Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
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GmatMathPro
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Statement 1: Let there be x red balls. We can select two red balls in xC2 ways, and the total number of ways to select two balls is 12C2=66. Thus, xC2/66=1/11 or xC2=6 so x=4. SUFFICIENTaditya.j wrote:Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
Statement 2: Again let there be x red balls. This means there are 12-x blue balls. Therefore there are x*(12-x) ways to choose one red ball and one blue ball, and again 12C2=66 ways to choose two balls. x(12-x)/66=9/22 or x(12-x)=27. By inspection we can see that x=3 or x=9 are both solutions. INSUFFICIENT.
Ans: A
However, on a real GMAT DS question, the statements wouldn't conflict with each other like this. What's the source on this?
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rijul007
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aditya.j wrote:Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
1) The probability of randomly selecting two red balls from the bag is 1/11.
lets say there are n red balls
Probability = nC2/12C2
nC2/66 = 1/11
nC2 = 6
n = 4
Sufficient
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
Probabilty = (No of ways of selecting one red and one blue ball)/(Total no of selections)
9/22 = x/12C2
x = 9/22 * 66
x = 27
lets say there are n red balls
no of blue balls = 12-n
no of ways of selecting one red and one blue = n(12-n) = x
n(12-n) = 27
n = 3 or 9
we dont no for sure whether no of red balls is 3 or 9
Insufficient
Option A
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Brent@GMATPrepNow
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Let R = # red ballsaditya.j wrote:Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
Let B = # blue balls
Given: R+B=12
Target: find the value of R
Statement 1: P(2 red) = 1/11
P(2 red) = P(1st ball is red) x P(2nd ball is red)
= (R/12)(R-1/11)
= (R^2-R)/132
Since we're told that P(2 red) = 1/11, we can write: (R^2-R)/132 = 1/11
Multiply both sides by 132 to get R^2 - R = 12
Set equal to zero: R^2 - R - 12 = 0
Factor: (R-4)(R+3)=0
R = -3, 4
Since R cannot be negative, R must equal 4
SUFFICIENT
Statement 2: P(1 red and 1 blue) = 9/22
P(1 red and 1 blue) = P(red then blue or blue then red)
= P(red then blue) + P(blue then red)
= (R/12)(B/11) + (B/12)(R/11)
= RB/66
Since we're told that P(1 red and 1 blue) = 9/22, we can write: RB/66 = 9/22
Multiply both sides by 66 to get RB=27
So, we know that RB=27 and R+B=12
case a: R=3 and B=9
case b: R=9 and B=3
Since R can have 2 different values, statement 2 is INSUFFICIENT
Answer = A
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LalaB
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1) (R/12)*((R-1)/11)=1/11aditya.j wrote:Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
R(R-1)=12
R=4 hence, (1) is suff
2) 2*(R/12)*((12-R)/11)=9/22
R(12-R)=27
Since R can be 3 or 9 , (2) is insuff
GmatMathPro wrote:Statement 1: Let there be x red balls. We can select two red balls in xC2 ways, and the total number of ways to select two balls is 12C2=66. Thus, xC2/66=1/11 or xC2=6 so x=4. SUFFICIENTaditya.j wrote:Q) A bag contains 12 balls some of which are red and rest are blue. How many red balls are there in the bag?
1) The probability of randomly selecting two red balls from the bag is 1/11.
2) When two balls are selected randomly from the bag the probability of selecting one red and one blue ball is 9/22
I really couldn't figure this one out!
Statement 2: Again let there be x red balls. This means there are 12-x blue balls. Therefore there are x*(12-x) ways to choose one red ball and one blue ball, and again 12C2=66 ways to choose two balls. x(12-x)/66=9/22 or x(12-x)=27. By inspection we can see that x=3 or x=9 are both solutions. INSUFFICIENT.
Ans: A
However, on a real GMAT DS question, the statements wouldn't conflict with each other like this. What's the source on this?
These are from the Jamboree Prep Tests.
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Matt@VeritasPrep
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It's (n choose 2) = 6, I think, which becomesBhamini wrote:I am not able to understand how NC2=4???Can you please break it down for me.
Is it n!/(n-2)!2!=6 and then after that?
n! / ((n-2)! * 2!) = 6, or
n! = 6 * (n-2)! * 2, or
n! = 12 * (n - 2)!, or
n * (n - 1) = 12
From which we find n = 4.
