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The surface area of a square tabletop was changed so that one of the dimensions was reduced by 1 inch and the other dimension was increased by 2 inches. What was the surface area before these changes were made?

(1) After the changes were made, the surface area was 70 square inches.

(2) There was a 25 percent increase in one of the dimensions.

 

guys please explain in detail

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I believe that the answer is D.

 

The stem tells us that a SQUARE (i.e. all sides equal) tabletop had one of its sides increased by 2 inches, and another decreased by 1. So, if we let "s" stand for the length of a side of the ORIGINAL tabletop, the formula for the surface area of the new tabletop is...

 

A = (s-1)(s+2) = s^2 + s - 2

 

So all we need to find is s and we can solve the problem.

 

(1) Tells us that the new SA is 70. So...

70 = s^2 + s - 2

=> s^2 + s - 72 = 0

=> (s-8)(s+9) = 0

So s can be either -9 or 8...it's a real length so it can't be negative, so I would go with 8, SUFFICIENT.

 

(2) Tells us that one of the sides increased by 25%. Only one side increased, and it increased by 2. So...

.25s=2, SUFFICIENT.

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