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Questions on 2 ps problems

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Questions on 2 ps problems

by xcise_science » Sat Dec 01, 2007 5:05 pm
For Q3, I just plugged in, I know there is a faster (& hopefully not complicated) way but I don't know it.

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Q24
here's what I did:
[1001^2 - 999^2] / [101^2 - 99^2]
==> [(1001+999)(1001-999)] / [(101+99)(101-99)]
==> (1001*1001)/(101*101) and I just didn't know where to go from here. I knew the answer had to be A or D.


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by Suyog » Sat Dec 01, 2007 6:02 pm
1. Let the first number be x so the a1 = x, a2 = 2x and so on.
The number are x, 2x, 4x, 8x, 16x

Given a5 - a2 = 12
so 16x - 2x = 12
14x = 12
x = 6/7

2. (999 + 2)^2 - 999^2 / (99+2)^2
=999^2 + 2(1998) + 4 - 999^2/ 99^2 + 2(198) + 4 - 99^2
=4000/400
=10
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