Player A dealt some cards to player B and himself from a full pack of playing cards and laid the rest aside. A then said to B "If you give me a certain number of your cards, I will have four times as many cards as you will have. If I give you the same number of cards, I will have thrice as many cards as you will have." Of the given choices, which could represent the number of cards with player A?
(a) 9
(b) 31
(c) 12
(d) 35
My work :
$52 = y + (52 -y)$ -----initially y cards were dealt in total.
$y = c1 + (y-c1)$ -----cards with players A and B respectively
Condition 1 : $c1 + a = 4 ( y - c1 - a )$
Condition 2 : $c1 - b = 3 ( y - c1 + b)$
Eliminating y, we get c1 = 15a + 16b
The possible combinations for a and b are $(1,1), (1,2), (2,1)$.
So for the given choices, it should be the option b but I am not convinced. Is the approach correct?