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A solution leaks out of a jar at the rate of p liters for every q hour

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A solution leaks out of a jar at the rate of p liters for every q hour  [#permalink]

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New post 27 Sep 2018, 04:23
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A solution leaks out of a jar at the rate of p liters for every q hours. If the solution costs 10 dollars per liter, what is the cost, in dollars, of the amount of the solution that will leak out in r hours?

(A) pr/(10q)

(B) 10q/(pr)

(C) 10p/(qr)

(D) 10pr/q

(E) 10qr/p

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Re: A solution leaks out of a jar at the rate of p liters for every q hour  [#permalink]

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New post 27 Sep 2018, 05:15
Bunuel wrote:
A solution leaks out of a jar at the rate of p liters for every q hours. If the solution costs 10 dollars per liter, what is the cost, in dollars, of the amount of the solution that will leak out in r hours?

(A) pr/(10q)

(B) 10q/(pr)

(C) 10p/(qr)

(D) 10pr/q

(E) 10qr/p


Leaking rate p liters for every q hours
i.e. in 1 hour, Leakage = p/q litres
i.e. in r hour, Leakage = r*(p/q) litres

Cost = $10 per litre

i.e. Cost of Total Leakage = 10*r*(p/q)

Answer: Option D
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Re: A solution leaks out of a jar at the rate of p liters for every q hour  [#permalink]

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New post 27 Sep 2018, 05:19
Algebraically.

Rate = p/q
Cost = 10 dollars

Now we are requested for the cost for r hours.

So it would be (10*p*r)/q

We can also test this by plugging numbers if we say p = 2, q = 1 and r = 4

The leaked solution is 8 liters in 4 hours.

The cost of the the leakage per liter is 10 so that is 10* 8 = 80.

Answer choice D

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Re: A solution leaks out of a jar at the rate of p liters for every q hour &nbs [#permalink] 27 Sep 2018, 05:19
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A solution leaks out of a jar at the rate of p liters for every q hour

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