If a and b are non zero integers is ab/(a + b) < ab?

Asked: If a and b are non zero integers is ab/(a + b) < ab?
Is ab/(a+b) < ab?
Is ab/(a+b) - ab < 0?
Is ab {1/(a+b) - 1} < 0?
Is 1/(a+b) - 1 < 0?
Is 1/(a+b) < 1?
Is (a+b) > 1?

(1) a = 6b
If b is -ve ; a is -ve and ab<0
But if b = 2; a=12; ab=24 > 1
NOT SUFFICIENT

(2) a + b < 0
a + b < 0
a + b is NOT >1
SUFFICIENT

IMO B

I took a simple approach to this one that led to a clear and fast answer.

Statement (1)

Substitute for a:

\(\frac{6b*b}{6b + b} < 6b*b\)

\(\frac{6b^2}{7b} < 6b^2\)

Now, we can plainly see that both sides contain 6b squared, and we know that since b must be a nonzero integer, this product must be positive. We can thus reimagine the inequality:

\(\frac{POS}{7b} < POS\)

Now, if b is positive, the right-hand side must be greater, since we would be taking the same value on the left-hand side and dividing by a multiple of 7. If, on the other hand, b is negative, the left-hand side will be negative, while the right-hand side will remain positive. Thus, regardless of whether b is positive or negative, the answer to the question is YES, and Statement (1) is SUFFICIENT.

Statement (2)

Substitute for (a + b), keeping things as simple as possible:

\(\frac{ab}{NEG} < ab\)

Now, we know nothing about a and b individually, so we can think of their product (as nonzero integers, remember) as either positive or negative.

a) ab is positive—

\(\frac{POS}{NEG} < POS\) True.

b) ab is negative—

\(\frac{NEG}{NEG} < NEG\) False.

Since we get conflicting results, we can rule out Statement (2) as sufficient on its own.

The answer must be (A).

Good luck with your studies, everyone.

- Andrew