Today’s GMAT challenge question comes from ManhattanGMAT. To help you with your GMAT studying, try to solve the problem on your own, and then read on for the explanation of its solution:
Problem
A rectangular solid has side lengths 3, 4, and z, where z is an integer. The total surface area of this solid is the same as that of a cube with side length s, where s is an integer. If z is the lowest possible integer that fits these conditions, what is s?
(A) 3
(B) 5
(C) 6
(D) 7
(E) 9
Solution
The surface area of a rectangular solid is the total area of all of its faces. Since the side lengths are 3, 4, and z, there will be two faces with area 12 (= 3×4), two faces with area 3z, and two faces with area 4z. Thus, the total surface area of this solid will be 2×(12 + 3z + 4z) = 24 + 14z.
Meanwhile, since a cube has 6 faces, its surface area is 6 times the area of one face, or 6s2.
We now set these expressions equal to each other.
24 + 14z = 6s2
12 + 7z = 3s2
We need to find two integers, z and s, that fit this equation, and z needs to be its lowest possible value. At this point, we should switch to testing numbers—although we can save ourselves some work if we notice that z must be a multiple of 3. (Why? Notice that the right side, 3s2, is a multiple of 3, and on the left side, 12 is a multiple of 3. That means 7z must be a multiple of 3, and since 7 is not a multiple of 3 itself, z must be a multiple of 3.) Also, z must be a positive integer (otherwise, the rectangular solid would not be a solid), so we can start with z = 3 and check.
If z = 3, 12 + 7z = 12 + 21 = 33 = 3s2, meaning that s2 = 11, but 11 is not a perfect square. No good.
If z = 6, 12 + 7z = 12 + 42 = 54 = 3s2, meaning that s2 = 18, but 18 is not a perfect square. No good.
If z = 9, 12 + 7z = 12 + 63 = 75 = 3s2, meaning that s2 = 25, so s = 5. This fits.
Since we are asked for s, we must pick 5. Be careful not to pick z (which would be answer E).
The correct answer is (B).
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Read the full article: GMAT Tip: Solid Surfaces