Aptitude - Pipes and Cistern - Discussion

Discussion Forum : Pipes and Cistern - General Questions (Q.No. 15)
15.
Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:
10
12
14
16
Answer: Option
Explanation:

Part filled in 2 hours = 2 = 1
6 3

Remaining part = 1 - 1 = 2 .
3 3

(A + B)'s 7 hour's work = 2
3

(A + B)'s 1 hour's work = 2
21

C's 1 hour's work = { (A + B + C)'s 1 hour's work } - { (A + B)'s 1 hour's work }

   = 1 - 2 = 1
6 21 14

C alone can fill the tank in 14 hours.

Discussion:
44 comments Page 2 of 5.

Soundharya saravanan said:   1 decade ago
A+B+C = 6 hrs after working 2 hrs.

A+B+C = 2 hrs.

So the remaining work is = 1-2/6 = 2/3.

A+B = 7 hrs.

(2/(3*7) = 2/21.

1/6-2/21 = 1/14.

Answer is: 14 hrs.

Arun said:   1 decade ago
Can't understand explain it clearly.

Sahil sharma said:   1 decade ago
How 6 hrs are calculated here?

Abhishek said:   1 decade ago
@Sahil sharma see the options in the answer. No one options are below 6 hours so first we need to find 6 hours work.

We have 2 hours work. Now we need to find out 6 hours work.

We know 2*3=6. So multiply by 3 on RHS to find out 6 hours work. If we wanted to know 8 hours work then we need to do 2*4=8. If 10 hours then 2*5=10 and so on.

Tapas said:   10 years ago
Explain clearly again.

Pooja said:   10 years ago
1/6-2/21 = 1/14 explain me in detail.

Suba peristina said:   9 years ago
A+B+C worked together for 2 hrs.

In 1hr they filled up to 1/6 parts thus for 2hrs they fill up to (1/6)*2 i.e.,2/6 parts. Which is 1/3.

Then we need to calculate remaining parts so total-filled parts give remaining part.

i.e., 1-1/3 = 2/3 part are yet remaining.

According to the question now A and B alone opened thus it works for 7 hrs to fill the remaining 2/3 part.

So (A+B)together fill (2/3)*(1/7)parts in 1hr i.e it together fill 2/21 parts in 1hr.

Now we need to calculate the time of C ;

So (A+B+C)'s 1hr work-(A+B)'s 1hr work will give C's 1hr work.

Therefore C's work in 1 HOUR = 1/6 - 2/21 = 1/14 parts.

So time taken by C to fill the tank is 14 hours.

Rahul Namdev said:   9 years ago
LCM of 1/6 . 1/2 . 1/7 = 84.
So, 6/84 = 14.

Namrata said:   9 years ago
As we know x/a + x/b + x/c = x/6.
Write equation as;

(X/6) * 2 + (x/6 - x/c) * 7 = x.

After solving this equation you will get answer as 14.

Sri vidhya said:   9 years ago
Why we need to calculate the final step for that one hour for A & B & C and A & B?


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