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

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.

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.

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

The easiest method is given in my first post.

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

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.

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

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

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

 

­
You can make a table:



However, you'd still need to do something else, say use the method shown here: 

https://gmatclub.com/forum/there-are-87 ... l#p1274151­