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A certain liquid leaks out of a container at the rate of k liters for

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A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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New post 14 Jan 2010, 07:01
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A certain liquid leaks out of a container at the rate of k liters for every x hours. If the liquid costs $6 per liter, what is the cost, in dollars, of the amount of the liquid that will leak out in y hours?

A) \(\frac {ky}{6x}\)

B) \(\frac {6x}{ky}\)

C) \(\frac {6k}{xy}\)

D) \(\frac {6ky}{x}\)

E) \(\frac {6xy}{k}\)
[Reveal] Spoiler: OA

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Re: A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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New post 14 Jan 2010, 07:27
lts lost per hour = k/x
lts lost in y hours = k*y/x
total cost = 6ky/x

hence D

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New post 14 Jan 2010, 12:33
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The way I see it

If it loses k liters in x hours, then in 1 hour it loses k/x. Therefore, in y hours it would lose (k/x) * y

Therefore total cost = 6ky/x

D is the answer

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Re: A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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zaarathelab wrote:
A certain liquid leaks out of a container at the rate of k liters for every x hours. If the
liquid costs $6 per liter, what is the cost, in dollars, of the amount of the liquid that will
leak out in y hours?

A) \(\frac {ky}{6x}\)

B) \(\frac {6x}{ky}\)

c) \(\frac {6k}{xy}\)

d) \(\frac {6ky}{x}\)

e) \(\frac {6xy}{k}\)



With this problem solving question it would be easier to plug in numbers.

Let's say that:
K=5
X=1
6=6
Y=10

So the liquid leaks out at a rate of 5l per hour, every litre cost 6$ and every hour costs 30$.
If Y= 10 the cost of the liquid per hour (30$) multiplied by Y(10) equals 300.
Our target is 300.

Plug in the numbers in the answer choices and the answer that yelds 300 will be the right one.

A \(\frac {ky}{6x}\) = 5(10)/6= 8.333333

B \(\frac {6x}{ky}\) = 6/5(10)= 0.12

C \(\frac {6k}{xy}\) = 6(5)/10 = 3

D \(\frac {6ky}{x}\) = 6(5)(10)/1 = 300 CORRECT

(we could stop at D but if you have time go ahead)

E \(\frac {6xy}{k}\) = 60/5 = 12

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A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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New post 31 Dec 2014, 07:48
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For these questions, start with the only value that is NOT a ratio and start canceling units. In this case, that's 'y' hours.

\(y \ hours * \frac{ k \ liters }{x hours} *\frac{$6}{liter} = \frac{6yk}{x}.\)

Answer: D

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Re: A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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New post 16 Jan 2016, 06:51
Let's say:
K=2
X=3
Y=5
leaking every an hour will be 6*2/3 , and after Y hours: (6*2)*5/3. Plugging in our variables we have (K*6*Y)/X. So answer choice is D.

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Re: A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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New post 31 Oct 2017, 16:39
zaarathelab wrote:
A certain liquid leaks out of a container at the rate of k liters for every x hours. If the liquid costs $6 per liter, what is the cost, in dollars, of the amount of the liquid that will leak out in y hours?

A) \(\frac {ky}{6x}\)

B) \(\frac {6x}{ky}\)

C) \(\frac {6k}{xy}\)

D) \(\frac {6ky}{x}\)

E) \(\frac {6xy}{k}\)


The leak rate is k/x; thus, in y hours, ky/x liters will leak out, which will cost (6)(ky/x) = 6ky/x dollars.

Answer: D
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Re: A certain liquid leaks out of a container at the rate of k liters for [#permalink]

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New post 02 Nov 2017, 20:53
Answer should be D

lost per hour = k/x
lost in y hours = k*y/x
total cost = 6ky/x
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Re: A certain liquid leaks out of a container at the rate of k liters for   [#permalink] 02 Nov 2017, 20:53
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