In the grid above, the variables a through p are each equal to 2, 3, 5

(1) bcehilno = \(35^4 = 5^4 * 7^4\)
so bc=35; eh=35 ,il=35 , no=35
hence fg=2*3=6 and kj=2*3=6

Sufficient.


(2) (2.25^2)afkp = dgjm
\(2.25^2 = 1.5*1.5*1.5*1.5 = \frac{3}{2}* \frac{3}{2}* \frac{3}{2}* \frac{3}{2}\)
so, d,g,j,m=3 and a,f,k,p=2
sufficient.

Answer D

MANHATTAN GMAT OFFICIAL SOLUTION:

Although there are 16 different variables in this 4×4 grid, the values of these variables are severely restricted: they can only be 2, 3, 5, or 7 (you should recognize these as primes), and only one of each can exist in any row or column. You are asked for the value of the product fgjk, which is the 2×2 grid at the center.

(1) SUFFICIENT. Note that 35^4 = 5^4*7^4, so four of the variables must be 5’s, while the other four must be 7’s. You are not allowed to put two of the same values in any row or column, so a possible arrangement (there are many) looks like this:


The central grid must be composed of two 2’s and two 3’s, since each row and each column must have exactly one of each possible value, and you’ve already used up the 5’s and 7’s. Although you cannot be sure which particular variable in the central 2×2 grid is which value, you know that the product is 36 (2×2×3×3).

(2) SUFFICIENT. First, use FDP connections: rewrite 2.25 as 9/4, or (3/2)^2. Thus, 2.25^2= (3/2)^4. Cross-multiplying by 2^4, you get \(3^4afkp = 2^4dgjm\).

Since the only possible values of the variables are 2, 3, 5, and 7, the only possibilities for the values of the variables are a = f = k = p = 2, while d = g = j = m = 3.

This tells us that fgjk = 36.

The correct answer is D.

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