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06 May 2026, 19:00
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Question Stats:
33% (00:10) correct
67%
(01:48)
wrong
based on 25
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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%.
The OA will be automatically revealed on Sunday 10th of May 2026 07:00:44 PM Pacific Time Zone
(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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The OA will be automatically revealed on Sunday 10th of May 2026 07:00:44 PM Pacific Time Zone
06 May 2026, 20:55
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Let:
B = number of blue tokens
G = number of green tokens
S = number of silver tokens
We need to know whether G > B.
Statement (1):
After removing 1/4 of blue tokens:
Blue remaining = 3B/4
Probability of green:
G / (3B/4 + G + S) = 3/5
Cross-multiply:
5G = 3(3B/4 + G + S)
5G = 9B/4 + 3G + 3S
2G = 9B/4 + 3S
Since S is unknown, we cannot determine whether G > B.
Not sufficient.
Statement (2):
After removing 2/5 of green tokens:
Green remaining = 3G/5
Probability of blue:
B / (B + 3G/5 + S) = 2/5
Cross-multiply:
5B = 2(B + 3G/5 + S)
5B = 2B + 6G/5 + 2S
3B = 6G/5 + 2S
Again S unknown, cannot determine relationship between G and B.
Not sufficient.
Together:
From (1):
2G = 9B/4 + 3S
→ 8G = 9B + 12S
From (2):
3B = 6G/5 + 2S
→ 15B = 6G + 10S
Solve simultaneously.
From second:
6G = 15B − 10S
→ 8G = 20B − (40/3)S
Set equal to first:
9B + 12S = 20B − (40/3)S
Bring terms together:
11B = (76/3)S
So:
B = 76S/33
Substitute into:
15B = 6G + 10S
15(76S/33) = 6G + 10S
380S/11 = 6G + 10S
270S/11 = 6G
G = 45S/11
Compare:
G = 45S/11
B = 76S/33
Convert:
G = 135S/33
Thus:
G > B
Sufficient together.
Answer: C
B = number of blue tokens
G = number of green tokens
S = number of silver tokens
We need to know whether G > B.
Statement (1):
After removing 1/4 of blue tokens:
Blue remaining = 3B/4
Probability of green:
G / (3B/4 + G + S) = 3/5
Cross-multiply:
5G = 3(3B/4 + G + S)
5G = 9B/4 + 3G + 3S
2G = 9B/4 + 3S
Since S is unknown, we cannot determine whether G > B.
Not sufficient.
Statement (2):
After removing 2/5 of green tokens:
Green remaining = 3G/5
Probability of blue:
B / (B + 3G/5 + S) = 2/5
Cross-multiply:
5B = 2(B + 3G/5 + S)
5B = 2B + 6G/5 + 2S
3B = 6G/5 + 2S
Again S unknown, cannot determine relationship between G and B.
Not sufficient.
Together:
From (1):
2G = 9B/4 + 3S
→ 8G = 9B + 12S
From (2):
3B = 6G/5 + 2S
→ 15B = 6G + 10S
Solve simultaneously.
From second:
6G = 15B − 10S
→ 8G = 20B − (40/3)S
Set equal to first:
9B + 12S = 20B − (40/3)S
Bring terms together:
11B = (76/3)S
So:
B = 76S/33
Substitute into:
15B = 6G + 10S
15(76S/33) = 6G + 10S
380S/11 = 6G + 10S
270S/11 = 6G
G = 45S/11
Compare:
G = 45S/11
B = 76S/33
Convert:
G = 135S/33
Thus:
G > B
Sufficient together.
Answer: C
07 May 2026, 02:46
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I think the answer is A
How i did it, but it took time. would be good to see an easier way given its an easy Q.
O= others silver + blue in 1 and silver + green in 2
we need to det if g>b
(1) If 1/4 of the blue tokens "b" were removed from the drawer, the probability of selecting a green token P(g)would be 60%.
so now we have 3/4b and P(g)=60/100 or 6/10=6g/(6g+4O)
so Blue of the 4 O used here can be min 1 and max 3. so originally blue would be min 4/3 or max 4. either ways green is 6 and hence b,g
so this is sufficient
(2) If 2/5 of the green tokens were removed from the drawer, the probability of selecting a blue token would be 40%. same as above
b/b+o=40% or 4/10...4(b)/4(b)+6O.. 6O can be with 1green or 5 greens..which is 3/5th of greens. so Green would have been min 5/3 or max 25/3.
In 5/3 case its lower than 4blue above and in 25/3 case is higher than blue above. so insufficent.
Hence A
How i did it, but it took time. would be good to see an easier way given its an easy Q.
O= others silver + blue in 1 and silver + green in 2
we need to det if g>b
(1) If 1/4 of the blue tokens "b" were removed from the drawer, the probability of selecting a green token P(g)would be 60%.
so now we have 3/4b and P(g)=60/100 or 6/10=6g/(6g+4O)
so Blue of the 4 O used here can be min 1 and max 3. so originally blue would be min 4/3 or max 4. either ways green is 6 and hence b,g
so this is sufficient
(2) If 2/5 of the green tokens were removed from the drawer, the probability of selecting a blue token would be 40%. same as above
b/b+o=40% or 4/10...4(b)/4(b)+6O.. 6O can be with 1green or 5 greens..which is 3/5th of greens. so Green would have been min 5/3 or max 25/3.
In 5/3 case its lower than 4blue above and in 25/3 case is higher than blue above. so insufficent.
Hence A
