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27 Dec 2013, 05:41
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D
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Difficulty:
5%
(low)
Question Stats:
84% (01:07) correct
16%
(01:21)
wrong
based on 976
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Some computers at a certain company are Brand X and the rest are Brand Y. If the ratio of the number of Brand Y computers to the number of Brand X computers at the company is 5 to 6, how many of the computers are Brand Y?
(1) There are 80 more Brand X computers than Brand Y computers at the company.
(2) There is a total of 880 computers at the company.
(1) There are 80 more Brand X computers than Brand Y computers at the company.
(2) There is a total of 880 computers at the company.
30 Dec 2013, 05:41
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SOLUTION
Some computers at a certain company are Brand X and the rest are Brand Y. If the ratio of the number of Brand Y computers to the number of Brand X computers at the company is 5 to 6, how many of the computers are Brand Y?
Given that \(\frac{y}{x}=\frac{5}{6}\) we need to find the value of y.
(1) There are 80 more Brand X computers than Brand Y computers at the company --> \(x = 80 + y\). We have two distinct linear equations (\(\frac{y}{x}=\frac{5}{6}\) and \(x = 80 + y\)) with two unknowns, thus we can solve for both. Sufficient.
(2) There is a total of 880 computers at the company --> \(x + y=880\). We have two distinct linear equations (\(\frac{y}{x}=\frac{5}{6}\) and \(x + y=880\)) with two unknowns, thus we can solve for both. Sufficient.
Answer: D.
Some computers at a certain company are Brand X and the rest are Brand Y. If the ratio of the number of Brand Y computers to the number of Brand X computers at the company is 5 to 6, how many of the computers are Brand Y?
Given that \(\frac{y}{x}=\frac{5}{6}\) we need to find the value of y.
(1) There are 80 more Brand X computers than Brand Y computers at the company --> \(x = 80 + y\). We have two distinct linear equations (\(\frac{y}{x}=\frac{5}{6}\) and \(x = 80 + y\)) with two unknowns, thus we can solve for both. Sufficient.
(2) There is a total of 880 computers at the company --> \(x + y=880\). We have two distinct linear equations (\(\frac{y}{x}=\frac{5}{6}\) and \(x + y=880\)) with two unknowns, thus we can solve for both. Sufficient.
Answer: D.
General Discussion
10 Jan 2014, 11:17
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albany09
We're given: X + Y = Total Computers and 5Y = 6X, with this we can solve for X or for Y.. SO, we only need to solve for ONE variable to answer the question
1) Solves for X and Y given our second equation. Thus, this is sufficient.
2) Solves for TC, thus, this is sufficient since we have a fixed relation between the # of X to Y computers.
So, the answer is D.
