7
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:

In a public bathroom there are n taps 1,2,3...n. Tap1 and Tap2 take equal time to fill the tank while Tap3 takes half the time taken by Tap2 and Tap4 takes half the time taken by Tap3. Similarly each next number of tap takes half the time taken by previous number of Tap i.e, K-th  Tap takes half the time taken by (K-1)th Tap.

If the 10th tap takes 2 hours to fill the tank alone then what is the ratio of  efficiency of 8th tap and 12th tap, respectively?

 A) 4:1 B) 5:3 C) 16:1 D) 1:16

Explanation:

Time taken by 8th tap = $\inline&space;2\times&space;2\times&space;2$ = 8 hours

and time taken by 12th tap = $\inline&space;2\times&space;\frac{1}{2}\times&space;\frac{1}{2}$ = $\inline&space;\frac{1}{2}$ hour

Ratio of time taken by 8th 8th tap and 12th tap = $\inline&space;8:\frac{1}{2}$ =16:1

Therefore, Ratio of efficiencies of 8th tap and 12th tap =1:16

5 670
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

4 1214
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

Answer & Explanation Answer: B) 3:09 pm

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 1135
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.

8 2289
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

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.