8
Q:

Adam and Smith are working on a project. Adam takes 6 hrs to type 36 pages on a computer, while Smith takes 5 hrs to type 40 pages. How much time will they take, working together on two different computers to type a project of 120 pages?

 A) 8 hrs 45 min B) 8 hrs 42 min C) 8 hrs D) 8 hrs 34 min

Answer:   D) 8 hrs 34 min

Explanation:

Number of pages typed by Adam in 1 hour = $\inline \fn_jvn \small \frac{36}{6}$ = 6
Number of pages typed by Smith in 1 hour = $\inline \fn_jvn \small \frac{40}{5}$ = 8
Number of pages typed by both in 1 hour = (6 + 8) = 14
Time taken by both to type 110 pages = (120 * 1/14) = 8 $\inline \fn_jvn \small \frac{4}{7}$ = 8 hrs 34 min.

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

2 1382
Q:

A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in 55% of the time. How many workers were there in the group?

 A) 50 B) 40 C) 45 D) 10

Explanation:

It can be solved easily through option.

$\inline&space;\left&space;(&space;10+9+8+....+1&space;\right&space;)=10\left&space;(&space;10\times&space;\frac{55}{100}&space;\right&space;)$

55 = 55     Hence correct

Alternatively:

$\inline&space;\frac{n(n+1)}{2}=n\times&space;\frac{55n}{100}$

=> n= 10

In Both cases total work is 55man-days.

10 2632
Q:

Mr. stenley employed a certain number of typist for his project. 8 days later 20% of the typist left the job and it was found that it took as much time to complete the rest work from then as the entire work needed with all the employed typists. The average speed of a typist is 20 pages/hour. Minimum how many typist could be employed?

 A) 10 B) 5 C) 15 D) 4

Explanation:

Since 20% $\left&space;(&space;i.e,\frac{1}{5}&space;\right&space;)$ typists left the job. So, there can be any value which is multiple of 5 i.e, whose 20% is always an integer. Hence, 5 is the least possible value.

2 679
Q:

A Contractor employed a certain number of workers  to finish constructing a road in a certain scheduled time. Sometime later, when a part of work had been completed, he realised that the work would get delayed by three-fourth of the  scheduled time, so he at once doubled the no of workers and thus he managed to finish the road on the scheduled time. How much work he had been completed, before increasing the number of workers?

 A) 10 % B) 14 2/7 % C) 20 % D) Can't be determined

Explanation:

Let he initially employed x workers which works for D days and he estimated 100 days for the whole work and then he doubled the worker for (100-D) days.

D * x +(100- D) * 2x= 175x

=>  D= 25 days

Now , the work done in 25 days = 25x

Total work = 175x

therefore, workdone before increasing the no of workers = $\frac{25x}{175x}\times&space;100=14\frac{2}{7}$ %

17 2683
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.