In a stack of cards, 9 cards are blue and the rest are red

Okay here is best bet to use plug in method

lets start with C
probability to get first blue is 9/12 x second blue 8/11

bingo! that turns out to be 6/11

IMO C

\(9/(9+x)*8/(8+x) = 6/11\)

\(0 = 17x+x^2-60\)

\(0 = (x+20)(x-3)\)

x=3

9+3 = 12

the answer is (C)

Total card x
so,
9/x*8/x-1 = 6/11
x^2 - x -132= 0
x = 12
Ans. C

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

9 - B
let total cards be n

(9/n) * (8/(n-1)) = 6/11
(72 * 11) / 6 = n (n-1)
12 * 11 = n (n-1)

n = 12

Hi All,

This question can be solved by TESTing THE ANSWERS.

We're told that we have 9 blue cards and an unknown number of red cards. We're also told that if 2 cards are to be chosen at random from the stack WITHOUT replacement, then the probability that the cards chosen will BOTH be BLUE is 6/11. We're asked for the TOTAL number of cards.

Normally, when TESTing THE ANSWERS, we should start with either B or D. Answer B looks easier, so let's start there.

Answer B: 11 total cards

With 11 total cards, and 9 blue cards, the probability of pulling two blue cards is...
(9/11)(8/10) = 36/55

Since 6/11 = 30/55, this is clearly NOT the answer. The probability that occurs with Answer B is a little TOO BIG, so we need an answer that lowers the probability (and thus, requires MORE total cards...).

Answer C: 12 total cards

With 12 total cards, and 9 blue cards, the probability of pulling two blue cards is...
(9/12)(8/11) = 24/44 = 6/11
This is an exact MATCH for what we were told, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Assume the numbe rof red cards = x
Total cards = 9 + x

P(Both blue cards) = \(\frac{9}{{x+9}} * \frac{8}{{x + 8}}\) = \(\frac{6}{11}\)

\(\frac{72}{{(x + 9)(x + 8)}}\) = 6/11
132 =\(x^2\) + 17x + 72
\(x^2\) + 17x - 60 =0
(x + 20)(x - 3) = 0
x can take only positive values, hence x = 3

Total cards = 9 + 3 = 12
Option C

Just adding another way to answer such a question.

\(b=9\)
\(x=b+r\)

\(\frac{b}{x} *\frac{b-1}{x-1}=b/x\)
\(\frac{9}{x} *\frac{8}{x-1}=6/11\)
\(\frac{72}{x(x-1)}=\frac{6}{11}\)

\(12*11=x(x-1)\)

\(x=12\)

So answer is c

The algebra on this was brutal...took me 4+ minutes which is absurd for a question of this difficulty. How do you when you should use the answers to solve?
_________________

BLUE cards = 9

Probability of drawing 2 blue cards =6/11

Let assume X be total no of cards

probability of drawing 1 blue card = 9/x

now total cards in deck are x-1 and blue card remaining are 8

Probability of drawing 2nd blue card = 8/x-1

Therefore probability of drawing 2 blue cards =6/11=9/x*8/x-1

solving equation, x=12

(9/9+x)*(8/8+x)=6/11

72/[(9+x)*(8+x)]=6/11

11*72=6(9+x)*(8+x)

11*12=(9+x)*(8+x)
so we want 9+x=12 and 8+x=11
hence x=3

Please how did you quickly figure out that 12 x 11 = 132

Hi Dufa,

Depending on how you prefer to do Arithmetic, you might sometimes find it faster/easier to break a calculation into 'pieces'

For example:

12 x 11 = (12 x 10) + (12 x 1) = 120 + 12 = 132

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

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