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VIC problem from Knewton

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by srcc25anu » Wed Mar 23, 2011 4:47 pm
a+b = p
a-b = q
adding the two we get 2a = p=q or a= p+q/2
now b = a-q or p+q-2q / 2 = p-q/2
ab = (p+q/2)*(p-q/2) = p^2-q^2 / 4
ans A (typo error)
Last edited by srcc25anu on Wed Mar 23, 2011 7:53 pm, edited 1 time in total.
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by ajmoney09 » Wed Mar 23, 2011 6:05 pm
Answer is B.

I plugged in easy values.

a+b = P & a-b=q

IF:
a=5
b=3

5+3=8
5-3=2
P = 8
Q = 2

a(b)=
5(3) = 15

Plug and chug now.... usually start from bottom to top because on problems like this they want you to waste time, plugging and chugging every where and it is sometimes on the bottom, but you can easily remove some answer choices.... E is obviously removed.
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by rohu27 » Wed Mar 23, 2011 7:38 pm
option A right?
srcc25anu wrote:a+b = p
a-b = q
adding the two we get 2a = p=q or a= p+q/2
now b = a-q or p+q-2q / 2 = p-q/2
ab = (p+q/2)*(p-q/2) = p^2-q^2 / 4
ans B
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by manpsingh87 » Wed Mar 23, 2011 7:46 pm
tonebeeze wrote:Please see attachment. Can someone walk me through the algebra. Thanks
a+b=p;---------1)
a-b=q;----------2)

squaring equations 1 and 2 we get;

(a+b)^2=p^2;----3)
(a-b)^2=q^2;------4)

Now subtracting 3 and 4 we have

4ab=p^2-q^2;

ab= p^2-q^2/4 ;

hence A
O Excellence... my search for you is on... you can be far.. but not beyond my reach!
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by srcc25anu » Wed Mar 23, 2011 7:54 pm
Yes, my mistake. its A indeed.
rohu27 wrote:option A right?
srcc25anu wrote:a+b = p
a-b = q
adding the two we get 2a = p-q or a= p+q/2
now b = a-q or p+q-2q / 2 = p-q/2
ab = (p+q/2)*(p-q/2) = p^2-q^2 / 4
ans B
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