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zaarathelab
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A certain liquid leaks out of a container at the rate of k liters for 14 Jan 2010, 07:01
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D
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87% (01:19) correct 13% (01:39) wrong based on 1349 sessions

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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) \(\frac {6xy}{k}\)
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D
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A certain liquid leaks out of a container at the rate of k liters for 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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A certain liquid leaks out of a container at the rate of k liters for 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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A certain liquid leaks out of a container at the rate of k liters for 14 Jan 2010, 07:27
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lts lost per hour = k/x
lts lost in y hours = k*y/x
total cost = 6ky/x

hence D
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A certain liquid leaks out of a container at the rate of k liters for 17 Dec 2014, 05:52
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zaarathelab
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 16 Jan 2016, 06:51
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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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A certain liquid leaks out of a container at the rate of k liters for 31 Oct 2017, 16:39
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zaarathelab
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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A certain liquid leaks out of a container at the rate of k liters for 02 Nov 2017, 20:53
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Answer should be D

lost per hour = k/x
lost in y hours = k*y/x
total cost = 6ky/x
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A certain liquid leaks out of a container at the rate of k liters for 05 Sep 2018, 22:50
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zaarathelab
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

Hi Scott,

I hope you're well. In an algebraic expression like the above, why wouldn't it be 6k/xy? Isn't y representative of hours? Sorry for the silly question.

Thanks in advance.
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A certain liquid leaks out of a container at the rate of k liters for 21 Nov 2019, 16:31
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Hi ScottTargetTestPrep
What is the difference between saying "k liters per x hours" and "k liters for every x hours"?

The wording in the later confused me a bit.

Thank You,
Dablu
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A certain liquid leaks out of a container at the rate of k liters for 10 Dec 2019, 09:03
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Any similar problems that only have variables in the answer choices?
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A certain liquid leaks out of a container at the rate of k liters for 13 Dec 2020, 06:48
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We're given the ratio \(\frac{k liters }{ x hours}\), which is the leak rate.

We're given the ratio \(\frac{$6 }{ liter}\), which is the cost.


\(leak rate\) * \(cost\) * \(y\) = cost of liquid that will leak out in y hours

\(\frac{$6 }{ liter}\) * \(\frac{k liters }{ x hours}\) = \(\frac{6ky}{x}\)

Answer is D.

Alternatively, we can choose easy numbers.

Let x = 1 hour
Let y = 2 hours

k = 1

Cost of the amount of liquid that will leak out in 1 hour: $6
Cost of the amount of liquid that will leak out in 2 hours: $12

We need an answer that is going to give us $12.

\(\frac{6ky}{x }= \frac{6(1)(2) }{ (1)} = 12\)

Answer is D.
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A certain liquid leaks out of a container at the rate of k liters for 16 Dec 2020, 02:25
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Bunuel, this is a Gmatprep question, please tag. Thanks!
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A certain liquid leaks out of a container at the rate of k liters for 16 Dec 2020, 02:35
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Bunuel, this is a Gmatprep question, please tag. Thanks!

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Added the tag. Thank you!
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A certain liquid leaks out of a container at the rate of k liters for 17 Dec 2021, 01:28
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A certain liquid leaks out of a container at the rate of k liters for every x hours.

Leak rate per hour = k/x
Amount of the liquid that will leak out in y hours =\(\frac{ k}{x }* y\)

Since the liquid cost $6/ liter, the total cost = \(\frac{ k}{x }* y * 6\) = 6ky/x

Option D is the answer.

Thanks,
Clifin J Francis,
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A certain liquid leaks out of a container at the rate of k liters for 20 Feb 2024, 15:07
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