SOLUTION: How to find at least two new dimensions of the new box that's 4"by12"by8" that both have volumes 20% more than 384. Please help, so lost.

Algebra ->  Volume -> SOLUTION: How to find at least two new dimensions of the new box that's 4"by12"by8" that both have volumes 20% more than 384. Please help, so lost.      Log On


   

Question 720366: How to find at least two new dimensions of the new box that's 4"by12"by8" that both have volumes 20% more than 384. Please help, so lost.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
That's a confusing question.
+4%2A12%2A8+=+384+
So, I guess they want 2 different boxes whose
volumes are 20% more than the given box
+384+%2B+.2%2A384+=+1.2%2A384+
+1.2%2A384+=+460.8+
-------------------
I think the thing to realize here is that if you
multiply any one of the 3 dimensions of the
original box by +1.2+, you will get a
volume that is 20% larger, so
+8%2A1.2+=+9.6+
4 x 12 x 9.6 is a possible box
-------------------------
+12%2A1.2+=+14.4+
4 x 14.4 x 8 is a possible box
-------------------------
+4%2A1.2+-+4.8+
4.8 x 12 x 8 is a possible box
--------------------------
Note all 3 give volume = 460.8 in3