Hi,

let's say \(\frac{a}{b}\) = \(x\)

And if we divide both the quantities with \(b\) , we will get
Qt A = \(\frac{a}{b}\) = \(x\) & Qt B = \(1\)

We are given that \(\frac{a^2}{b^2}=\frac{a}{b}\)
put \(\frac{a}{b}\) = \(x\)

\(x^2 = x\) :::: \(x = 1\)

By putting the value in Qt A, we will get \(Qt A = Qt B\)

Answer C

rx10 sir,
Whats wrong in my approach? THANKS IN ADVANCE

Sir,
Can you please tell me how Qt A will be \(x^2\) ?
Go through your solution again.

Consider this : rather than \(a/b\) take \(p/q\)

You'll find it.

Given: (a^2/b^2).....(a/b)^2......i assume (a/b) as x so, this (a/b)^2 becomes x^2

Qt A is \(a\) only.

So that can't be \(x^2\) , for that we will need \(\frac{a^2}{b^2}\)

Please ask if the doubt remains.

HAHA. I made silly mistake. NOw I got it sir. Thanks

a and b must be different from zero.

From this we infer that the fraction a/b b cannot be zero otherwise the fraction is indefinite

Now, for being a^2/b^2=a/b

a and b must be one

but also 2, 3 four. In any case we will end up that 1=1

C is the answer

Now i edit my answer and it goes like this.

I put \(\frac{(a}{b)} =x\).
Now, quantity \(a= x^2 \)and quantity b =x




subtracting x in both quantity,\( x^2 -x\) in a and 0 in b,


Solving a getting x=1 and x =0

if x=0,
a/b =0, ....


or, x =1

a/b = 1 implies a = b.

Either / or signifies it must be a =b.
Hence answer is c.

Multiply both the denominators to the opposite side. We can do this because the denoms can not be zero.
We are given that ab is not equal to zero so both a and b is not zero.
So we can cancel them out from the equation with the equation remaining to be a=b.