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A box contains red and blue balls only. If there are 8 balls

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A box contains red and blue balls only. If there are 8 balls  [#permalink]

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New post 27 Nov 2010, 20:36
1
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A
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D
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  85% (hard)

Question Stats:

59% (01:11) correct 41% (01:28) wrong based on 134 sessions

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A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

Please explain your appraoch
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Re: Balls in the box  [#permalink]

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New post 27 Nov 2010, 21:01
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1
udaymathapati wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

Please explain your appraoch

Lets say there are R red balls and 8-R blue balls.
S1: P(Both-Red) = \(\frac{R}{8}*\frac{R-1}{7} = \frac{5}{14}\)
so R(R-1) = 5*4, hence R = 5 (considering only positive value). Sufficient

S2: P(Red-Blue) = \(\frac{R}{8}*\frac{8-R}{7} = \frac{15}{56}\)
so R(8-R) = 15, hence R = 5 or 3. Not Sufficient

Answer: A
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Re: A box contains red and blue balls only. If there are 8 balls  [#permalink]

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New post 28 Aug 2018, 17:49
udaymathapati wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.

Please explain your appraoch


I was able to deduce statement 1 to \(R(R-1)/8*7\) and statement 2 to \(R(8-R)=15\). I didn't solve it further as I thought R can be calculated but end up getting it wrong :P

I chose D as answer. :(
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A box contains red and blue balls only. If there are 8 balls  [#permalink]

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New post 28 Aug 2018, 18:39
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
Let the number of red balla be R and blue balla be B

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
if there are R red balls , the first can be picked in R ways and the next in R-1 ways
Therefore probability = \(\frac{R}{8}*\frac{R-1}{8-1}=\frac{5}{14}.......R(R-1)=\frac{8*7*5}{14}=5*4\)
Thus R =5
Sufficient

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.
Probability = \(\frac{R}{8}*\frac{B}{8-1}=\frac{15}{56}.......R*B=\frac{8*7*15}{56}=15=5*3\)
5*3 is the only possibility as 5+3=8 but R could be 3 or 5
Insufficient

A
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A box contains red and blue balls only. If there are 8 balls  [#permalink]

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New post 28 Aug 2018, 19:57
1
chetan2u wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
Let the number of red balla be R and blue balla be B

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
if there are R red balls , the first can be picked in R ways and the next in R-1 ways
Therefore probability = \(\frac{R}{8}*\frac{R-1}{8-1}=\frac{5}{14}.......R(R-1)=\frac{8*7*5}{14}=5*4\)
Thus R =5
Sufficient

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.
Probability = \(\frac{R}{8}*\frac{B}{8-1}=\frac{15}{56}.......R*B=\frac{8*7*15}{56}=15=5*3\)
5*3 is the only possibility as 5+3=8
Sufficient

D


chetan2u,
u also made same mistake which I made. Only statement 1 is sufficient. Statement 2 gives, two values for R=3,5.
Answer should be A.
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Re: A box contains red and blue balls only. If there are 8 balls  [#permalink]

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New post 28 Aug 2018, 20:05
ammuseeru wrote:
chetan2u wrote:
A box contains red and blue balls only. If there are 8 balls in total, how many red balls are in the box?
Let the number of red balla be R and blue balla be B

(1) If two balls are randomly selected without replacement, the probability that both balls are red is 5/14
if there are R red balls , the first can be picked in R ways and the next in R-1 ways
Therefore probability = \(\frac{R}{8}*\frac{R-1}{8-1}=\frac{5}{14}.......R(R-1)=\frac{8*7*5}{14}=5*4\)
Thus R =5
Sufficient

(2) If two balls are randomly selected without replacement, the probability is 15/56 that the first ball is red and the second ball is blue.
Probability = \(\frac{R}{8}*\frac{B}{8-1}=\frac{15}{56}.......R*B=\frac{8*7*15}{56}=15=5*3\)
5*3 is the only possibility as 5+3=8
Sufficient

D


chetan2u,
u also made same mistake which I made. Only statement 1 is sufficient. Statement 2 gives, two values for R=3,5.
Answer should be A.


Thanks..
I for some reason read Red balls are more than blue although it is not mentioned. May be I had a similar question earlier.

Kudos
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A box contains red and blue balls only. If there are 8 balls &nbs [#permalink] 28 Aug 2018, 20:05
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