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Suppose y=k/x, where k is a constant
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09 Jul 2021, 08:31
09 Jul 2021, 08:31
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Question Stats:
94% (00:50) correct
5% (03:16) wrong based on 18 sessions
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Suppose \( y=\frac{k}{x}\), where k is a constant. If \(y=\frac{2}{9}\) when x=540, what is the value of y when x=360 ?
Show: :: OA
1/3
Intern
Joined: 11 Jan 2021
Posts: 27
Given Kudos: 1275
GRE 1: Q167 V159
Re: Suppose y=k/x, where k is a constant
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09 Jul 2021, 09:36
09 Jul 2021, 09:36
1
\(y = \frac{k}{x}\)
At, \(y = \frac{2}{9}, x = 540\)
Thus, \(k = \frac{2}{9} * 540 = 120\)
At \(x = 360\),
\(y = \frac{120}{360} = \frac{1}{3}\)
At, \(y = \frac{2}{9}, x = 540\)
Thus, \(k = \frac{2}{9} * 540 = 120\)
At \(x = 360\),
\(y = \frac{120}{360} = \frac{1}{3}\)
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
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Re: Suppose y=k/x, where k is a constant
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11 Nov 2021, 12:11
11 Nov 2021, 12:11
1
Carcass wrote:
Suppose \( y=\frac{k}{x}\), where k is a constant. If \(y=\frac{2}{9}\) when x=540, what is the value of y when x=360 ?
Show: :: OA
1/3
Given: The pair of values \(x = 540\) and \(y = \frac{2}{9}\) is a solution to the equation \(y=\frac{k}{x}\)
Plug those values into the equation to get: \(\frac{2}{9} =\frac{k}{540}\)
Cross multiply: \(9k = 1080\)
Solve: \(k = \frac{1080}{9}=120\)
So, the original equation is \(y=\frac{120}{x}\)
Finally, to answer the question, we'll plug in \(x = 360\) to get: \(y=\frac{120}{360}=\frac{1}{3}\)
Answer: 1/3
