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It is observed that 2/7 f the balls that have red color also have

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VP
Joined: 31 Oct 2013
Posts: 1493
Concentration: Accounting, Finance
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It is observed that 2/7 f the balls that have red color also have  [#permalink]

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28 Jan 2018, 05:03
1
14
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Difficulty:

85% (hard)

Question Stats:

52% (02:56) correct 48% (02:54) wrong based on 105 sessions

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There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f 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 has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT
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Joined: 24 Jun 2013
Posts: 142
Location: India
Schools: ISB '20, GMBA '20
It is observed that 2/7 f the balls that have red color also have  [#permalink]

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03 Feb 2018, 22:32
9
2
poul_249 wrote:
Still not able to understand. Could someone explain ?

You have to visualise the problem as set problem...
Given 87 Balls with Red and Green colour
So the set equation becomes; Total (Red & Green Balls) = #Red Balls + # Green Balls - Both (Red & Green)

Given is 2/7 of Red balls also are Green ==> Both ( Red and green) = $$\frac{2}{7}$$ * # Red Balls
Also it is stated that 3/5 of Green balls are red, which is again stating that ==> Both ( Red and green) = $$\frac{3}{7}$$ * # Green Balls

Essentially it means that #Red Balls = $$\frac{7}{2}$$* Both (Red and Green)
Also it means that #Green Balls = $$\frac{7}{3}$$* Both (Red and Green)

Now coming back to original equation Total (Red & Green Balls) = #Red Balls + # Green Balls - Both (Red & Green)

Substituting the values we got above

87= $$\frac{7}{2}*Both + \frac{7}{3}*Both - Both$$
So Both = 18

We are asked to find Both/Total =$$\frac{18}{87}$$ = $$6/29$$

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It is observed that 2/7 f the balls that have red color also have  [#permalink]

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Updated on: 03 Feb 2018, 22:34
5
2
ven diagram..
87=R+G-Both RG

given that ...Both = 2R/7 = 3G/7; we need to find Both/total

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

Both = 18

therefore , 18/87 = 6/29 D

Corrected G in 3G/7

Originally posted by doomedcat on 28 Jan 2018, 05:29.
Last edited by doomedcat on 03 Feb 2018, 22:34, edited 1 time in total.
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Re: It is observed that 2/7 f the balls that have red color also have  [#permalink]

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02 Feb 2018, 09:45
Still not able to understand. Could someone explain ?
Intern
Joined: 10 Sep 2017
Posts: 4
Re: It is observed that 2/7 f the balls that have red color also have  [#permalink]

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15 Feb 2018, 22:07
1
selim wrote:
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f 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 has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT

R Red
G Green
X Both

Given:
Total Balls = 87
(2/7)*R = X ----> R = (7/2)*X ---->(1)
(3/7)*G = X ----> G = (7/3)*X ---->(2)

formula : T = R+G - both(X)
Applying 1 and 2 in formula
87= (7/2)*X + (7/3)*X - X
Solving, we get X= 18

Now fr fraction X/T = 18/87 = 6/29

VP
Joined: 31 Oct 2013
Posts: 1493
Concentration: Accounting, Finance
GPA: 3.68
WE: Analyst (Accounting)
Re: It is observed that 2/7 f the balls that have red color also have  [#permalink]

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28 Aug 2019, 10:40
KSBGC wrote:
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f 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 has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT

Bunuel's Solution:

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.

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Re: It is observed that 2/7 f the balls that have red color also have  [#permalink]

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02 Sep 2019, 19:21
KSBGC wrote:
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f 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 has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT

We can create the equations:

r + g - both = 87

and

(2/7)r = (3/7)g (note: either side of the equation = both)

2r = 3g

r = 3g/2

Substituting, we have:

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

5g/2 - 3g/7 = 87

Multiplying by 14, we have:

35g - 6g = 87 x 14

29g = 87 x 14

g = 3 x 14 = 42

Therefore, both = 3(42)/7 = 3(6) = 18, and balls of both colors are 18/87 = 6/29 of all the balls.

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Re: It is observed that 2/7 f the balls that have red color also have  [#permalink]

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03 Dec 2019, 16:42
KSBGC wrote:
There are 87 balls in a jar. Each ball is painted with at least one of the two colors, red and green. It is observed that 2/7 f 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 has both red and green colors?

a) 6/14
b)2/7
c) 6/35
d) 6/29
e) 1/5

source: NOVA GMAT

r+g-both+none=87…r+g-both=87
2/7*r=3/7*g…r=3/2*g
3/2*g+g-3/7*g=87…5g/2-3g/7=87
35g-6g/14=87…g=87*14/29
both/total=(3/7*g)/87=(3/7*87*14/29)/87=6/29

Ans (D)
Re: It is observed that 2/7 f the balls that have red color also have   [#permalink] 03 Dec 2019, 16:42
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