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An jar is filled with red, white, and blue tokens that are e

by hazelnut01 » Tue Apr 04, 2017 5:58 pm
A jar is filled with red, white, and blue tokens that are equivalent except for their color. The chance of randomly selecting a red token, replacing it, then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21

Source: Manhattan Prep
Official Answer : D
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by Jay@ManhattanReview » Wed Apr 05, 2017 12:50 am
ziyuenlau wrote:A jar is filled with red, white, and blue tokens that are equivalent except for their color. The chance of randomly selecting a red token, replacing it, then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21

Source: Manhattan Prep
Official Answer : D
Hi ziyuenlau,

Say there are r red, w white, and b blue tokens.

As per the question,

[r/(r+w+b)] * [w/(r+w+b)] = [b/(r+w+b)]

=> [(rw)/(r+w+b)] = b

=> rw = b(r+w+b)

Since r, w and b are multiples of 3, let's assume that r = 3a, w = 3b and b = 3c; where a, b and c are positive integers.

Thus, rw = b(r+w+b) => 3a*3b = 3c(3a+3b+3c)

=> ab = c(a+b+c)

We see that in RHS, c is multiplied to (a+b+c), the sum of a, b, and c, thus, the value of c must be the least among a, b and c.

Say c = 1

Thus, ab = (a+b+1)

Now since a and b must be greater than c = 1, the minimum value they can take = 2

Say a = b = 2

Thus, ab = (a+b+1) => LHS = ab = 2*2 = 4 and RHS = (a+b+1) = (2+2+1) = 5. They are not equal. So this is not possible!

Say a = 2 and b = 3

Thus, ab = (a+b+1) => LHS = ab = 2*3 = 6 and RHS = (a+b+1) = (2+3+1) = 6. They are equal. Thus works.

Thus, a = 2, b = 3 and c = 1.

=> # of red token = 3a = 3*2 = 6
=> # of white token = 3b = 3*3 = 9
=> # of blue token = 3c = 3*1 = 3

=> Total number of tokens = 6 + 9 + 3 = 18.

The correct answer: D

Hope this helps!

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide

-Jay
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by GMATGuruNY » Wed Apr 05, 2017 3:22 am
ziyuenlau wrote:A jar is filled with red, white, and blue tokens that are equivalent except for their color. The chance of randomly selecting a red token, replacing it, then randomly selecting a white token is the same as the chance of randomly selecting a blue token. If the number of tokens of every color is a multiple of 3, what is the smallest possible total number of tokens in the jar?

(A) 9
(B) 12
(C) 15
(D) 18
(E) 21
We can PLUG IN THE ANSWERS, which represent the total number of tokens.
When the correct answer choice is plugged in, P(red then white) = P(blue).
Since the question stem asks for the SMALLEST POSSIBLE TOTAL, start with the smallest answer choice.

A: 9 tokens, implying 3 red, 3 white, 3 blue
P(red then white) = 3/9 * 3/9 = 1/9.
P(blue) = 3/9 = 1/3.
Since P(red then white) < P(blue), eliminate A.

To increase the value of P(red then white), R+W must constitute a greater portion of the total.
Thus, as we test the remaining answer choices, we should keep increasing the value of R+W.

B: 12 tokens, allowing for 6 red, 3 white, 3 blue
P(red then white) = 6/12 * 3/12 = 1/8.
P(blue) = 3/12 = 1/4.
Since P(red then white) < P(blue), eliminate B.

C: 15 tokens
Case 1: 9 red, 3 white and 3 blue, in which case P(red then white) = 9/15 * 3/15 = 3/25.
Case 2: 6 red, 6 white and 3 blue, in which case P(red then white) = 6/15 * 6/15 = 4/25.
P(blue) = 3/15 = 1/5.
In each case, P(red then white) < P(blue).
Eliminate C.

Notice that Case 2 -- in which R=W -- yields a greater result than Case 1.
Thus, in D and E, we should test cases in which the values for R and W are as close as possible.

D: 18 tokens, allowing for 9 red, 6 white, 3 blue
P(red then white) = 9/18 * 6/18 = 1/6.
P(blue) = 3/18 = 1/6.
Success!

The correct answer is D.
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