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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1055
GPA: 3.39
A certain liquid leaks out of a container at the rate of k liters for
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06 Feb 2023, 21:49
06 Feb 2023, 21:49
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Expert Reply
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Question Stats:
80% (01:45) correct
20% (02:15) wrong based on 5 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) \(\frac {6xy}{k}\)
A) \(\frac {ky}{6x}\)
B) \(\frac {6x}{ky}\)
C) \(\frac {6k}{xy}\)
D) \(\frac {6ky}{x}\)
E) \(\frac {6xy}{k}\)
Re: A certain liquid leaks out of a container at the rate of k liters for
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08 Feb 2023, 14:28
08 Feb 2023, 14:28
1
Think of this one in terms of the rate in liters per hour: k/x
So, in y hours, the number of liters would be k/x * y
SInce we're looking at the total cost per liter, simply multiply the whole thing times 6.
6ky/x
Answer: D
So, in y hours, the number of liters would be k/x * y
SInce we're looking at the total cost per liter, simply multiply the whole thing times 6.
6ky/x
Answer: D
Re: A certain liquid leaks out of a container at the rate of k liters for
[#permalink]
08 Feb 2023, 15:40
08 Feb 2023, 15:40
1
Expert Reply
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
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
