Question

A, B, C, D, and E play a game of cards. B has 6 cards less than E and A has 6 cards less than D. A and B together have a third of the number of cards D and E have. If B has two cards less than what C has and the total number of cards is 30, how many cards does B have?

• A. 5
• B. 3
• C. 4
• D. 2

Answer 1

The correct option is C 4

We denote:

$Person$	$A$	$B$	$C$	$D$	$E$
$Numberofcards$	$a$	$b$	$c$	$d$	$e$

Now, we know that B has 6 cards less than E and A has 6 cards less than D.

$⟹b=e–6a=d–6$

We also know that A and B together have a third of the number of cards D and E have.

$⟹a+b=\frac{d+e}{3}$

Combining the two equations, we find:

$(e–6)+(d–6)=\frac{d+e}{3}⟹(d+e)–12=\frac{d+e}{3}⟹2\times \frac{(d+e)}{3}=12⟹d+e=\frac{12\times 3}{2}⟹d+e=18$

Using this, we can caluclate the total number of cards A and B together have:

$a+b=\frac{d+e}{3}=\frac{18}{3}=6$

Next, we know that B has 2 cards less than what C has.

$b=c–2⟹c=b+2$

And finally, we know that the total number of cards they have together is 30.

$⟹a+b+c+d+e=30⟹(a+b)+\left(c\right)+(d+e)=30⟹\left(6\right)+(b+2)+\left(18\right)=30⟹(b+2)+24=30⟹b=30–26⟹b=4$

Hence, B has 4 cards.