A, B, C and D play a game of cards. A says to B, ‘If I give you 8 cards, you will have as many as C has and I shall have 3 less than what C has. Also if I take 6 cards from C, I shall have twice as many as D has’. If B and D together have 50 cards, how many cards has A got?

Let number of cards with A be $$x$$, with B = $$y$$, with C = $$z$$ and with D = $$(50-y)$$

According to ques,

=> $$y+8=z$$ and $$x-8=z-3$$

Comparing above equations, => $$y+8=x-5$$

=> $$x-y=13$$ -------------(i)

Also, $$x+6=2(50-y)$$

=> $$x+6=100-2y$$

=> $$x+2y=94$$ -------------(ii)

Subtracting equation (i) from (ii), => $$3y=81$$

=> $$y=\frac{81}{3}=27$$

Substituting it in equation (i), we get : $$x=13+27=40$$

=> Ans - (D)