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Two pipes can fill an empty tank separately in 24 minutes and 40 minutes respectively and a third pipe can empty 30 gallons of water per minute. If all three pipes are open, empty tanks become full in one hour. The capacity of the tank (in gallons) is:
Answer & Solution
Correct Answer:
Option
B
Let capacity of the tank = x gallons;
Part of the tank filled in 1 minute
$$\eqalign{ & = {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 \cr & {\text{Or}}, {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 = \frac{x}{{60}} \cr & {\text{Or}}, \frac{x}{{24}} + \frac{x}{{40}} - \frac{x}{{60}} = 30 \cr & {\text{Or}}, {\frac{{ {10x + 6x - 4x} }}{{240}}} = 30 \cr & {\text{Or}},\,12x = 30 \times 240 \cr & {\text{Or}},\,x = 600\,{\text{gallons}} \cr} $$
$$\eqalign{ & = {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 \cr & {\text{Or}}, {\frac{x}{{24}}} + {\frac{x}{{40}}} - 30 = \frac{x}{{60}} \cr & {\text{Or}}, \frac{x}{{24}} + \frac{x}{{40}} - \frac{x}{{60}} = 30 \cr & {\text{Or}}, {\frac{{ {10x + 6x - 4x} }}{{240}}} = 30 \cr & {\text{Or}},\,12x = 30 \times 240 \cr & {\text{Or}},\,x = 600\,{\text{gallons}} \cr} $$
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Loginwork done by all three pipe together is 601
Work done by third pipe to empty tha tank =601−(241+401)
=601−1208
=−201 [-ve sign means emptying]
∴ Volume of 201 part =30 gallons
∴ Volume of whole tank =30×20=600 gallons
Answer is 600 gallons.