The volume of a rectangular box x inches wide, y inches long, and z inches high is 108 in^3. If x≤y≤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 the option A .
Experts, could you show me how to show that the A is the correct answer? I'd be thankful for your help.
The volume of a rectangular box x inches wide
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We are given that the volume of a rectangular box x inches wide, y inches long, and z inches high = 108 in^3 if x ≤ y ≤ z.VJesus12 wrote:The volume of a rectangular box x inches wide, y inches long, and z inches high is 108 in^3. If x≤y≤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 the option A .
Experts, could you show me how to show that the A is the correct answer? I'd be thankful for your help.
Thus, xyz = 108
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.
The largest face of the box would be yz since we are given that x ≤ y ≤ z.
Thus, yz = 36. We already know that xyz = 108. Thus, x = (xyz) / (yz) = 108/36 = 3 in
Thus, x + yz = 3 + 36 = 39. Sufficient.
(2) The longest edge of the box is 9 inches.
=> z = 9. This can't help. We can't the value of x + yz. Insufficient.
The correct answer: A
Hope this helps!
-Jay
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