Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 1)
1.
Three pipes A, B and C can fill a tank from empty to full in 30 minutes, 20 minutes, and 10 minutes respectively. When the tank is empty, all the three pipes are opened. A, B and C discharge chemical solutions P,Q and R respectively. What is the proportion of the solution R in the liquid in the tank after 3 minutes?
Answer: Option
Explanation:
| Part filled by (A + B + C) in 3 minutes = 3 |  |
1 | + | 1 | + | 1 |  |
= | |
3 x | 11 | |
= | 11 | . |
| 30 | 20 | 10 | 60 | 20 |
| Part filled by C in 3 minutes = | 3 | . |
| 10 |
 Required ratio = |
|
3 | x | 20 | |
= | 6 | . |
| 10 | 11 | 11 |
Discussion:
75 comments Page 8 of 8.
Akshay Kumar Roy said:
1 year ago
First take the LCM of all the time taken by A, B, and C which we will give us total work = 60, and then take out the efficiency by the formula (total work/time) A-> 2 ,B-> 3 and C-> 6 and then take out the Ratio of C's R liquid proportion (6/11) by multiplying both the numerator and denominator by 3.
And the result we will get is 6/11.
And the result we will get is 6/11.
(5)
Saikat said:
1 year ago
Total 60 unit.
A = 1 minute 2 unit
B =1 min 3 unit
C =1 min 6 unit.
So, Total (6+3+2)=11 unit.
C contribution is 6/11 unit.
A = 1 minute 2 unit
B =1 min 3 unit
C =1 min 6 unit.
So, Total (6+3+2)=11 unit.
C contribution is 6/11 unit.
(53)
DVJS said:
4 weeks ago
Total 60 unit. (LCM of 10 20 30).
A in 1 min = 2 units.
B in 1 min = 3 units.
C in 1 min = 6 units.
A + B + C = (6 + 3 + 2) = 11 units.
So, C's contribution is 6/11 units.
A in 1 min = 2 units.
B in 1 min = 3 units.
C in 1 min = 6 units.
A + B + C = (6 + 3 + 2) = 11 units.
So, C's contribution is 6/11 units.
(5)
APA said:
2 weeks ago
A in 1 min = 1/30.
B in 1 min = 1/20.
C in 1 min = 1/10.
1/30 + 1/20 + 1/10 => (2 + 3 + 6)/60 => 11/60.
Now, three pipes in 3 min = 3 * 11/60 => 11/20.
C share in 3 min = 3 * 1/10 => 3/10.
Required ratio = (3/10 * 20/11) => 6/11.
B in 1 min = 1/20.
C in 1 min = 1/10.
1/30 + 1/20 + 1/10 => (2 + 3 + 6)/60 => 11/60.
Now, three pipes in 3 min = 3 * 11/60 => 11/20.
C share in 3 min = 3 * 1/10 => 3/10.
Required ratio = (3/10 * 20/11) => 6/11.
(3)
Dinesh said:
2 weeks ago
Good, Thanks for explaining.
(1)
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