Here's my approach:
Let "k" be the number of kings (black) that may be part of the 8 BLACK cards added to Deck 1.
No. of red aces in the new deck remains unchanged i.e. 2
No. of cards in the new deck = 52+8 = 60
No. of kings in the new deck = 4+k (2 Red kings + 2 Black kings + k Black kings)
Thus, probability of picking a red ace or a king is :
2/(52+8) + (4+k)/(52+8) = (6+k)/60
Now, since it's given that the above probability is greater than 1/8
We get,
(6+k)/60 > 1/8
i.e. k > 3/2
i.e. k > 1
k can only take 3 values (0 black kings or 1 black king or 2 black kings)
Thus, since k>1, k=2.
Now, the required probability of Amy picking a black king or a red jack from the new pack can be calculated as follows:
Total no. of black kings in the new pack = 4 (2 from Deck 1 + 2 from the 8 cards added to Deck 1)
Total no. of red Jacks in the new pack = 2
Thus,
4/60 + 2/60 = 6/60 = 1/10.
I got (D) as the answer.
Did anyone else get the same answer???
Cheers,
Taz
_________________
If this post helped you in your GMAT prep in anyway, please take a moment and hit the "Kudos" button.
It'll make my day