SOLUTION: A rectangle has integer side lengths. One pair of opposite sides is increased by 30% and the other pair of sides is decreased by 20%. The new side lengths are also integers. What i

Algebra ->  Rectangles -> SOLUTION: A rectangle has integer side lengths. One pair of opposite sides is increased by 30% and the other pair of sides is decreased by 20%. The new side lengths are also integers. What i      Log On


   

Question 203191: A rectangle has integer side lengths. One pair of opposite sides is increased by 30% and the other pair of sides is decreased by 20%. The new side lengths are also integers. What is the smallest possible area, in square units, of the new
rectangle?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
Let x and y represent the sides of the original rectangle. And we are given that they are both integers. (And, since they are sides of a rectangle, we must assume that they are positive integers.)

If we increase a side by 30% then the new side is 130% of the original side. 130% = 130/100 = 13/10. So if the original side is x the increased side is %2813%2F10%29x.
The problem tells us that this new side, %2813%2F10%29x is also an integer. The only way %2813%2F10%29x could be an integer is if x is a multiple of 10. And since x must be positive and we want the smallest possible area, x must be 10. And the increased side would be %2813%2F10%29%2A10+=+13

Similarly, if we decrease y by 20% the new side is 80% of the original side. 80% = 80/100 = 4/5. So the decreased side is %284%2F5%29y. This also must be an integer so y must be a multiple of 5. And since y must be positive and since we are interested in the smallest area, y must be 5. And the decreased side is %284%2F5%29%2A5+=+4.

With the new sides of 13 and 4, the smallest possible are is 13*4 = 52.