Aptitude - Pipes and Cistern - Discussion
Discussion Forum : Pipes and Cistern - General Questions (Q.No. 4)
4.
Two pipes A and B can fill a cistern in 37
minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
minutes and 45 minutes respectively. Both pipes are opened. The cistern will be filled in just half an hour, if the B is turned off after:
Answer: Option
Explanation:
Let B be turned off after x minutes. Then,
Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.
 x |
 |
2 | + | 1 |  |
+ (30 - x). | 2 | = 1 |
| 75 | 45 | 75 |
 |
11x | + | (60 -2x) | = 1 |
| 225 | 75 |
11x + 180 - 6x = 225.
x = 9.
Discussion:
90 comments Page 3 of 9.
OmkarR said:
5 years ago
@All.
According to me, the solution is;
The cistern will be filled in just half an hour, if the B is turned off after:
i.e A work 30 min so,
A total work 75/2 minutes. 30 /75/2=4/5.
Remaining work is 1-4/5=1/5,
So,
B can fill in 45 minute. Then 45/5 =9.
According to me, the solution is;
The cistern will be filled in just half an hour, if the B is turned off after:
i.e A work 30 min so,
A total work 75/2 minutes. 30 /75/2=4/5.
Remaining work is 1-4/5=1/5,
So,
B can fill in 45 minute. Then 45/5 =9.
(1)
Devi said:
6 years ago
For this type of question, we can approach like;
Here pipe A fill-in 75/2min(let it be 'x'), pipe B in 45 min(let it be 'y').
At what time B pipe should be close to fill a total cistern in 30 min (let it be 't'min)is,
y(1-(t/x))= 45(1-((30*2)/75)) = 9min.
Here pipe A fill-in 75/2min(let it be 'x'), pipe B in 45 min(let it be 'y').
At what time B pipe should be close to fill a total cistern in 30 min (let it be 't'min)is,
y(1-(t/x))= 45(1-((30*2)/75)) = 9min.
Shekhar ssc said:
10 years ago
Let A one min work = 1/37.5 = 2/75.
Let B one min work = 1/45.
Let B turn off after x min.
And so tank filled in 30 min.
So Tank full = Part filled by A and B + Part filled by only A.
1 = (A+B)x+(30-x)A.
1 = 11x/225+(30-x)2/75.
That's way x = 9.
Let B one min work = 1/45.
Let B turn off after x min.
And so tank filled in 30 min.
So Tank full = Part filled by A and B + Part filled by only A.
1 = (A+B)x+(30-x)A.
1 = 11x/225+(30-x)2/75.
That's way x = 9.
Hardik said:
10 years ago
Both together can do a work in 225/11 minute.
Now pipe A can do a work in 75/2 minute. He alone did work for 30 minutes in last.
30*2/75=4/5.
Now 1/5 had been filled by a and b together. 1 cistern in 225/11 minutes then 1/5 in how many minutes ?
Now pipe A can do a work in 75/2 minute. He alone did work for 30 minutes in last.
30*2/75=4/5.
Now 1/5 had been filled by a and b together. 1 cistern in 225/11 minutes then 1/5 in how many minutes ?
Swapnil said:
7 years ago
Pipe A fills the tank in 75/2 Min.
As pipe A runs 30 mins it fills,
75/2 min = 1 tank.
30 min = 1*2/75*30 = 4/5,
Pipe A fills 4/5 in 30 mins, remaining 1/5.
Pipe B fills 1 full tank in 45 mins,so.
1 tank = 45 min.
1/5 tank = 45/5 = 9 min.
As pipe A runs 30 mins it fills,
75/2 min = 1 tank.
30 min = 1*2/75*30 = 4/5,
Pipe A fills 4/5 in 30 mins, remaining 1/5.
Pipe B fills 1 full tank in 45 mins,so.
1 tank = 45 min.
1/5 tank = 45/5 = 9 min.
(1)
Deepak said:
8 years ago
Best solution:
Parts filled by A in 1 min = 2/75
Parts filled by B in 1 min = 1/45
Now;
In 30 min tank is full so,
Parts filled by A in 30 min = 30*2/75 =60/75
Parts filled by B in X min = X/45
So, 60/75 + X/45 = 1
Therefore ,X = 9 min.
Parts filled by A in 1 min = 2/75
Parts filled by B in 1 min = 1/45
Now;
In 30 min tank is full so,
Parts filled by A in 30 min = 30*2/75 =60/75
Parts filled by B in X min = X/45
So, 60/75 + X/45 = 1
Therefore ,X = 9 min.
Nehal said:
1 decade ago
Hey,
After first step just open the brackets i.e,
2x/75 + x/45 + 60/45 - 2x/75 = 1.
(2x/75 - 2x/75) + x/45 + 60/45 = 1.
0 + x/45 + 4/5 = 1.
Take 1/5 as common,
x/9 + 4 = 5.
x/9 = 1.
x = 9.
No need of complex calculations.
After first step just open the brackets i.e,
2x/75 + x/45 + 60/45 - 2x/75 = 1.
(2x/75 - 2x/75) + x/45 + 60/45 = 1.
0 + x/45 + 4/5 = 1.
Take 1/5 as common,
x/9 + 4 = 5.
x/9 = 1.
x = 9.
No need of complex calculations.
MERAJ HUSAIN said:
1 decade ago
I can get you understood here:
Since cistern is one which is to be filled by A & B.
Hence 1 = work of 1 min of A *30+work in 1 min of B*x.
1 = (2/75)*30+(1/45)*x.
x/45 = 1-60/75 = 15/75 = 1/5.
x = (1/5)*45 = 9 mins.
Since cistern is one which is to be filled by A & B.
Hence 1 = work of 1 min of A *30+work in 1 min of B*x.
1 = (2/75)*30+(1/45)*x.
x/45 = 1-60/75 = 15/75 = 1/5.
x = (1/5)*45 = 9 mins.
Aka said:
9 years ago
Yet simpler approach:
A's 1 minute work = 2/75,
B's 1 minute work = 1/45,
Let B be opened for first x minutes while A is opened for entire 30 mins.
=> 2/75 * 30 + 1/45 * x = 1.
By solving it, we get the answer.
A's 1 minute work = 2/75,
B's 1 minute work = 1/45,
Let B be opened for first x minutes while A is opened for entire 30 mins.
=> 2/75 * 30 + 1/45 * x = 1.
By solving it, we get the answer.
(1)
Anitha said:
3 years ago
A = 371/2
= 75/2 LCM of A & B(225) A=6, B=5
B=45.
Pipe is opened for 30 minutes so,
A * 30 = 6 * 30 = 180.
We need B's time,
Total time - A's time = 225 - 180.
= 45.
Time = C/E = 45/5.
=9 mins.
= 75/2 LCM of A & B(225) A=6, B=5
B=45.
Pipe is opened for 30 minutes so,
A * 30 = 6 * 30 = 180.
We need B's time,
Total time - A's time = 225 - 180.
= 45.
Time = C/E = 45/5.
=9 mins.
(88)
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