Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 6)
6.
Two pipes can fill a tank in 20 and 24 minutes respectively and a waste pipe can empty 3 gallons per minute. All the three pipes working together can fill the tank in 15 minutes. The capacity of the tank is:
Answer: Option
Explanation:
| Work done by the waste pipe in 1 minute = | 1 | - |  |
1 | + | 1 |  |
| 15 | 20 | 24 |
| = | |
1 | - | 11 | |
| 15 | 120 |
| = - | 1 | . [-ve sign means emptying] |
| 40 |
 Volume of |
1 | part = 3 gallons. |
| 40 |
Volume of whole = (3 x 40) gallons = 120 gallons.
Discussion:
48 comments Page 5 of 5.
Jeevitha said:
5 years ago
Thanks a lot @Gohor. Your method is simple to understand.
(2)
Apoorv said:
5 years ago
Pipe 1 can fill the tank in 20 minute and pipe 2 can fill it in 24 min.
Assume capacity of Tank is 120 gallons (LCM of 20,24).
Now,
Rate of pipe 1 = 120/20 = +6 gallons/minute.
Rate of pipe 2 = 120/24 = +5 gallons/minute.
Rate of drainage pipe = -3 gallons/minute (Note: -ve sign be because it is draining water).
Rate * time = Work.
(6+5+(-3)) * 15 min = 120 gallons.
Assume capacity of Tank is 120 gallons (LCM of 20,24).
Now,
Rate of pipe 1 = 120/20 = +6 gallons/minute.
Rate of pipe 2 = 120/24 = +5 gallons/minute.
Rate of drainage pipe = -3 gallons/minute (Note: -ve sign be because it is draining water).
Rate * time = Work.
(6+5+(-3)) * 15 min = 120 gallons.
(13)
Abhi said:
5 years ago
Amount filled by the tank in 1 minute = (1/20)+(1/24),
Assuming x= capacity of the tank,
Amount emptied by the tank in 1 minute = (3/x).
Total amount filled in 1 minute when all 3 pipes are on = (1/15).
So,
(1/20+1/24)-(3/x) = (1/15),
Hence, x = 120 gallons.
Assuming x= capacity of the tank,
Amount emptied by the tank in 1 minute = (3/x).
Total amount filled in 1 minute when all 3 pipes are on = (1/15).
So,
(1/20+1/24)-(3/x) = (1/15),
Hence, x = 120 gallons.
(8)
Preethi said:
5 years ago
@Krish.
Your explanation is good. Thanks.
Your explanation is good. Thanks.
Ram said:
4 years ago
Thanks all for explaining.
(1)
Jamshaid said:
3 years ago
Not agreed.
The Suggested Solution is;
Let Tank be of X Gallons.
Speed of filling X/20+X/24; Emptying in 15 Min = 15 * 3 Gal/Mon = 45 G, then;
15 min filling eq. 15 * (X/20 + X/24) - 45 = X.
X = 25 (15/21) G.
The Suggested Solution is;
Let Tank be of X Gallons.
Speed of filling X/20+X/24; Emptying in 15 Min = 15 * 3 Gal/Mon = 45 G, then;
15 min filling eq. 15 * (X/20 + X/24) - 45 = X.
X = 25 (15/21) G.
(2)
Lingaraj Behera said:
3 years ago
Let A take 20 min B take 24 and C 3g(g for gallons).
So, the LCM of (20,24,3g) is 120 gallons, so A fills 6g per min, B fills 5 g per min and C -3g per min.
In 1 min the tank fill by 5+6-3 = 8 gallons.
It takes 15 min to fill, so tank capacity is 8*15 = 120 gallons.
So, the LCM of (20,24,3g) is 120 gallons, so A fills 6g per min, B fills 5 g per min and C -3g per min.
In 1 min the tank fill by 5+6-3 = 8 gallons.
It takes 15 min to fill, so tank capacity is 8*15 = 120 gallons.
(29)
Aniket more said:
2 years ago
@All.
We all know that lcm is the total capacity of the tank now here two values are already given which are 20 and 24 and 3rd value of the outlet pipe is not given so the logic here is the lcm of 20 24 and the unknown 3rd value will be divisible by all the three numbers right?
Hence check out all four options which option is completely divisible by 24 and 20 and that is your answer.
We all know that lcm is the total capacity of the tank now here two values are already given which are 20 and 24 and 3rd value of the outlet pipe is not given so the logic here is the lcm of 20 24 and the unknown 3rd value will be divisible by all the three numbers right?
Hence check out all four options which option is completely divisible by 24 and 20 and that is your answer.
(9)
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