Avas desk drawer contains only blue, green, and silver tokens

Question

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%.

Bunuel · Accepted Answer

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.

AdarshSambare · Answer

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

SS990 · Answer

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

BongBideshini · Answer

Let blue = B, green = G, silver = S.
Need to know: Is G > B
 Statement (1) 
After removing 1/4 of blue tokens:
G /   = 3/5
5G = (9/4)B + 3G + 3S
2G = (9/4)B + 3S
G = (9/8)B + (3/2) S

Since S > 0, G > B always.
 Sufficient

Statement (2) 
B/  = 2/5
5B = 2B + (6/5)G + 2S
3B = (6/5)G + 2S
G = (5/2)B − (5/3) S
Since S can vary, G > B and G < B, both cases are possible.
 Insufficient

Ans (A)

Avas desk drawer contains only blue, green, and silver tokens

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