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There are 87 balls in a jar. Each ball is painted with at le
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04 Oct 2013, 04:44
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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 Bunel can you please solve it with some excellent method?
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Re: There are 87 balls in a jar. Each ball is painted with at le
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04 Oct 2013, 04:56
Bunuel wrote: honchos 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
Bunel can you please solve it with some excellent method? 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. 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.
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Re: There are 87 balls in a jar. Each ball is painted with at le
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Re: There are 87 balls in a jar. Each ball is painted with at le
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04 Oct 2013, 04:52
Bunuel wrote: honchos 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
Bunel can you please solve it with some excellent method? 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. 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.
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Re: There are 87 balls in a jar. Each ball is painted with at le
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04 Oct 2013, 04:59
Bunuel wrote: Bunuel wrote: honchos 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
Bunel can you please solve it with some excellent method? 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. 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.
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Re: There are 87 balls in a jar. Each ball is painted with at le
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04 Oct 2013, 05:00
honchos wrote: Bunuel wrote: Bunuel 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
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. 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.
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Re: There are 87 balls in a jar. Each ball is painted with at le
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04 Oct 2013, 05:01
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.
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Re: There are 87 balls in a jar. Each ball is painted with at le
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30 Aug 2014, 06:20
Hello, I understand the solution to this question provided by you, however, I am not able to understand the flaw in the method I used to solve the question earlier, which gave me wrong answer. Please help me find out what wrong I did, else I may keep repeating the mistake. Thank you in advance! My solution : Balls with both green and red color: 2/7*R = 3/7*G solving the equation: R/G = 3/2 , which means R = 87*3/5 or G = 87*2/5 Fraction of the balls in the jar have both red and green colors= In terms of R: 3/5*2/7 = 6/35 or, in terms of G: 2/5*3/7 = 6/35. I don't know what wrong I did here. Sigh! Please help. Bunuel wrote: Bunuel wrote: honchos 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
Bunel can you please solve it with some excellent method? 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. 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.



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Re: There are 87 balls in a jar. Each ball is painted with at le
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01 Sep 2014, 06:47
To be more academic: 87=r+g2/7*r 2/7*r=3/7*g, so g=2/3*r
substitute to get 87=r+2/3*r2/7*r => 29/21*r => r=63 =>2/7*63=18=>18/87=6/29



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Re: There are 87 balls in a jar. Each ball is painted with at le
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01 Sep 2014, 07:18
mneeti wrote: Hello, I understand the solution to this question provided by you, however, I am not able to understand the flaw in the method I used to solve the question earlier, which gave me wrong answer. Please help me find out what wrong I did, else I may keep repeating the mistake. Thank you in advance! My solution : Balls with both green and red color: 2/7*R = 3/7*G solving the equation: R/G = 3/2 , which means R = 87*3/5 or G = 87*2/5Fraction of the balls in the jar have both red and green colors= In terms of R: 3/5*2/7 = 6/35 or, in terms of G: 2/5*3/7 = 6/35. I don't know what wrong I did here. Sigh! Please help. Bunuel wrote: Bunuel 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
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. 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. How can R be 3/5th of 87 if this even not an integer? The same for G. The point is that you cannot write R = 87*3/5 from R/G = 3/2 because there is an overlap between R and G. Total number is not 5 parts because 3 parts of R and 2 parts of G have overlap. Hope it's clear.
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Nova  Fractions question
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21 Jun 2017, 02:49
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 RB, the number of green balls is GB, and the number of balls with both red and green balls is B. Now, the total number of balls is T = (RB) + (GB) + 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+GB yields T = 7B/2 + 7B/3B. Solving for B yields B= 6T/29.



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