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 1 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)

Ankitkumar Dandiwala said:   2 years ago
Cross-check:

TA = 60 min.
TB = 40 min.
As per the given statement, A and B run together for 15 min (=Answer/2=30/2), and B runs alone for the next 15 mins (=Answer/2=30/2).
P1 = Part filled by first 15 min = 15*(1/TA+1/TB) = 15*(1/60+1/40) = 15*5/120=5/8.
P2 = Part filled by next 15 min = 15*(1/TB) = 15*(1/40) = 3/8.
Full tank = summation of both parts = P1+P2 = 5/8+3/8 = 1.
Hence it is proved that the given explanation is correct.
(2)

Pratham said:   4 years ago
I agree The right answer is 32.
(14)

Sagar P. said:   5 years ago
LCM(60,40) = 120.
Suppose the capacity of the tanker is 120litre.
Then, the quantity filled by pipe A in 1 min=120/60= 2 litres.
The quantity filled by pipe B in 1 min=120/40=3 litre.

The quantity filled by pipe A and B together in 1 min= 2 + 3 = 5 litre.

Suppose the tanker is filled in x minutes.

Then, to fill the tanker from the empty state, B is used for x/2 minutes and A &B together is used for the rest x/2 minutes.

3x/2 + 5x/2 = 120;
x/2(3+5) = 120;
8x/2 = 120
X = 30.

Required time =30 minutes.
(26)

Mehedi Alam said:   6 years ago
Total = 240 units.
Efficiency of A = 4 unit.
Efficiency of B = 6.

So required time = (240 *1/2) /6+(120/10) = 20+12 = 32.
Why ans is 30?
(7)

Md Harej Alam said:   6 years ago
Why not 32?
(7)

Ram Deen said:   6 years ago
A = 60 min.
B = 40 min.

Total work= 120.

The capacity of A and B is 2 and 3 respectively.

In Q, is clear that B is used for half the time, so we will take 20 min.

So, in 20 min B will fill = 60 work.
Remaining work = 60.
Now, both A and B is used for other half means is clear that for other half work that is 60.

So the capacity of A+B=5.

Time = 60/5 = 12 min.
Total time = 20 + 12 = 32 min.
(15)

Rashiqr Rahuman said:   6 years ago
Thanks @Sanjay Pahade.

Santosh said:   6 years ago
Thanks @Viz.

Chinmay said:   7 years ago
Question is asked in terms of time, not volume (work). So, 30 mins is the answer. LCM method assumes half volume not half time.
(1)


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