A tank is emptied everyday at a fixed time point. Immediately thereafter, either pump A or pump B or both start working until the tank is full. On Monday, A alone completed filling the tank at 8 pm. On Tuesday, B alone completed filling the tank at 6 pm. On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the tank. At what time was the tank filled on Thursday if both pumps were used simultaneously all along?
Started 10 months ago by Shashank in
Explanatory Answer
Given that a tank is emptied everyday at a fixed time point and after that either pump A or B or
both start working to fill the tank.
On Monday, A alone completed filling the tank at 8 pm.
On Tuesday, B alone completed filling the tank at 6 pm.
On Wednesday, A alone worked till 5 pm, and then B worked alone from 5 pm to 7 pm, to fill the
tank.
We have to find at what time will the tank filled on thursday if both were used simultaneously all
along
A is doing 3 hours less work on Wednesday which B completes in 2 hours therefore B = 1.5 A
⟹ (n)/(n-2) = 3/2
⟹ 2n = 3n – 6
⟹ n = 6
A takes 6 hours and B takes 4 hours or the tank is closed at 2 pm
When they are together open
⟹ 1/6 + 1/4 = (2+3)/12
⟹ Rate is 5/12 or they can fill the entire tank in 12/5 hours which is 2 hours 24 minutes
Starting from 2 pm, they fill the entire tank by 4:24 pm
Here the break through is figuring out the ratio or efficiency of the rate at which they fill is B : A
= 3 : 2
The conventional way is
⟹ (n−3)/n + 2/(n-2) = 1
n-3 hours at the rate of 1/𝑛 per hour
2 hours at the rate 1/(𝑛−2) per hour
Solving this we get n = 6 and we can find solution using this n.
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