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?
15 min
20 min
27.5 min
30 min
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 6 of 6.

Asik Ahamed said:   2 months ago
Given :
A = 1/60 min
B = 1/40 min.

B works alone half time means 1/2, no confusion, put x/2.
A and B both work together that also one half puts another x/2.

Step 1 :
First, we do B alone in one half the value is ×/2 and the actual time of B is 1/40, so, x/2×1/40 = ×/80.
B half value = x/80.

Step 2 :
A and B work, take lcm of 60,40 = 120.
It looks like 2/120 + 3/120 = 5/120 => 1/24.
Then implement the A and B work half x/2 × 1/24 = x/48.
A and B work half value = x/48.

Step 3 :
Adding your half's answers B half = x/80 and A&B half = x/48.
Total work done = 1,
So, x/80 + x/48 =1,
Take lcm on both sides, don't forget to take lcm, lcm of 80,48 = 240
3x/120 + 5x/240 = 8x/240 = 1.

Then simplify,
8x/240 = 1.
x = 240/8 => 30 minutes.
(2)


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