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05 Aug 2026, 12:10
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Ava’s desk drawer contains only blue, green, and silver tokens, with at least one token of each color. Are there more green tokens than blue tokens in the drawer?
(1) If \(\frac{1}{4}\) of the blue tokens were removed from the drawer, the probability of selecting a green token would be \(60\%\).
(2) If \(\frac{2}{5}\) of the green tokens were removed from the drawer, the probability of selecting a blue token would be \(40\%\).
(1) If \(\frac{1}{4}\) of the blue tokens were removed from the drawer, the probability of selecting a green token would be \(60\%\).
(2) If \(\frac{2}{5}\) of the green tokens were removed from the drawer, the probability of selecting a blue token would be \(40\%\).
05 Aug 2026, 12:10
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Official Solution:
Ava’s desk drawer contains only blue, green, and silver tokens, with at least one token of each color. Are there more green tokens than blue tokens in the drawer?
Let \(B\) be the number of blue tokens, \(G\) be the number of green tokens, and \(S\) be the number of silver tokens.
The question asks whether \(G > B\).
(1) If \(\frac{1}{4}\) of the blue tokens were removed from the drawer, the probability of selecting a green token would be \(60\%\).
After \(\frac{1}{4}\) of the blue tokens are removed, \(\frac{3B}{4}\) blue tokens remain. So this statement gives:
\(\frac{G}{\frac{3B}{4} + G + S} = \frac{3}{5}\)
\(5G = \frac{9B}{4} + 3G + 3S\)
\(2G = \frac{9B}{4} + 3S\)
\(G = \frac{9B}{8} + \frac{3S}{2}\)
Since \(\frac{9B}{8}\) is already greater than \(B\), and since \(S\) is positive, \(G\) must be greater than \(B\).
Sufficient.
(2) If \(\frac{2}{5}\) of the green tokens were removed from the drawer, the probability of selecting a blue token would be \(40\%\).
After \(\frac{2}{5}\) of the green tokens are removed, \(\frac{3G}{5}\) green tokens remain. So this statement gives:
\(\frac{B}{B + \frac{3G}{5} + S} = \frac{2}{5}\)
\(5B = 2B + \frac{6G}{5} + 2S\)
\(3B = \frac{6G}{5} + 2S\)
\(15B = 6G + 10S\)
This is not enough to say whether \(G > B\).
For example, if \(B = 4\), \(G = 5\), and \(S = 3\), then \(15B = 60\) and \(6G + 10S = 60\), so the statement is satisfied and \(G > B\).
But if \(B = 6\), \(G = 5\), and \(S = 6\), then \(15B = 90\) and \(6G + 10S = 90\), so the statement is satisfied and \(G < B\).
Not sufficient.
Answer: A
Ava’s desk drawer contains only blue, green, and silver tokens, with at least one token of each color. Are there more green tokens than blue tokens in the drawer?
Let \(B\) be the number of blue tokens, \(G\) be the number of green tokens, and \(S\) be the number of silver tokens.
The question asks whether \(G > B\).
(1) If \(\frac{1}{4}\) of the blue tokens were removed from the drawer, the probability of selecting a green token would be \(60\%\).
After \(\frac{1}{4}\) of the blue tokens are removed, \(\frac{3B}{4}\) blue tokens remain. So this statement gives:
\(\frac{G}{\frac{3B}{4} + G + S} = \frac{3}{5}\)
\(5G = \frac{9B}{4} + 3G + 3S\)
\(2G = \frac{9B}{4} + 3S\)
\(G = \frac{9B}{8} + \frac{3S}{2}\)
Since \(\frac{9B}{8}\) is already greater than \(B\), and since \(S\) is positive, \(G\) must be greater than \(B\).
Sufficient.
(2) If \(\frac{2}{5}\) of the green tokens were removed from the drawer, the probability of selecting a blue token would be \(40\%\).
After \(\frac{2}{5}\) of the green tokens are removed, \(\frac{3G}{5}\) green tokens remain. So this statement gives:
\(\frac{B}{B + \frac{3G}{5} + S} = \frac{2}{5}\)
\(5B = 2B + \frac{6G}{5} + 2S\)
\(3B = \frac{6G}{5} + 2S\)
\(15B = 6G + 10S\)
This is not enough to say whether \(G > B\).
For example, if \(B = 4\), \(G = 5\), and \(S = 3\), then \(15B = 60\) and \(6G + 10S = 60\), so the statement is satisfied and \(G > B\).
But if \(B = 6\), \(G = 5\), and \(S = 6\), then \(15B = 90\) and \(6G + 10S = 90\), so the statement is satisfied and \(G < B\).
Not sufficient.
Answer: A
