Problem Solving

We must remember that on the GMAT there are no points for style. Our goal is not to solve the problem but to select the correct answer choice.

We can start by using common sense. Mark has fewer cards than does John. Accordingly, Mark must have fewer than 60 cards. Answer choices (A) and (B) should be eliminated at the outset.

Since we are left with 3 choices, we can easily backsolve the problem by selecting the middle of the remaining three choices, aka (D). If Mark has 46 cards before the swap, then John must have 120-46 = 74 before the swap. When John gives 5 to Mark and gets 2 in return, he has a net loss of 3 cards, bringing his count to 71 whereas Mark gains 3 and has 49 after the swap. This is a 22-card difference. Accordingly (D) must be the best answer. Had (D) not been the best answer, we would have known whether to go for a higher or lower number. Accordingly, we would not have had to retest the answers.

However, perhaps there are some of you out there who are not interested in go-for-the-throat solutions. You want something elegant to wow your friends. Very well, try this:

Since after the swap, the difference between the number of John's cards and that of Mark's cards was 22, we can conclude that if John had 22 fewer cards that Mark and John would have the same number of cards. So we can subtract 22 from 120 (the number of total cards), leaving 98. Thus, Mark has 98/2 = 49 cards after the swap. From there we need only "undo" the swap by returning 5 cards to John and returning 2 cards to Mark, leaving Mark with 3 fewer cards. Thus Mark must have had 46 cards before the swap. This is answer choice (D).

lheiannie07 wrote:John and Mark each own a collection of baseball cards. The two collections combined contain 120 cards. If John were to trade 5 cards to Mark and receive 2 of Mark's cards in return, John would have 22 more cards than Mark does. How many cards does Mark possess before the proposed trade?

A. 85
B. 74
C. 52
D. 46
E. 35

I'm confused how to set up the formulas here. Can any experts help?

OA D

Hello lheiannie07.

I would sove it like this: $$J+M=120.$$ Also $$J-3=M+3+22\ \ \ \Rightarrow\ \ J-M=28.$$ Now, if we add this two expressions we get
$$2J=148\ \Rightarrow\ \ J=74.$$ This implies that Mark would have 46 cards before the trade. Therefore, the correct answer is the option D.

let 'j', 'm' represent the amount of cards in John and mark's collections respectively.
$$j+m=120\ ------\left(i\right)$$
If john gives 5 cards to Mark and receive 2 in return, John now have $$j-5+2=j-3\ cards\ and\ mark\ has\ m+3\ cards$$
$$j-3=22+\left(m+3\right)$$
$$j-m=28----\left(ii\right)$$
$$solving\ \left(i\right)\ and\ \left(ii\right),\ $$
$$j+m=120$$
$$j-m=28$$
$$we\ have\ 2m=92......\ Hence,\ m=46$$
$$Therefore,\ mark\ has\ 46cards\ before\ the\ trade\ \left(option\ D\right)$$