Bunuel wrote:
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
total combinations: 16 rectangles = 4 choices length • 4 choices of width = 16 combinations
4 ≤ area ≤ 8 and 10 ≤ perimeter ≤ 12 are made of integers 1 ≤ length • width ≤ 4
if area = 4, then lw={4,1}{2,2}{1,4}, but lw={2,2} gives perimeter=2(2+2)=8 outside range, lw={4,1}{1,4}=2 options
if area = 6, then lw={2,3}{3,2}{6,1}{1,6}, but {6} outside range, lw={2,3}{3,2}=2 options
if area = 8, then lw={4,2}{2,4}{8,1}{1,8}, but {8} outside range, lw={4,2}{2,4}=2 options
if area = 5, then lw={5,1}{1,5}, but {5} is outside range, 0 options (notice this happens for all "prime" areas)
if area = 7, then 0 options;
favorable combinations: {2,3}{3,2}{4,2}{2,4}{4,1}{1,4}=2+2+2=6
probability(fav/total): 6/16=3/8
Answer (C)