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 2 of 9.
Clinton said:
5 years ago
((A - Time of closing)/A) * B = Answer.
Here,
A=75/2.
Time of Closing the pipe=30.
B = 45.
So, Apply in the formula.
(((75/2)-30)/(75/2)) * 45 = 9 (Answer).
Here,
A=75/2.
Time of Closing the pipe=30.
B = 45.
So, Apply in the formula.
(((75/2)-30)/(75/2)) * 45 = 9 (Answer).
(2)
Narula said:
7 years ago
The much simple way is the LCM method.
LCM of (75/2 and 45) = 225, assume this is a total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
LCM of (75/2 and 45) = 225, assume this is a total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
(1)
Abhijeet Pol said:
8 years ago
Much simple way is the LCM method.
LCM of (75/2 and 45) = 225, assume this is total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
LCM of (75/2 and 45) = 225, assume this is total unit of work.
Thus, efficiency of A = [(225)/(75/2)] = 6 units/min and B = 225/45 = 5 units/min.
Since A works for 30 mins, he will finish = 6 x 30 = 180 units.
Remaining = 225 - 180 = 45 units to be completed by B in = (45/5) = 9 mins.
(1)
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)
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)
Nilesh12 said:
6 years ago
Formula(1 - total time/2nd pipe time)*1st pipe time.
then,
(1-30/75/2)*45.
Means,
(1-60/75)*45.
(75-60/75)*45.
(1/5)*45.
Answer 9.
then,
(1-30/75/2)*45.
Means,
(1-60/75)*45.
(75-60/75)*45.
(1/5)*45.
Answer 9.
(1)
Karansingh said:
6 years ago
There is a shorcut trick here.
Find lcm of 75 and 45.
Lcm is 225.
Divide by 75 and 45 to lcm.
Hereafter dividing lcm you will get 3 and 5.
Now given is that cistern will be filled in half hour means 30 min if b turned off.
They do 30/5 = 6.
Now add 3+6 = 9 (why to add this because we asked that if it is closed then we have to add that 3+6).
Find lcm of 75 and 45.
Lcm is 225.
Divide by 75 and 45 to lcm.
Hereafter dividing lcm you will get 3 and 5.
Now given is that cistern will be filled in half hour means 30 min if b turned off.
They do 30/5 = 6.
Now add 3+6 = 9 (why to add this because we asked that if it is closed then we have to add that 3+6).
(1)
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)
Hemant said:
9 years ago
In 1min=2/75 part of the cistern is filled by pipe A.
And In 1 min=1/45 part of the cistern is filled by pipe B.
Now,
Pipe A worked for 30 min and pipe B work for x min we suppose,
To fill complete cistern = 1 part.
2/75 * 30 + 1/45 * x = 1,
12/15 + x/45 = 1,
(36 + x)/45 = 1,
x = 9 min.
And In 1 min=1/45 part of the cistern is filled by pipe B.
Now,
Pipe A worked for 30 min and pipe B work for x min we suppose,
To fill complete cistern = 1 part.
2/75 * 30 + 1/45 * x = 1,
12/15 + x/45 = 1,
(36 + x)/45 = 1,
x = 9 min.
Naidu said:
9 years ago
How 30 came after 2/75?
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