Two pumps, working alone, can fill an empty pool in 10 hours and 15 ho

Question

Two pumps, working alone, can fill an empty pool in 10 hours and 15 hours, respectively. The first pump initially started alone for h hours; after which the second pump was also started. If it took a total of 7 hours for the pool to be filled completely by the both the pumps, what is the value of h?

(A) 2.00
(B) 2.50
(C) 3.00
(D) 3.30
(E) 4.00

(A) 2.00
(B) 2.50
(C) 3.00
(D) 3.30
(E) 4.00

ramlala · Accepted Answer

Pump X rate to fill the pool is  \frac{1}{10}  or  \frac{3}{30}

Pump Y rate to fill the pool is  \frac{1}{15}  or  \frac{2}{30}

Both Pump X & Y combined  \frac{3}{30}+\frac{2}{30} = \frac{5}{30}

Start with Choice A
h=2 hours & Both = 5 hours
i.e.  \frac{3}{30}*2 + \frac{5}{30}*5 = \frac{31}{30}   Eliminate

Choice B 
h=2.5 hours & Both = 4.5 hours
i.e.  \frac{3}{30}*2.5 + \frac{5}{30}*4.5 = \frac{30}{30}   That is our answer

Mislead · Answer

The timing of the 2 pumps are 10hr and 15hr, lets convert them in ratio:
2:3 and hence the efficiency will be in the ratio of 3:2, both combined there efficiency will be of 5units.
Timing = 2:3
Efficiency= 3:2:5 (5 being the combined efficiency)
Timing = 5:3 (choosing the efficiency of the 1st pump since h is related to 1st pump)
since the timing of 1st pump is 10, both working together will take 6hr but its mentioned that it takes 7hours to complete the work so additional 5unit of work has to be completed in 1hr and working alone for 2.5hr 1st pump will create a buffer of 5unit of work.

IMO B

hykhan · Answer

I found that smart plugging of numbers helped. If we started with A (2 hours), then pump A completed 2/10 = 1/5 of the work in 2 hours. So there is 4/5 work remaining. the combined rate is 1/10 + 1/15 = 1/6. So at a rate of 1/6, it would take 6*4/5 = 24/5 = 4.8 hours to complete the remaining work. 4.8 + 2 = 6.8 which is pretty close to our required number of 7. I would be pretty comfortable with guessing B at this point, but if I had time I'd probably test that value next.

2.5 hours means 25% of the work is done in 2.5 hours by pump A, and 75% of the work remains.

At a combined rate of 1/6, 3/4 of the work is accomplished at 6*3/4 = 4.5 hours. 2.5 + 4.5 = 7 and our math checks out.

B

2.5 hours means 25% of the work is done in 2.5 hours by pump A, and 75% of the work remains.

At a combined rate of 1/6, 3/4 of the work is accomplished at 6*3/4 = 4.5 hours. 2.5 + 4.5 = 7 and our math checks out.

B

ace316 · Answer

Solution:

(Here the unit of Rate is Pool / hour)

Pump 1

1 Pool ⇌ 10 hours

1 hour ⇌ 1/10 Pool => R1 = (1/10 Pool/hr)

Pump 2

1 Pool ⇌ 15 hours

1 hour ⇌ 1/15 Pool => R2 = (1/15 Pool/hr)

Pump 1 + Pump 2 (Combined rate) 
i.e. R12 = R1 + R2 = 1/10 + 1/15 =>R12 = 1/6 Pool/hr
=> 1 Pool ⇌ 6 hrs

Pump 1 worked for h hours alone.

=> Pool Filled = 1/10 * h

Pump 1 and Pump 2 worked together for (7-h) hours.

=> Pool Filled = 1/6 * 7-h

Therefore total work,

{h/10} + {(7-h)/6} = 1 (Pool)

=> h = 5/2 or 2.5 hours (Answer Choice B)

AJSIPA · Answer

Pump A works for full 7 hours and pump 2 works for only 7-h hours 
pump A rate in 1 hr=1/10 pool/hr filled
pump B rate in 1 hr=1/15 pool/hr filled
1/10*7+1/15*(7-h)=1
h=2.5

Sabuj40040 · Answer

LCM(10,15)=30
First Pump efficiency 30/10=3 unit 
Second Pump efficiency 30/15=2 unit
Time taken by first pump x hours, therefore second pump takes (7-x) hours
ATQ, 
 3*x+(3+2)*(7-x)=30
 or, x= 2.5

Two pumps, working alone, can fill an empty pool in 10 hours and 15 ho

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Two pumps, working alone, can fill an empty pool in 10 hours and 15 ho

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