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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?

if we say the cookies are in either the blue tin or the green tin, it means that they can only be in one of them, not both or not shared. So how come you can get 1/4 later in the blue tin ? i don't get this one

The question said 2/3 of all the cookies were placed in either the blue tin or the green tin. Not ALL the cookies were placed in blue or green.

Also, it was stated in the first sentence that "A batch of cookies was divided among 3 tins...". Hope this helps!

Saw this in Kaplan 800 (2007-2008 edition page 286). I was confused as well, but now I see that the confusion comes from not reading the question carefully. The confusion is based on what "other tins" refers to. This is an example of how reading comprehension comes to play in the quantitative part!

P.S. Wow! I just realized it's 4 days away from the 4th anniversary since the previous post!

Hi everyone, I am having understanding part of the solution to this problem.

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 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

I realized that all the numerators are greater that their denominators, but my questions is, when I read kaplan's explanations, why it used this formula, green cookies/ green and red cookies, and it did not consider the blue cookies in the formula?

The answer is C. The question asks “what fraction of the cookies that were placed in the other tins were placed in the green tin?” This means what fraction of the cookies not placed in the blue tin were in the green tin. To solve subtract ¼ from 2/3. (This will give you the percentage of cookies in the green tin over all the cookies (blue, green and red). This equals 5/12. Then divide this number by the total number of cookies in both the green and red tin (or not the blue tin). This number equals (total number of cookies in both the green and red tin) 9/12. (5/12 +1/3). Then divide 5/12 by 9/12 and you get 5/9

if you read the highlighted sentence carefully , you will see "blue tin" is the main topic of the sentence . hence other over there refers to anything but blue(i.e R & G)

so basically what they are trying to ask is what fraction of R & G is G

B+G = 2/3 B = 1/4 => G = 2/3 - 1/4 = 5/12 R = 1/3

=> G/(R+G) = (5/12)/(1/3+5/12) = 5/9

Answer is C.

manalq8 wrote:

Hi everyone, I am having understanding part of the solution to this problem.

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 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

I realized that all the numerators are greater that their denominators, but my questions is, when I read kaplan's explanations, why it used this formula, green cookies/ green and red cookies, and it did not consider the blue cookies in the formula?

Hi, This is my first post on GMATclub. I have joined mastergmat's online course and it seems to be working fine. Solved this question using the Pluggin In method (one of Mastergmat's many clever ways of tackling such questions).

Assuming that the total number of cookies is 24, the number of cookies of each color can be found out. Thus, we have 6 for Blue, 10 for green and 8 for red. The answer thus can be calculated as (R+B)/G= 14/10=7/5.

Done in 34 seconds.

gmatclubot

Re: A batch of cookies
[#permalink]
11 Sep 2011, 09:17

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