PS-Probability-# of cards

Question

Guys,

Except for POE (process of elimination), is there any easier way to solve the problem below?

Q30. In a stack of cards, 9 cards are blue and the rest are red. If 2 cards are to be chosen at random from the stack without replacement, the probability that the cards chosen will both be blue is 6/11. What is the number of cards in the stack?

A.10 B.11 C.12 D.15 E.18

grad_mba · Answer

Let there be n cards, 

9B => (n-9)R

given, 9C2/nC2 = 6/11

now solving, n(n-1) = 132 => n = 12

andrealittrell · Answer

hey.. i am new and just started studying.. i WAS good at finite math.. but i completely forget it!

i have a question.. what are the "C;s" ur referring to??

also, i thought u could do this.
 
i would try with 11 cards and do

9/11 x 8/11 (bc u would have only 8 left the next time)
that was nto right, and it needed to be less so then i tried

9/12 x 8/12 and it seemed correct for me. so therefore, 12 cards???

please, correct me if i am wrong! also, if anyone has sheets on probability, that be sweet!

andrealittrell · Answer

woops! i am awesome!!

basically.. for the first card u have 9/12 x 8/11 (bc now 1 is gone.. forgot about that!)

and that works.. u can factor it down.. i guess that is the easiest for me.. 


i also need clarification on what the C means.. sitll do not get it!

javed · Answer

Let there be x red cards so the total no. of cards are 9+x. there for probability of choosing 2 red cards without replacement is (9/9+x)* (8/8+x)=6/11.
Therefore x^2 + 17x - 60 = 0.
(x-3)(x-20)=0.
Therefore x=3 or x=20. x cannot be negative so x=3. Therefore the total no.s of cards are = 9+3=12. Hence the answere is C.

Javed.

Cheers!

Caas · Answer

x-number of cards
9/x - probability of choosing one blue card
8/(x-1) - probability of choosing the secong blue card without the first having been replaced
Prob1 AND Prob2 = 9/x * 8/(x-1)=6/11
Solving this equation we get 12
The answer is C

dvtohir · Answer

Thanks guys,

IMO is also C.

andrealittrell: 

C in 9C2 stands for the term "combination". In this case: # of  different  2-card combinations taken out of 9 cards. The formula: n!/((n-r)!*r!), where n is the total number of objects and r is # of objects in the combination taken out of the total. In this problem n=9 and r=2.

F75 · Answer

I do it the same way as Caas' 
6*12 / (x(x-1)) = 6/11
12*11 = x(x-1)

but instead of solving I see what solution fit in the equation.
Potential solutions are A.10 B.11 C.12 D.15 E.18

Then answer C.

querio · Answer

Hi Caas,

I also followed the same method as you did, but I got stucked at the moment of calculating the roots of x^2-x-132. I didn't know how to calculate the square root of 529... Is there any quick way to get this number?

thanks!

PS-Probability-# of cards

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