Last week, Manhattan GMAT posted a GMAT question on our blog. Today, they have followed up with the answer:
First, get a sense of the sixteen rectangles. Here are some of the possibilities:
1×1
1×2
1×3
1×2
1×3
1×4
2×1 (notice that this is a different rectangle from the 1×2)
2×2
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.
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:
2×3 and 3×2
1×4 and 4×1
2×4 and 4×2
3×3
Total: 7 rectangles
Area between 4 and 8, inclusive:
2×2
2×3 and 3×2
1×4 and 4×1
2×4 and 4×2
Total: 7 rectangles
There are only 6 triangles on both lists:
2×3 and 3×2
1×4 and 4×1
2×4 and 4×2
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.