There are three inlet taps A, B and C in a tank. They can fill the tank in 5 h, 30 h and 105 h respectively, if
opened individually. The tank is empty. At first, all the 3 taps are opened simultaneously. After one hour,
tap C is closed and A and B are kept running. After another one hour, tap B is also closed. The remaining
portion of the tank was filled by tap A alone. What percentage of the tank was filled by tap A in this process
(answer to the nearest percentage)?
1. 92%
2. 82%
3. 10%
4. 89%
Answers
A fills 1/5 part in 1 hr
A fills 2/5 part in 2 hrs
B fills full in 30 hrs
B fills 1/30 part in 1 hr
B fills 1/15 part in 2 hrs
C fills full in 105 hrs
C fills 1/105 part in 1 hour
total filled part = 2/5 + 1/15 + 1/105
= (42 + 7 + 1)/105 = 50/105 = 10/21
empty part = 1 - 10/21 = (21 - 10)/21 = 11/21
the part filled by A = 2/5 + 11/21 = (42 + 55)/105 = 97/105
percent part filled by A alone = (97/105) × 100 = 92.38%
there option 1 is correct.
Answer:calculate the fraction of the tank to be filled by the taps respectively,
Tap A=5hours,in one hour the fraction of the tank filled is 1/5
Tap B=30 hours
1/30 of the rank is filled
Tap C=105hours
1/1050of the tank will be filled in one hour
When both taps are opened in one hour, the fraction of water in the tank will be
1/5+1/30+1/105
=51/210
After tap C is closed, fraction of the tank to be filled in one hour us
1/5+1/30=7/30
Thus after two hours the fraction of the tank already filled will be,,
51/210+7/30
=100/210
=10/21
Remaining portion of the tank to be filled will be,
1-10/21
=11/21,this is to be filled by tap A.
Total fraction filled by Tap A is
1/5+1/5+11/21
=97/105
Percentage is 97/105x100
=92.38
=92%
The answer is
1.92%
Step-by-step explanation: