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:
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 4 of 5.
Naveen said:
1 decade ago
(A+B)'s 7 hour work is 2/3.
(A+B)'s 1 hour work is(2/3)/7 = 2/3*7 = 2/21.
(A+B)'s 1 hour work is(2/3)/7 = 2/3*7 = 2/21.
Saranya said:
1 decade ago
Why 2/6?
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.
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.
Rechu said:
9 years ago
Assume total work = 6.
So, efficiency of a + b + c = 1.
we know w = e * t.
w = 1 * 2 {given}
= 2.
Balnce = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.
Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.
So, efficiency of a + b + c = 1.
we know w = e * t.
w = 1 * 2 {given}
= 2.
Balnce = 4 work.
a + b = 4 work.
w = e * t.
4 = e * 7.
e = 4/7.
Then c = 1 - 4/7 = 3/7.
c alone = 6/3/7 = 14 hrs.
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.
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.
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.
So, 6/84 = 14.
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