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

Question

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

Bunel can you please solve it with some excellent method?

(A) 6/14
(B) 2/7
(C) 6/35
(D) 6/29
(E) 6/42

Bunuel · Accepted Answer

The easiest way would be to spot that the correct answer must have the denominator which is a factor of 87=3*29. Only D fits. You don't even need to solve anything.

Answer: D.

honchos · Answer

Let T be the total number of balls, R the number of balls having red color, G the number having green color, and B the number having both colors.

So, the number of balls having only red is R – B, the number having only green is G – B, and the number having both is B. Now, the total number of balls is T = (R – B) + (G – B)+ B = R + G – B. We are given that 2/7 of the balls having red color have green also. This implies that B = 2R/7. Also, we are given that 3/7 of the green balls have red color. This implies that B = 3G/7. Solving for R and G in these two equations yields R = 7B/2 and G = 7B/3. Substituting this into the equation T = R + G – B yields
T = 7B/2 + 7B/3 – B. Solving for B yields B = 6T/29. Hence, 6/29 of all the balls in the jar have both colors.
The answer is (D). Note that we did not use the information: “There are 87 balls.” Sometimes, not all information in a problem is needed.

honchos · Answer

I was visualizing it this way 3/7 = 2/7 + 1/7
  Therefore total # that have both is finally 3/7, but still couldn't get the answer by this method.

May be I have incorrect comprehension of the question.

Bunuel · Answer

Conventional method:
2/7 of the balls that have red color also have green color --> {both} = x = 2/7*R;
3/7 of the balls that have green color also have red color --> {both} = x = 3/7*G.

87 = R + G - Both;
87 = 7x/2 + 7x/3 - x;
x = 18;
x/total = 18/87 = 6/29.

Answer: D.

honchos · Answer

Thanks actually I solved it this way, i was looking for som alternatively easy method.

Bunuel · Answer

The easiest method is given in my first post.

Temurkhon · Answer

To be more academic:
87=r+g-2/7*r
2/7*r=3/7*g, so g=2/3*r

substitute to get 87=r+2/3*r-2/7*r => 29/21*r => r=63 =>2/7*63=18=>18/87=6/29

Skurpis · Answer

Hello - I'm working through the NOVA book in addition to several others. I came across an explanation that I didn't quite understand. Can someone help me work through how the simplified equation gets from  T = 7B/2+7B/3 - B   to  6T/29 ?

Question 
There are 87 balls in a jar. Each ball is painted with at least one of the 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?

Answer 
Let T be the total number of balls, R the number of red balls, G the number of green balls, B the number of balls with both red and green.

So the number of red balls is  R-B , the number of green balls is  G-B , and the number of balls with both red and green balls is  B . Now, the total number of balls is  T = (R-B) + (G-B) + B or T = R + G - B

We are given that 2/7 of the balls that are red also have green. This implies that B = 2R/7. We are also given that 3/7 of the green balls have red color. This implies that B = 3G/7. Solving for R and G in these two equations yields R = 7B/2 and G = 7B/3. Substituting this into the equation  T = R+G-B   yields  T = 7B/2 + 7B/3-B . Solving for B yields  B= 6T/29 .

MathRevolution · Answer

Let T = Toal number of Balls (Given : 87), R = red color balls, G = Green color balls, B = Both color balls

T = R + G - B

*  \frac{2}{7}  of the balls that have red color also have green color => B(both) =  \frac{2}{7} *R => R =  \frac{7}{2} *B

*  \frac{3}{7}  of the balls that have green color also have red color => B(both) =  \frac{3}{7} *G => G =  \frac{7}{3} *B

T =  \frac{7}{2} *B +  \frac{7}{3} *B - B
=> 87 = B (  \frac{7}{2} + \frac{7}{3} -1)
=> 87 = B * (  \frac{21+14-6}{6} )
=> 87 = B * (  \frac{29}{6} )
=> B =  \frac{87*6}{29} 
=> B = 18

\frac{B}{Total}  =  \frac{18}{87}  =  \frac{6 }{ 29}

Answer D

adewale223 · Answer

Since RnG = 2R/7

And GnR = 3G/7

Therefore 2R/7 = 3G/7

2R = 3G

R = 3G/2

Now since RUG = R + G - RnG

87 = 3G/2 + G - 3G/7

1218 = 21G + 14G - 6G

29G = 1218

G = 42

RnG = 3G/7

= 18

Fraction = 18/87 = 6/29 Answer choice D

Nsp10 · Answer

Hi @Bunnel,
how can we do this question using set matrix 
Thank you in advance.

Bunuel · Answer

You can make a table: https://gmatclub.com/forum/download/file.php?id=81896 However, you'd still need to do something else, say use the method shown here: there-are-87-balls-in-a-jar-each-ball-is-painted-with-at-le-161001.html#p1274151 GMAT-Club-Forum-kvgwsvee.png

bumpbot · Answer

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