Consider a = 3 and b = 4.
Condition A:
4 / 7 + 3

Condition B:
3/7 - 4

Hence, A will be greater.

Consider a = -4 and b = -3
Condition A:
3 / 7 - 4

Condition B:
4 / 7 + 3

Hence B is greater.

Thus, answer must be D.

Picking numbers could be misleading

better to use a general approach

QA simplified becomes \(\frac{-(a^2+b^2)}{a^2-b^2}\)

Which means that the quantity is indeed positive

QB is \(\frac{a^2+b^2}{a^2-b^2}\) which is negative

Positive > negative

A is the answer

But the restriction is that |a| < |b| so your second assumption is not correct.

If you solve it algebraically then, it becomes [b (a-b) - a(a+b)]/(a^2 - b^2) is the condition on Quantity A

then if you expand on it it becomes ab - b^2 - a^2 - ab which reduces to - a^2 - b^2 = - (a^2 + b^2)/(a^2 - b^2)

If you do the same for the other side you will end up with the same numerator - (a^2 + b^2)/(b^2 - a^2)

Since both sides are negative and when you compare quantity A to B the direction is usually A > B

However, since both are negative it becomes is a < b (For answer choice A) which then becomes

is 1/(a^2 - b^2) < 1/(b^2 - a^2)

Since |b| > |a| then the b^2 > a^2

the left side ends being negative and the right side ends being positive answering the inequality of whether A < B, hence the answer choice is A.

I hope this help @esaktasynov

Good catch, really appreciate it.

I did the exact same steps till \(\frac{-(a^2+b^2)}{a^2-b^2}\) and \(\frac{a^2+b^2}{a^2-b^2}\)

However I multiplied both the fractions by \(a^2-b^2\) to cancel the denominator on both the quantities before comparing

This led me to compare the two quantities as \(-(a^2+b^2)\) which is always negative and \(a^2+b^2\) from which is always positive

So I got my answer as B.

Was my last step wrong? Is it not allowed to cancel the denominators to simplify the quantities further?

You're right to say that your last step was wrong.

If |a| < |b|, then we know that a² < b², which means a² - b² is NEGATIVE
So when you multiplied both quantities by a²-b², you were inadvertently multiplying both quantities by a NEGATIVE value, which caused unintended results.

For example consider this question:
QUANTITY A: 2
QUANTITY B: 3
Here it's obvious that Quantity B is greater

However, if we multiply both quantities by -1, we get:
QUANTITY A: -2
QUANTITY B: -3
Now, we see that Quantity A is greater

Cheers,
Brent

Thank you so much, its clear now

you said QB is -ve whereas we are getting negative sign in statement A

I manipulated both sides to get to an answer:

\(\frac{b}{a+b} - \frac{a}{a-b} ? \frac{a}{b+a} - \frac{b}{b-a }\)

\(\frac{b-a}{a+b} - \frac{a}{a-b} ? - \frac{b}{b-a }\)

\(\frac{b-a}{a+b} ? - \frac{b}{b-a } + \frac{a}{a-b}\)

\(\frac{b-a}{a+b} ? - \frac{b}{b-a } + \frac{(-1)-a}{(-1)(b-a)}\)

\(\frac{b-a}{a+b} ? - \frac{b}{b-a } + \frac{-a}{b-a}\)

\(\frac{b-a}{a+b} ? \frac{-b-a}{b-a}\)

\(\frac{b-a}{a+b} ? \frac{(-1)(a+b)}{b-a}\)

Here I crossed multiplied. You usually wouldn't want to do this because you run the risk of changing the sign (the '?' in our algebra above). However, this is a unique situation, and can be better seen in the result of the cross multiplication:

\((b-a)^2 ? (-1)(a+b)^2\)

We don't have to worry about the cases when \(b-a\) and \(a+b\) are positive or negative since they are being squared, so they are always positive.

Given that \((a+b)^2\) is being multiplied by \(-1\), it must be the case that:

\((b-a)^2 > (-1)(a+b)^2\)

So A is greater.


The only case where this wouldn't be true is if they were equal, or Choice C. In other words:

\(b-a = a + b\)

But quick algebra rules this possibility out:

\(b-a = a + b\)

\(-a = a\)

\(0 = 2a\)

\(0 = a\)

Which would imply that \(b = 0\), but we know that both cannot be \(0\) because of the initial restriction given in the question: \(|a|<|b|\) (or by recognizing that the denominators in the fraction of both A and B cannot be 0).

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