VIC problem from Knewton

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)

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.

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

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

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