Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 12)
12.
A large tanker can be filled by two pipes A and B in 60 minutes and 40 minutes respectively. How many minutes will it take to fill the tanker from empty state if B is used for half the time and A and B fill it together for the other half?
Answer: Option
Explanation:
| Part filled by (A + B) in 1 minute = |  |
1 | + | 1 |  |
= | 1 | . |
| 60 | 40 | 24 |
Suppose the tank is filled in x minutes.
| Then, | x | |
1 | + | 1 | |
= 1 |
| 2 | 24 | 40 |
 |
x | x | 1 | = 1 |
| 2 | 15 |
x = 30 min.
Discussion:
51 comments Page 4 of 6.
Suji said:
1 decade ago
Nothing to confuse last 2 lines of question is changed into eqn.
Given half time to fill. As we don't know min. We take as x ie. X/2.
A and B together+only B.
Given half time to fill. As we don't know min. We take as x ie. X/2.
A and B together+only B.
Manasa said:
1 decade ago
Thanks ishan. Well explained.
Santhosh said:
1 decade ago
Thank You Pankaj
Shubham said:
1 decade ago
But as calculated so both A and B will fill tank in 24min?
So what about this?
So what about this?
Viz said:
1 decade ago
x/2(only B) + x/2(A and B together) = 1.
x/2(1/40) + x/2 (1/60+1/40) = 1.
x/80 + x/2(5/120) = 1.
x/80 + 5x/240 = 1.
(3x+5x)/240 = 1.
8x = 240.
x=30 mins.
x/2(1/40) + x/2 (1/60+1/40) = 1.
x/80 + x/2(5/120) = 1.
x/80 + 5x/240 = 1.
(3x+5x)/240 = 1.
8x = 240.
x=30 mins.
SANJAY PAHADE said:
1 decade ago
A simple method for solving this problem without formula :
Firstly, For a 1/2 Minute i.e half-minute Pipe B is used so,
Now, remaining part of tank should be filled by both A and B in remaining 1/2 i.e half minute.
For which combined capacity of A and B have to used as both are working simultaneously.
Finally,
We have to add tank filled by B alone and A+B together.
So, required answer is 30 mins respectively.
Firstly, For a 1/2 Minute i.e half-minute Pipe B is used so,
1 1 1
- * - = -
2 40 80
Now, remaining part of tank should be filled by both A and B in remaining 1/2 i.e half minute.
For which combined capacity of A and B have to used as both are working simultaneously.
1 [ 1 1 ] 1
- * [ - + - ] = -
2 [ 40 80 ] 48
Finally,
We have to add tank filled by B alone and A+B together.
1 1 128
- + - = ---
80 48 3840
1
= -
30
So, required answer is 30 mins respectively.
M ramesh said:
1 decade ago
This problem solving easily this type.
Total time is = x.
b time = a+b time.
b one minute capacity = 2.5.
a+b one minute capacity = 2.5+1.66 = 4.16.
b+(a+b) = 1.
2.5*15m = 37.5.
4.16*15m = 62.5.
37.5+62.5 = 100, 15+15 = 30.
Total time is = x.
b time = a+b time.
b one minute capacity = 2.5.
a+b one minute capacity = 2.5+1.66 = 4.16.
b+(a+b) = 1.
2.5*15m = 37.5.
4.16*15m = 62.5.
37.5+62.5 = 100, 15+15 = 30.
Tom said:
1 decade ago
I don't understand that how 1/40 is for one minutes?
Sri said:
1 decade ago
Please check the answer is 32 minutes,
B can fill a tank in 40 minutes means it can fill half tank by 20 minutes
Half tank B = 20 minutes.
A and B can fill a tank in 24 minutes means it can fill half tank by 12 minutes
Half tank A+B = 12 minutes
So, the answer is 32 minutes.
B can fill a tank in 40 minutes means it can fill half tank by 20 minutes
Half tank B = 20 minutes.
A and B can fill a tank in 24 minutes means it can fill half tank by 12 minutes
Half tank A+B = 12 minutes
So, the answer is 32 minutes.
Bhushan said:
9 years ago
By LCM method answer is 32.
A B A+B
60 40 24 ---> Time.
2 3 5(3+2) ---> lit/min.
___ ___ _____
120 120 120 ---> capacity
For 60 lit, B will continue so it will take 20min.
And remaining 60 lit both A+B will work i.e will take 12min.
Total time = 20 + 12 = 32min.
A B A+B
60 40 24 ---> Time.
2 3 5(3+2) ---> lit/min.
___ ___ _____
120 120 120 ---> capacity
For 60 lit, B will continue so it will take 20min.
And remaining 60 lit both A+B will work i.e will take 12min.
Total time = 20 + 12 = 32min.
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