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11 Nov 2009, 07:42
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
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Question Stats:
57% (02:12) correct
43%
(02:26)
wrong
based on 593
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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
A. 15/2
B. 9/4
C. 5/9
D. 7/5
E. 9/7
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)
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)
11 Nov 2009, 08:19
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zaarathelab
# of cookies - X
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 \(B=\frac{1}{4}*X\); --> \(\frac{1}{4}*X+G=\frac{2}{3}*X\); --> \(G=\frac{5}{12}*X\).
\(\frac{G}{G+R}=(\frac{5}{12}*X)/(\frac{5}{12}*X+\frac{1}{3}*X)=\frac{5}{9}\)
Answer: C (5/9)
