Q-26) Three pipes A, B, and C can fill the tank in 10 hours, 20 hours and 40 hours respectively. In the beginning all of them are opened simultaneously. After 2 hours, tap C is closed and A and B are kept running. After the 4th hour, tap B is also closed. The remaining work is done by tap A alone. What is the percentage of the work done by tap A alone?
solve with equation please
Answers
Answer:
pipes 1: . 3_9/17 hour
pipes 2 :9 min
pipes 3 : 15 min
Step-by-step explanation:
1. Pipes A and B can fill a tank in 5 and 6 hours respectively. Pipe C can empty it in 12 hours. If all the three pipes are opened together, then the tank will be filled in:
A.
4
1
2
hours B.
1
13
17
hours
C.
3
9
17
hours D.
2
8
11
hours
Answer: Option C
Explanation:
Solution 1
Pipes A and B can fill the tank in 5 and 6 hours respectively. Therefore,
part filled by pipe A in 1 hour
=
1
5
part filled by pipe B in 1 hour =
1
6
Pipe C can empty the tank in 12 hours. Therefore,
part emptied by pipe C in 1 hour
=
1
12
Net part filled by Pipes A,B,C together in 1 hour
=
1
5
+
1
6
−
1
12
=
17
60
i.e., the tank can be filled in
60
17
=
3
9
17
hours.
Solution 2
LCM
(
5
,
6
,
12
)
=
60
Suppose capacity of the tank is
60
litre. Then,
Quantity filled by pipe A in 1 hour
=
60
5
=
12
litre.
Quantity filled by pipe B in 1 hour
=
60
6
=
10
litre.
Quantity emptied by pipe C in 1 hour
=
60
12
=
5
litre.
Quantity filled in 1 hour if all the pipes are opened together
=
12
+
10
−
5
=
17
litre.
Required time
=
60
17
=
3
9
17
hours.
Pipe A alone can fill the cistern in
37
1
2
=
75
2
minutes. Since it was open for
30
minutes, part of the cistern filled by pipe A
=
2
75
×
30
=
4
5
So the remaining
1
5
part is filled by pipe B.
Pipe B can fill the cistern in 45 minutes. So, time required to fill
1
5
part
=
45
5
=
9
minutes
Suppose the first pipe alone can fill the tank in
x
hours. Then,
second pipe alone can fill the tank in
(
x
−
5
)
hours,
third pipe alone can fill the tank in
(
x
−
5
)
−
4
=
(
x
−
9
)
hours.
Part filled by first pipe and second pipe together in 1 hr
= Part filled by third pipe in 1 hr
⇒
1
x
+
1
x
−
5
=
1
x
−
9