Aptitude - Pipes and Cistern - Discussion

Discussion Forum : Pipes and Cistern - General Questions (Q.No. 10)
10.
Two pipes A and B can fill a tank in 15 minutes and 20 minutes respectively. Both the pipes are opened together but after 4 minutes, pipe A is turned off. What is the total time required to fill the tank?
10 min. 20 sec.
11 min. 45 sec.
12 min. 30 sec.
14 min. 40 sec.
Answer: Option
Explanation:

Part filled in 4 minutes = 4 1 + 1 = 7 .
15 20 15

Remaining part = 1 - 7 = 8 .
15 15

Part filled by B in 1 minute = 1
20

1 : 8 :: 1 : x
20 15

x = 8 x 1 x 20 = 10 2 min = 10 min. 40 sec.
15 3

The tank will be full in (4 min. + 10 min. + 40 sec.) = 14 min. 40 sec.

Discussion:
37 comments Page 1 of 4.

Rkay said:   2 years ago
A = 15.
B = 20.
Lcm = 60.

Capacity A - 4
B - 3
(4+3)*4min = 28.
Remain 32.
By b only 32/3 = 10 2/3 ie 10m 40sec.
T time 14min 40sec.
(11)

Sidra Rayees said:   3 years ago
@Vinay Kumar Gupta is absolutely Right.

The Answer is 10 Minutes And 40 Seconds.

Sol:
A=15minutes
B=20 minutes.

Taking LCM of Both Which comes 60.
Therefore Total capacity of the Tank is 60 litres,

For A: 15 * 4 = 60.
For B: 20 * 3 = 60.

It Means Pipe A Fills 4 Litres In 1 Minute.
And Pipe B Fills 3 Litres In 1 Minute,
Both ate filling 3+4=7 litres In 1 minute.
Both are Working For 4 minutes together.
Therefore, 7*4=28 litres.
Now total capacity - 28litres
60-28=32.

Now Pipe B Have To Fill These Remaining 32 litres,
As pipe B Fills 3 Litres in 1 minute.
So 3*10 = 30.
30 litres in 10 minutes.
As Pipe B Fills 3 litres in one minute means 1 litre in 20 seconds.
So, the remaining 2 litres Will take 40 seconds.
Therefore The Answer will be 10 minute's and 40 seconds.
(10)

Kailash sharma said:   8 years ago
a=15;
b=20;
l.c.m of 15,20=60;
a's 1Min=60/15 => 4;
b's 1Min=60/20 => 3;
(a+b)'s 4Min= (4+3)*4=28;
Now 60-28 = 32;
b remaining time = 32/3 => 10.66.
Now add 10.66+4 =>14.66.
(9)

Pooarasu said:   2 years ago
Let x be the minutes that B opened,
4(1/15 +1/20) + x (1/20) = 1.
x=10 2/3 => 10min 40sec.
After solving x = 10min 40sec +4 min =>14min 40 =>answer.
(8)

Ezhil said:   5 years ago
@Sriganesh.

B can fill the tank in 1 min = 1/20 parts or work done of the tank.
After 4 min, B can alone to fill the remaining part of the tank?
min = 8/15 remaining parts or work done of the tank.
By cross multiply
1/20 parts = 1 min.
8/15 parts = ? min.
(8/15 * 1)/1/20 = ?
8/15 * 1 * 20 = ?
8/3*4 = ?
?=32/3=10 (2/3) min =10 min 40 sec (2/3 min convert to seconds 2/3 * 60=40 sec).

Thus, Total Time Required to fill the tank = 4 min + 10 min 40 sec.
= 14 min 40 sec.
(4)

Vipul Chaudhary said:   2 years ago
Let x be the minutes that B opened,
4 (1/15) + x (1/20) = 1.
After solving x = 14min 40sec => answer.
(3)

Debalina Roy said:   7 years ago
"1/20 :8/15 :: 1:x " I can't understand this. Please explain me.
(2)

Deepak verma said:   1 decade ago
Its so simple:
After 4 minute the a was turned off means till 4 minutes a and b work together.
So,
4(1/15+i/20) = 7/15.

Let total time take = x.

Than 4 minutes after a is turned off so,

7/15+(x-4)*i/20 = 1.

x = 14 min. 40sec.
(2)

Priya P. said:   1 decade ago
@ kalpna.

Thanks. Really less time consuming.
(1)

Revant said:   5 years ago
@Preeti.

Very clear solution. Thanks for explaining.
(1)


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