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# There are 87 balls in a jar. Each ball is painted with at le

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There are 87 balls in a jar. Each ball is painted with at le [#permalink]  03 Dec 2016, 11:31
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There are 87 balls in a jar. Each ball is painted with at least one of two colors, red or green. It is observed that 2/7 of the balls that have red color also have green color, while 3/7 of the balls that have green color also have red color. What fraction of the balls in the jar have both red and green colors?

(A) 6/14
(B) 2/7
(C) 6/35
(D) 6/29
(E) 6/42
[Reveal] Spoiler: OA

Last edited by Carcass on 05 Dec 2016, 08:47, edited 1 time in total.
Edited the question
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Re: There are 87 balls in a jar. Each ball is painted with at le [#permalink]  05 Dec 2016, 14:18
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Let number of red only balls be $$a$$ and green only ball be $$b$$ and both red and green be $$c$$.

So we can say $$a+b+c=87$$.

2/7 of the balls that have red color also have green color

So $$\frac{c}{(a+c)}= \frac{2}{7}$$.

3/7 of the balls that have green color also have red color

So $$\frac{c}{(b+c)}=\frac{3}{7}$$.

Inverting both fractions and adding them

$$\frac{(a+c)}{c}+\frac{(b+c)}{c}=\frac{7}{3}+\frac{7}{2}$$

Subtracting 1 from both sides

$$\frac{a+b+2c}{c}-1= \frac{35}{6}-1$$.

$$\frac{a+b+c}{c}=\frac{29}{6}$$

Inverting again

$$\frac{c}{a+b+c}=\frac{6}{29}$$.

Our objective was always to find $$\frac{c}{a+b+c}$$.

Hence option D is correct.
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Re: There are 87 balls in a jar. Each ball is painted with at le   [#permalink] 05 Dec 2016, 14:18
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