TestPrepUnlimited wrote:
Bunuel wrote:
If s is an integer greater than 1, how many 1-inch cubes can be packed into a rectangular box having sides \(s\), \(s + \frac{3}{2}\), and \(s - 1\), measured in inches?
Since we cannot liquidate the cubes, we have to take the biggest integer of each side that we can use. Among the length \(s + \frac{3}{2}\) we can only use \(s + 1\) of that length. Since the rest of the lengths are in integers we can use them directly.
Then the area is \(s * (s + 1) * (s - 1) = s * (s^2 - 1) = s^3 - s\).
Ans: A
Hi
TestPrepUnlimited and
Bunuel,
I am so confused... Here is what I had:
Assume S=3, integer less than 1, thus sides of the rectangular would be 3, 3+3/2= 9/2, 3-1=2, the volume of the rectangular 3*9/2*2= 27 inches, divided by 1-inches cube equals to 27 cubes.. plug the sides of the rectangular into all the options:
A. S^3-S = 27-3= 24
B. S^3 + S^2/2 + S/2 = 27+9/2+3/2 =36
C. S^3 - 2S + S = 27-6+3 = 24
D. S^3+S^2 -S = 27+9-3=33
E. S^3 = 27
Only E has the same value as 27 cubes.. Can you correct me what went wrong?
Thanks in advance!