Think theoretically, it is much better than plug-in number this time. It is cumbersome.

QA \(\frac{x}{x} + \frac{5}{x}\)

QB \(\frac{(x-1)}{(x-1)} + \frac{5}{(x-1)}\)

\(\frac{x}{x}\) and \(\frac{(x-1)}{(x-1)}\) are both equal to one

so we will end up with \(\frac{5}{x}\) and \(\frac{5}{x-1}\)

Now, if you plug an integer B > A but if you plug a fraction A > B. So the answer is D but the OE in the book is B.

Clearly a mistake. Hope this helps.

Regards

Great-- looks like I am learning now ..Thanks to this forum .what is OE book?

Actually the explanation I gave you. It takes into account only integers greater than one, which is impossible because you must take in account even fractions.

Hope this helps.
Regards

Thanks

I can't seem to make a clear decision. Everytime I am trying out any value I am getting B>A. Now, let's say after simplification we are dealing with

Quantity A Quantity B

1/X on side 1/ (X-1)

But if I take let's say X to be a fraction i.e. 8/3 (X has to be greater than 1)
then on side A we get 3/8= 0.375
and on side B we get 3/5= 0.6. in this case B>A

Again, when I try out an integer, say, 5,
I get 1/5 on side A
and 1/4 on side B.

In fact, when I apply the theory I get, that quantity B should be greater. Since X>1 that means X is a positive number. So, when X is in the denominator the result should be less than when a value less than X is in the denominator the result should be greater, provided that the numerator remains the same.

Could please explain at length. Cause I can't figure out a case where A>B.

Sincerely,
Scrat

This explanation is not right. Clearly, with x > 1, B is always greater, regardless of what you take for x, an integer or fraction>1

even after reducing it to

\frac{5}{x} and \frac{5}{(x-1)}, given that x>1 it can be clearly observed that \frac{5}{(x-1)} has a smaller denominator than \frac{5}{x} while the numerator in both cases is same. Hence \frac{5}{(x-1)} has to be greater. Correct answer is B.

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