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# Overlapping Set Problem

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Manager
Joined: 06 Apr 2011
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03 Aug 2011, 22:11
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33% (03:42) correct 67% (04:11) wrong based on 11 sessions

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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

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Asher

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03 Aug 2011, 22:44
1
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Asher wrote:
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

$$n(R \cup G)=n(R)+n(G)-n(R \cap G)$$

$$87=R+G-\frac{2}{7}R$$--------------1

$$87=R+G-\frac{3}{7}G$$--------------2

Solve these equations to find either R or G.

Both=(2/7)R OR (3/7)G

I got 6/29.

Ans: "D"
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03 Aug 2011, 22:57
Let x be the number of red balls in the jar and y be the number of green balls in the jar.

Then we know that n(red U green) = number of total balls in the jar = 87 (as it is given that each ball has at least one of two colors)

=> 87 = x + y - (3/7)*y [because 3/7 of the balls that have green color also have red color]
Also (2/7) * x = (3/7) * y [because 2/7 of the balls that have red color also have green color and it is given that 3/7 of the balls that have green color also have red color. These two numbers have to be equal].

Solving these two equations, we get y = 42 and x = 63.

So there are 63 red balls and 42 green balls in the jar.

Of these, number of only red balls = 63 - (2/7)*63 = 45
Number of only green balls = 42 - (3/7) * 42 = 24
Number of balls that are both green and red = 87 - 45 - 24 = 18

Therefore fraction of balls in the jar that have both red and green colors = 18/87 = 6/29

The correct answer is therefore (D).
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Manager
Joined: 06 Apr 2011
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Kudos [?]: 14 [0], given: 22

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05 Aug 2011, 01:11
Thanks fluke and gyanone.

i found this solution in another forum.. is it ok to do it this way as well

Quote:
So, Total = R + G - R&G
87 = R + G - R&G

We also now that, R*2/7 = G*3/7 = R&G
therefore, G = R* 2/3

Let's go back to the Total equation:
87 = R + R*2/3 - R*2/7
In the end you get: R = 87 * (21/29)
R&G = R * 2/7 = 87 * 21/29 * 2/7 = 87 * 6/29
R&G/87 = 6/29

In other words, 6/29 of the total number of balls have both the colors red & green

is it ok to say that r&g/87 = 6/29. hence ans. is 6/29.
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Asher

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Joined: 20 Dec 2010
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Kudos [?]: 1523 [1] , given: 376

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05 Aug 2011, 01:20
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Asher wrote:
Thanks fluke and gyanone.

i found this solution in another forum.. is it ok to do it this way as well

Quote:
So, Total = R + G - R&G
87 = R + G - R&G

We also now that, R*2/7 = G*3/7 = R&G
therefore, G = R* 2/3

Let's go back to the Total equation:
87 = R + R*2/3 - R*2/7
In the end you get: R = 87 * (21/29)
R&G = R * 2/7 = 87 * 21/29 * 2/7 = 87 * 6/29
R&G/87 = 6/29

In other words, 6/29 of the total number of balls have both the colors red & green

is it ok to say that r&g/87 = 6/29. hence ans. is 6/29.

Yes, this is good as well. It uses substitution rather than simultaneous solution of equation in 2-variables.
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05 Aug 2011, 08:09
Asher,

The way you have solved it is fine as well. As fluke has pointed out, your method just uses substitution instead of solving the two equations to get the values of the variables.

The point to note here is that if you can use substitution instead of solving the two equations for the two variables, do that on the exam -> might save you those extra precious 5-10 seconds.
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Joined: 27 Feb 2011
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Kudos [?]: 3 [0], given: 9

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05 Aug 2011, 12:50
fluke wrote:

$$n(R \cup G)=n(R)+n(G)-n(R \cap G)$$

$$87=R+G-\frac{2}{7}R$$--------------1

$$87=R+G-\frac{3}{7}G$$--------------2

Solve these equations to find either R or G.

Both=(2/7)R OR (3/7)G

I got 6/29.

Ans: "D"

Nice way of solving.. thanks!
Manager
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06 Aug 2011, 13:40
I took a shortcut to do this one if a few seconds. I said that the denominator had to be a factor of 87. Only answer choice D had a factor of 87--29.

This method wouldn't have worked if there were another answer choice with 29 or 3.
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10 Aug 2011, 02:06
good ways of solving them ..

but a very quick soln (for this prbm )

87 is divisible by 29 and 3

since 29 is one of the deno in all the choices... it has to be tht choice ... !!!

(soln by a trick)
Re: Overlapping Set Problem   [#permalink] 10 Aug 2011, 02:06
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