I thought the OA was confusing until I realized that the trick to this question is in analyzing the answer choices.

There are several possible combinations of integers multiplied together that produce values bet'n 15 and 200. But the Q is to seek a low card and a high card contained in the deck.

Now, observe the answer choices.

A: 3 and 15 - While the lower extreme of 3 could certainly satisfy, the upper extreme falls outside of the requirements of the question stem. 15 multiplied by any greater integer will be >200.

B: 3 and 20 - See reasoning for 'A' above (i.e. 20 multiplied by any greater integer will be >200)

D: 4 and 14 - See reasoning for 'A' above (i.e. 14 multiplied by any greater integer will be >200)

E: 5 and 14 - See reasoning for 'A' above (i.e. 14 multiplied by any greater integer will be >200)

That only leaves choice C. 13 by 14 results in 182, which lies between 15 and 200.

I hope my explanation is clear....

Cheers,
Sandra

Just to add to beeparoo's explanation...

The problem states that the children multiply the card's value by a number greater than that card. So for example, you had a card value "5". Now you could multiply this by any number greater than 5, say 6/7/8/9... to get a result 30/35/40/45...

But the question also mentions that the range of these multiplications is between 15 and 200. So if you had a card 3, you could multiply that by 4, but that would give you 12 (<15). So the minimum card would have to be 4 so that the min possible value is 4*5 = 20.

This leaves you with option (c) and (d) only. Now for the higher range, take the answer and plug it in to arrive at answer (c).