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The surface area of a square tabletop was changed so that

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The surface area of a square tabletop was changed so that [#permalink] New post 31 Aug 2003, 08:39
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.

Answer is D. Please explain how? It think it is A
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 [#permalink] New post 31 Aug 2003, 10:41
It is D, because the only dimension was reduced by 1 inch. The other one was increased by 2.
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 [#permalink] New post 31 Aug 2003, 18:13
Stolyar, could you please give a more detailed explanation. I still do not understand.
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 [#permalink] New post 05 Sep 2003, 07:08
The answer is D

Assuming the length of the Square table is L. this should also be the Breadth. Now one is incremented and the other decremented

So the new dimensions will be L-1 and L+2. The new surface area would be
( L -1 ) (L + 2 ) = 70

Solving for L will give you 1 +ve val . Using that area of the sqaure can be determined


Statement 2 says that one dimension had a 25 % inc

Therefore %inc = ( amt of inc * 100 ) / Actual amt
in this case 25 = ( (L + 2 - L) * 100 ) / L
This should help you solve for L and therfore get the area of the square

Answer is D both statements satisfy the condition
  [#permalink] 05 Sep 2003, 07:08
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