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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]

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New post 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  [#permalink]

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New post 04 Oct 2013, 04:56
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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  [#permalink]

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New post 04 Oct 2013, 04:49
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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.
_________________

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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: There are 87 balls in a jar. Each ball is painted with at le  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 04 Oct 2013, 05:00
1
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: There are 87 balls in a jar. Each ball is painted with at le  [#permalink]

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New post 04 Oct 2013, 05:01
2
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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  [#permalink]

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New post 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  [#permalink]

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New post 01 Sep 2014, 06:47
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
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Re: There are 87 balls in a jar. Each ball is painted with at le  [#permalink]

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New post 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*2/5

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:
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Nova - Fractions question  [#permalink]

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New post 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 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.
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Re: There are 87 balls in a jar. Each ball is painted with at le  [#permalink]

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