Manhattan GMAT Challenge Problem of the Week - 4 April 2011

by Manhattan Prep, Apr 4, 2011

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Question

Sixteen rectangles are formed by choosing all possible combinations of integer length l and integer width w such that 1 l 4 and 1 w 4. The probability that a rectangle chosen at random from this set of rectangles has perimeter p and area a such that both 10 p 12 and 4 a 8 is

(A) 9/256

(B) 49/256

(C) 3/8

(D) 7/16

(E) 7/8

Answer

First, get a sense of the sixteen rectangles. Here are some of the possibilities:

11

12

13

14

21 (notice that this is a different rectangle from the 12)

22

etc.

To compute the desired probability, count the rectangles that meet both conditions. Note that some rectangles will meet one condition but not both. If necessary, make separate lists, then cross-reference.

Perimeter between 10 and 12, inclusive:

23 and 32

14 and 41

24 and 42

33

Total: 7 rectangles

Area between 4 and 8, inclusive:

22

23 and 32

14 and 41

24 and 42

Total: 7 rectangles

There are only 6 triangles on both lists:

23 and 32

14 and 41

24 and 42

Thus, the probability of choosing one of these triangles is 6/16, or 3/8.

One trap in this problem is that you might think that you should calculate the separate probabilities of meeting either condition and then multiply those probabilities together, because there is an and condition at work. However, these probabilities are not independent. The perimeter and the area of the rectangles in question are not independently determined. Instead, what you should do is simply count the rectangles that meet both conditions, then divide by the total number of rectangles.

The correct answer is C.

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