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A
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Difficulty:
95%
(hard)
Question Stats:
35% (02:46) correct
65%
(02:25)
wrong
based on 65
sessions
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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 1/4 of the blue tokens were removed from the drawer, the probability of selecting a green token would be 60%.
(2) If 2/5 of the green tokens were removed from the drawer, the probability of selecting a blue token would be 40%.
(1) If 1/4 of the blue tokens were removed from the drawer, the probability of selecting a green token would be 60%.
(2) If 2/5 of the green tokens were removed from the drawer, the probability of selecting a blue token would be 40%.
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10 May 2026, 02:30
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GMAT Club 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 1/4 of the blue tokens were removed from the drawer, the probability of selecting a green token would be 60%.
After 1/4 of the blue tokens are removed, 3B/4 blue tokens remain. So this statement gives:
G/(3B/4 + G + S) = 3/5
5G = 9B/4 + 3G + 3S
2G = 9B/4 + 3S
G = 9B/8 + 3S/2
Since 9B/8 is already greater than B, and since S is positive, G must be greater than B.
Sufficient.
(2) If 2/5 of the green tokens were removed from the drawer, the probability of selecting a blue token would be 40%.
After 2/5 of the green tokens are removed, 3G/5 green tokens remain. So this statement gives:
B/(B + 3G/5 + S) = 2/5
5B = 2B + 6G/5 + 2S
3B = 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 1/4 of the blue tokens were removed from the drawer, the probability of selecting a green token would be 60%.
After 1/4 of the blue tokens are removed, 3B/4 blue tokens remain. So this statement gives:
G/(3B/4 + G + S) = 3/5
5G = 9B/4 + 3G + 3S
2G = 9B/4 + 3S
G = 9B/8 + 3S/2
Since 9B/8 is already greater than B, and since S is positive, G must be greater than B.
Sufficient.
(2) If 2/5 of the green tokens were removed from the drawer, the probability of selecting a blue token would be 40%.
After 2/5 of the green tokens are removed, 3G/5 green tokens remain. So this statement gives:
B/(B + 3G/5 + S) = 2/5
5B = 2B + 6G/5 + 2S
3B = 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.
General Discussion
Updated on: 10 May 2026, 22:40
Originally posted by AdarshSambare on 06 May 2026, 20:55.
Last edited by AdarshSambare on 10 May 2026, 22:40, edited 1 time in total.
Last edited by AdarshSambare on 10 May 2026, 22:40, edited 1 time in total.
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We need to determine whether green tokens (G) are more than blue tokens (B).
Prethinking:
Try to translate each condition into equations and see if we can uniquely compare G and B.
Let blue = B, green = G, silver = S.
Statement (1):
After removing 1/4 of blue, blue becomes 3B/4.
Probability of green = 60% = 3/5:
G / (3B/4 + G + S) = 3/5
Cross-multiply:
5G = 3(3B/4 + G + S)
5G = 9B/4 + 3G + 3S
2G = 9B/4 + 3S
This does not uniquely compare G and B because S is still unknown. Not sufficient alone.
Statement (2):
After removing 2/5 of green, green becomes 3G/5.
Probability of blue = 40% = 2/5:
B / (B + 3G/5 + S) = 2/5
Cross-multiply:
5B = 2(B + 3G/5 + S)
5B = 2B + 6G/5 + 2S
3B = 6G/5 + 2S
Still not sufficient alone.
Now combine (1) and (2):
From (1): 2G = 9B/4 + 3S
From (2): 3B = 6G/5 + 2S
We can eliminate S and solve relationship between G and B. Doing that gives a fixed inequality:
After simplification, it leads to G > B always.
So the question can be answered definitively with both statements.
Answer: A
Prethinking:
Try to translate each condition into equations and see if we can uniquely compare G and B.
Let blue = B, green = G, silver = S.
Statement (1):
After removing 1/4 of blue, blue becomes 3B/4.
Probability of green = 60% = 3/5:
G / (3B/4 + G + S) = 3/5
Cross-multiply:
5G = 3(3B/4 + G + S)
5G = 9B/4 + 3G + 3S
2G = 9B/4 + 3S
This does not uniquely compare G and B because S is still unknown. Not sufficient alone.
Statement (2):
After removing 2/5 of green, green becomes 3G/5.
Probability of blue = 40% = 2/5:
B / (B + 3G/5 + S) = 2/5
Cross-multiply:
5B = 2(B + 3G/5 + S)
5B = 2B + 6G/5 + 2S
3B = 6G/5 + 2S
Still not sufficient alone.
Now combine (1) and (2):
From (1): 2G = 9B/4 + 3S
From (2): 3B = 6G/5 + 2S
We can eliminate S and solve relationship between G and B. Doing that gives a fixed inequality:
After simplification, it leads to G > B always.
So the question can be answered definitively with both statements.
Answer: A
