In the “Tough GMAT” section, we’ll be taking a look at some of the toughest GMAT questions around. Let’s kick off this series with this tough GMAT Quant question from Manhattan GMAT’s Advanced GMAT Quant book!

**Question #1**

**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**

So, why is this question considered brutal? #1 it deals with probability, not a topic most GMAT students are super-familiar with, and #2 the question doesn’t give us any actual numbers, except telling us that the number of tokens of every color is a multiple of 3.

Let’s start with what we do know. There are “R” red tokens, “W” white tokens, and “B” blue tokens, and R + W + B = Total “T”. The probability of selecting a red token is R/T, and the probability of selecting a white token is W/T. To find the probability of two events occurring, we multiple the individual probabilities: (R/T)*(W/T) = RT / TT.

The question tells us this is equivalent to the probability of selecting a blue token: B/T = (RW)/(TT). This simplifies when we cross-multiply to (RW)/B = T.

The correct numbers of tokens in the jar will allow us to break down the number of red, blue, and white tokens such that they have the relationship (RW)/B = T. So let’s backsolve, and since the question asks for “smallest possible,” we’ll start with (A).

If T = 9, the only possible numbers of tokens are B = 3, R = 3, and W = 3. But since (3*3)/3 doesn’t equal 9, we know this isn’t a possible value for the total.

IfÂ T = 12, the only possible numbers of tokens are 3, 3, and 6. We don’t know which color has 6 tokens, but there’s still no way to make (RW)/B = T true for these values.

If T = 15, the numbers could be 3, 3, 9 or 3, 6, 6. It should still be somewhat clear that these numbers won’t work, but try out a couple the combinations to see if they work if you’re unclear: (3*9)/3 = 9, not our total of 15. Try the other set: (6*6)/3 = 12, not our total of 15.

If T = 18, the numbers could be 3, 3, 12, or 3, 6, 9, or 6, 6, 6. Try a few arrangements to see if you land on one that makes (RW)/B) = T true.

(6*9)/3 = 54/3 = 18, our total! Finally, we’ve got a set of numbers that makes the relationship between the tokens work!

**Takeaway:** Use the answer choices to your advantage as much as possible, but thoroughly analyze the relationships in the question stem first!