Bunuel wrote:
In the figure above, all angles are right angles. If the lengths a, b, u, and v are integers, what is the perimeter of the six-sided figure?
(1) ab = 77
(2) u/v = 3/5
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Official Solution (Credit: Manhattan Prep)
Before tackling the statements, note that the desired perimeter is the same as that of a rectangle measuring a by b units. This fact can be seen perhaps most easily by completing the rectangle:
See attachment!
The large rectangle can be formed by relocating the segments with lengths u and v to the locations of the dotted lines, without changing anything else in the figure, so the two perimeters are the same.
The desired perimeter is thus the same as the perimeter of the large rectangle, i.e., 2a + 2b.
This expression can also be found by filling in the lengths of the two missing sides. By subtraction, the unlabeled horizontal side has length b – v, and the unlabeled vertical side has length a – u. Therefore, the perimeter of the figure is a + b + u + v + (b – v) + (a – u) = 2a + 2b.
(1) SUFFICIENT: The lengths a and b are positive integers whose product is 77, so a = 11 and b = 7, or else a = 7 and b = 11. (Note that a and b cannot be 1 and 77. If a = 1 then u cannot be an integer; likewise, if b = 1 then v cannot be an integer.)
In either case, the desired perimeter is 2a + 2b = 2(11) + 2(7) = 36.
(2) INSUFFICIENT: This statement gives no information about the values of a or b, so the desired perimeter cannot be found.
The correct answer is A.
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