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mozerng
A pump takes 4 hours to fill an empty pool. If the pump increased the water level of the pool to 80 percent of its total capacity after pumping water for 1 hour and 30 minutes, what percent of the pool was empty before the pump started working?

A: 37.5%
B: 42.5%
C: 57.5%
D: 62.5%
E: 70.0%

Given:
1. A pump takes 4 hours to fill an empty pool.
2. The pump increased the water level of the pool to 80 percent of its total capacity after pumping water for 1 hour and 30 minutes.

Asked: What percent of the pool was empty before the pump started working?

1. A pump takes 4 hours to fill an empty pool.
In 1 hour pump fills = 1/4 = 25% of pool

2. The pump increased the water level of the pool to 80 percent of its total capacity after pumping water for 1 hour and 30 minutes.
In 1 hour 30 minutes pump fills = 25% * 1.5 = 37.5% of pool
The percent of the pool was fill before the pump started working = 80% - 37.5% = 42.5%
The percent of the pool was empty before the pump started working = 100% - 42.5% = 57.5%

IMO C
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Let , initially pump had = x parts filled

In 3/2 hours pump filled (4/5 - x ) parts

In 1 hour pump filled = (4-5x) *2 / 15

Now pump 's rate = 1/4 parts/hour

(4-5x) *2 / 15 = 1/4
x= 17/40

Hence , the percent of the pool that was empty before the pump started working = 1 - 17/40 = 23/40 * 100 = 57.5%

C is the answer.
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