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27 Mar 2017, 01:44
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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.
(1) The area of the largest face of the box is 36 \(in^2\).
(2) The longest edge of the box is 9 inches.
27 Mar 2017, 01:55
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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?
The volume = (width)*(length)*(height) = xyz = 108.
(1) The area of the largest face of the box is 36 \(in^2\) --> since \(x ≤ y ≤ z\), the area of the largest face is yz, thus given that yz = 36. Substitute in the volume formula: x*yz = 36x = 108 --> x = 3. Therefore, x + yz = 3 + 36 = 39. Sufficient.
(2) The longest edge of the box is 9 inches --> z = 9 --> xy = 12. Multiple values are possible: for example, x = 2 and y = 6, OR x = 3 and y = 4. Not sufficient.
Answer: A.
Hope it's clear.
The volume = (width)*(length)*(height) = xyz = 108.
(1) The area of the largest face of the box is 36 \(in^2\) --> since \(x ≤ y ≤ z\), the area of the largest face is yz, thus given that yz = 36. Substitute in the volume formula: x*yz = 36x = 108 --> x = 3. Therefore, x + yz = 3 + 36 = 39. Sufficient.
(2) The longest edge of the box is 9 inches --> z = 9 --> xy = 12. Multiple values are possible: for example, x = 2 and y = 6, OR x = 3 and y = 4. Not sufficient.
Answer: A.
Hope it's clear.
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27 Mar 2017, 02:09
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Statement 1 :
yz=36
So, x=3
So, after putting the value of x and yz, we get a definite result, so SUFFICIENT.
Statement 2:
here, z=9
we know that xyz=108
So, xy=12
for different values of xy, it gives 12. So, the value of x+yz vary time to time.=>Insufficient.
So, The correct choice is .
Thank you...
