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Manager  Joined: 06 Apr 2010
Posts: 112
A box contains red and blue balls only. If there are 8 balls  [#permalink]

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6 00:00

Difficulty:   75% (hard)

Question Stats: 61% (02:03) correct 39% (02:05) wrong based on 201 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
Manager  Status: I rest, I rust.
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Re: Balls in the box  [#permalink]

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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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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 I chose D as answer. Math Expert V
Joined: 02 Aug 2009
Posts: 7686
A box contains red and blue balls only. If there are 8 balls  [#permalink]

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

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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.
Math Expert V
Joined: 02 Aug 2009
Posts: 7686
Re: A box contains red and blue balls only. If there are 8 balls  [#permalink]

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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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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A box contains red and blue balls only. If there are 8 balls  [#permalink]

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Hi All,

We're told that a box contains red and blue balls only and that there are 8 balls in total. We're asked for the number of 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

To start, since we're selecting balls from the box WITHOUT replacement, the probability that the first ball will be red is X/8 (where X is the number of red balls). If we pull a red ball on the first try, then there will be one FEWER red ball for the second try, so the probability of pulling a red ball then is (X-1)/7.

Thus, the overall probability will be (X)(X-1)/(8)(7) = (X)(X-1)/56. The given fraction is 5/14, which has clearly been reduced. 5/14 = 20/56, so we're looking for a value of X such that (X)(X-1) = 20. There's only one positive value that fits here... X= 5 --> since (5)(4) = 20
Fact 1 is 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.

Based on the work that we did in Fact 1, we can look for a similar pattern in Fact 2. The numerator of 15/56 is '15', so we need two numbers that SUM to 8 (one for the number of red balls and one for the number of blue balls) that when multiplied together give us 15. Those would clearly be 3 and 5. However, we do NOT know which one is which. We could have 3 red balls and 5 blue balls OR 5 red balls and 3 blue balls.
Fact 2 is INSUFFICIENT

Final Answer:

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www.empowergmat.com/ Re: A box contains red and blue balls only. If there are 8 balls   [#permalink] 29 Nov 2018, 16:42
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