manikdhingra Posted June 30, 2008 Share Posted June 30, 2008 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 Quote Link to comment Share on other sites More sharing options...
Creslin Posted June 30, 2008 Share Posted June 30, 2008 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. Quote Link to comment Share on other sites More sharing options...
Test Taker Posted June 30, 2008 Share Posted June 30, 2008 D it is. Let s be the length of each side. After changing,the sides would be s-1 and s+2. 1.This gives (s-1)(s-2)=70.Solving this gives s=8.suff 2 s*1/4 = 2 =>s=8 suff Quote Link to comment Share on other sites More sharing options...
manikdhingra Posted July 1, 2008 Author Share Posted July 1, 2008 yeah,.. u rite... ans is d...thanks guys......actually i was assuming the table top as a cuboid.....thats where i went wrong....... Quote Link to comment Share on other sites More sharing options...
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