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A school equipment closet contains only red, blue, and gray practice jerseys, with different numbers of jerseys of each color. If the closet contains 90 jerseys in total, with at least 20 jerseys of each color, are there more red jerseys than blue jerseys?
(1) The number of red jerseys is \(\frac{2}{7}\) of the total number of blue and gray jerseys.
(2) The total number of red and blue jerseys is equal to the number of gray jerseys.
(1) The number of red jerseys is \(\frac{2}{7}\) of the total number of blue and gray jerseys.
(2) The total number of red and blue jerseys is equal to the number of gray jerseys.
06 Aug 2026, 05:59
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Official Solution:
A school equipment closet contains only red, blue, and gray practice jerseys, with different numbers of jerseys of each color. If the closet contains 90 jerseys in total, with at least 20 jerseys of each color, are there more red jerseys than blue jerseys?
Let \(R\), \(B\), and \(G\) be the numbers of red, blue, and gray jerseys, respectively.
Given:
• \(R + B + G = 90\)
• Each of \(R\), \(B\), and \(G\) is at least 20, and the three numbers are all different.
The question asks whether \(R > B\).
(1) The number of red jerseys is \(\frac{2}{7}\) of the total number of blue and gray jerseys.
So:
\(R = \frac{2}{7} * (B + G)\)
Since \(B + G = 90 - R\):
\(R = \frac{2}{7} * (90 - R)\)
\(7R = 180 - 2R\)
\(R = 20\)
Since each color has at least 20 jerseys and the three numbers are all different, \(B\) cannot be 20. Therefore, \(B\) must be greater than 20.
So \(R < B\), and the answer to the question is NO.
Sufficient.
(2) The total number of red and blue jerseys is equal to the number of gray jerseys.
So:
\(R + B = G\)
Since \(R + B + G = 90\):
\(G + G = 90\)
\(G = 45\)
So:
\(R + B = 45\)
This does not determine whether \(R > B\).
For example, if \(R = 24\) and \(B = 21\), then \(R > B\).
But if \(R = 21\) and \(B = 24\), then \(R < B\).
Not sufficient.
Answer: A
A school equipment closet contains only red, blue, and gray practice jerseys, with different numbers of jerseys of each color. If the closet contains 90 jerseys in total, with at least 20 jerseys of each color, are there more red jerseys than blue jerseys?
Let \(R\), \(B\), and \(G\) be the numbers of red, blue, and gray jerseys, respectively.
Given:
• \(R + B + G = 90\)
• Each of \(R\), \(B\), and \(G\) is at least 20, and the three numbers are all different.
The question asks whether \(R > B\).
(1) The number of red jerseys is \(\frac{2}{7}\) of the total number of blue and gray jerseys.
So:
\(R = \frac{2}{7} * (B + G)\)
Since \(B + G = 90 - R\):
\(R = \frac{2}{7} * (90 - R)\)
\(7R = 180 - 2R\)
\(R = 20\)
Since each color has at least 20 jerseys and the three numbers are all different, \(B\) cannot be 20. Therefore, \(B\) must be greater than 20.
So \(R < B\), and the answer to the question is NO.
Sufficient.
(2) The total number of red and blue jerseys is equal to the number of gray jerseys.
So:
\(R + B = G\)
Since \(R + B + G = 90\):
\(G + G = 90\)
\(G = 45\)
So:
\(R + B = 45\)
This does not determine whether \(R > B\).
For example, if \(R = 24\) and \(B = 21\), then \(R > B\).
But if \(R = 21\) and \(B = 24\), then \(R < B\).
Not sufficient.
Answer: A
