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Error in Kaplan 800 - Word Problem # 5

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by bml1105 » Thu May 29, 2014 4:49 pm
A batch of cookies was divided among three tins: 2/3 of all the cookies were placed in either the blue tin or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction of the cookies that were placed in the other tins were placed in the green tin?

(A) 15/2
(B) 9/4
(C) 5/9
(D) 7/5
(E) 9/7


OA: C
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Source: — Problem Solving |

by bml1105 » Thu May 29, 2014 4:51 pm
When reading the explanation in the book it says what we are looking for is as follows:

(# of cookies in green tin)/{(# of cookies in green tin)+(# of cookies in red tin)}

Why would the number of cookies in the green tin be in the denominator? Shouldn't it be the # of cookies in the blue tin + the # of cookies in the red tin, since the question asks "what fraction of the cookies that were placed in the other tins were placed in the green tin"
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by Brent@GMATPrepNow » Thu May 29, 2014 5:49 pm
bml1105 wrote:A batch of cookies was divided among three tins: 2/3 of all the cookies were placed in either the blue tin or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction of the cookies that were placed in the other tins were placed in the green tin?

(A) 15/2
(B) 9/4
(C) 5/9
(D) 7/5
(E) 9/7


OA: C
Let's start with a certain number of cookies that works well with all of the fractions.
Let's say that there are 36 cookies altogether.

2/3 of all the cookies were placed in either the blue tin or the green tin
So, 24 cookies were placed in either the blue tin or the green tin

The rest were placed in the red tin
So, 12 cookies were placed in the red tin

1/4 of all the cookies were placed in the blue tin
So, 9 cookies were placed in the blue tin

So, the number of cookies that were placed in the green tin = 36 - 12 - 9 = 15

If 1/4 of all the cookies were placed in the blue tin, what fraction of the cookies that were placed in the other (non blue) tins were placed in the green tin?

There are 27 cookies in the non-blue tins.
15 of them are in the green tin
Fraction = 15/27 = [spoiler]5/9[/spoiler]

Answer: C

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by bml1105 » Thu May 29, 2014 8:09 pm
Ohhh ok. Thanks! That does make sense. I was definitely reading the question wrong
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by [email protected] » Thu May 29, 2014 9:21 pm
Hi bml1105,

Are these answer choices formatted correctly? I agree with Brent's approach (TESTing VALUES works really well on this question), but these answers, as written, imply that the only possible answer is the correct one.

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
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by theCodeToGMAT » Thu May 29, 2014 9:37 pm
B/G = 2/3 * C

R = 1/3 * C

B = 1/4 * C

Let Cookies be 60

B/G = 40
R = 20
B = 15
So, G = 25

Fraction: 25/45 = 5/9
R A H U L
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by bml1105 » Fri May 30, 2014 8:19 pm
Rich, I just double checked the book and it is written correctly. It's Kaplan 800 - Arithmetic and Algebra Word Problems #5.

I was reading the question as what fraction of cookies is in green compared to the cookies in blue and red OR (# in green tin)/[(# in red tin) + (# in blue tin)] and wondering why I was getting it wrong.
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bml1105 wrote:
Thu May 29, 2014 4:49 pm
A batch of cookies was divided among three tins: 2/3 of all the cookies were placed in either the blue tin or the green tin, and the rest were placed in the red tin. If 1/4 of all the cookies were placed in the blue tin, what fraction of the cookies that were placed in the other tins were placed in the green tin?

(A) 15/2
(B) 9/4
(C) 5/9
(D) 7/5
(E) 9/7
Let the total number of cookies = the LCM of the two denominators in the prompt = 12
Since 2/3 of the cookies are placed in the blue or green tin, B+G = 2/3* 12 = 8
Since 1/4 of the cookies are placed in the blue tin, B = 1/4 * 12 = 3

What fraction of the cookies that were placed in the other tins were placed in the green tin?
other tins = total - B = 12-3 = 9
G = (B+G) - B = 8-3 = 5
green/other = 5/9

The correct answer is C.
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