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A jar is filled with red, white, and blue tokens that are eq
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11 Nov 2013, 20:23
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40% (02:48) correct 60% (02:35) wrong based on 149 sessions
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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
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Re: An jar is filled with red, white, and blue tokens that are e
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11 Nov 2013, 20:37
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 333. 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. 1233 wouldn't work, because again you need both r&w to be larger than b. 666 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



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Re: An jar is filled with red, white, and blue tokens that are e
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07 Apr 2014, 01:31
Oh!took me 10 minutes to figure out!! And that too by options. All of these options are are multiples of \(3\).So dividing them equally will not help because in that case we will never have equal probability of selecting two tokens on the one hand and selecting just one on the other. By same logic,we can never have two same no. of differently colored tokens. To fulfill the condition,we need to have all different nos. of differently colored tokens. Then I checked the first three multiples of \(3:3,6\) & \(9\) added up to \(18\). With all mental capacity exhausted,clicked \(18\)! (Now I realize the question asks for smallest no. of tokens.I know this is shoddy logic but still got it right!)



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Re: A jar is filled with red, white, and blue tokens that are eq
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07 Apr 2014, 03:42
AccipiterQ 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 Let us say that there are r, w and b tokens. The question says that (r + w)/(r + w + b) x 2! = (b)/(r + w + b) Hence 2 r+ 2 w = b Number of tokens is a multiple of 3 hence the minimum number will be r, b, w = 3, 3, 2(3) + 2(3) = 3, 3, 12 = 18 (The key is 2!  which is the number of ways r and w can arrange itself)
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Re: A jar is filled with red, white, and blue tokens that are eq
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08 Apr 2014, 06:04
AccipiterQ 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 Say, no of red balls is R, white balls is W and blue balls is B and total number of balls is T (= R + W + B) (R/T) * (W/T) = B/T (Select a red token, replacing and then white token = Select a blue token) RW = BT 3a*3b = 3c*(3a + 3b + 3c) (Since R, W and B must be multiples of 3, a, b and c must be positive integers) ab/c = a+b+c Since a, b and c must be 1 at least, a+b+c must be at least 3. But that will make the left hand side 1. Let's try to keep c as 1 since it reduces complexity and keeps the right hand side low. Now, which two numbers when multiplied give 1 more than when added. Think of small numbers like 1, 2 and 3 since the total is small as is apparent from the options. The values that satisfy this are 2 and 3. 2*3 = 6 which is 1 more than 2+3 = 5. Hence total number of balls = 3 ( 1 + 2 + 3) = 18
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Re: A jar is filled with red, white, and blue tokens that are eq
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19 Jul 2015, 22:53
AccipiterQ 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 (Red / Total Coins)*(White / Total Coins) = (Blue / Total Coins) i.e. Red*White = Blue*Total Coins Let, Red = 3a White = 3b Blue = 3c Total Coins = 3(a+b+c) i.e. 3a * 3b = 3c*3(a+b+c) i.e. a*b = c*(a+b+c) For smallest values of a, b and c 2*3 = 1*(1+2+3) i.e. Minimum Total Coins = 3*(1+2+3) = 18 Answer: Option D
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Re: A jar is filled with red, white, and blue tokens that are eq
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07 Oct 2015, 12:09
So as stated above after some manipulations we have: r*w=b*(r+w+b), If we consider a hint about probability, it says that there are less white balls than white or red ones. Let's look at the answer choices: (A) 9: Is out, because than we have r=w=b and we know that b is less (B) 12: Is out, smallest multiple of 3 is 3=Blue, than we have B=3, W(>B)=6 and r=b (C) 15: b=3, r=w=6 let's plug this values in the equation above > 6*6=3*(6+6+3) is not true OUT (D) 18: b=r=w=6 wrong combination, we need b=3, w=6, r=9 > 6*9=3*(9+6+3) TRUE (E) 21: b=3, w=r=9 > 9*9=3*(9+9+3) is not true OUT Answer (D)
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Re: A jar is filled with red, white, and blue tokens that are eq
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18 Feb 2017, 07:49
So, why is this question considered brutal? #1 it deals with probability, not a topic most GMAT students are superfamiliar 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 crossmultiply 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. If 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!
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A jar is filled with red, white, and blue tokens that are eq
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09 Mar 2019, 08:26
A lot of good algebraic solutions have been posted, it's not my strongest point so the fastest way for me was just playing with the numbers... hopefully this approach makes sense:
From the prompt are given a ratio of r:w:b that are all multiples of 3. We also get that p(Choose r) AND p(Choose w) = p(Choose b). We need the smallest Total, so minimizing the amount of each token while fulfilling the ratio requirement.
A) I could see that there's no way to break it down given the constraints. Basically, if r,w,b have to to be multiples of 3 the only way is 3/9 * 3/9 = 3/9 > 1/9 ≠ 1/3. This made me realize that b has to be a much smaller amount and that r and w can never be equal. Looking at the answers I saw 3 odd numbers and 2 even numbers  I thought that C and E are out because there won't be factors of 5, 7 in the numerator to reduce the fractions on the LHS). Must be B or D.
B) I tried 3/12 * 6/12 = 3/12 > 1/4*1/2 ≠ 1/4. This led me to think that p(Choose b) has to be an even smaller amount of the total.
D) Tried 3/18 = 9/18 * 6/18 since it seemed correct and picked D. Looking back on it, I kind of intuitively got that you need a certain number of factors to reduce the fraction on the LHS. For example, 12/18 * 3/18 = 3/18 won't work because now we have 12*6 at the top to reduce 6*3*3 so it won't equal 1/6, while in the correct version we have 6*9 to reduce 6*3*3 to make it equal 1/6.
And the reason C,E don't work is exactly that  no matter what combination of factors of 3 you try, you will get fractions which don't reduce the denominator enough.



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Re: A jar is filled with red, white, and blue tokens that are eq
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09 Mar 2019, 08:38
It took 8 min 23sec to solve this question...it feels bad




Re: A jar is filled with red, white, and blue tokens that are eq
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09 Mar 2019, 08:38






