AccipiterQ wrote:
An 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
This is from the Manhattan Advanced Quant Guide. Here's how I solved it:
Since we know that the probability of drawing R then W = probability of drawing b we have an equation
\(\frac{r}{(b+w+r)} * \frac{w}{(b+w+r)}=\frac{b}{(b+w+r)}\)
\(\frac{rw}{(b+w+r)^2}=\frac{b}{(b+w+r)}\)
rw*(b+w+r)=b*(b+w+r)^2
rw=b(b+w+r)
now at this point I paid attention to the fact that b, w, and r have to be multiples of 3. Also, there has to be fewer blue chips than red or white. So (A) was out, since the only way to break 9 into 3's is 3-3-3. So now looking at it again I realized the answer had to be able to be broken down into factors of 3, which could add back up to the answer (since b+w+r=total chips). So (B) wouldn't work, since the only way to break it into 3 factors of 3 that add up to 12 is 3,3,6, and both R AND W have to be larger than B. Next I looked at (C), the only way for that to work is 3, 6, 6, with b=3, r=6, w=6. So I plugged that into the equation
6*6=3(3+6+6)
36=45
So C was out. Next I looked at (D). With D the only way to break down 18 into 3 factors of 3 where at least 2 of them were larger than the third is 3,6, and 9. 12-3-3 wouldn't work, because again you need both r&w to be larger than b. 6-6-6 is also out for the same reason. So I checked it out:
6*9=3(3+6+9)
54=54
YAAAY! The answer is D.
This is probably the hardest problem I've solved on my own, so I was quite happy