The volume of a rectangular box x inches wide, y inches long, and z inches high...

[BTGmoderatorLU]

Source: Manhattan Prep

The volume of a rectangular box x inches wide, y inches long, and z inches high is 108 in^3. If \(x \leq y \leq z\), what is x+yz?

1) The area of the largest face of the box is 36 in^2.
2) The longest edge of the box is 9 inches.

The OA is A

[Jay@ManhattanReview]

Source: Manhattan Prep

The volume of a rectangular box x inches wide, y inches long, and z inches high is 108 in^3. If \(x \leq y \leq z\), what is x+yz?

1) The area of the largest face of the box is 36 in^2.
2) The longest edge of the box is 9 inches.

The OA is A

So we have xyz = 108 and \(x \leq y \leq z\).

We have to get the value of x + yz.

Let's take each statement one by one.

1) The area of the largest face of the box is 36 in^2.

Since \(x \leq y \leq z\), the area of the largest face would be yz. Thus, yz = 36. Thus, x = xyz/yz = 108/36 = 3. And the value of x + yz = 3 + 36 = 39. Sufficient.

2) The longest edge of the box is 9 inches.

=> z = 9. Thus, xy = xyz/z = 108/9 = 12.

Case 1: Say y = 4. Thus, x = xy/y = 12/4 = 3. Thus, the value of x + yz = 3 + 4*9 = 39.
Case 2: Say y = 2. Thus, x = xy/y = 12/2 = 6. Thus, the value of x + yz = 6 + 2*9 = 24.

The correct answer: A

Hope this helps!

-Jay
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