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A batch of cookies was divided amomg 3 tins: 2/3 of all the

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Joined: 17 Aug 2009
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A batch of cookies was divided amomg 3 tins: 2/3 of all the  [#permalink]

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11 Nov 2009, 07:42
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75% (hard)

Question Stats:

56% (02:15) correct 44% (02:27) wrong based on 373 sessions

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A batch of cookies was divided amomg 3 tins: 2/3 of all the cookies were placed in either the blue 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
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Joined: 02 Sep 2009
Posts: 58465

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11 Nov 2009, 08:19
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1
zaarathelab wrote:
A batch of cookies was divided amomg 3 tins: 2/3 of all the cookies were placed in either the blue 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

2/3 of all the cookies were placed in either the blue or the green tin: $$B+G=\frac{2}{3}*X$$
The rest were placed in the red tin: $$R=\frac{1}{3}*X$$
1/4 of all the cookies were placed in the blue tin: $$B=\frac{1}{4}*X$$.

What fraction of the cookies that were placed in the other tins (other then blue, or in Green and Red) were placed in the green tin: $$\frac{G}{G+R}=?$$

$$B+G=\frac{2}{3}*X$$ and $$B=\frac{1}{4}*X$$; --> $$\frac{1}{4}*X+G=\frac{2}{3}*X$$; --> $$G=\frac{5}{12}*X$$.

$$\frac{G}{G+R}=(\frac{5}{12}*X)/(\frac{5}{12}*X+\frac{1}{3}*X)=\frac{5}{9}$$

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11 Nov 2009, 08:46
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zaarathelab wrote:
A batch of cookies was divided amomg 3 tins: 2/3 of all the cookies were placed in either the blue 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

Not a clear question! How one can infer/imply "other tins" mean "green + red tins"? Is that the massage the question supplying? How "other tins" is not mean "blue + red tins"?

Just to get the idea about the question i..e. how precisely and correctly the quesion is designed? Is it a question unnecessarily designed to be a tough or is it really tough?

For me, its not a standard GMAT level question.
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11 Nov 2009, 08:57
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GMAT TIGER wrote:
Not a clear question! How one can infer/imply "other tins" mean "green + red tins"? Is that the massage the question supplying? How "other tins" is not mean "blue + red tins"?

Just to get the idea about the question i..e. how precisely and correctly the quesion is designed? Is it a question unnecessarily designed to be a tough or is it really tough?

For me, its not a standard GMAT level question.

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

I think that the question implies that other tins then blue, as previously the stem in the same sentence was describing the fraction in blue tin.

I don't know the source but for me this could be the GMAT question.
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24 Feb 2010, 02:05
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Let Total no. of cookies = 12 (LCM(3,4))
Cookies in Blue +Green Tin = (2/3)*12 = 8
Cookies in Red Tin = 12-8 = 4
Cookies in Blue tin = (1/4)*12 = 3
Cookies in Green Tin = 8-3 = 5
Green/other(other than Blue) = 5/(4+5) = 5/9 (C)
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Re: PS Word Problem Kaplan 600  [#permalink]

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21 Sep 2010, 07:07
1
ezinis 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

I was frustrated at this question by the way it worded. The OA is C, what the hell.

B + G = 2/3
R = 1 - 2/3 = 1/3
B = 1/4
So G = 2/3 - 1/4 = 5/12
G/(G+R) = G/(1-B) = (5/12)/(3/4) = 5/9
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Re: PS Word Problem Kaplan 600  [#permalink]

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23 Sep 2010, 14:42
shrouded1 wrote:
ezinis 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

I was frustrated at this question by the way it worded. The OA is C, what the hell.

B + G = 2/3
R = 1 - 2/3 = 1/3
B = 1/4
So G = 2/3 - 1/4 = 5/12
G/(G+R) = G/(1-B) = (5/12)/(3/4) = 5/9

How did u assumed that it has to be G/(G+R) instead of (B+R) ?
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Re: PS Word Problem Kaplan 600  [#permalink]

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23 Sep 2010, 15:03
onedayill wrote:
How did u assumed that it has to be G/(G+R) instead of (B+R) ?

The question says 1/4th were placed in the blue tin, what percentage of the ones placed in the other tins ....

So other tins ==> Green + Red
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Re: A batch of cookies was divided amomg 3 tins: 2/3 of all the  [#permalink]

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27 Mar 2016, 20:16
zaarathelab wrote:
A batch of cookies was divided amomg 3 tins: 2/3 of all the cookies were placed in either the blue 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

my approach:
8 were in B and Green, and 4 were in red.
we need to put only 1/4 in blue, or 3 cookies. that leaves us with 9 remaining cookies.
since we need a fraction X/9, we can easily eliminate A, B, D, and E.
moreover, since in green tin maximum can be 9 remaining cookies, we would get 9/9, or 1.
A - 15/2 = 7.5 way more than needed.
B - 9/4 - 2.25 - no.
C - looks ok
D - >1 so no
E - > 1 so no.
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Re: A batch of cookies was divided among three tins: 2/3 of all  [#permalink]

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19 Sep 2019, 03:29
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Re: A batch of cookies was divided among three tins: 2/3 of all   [#permalink] 19 Sep 2019, 03:29
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