:

Total work = 100+50 = 150man-days

In 8 days 100 man-days work  has been completed. Now on 9th and 10th day there will be 25 workers. So in 2 days they wll complete additional 50 man- days work. Thus the work requires 2 more days.

4 1577
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

3 1551
Q:

A single reservoir supplies the petrol to the whole city, while the reservoir is fed by a single pipeline filling the reservoir with the stream of uniform volume. When the reservoir is full and if 40,000 liters of petrol is used daily, the suply fails in 90 days.If 32,000 liters of petrol is used daily, it fails in 60 days. How much petrol can be used daily with out the supply ever failing?

 A) 64000 liters B) 56000 liters C) 78000 liters D) 60000 liters

Explanation:

Let x liter be the per day filling and v litr be the capacity of the reservoir, then

90x + v = 40000 * 90     -----(1)

60x + v= 32000 * 60     ------(2)

solving eq.(1) and (2) , we get

x = 56000

Hence , 56000 liters per day can be used without the failure of supply.

3 1409
Q:

Two pipes A and B can fill a cistern in 4 minutes and 6 minutes respectively . If these pipes are turned on alternately for 1 minute each how long will it take to the cistern to fill?

As the pipes are operating alternatively, thus their 2 minutes job is =$\inline \frac{1}{4}+\frac{1}{6}=\frac{5}{12}$

In the next 2 minutes the pipes can fill another $\inline \frac{5}{12}$ part of cistern.

$\inline \therefore$ In 4 minutes the two pipes which are operating alternatively will fill $\inline \frac{5}{12}+\frac{5}{12}=\frac{5}{6}$

Remaining part = $\inline 1-\frac{5}{6}=\frac{1}{6}$

Pipe A can fill $\inline \frac{1}{4}$ of the cistern in 1 minute

Pipe A can fill $\inline \frac{1}{6}$ of the cistern in =$\inline 4\times \frac{1}{6}=\frac{2}{3}$  min

$\inline \therefore$Total time taken to fill the Cistern

$\inline 4+\frac{2}{3}=4\frac{2}{3}$ minutes