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by mendiratta » Wed Oct 17, 2007 11:03 pm
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.

OA given is D. But I think its wrong.
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by josephcho77 » Wed Oct 17, 2007 11:44 pm
Hi!
D is the correct answer.

Since a tabletop is a square, let x be the length and width.

After changes were made, the new area would be (x-1)*(x+2)

(1) says that (x-1)*(x+2)=70 which is x^2+x-72=0
(x+9)(x-8 )= 0
x can be -9 or 8
but since length or width can't be negative x is 8 (sufficient)

(2) says that 1.25x=x+2 thus x=8 (sufficient)
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by arora007 » Sat Jan 29, 2011 11:05 am
Oh! missed on the single world "square" and lost it!!
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