Nice question! (I think it's harder than medium)

If $xy < 0$, then there are two possible cases: EITHER x is positive and y is negative, OR x is negative and y is positive
Let's consider both possible cases.


case i: x is positive and y is negative

Take: $\frac{a}{x} < \frac{b}{y}$

Multiply both sides by x to get: $a < \frac{bx}{y}$ [since x is POSITIVE, the inequality symbol REMAINS facing the same direction]

Now multiply both sides by y to get: $ay > bx$ [since y is NEGATIVE, the direction of the inequality symbol is REVERSED]

In this case, we get $ay > bx$, which means Quantity A is greater


case ii: x is negative and y is positive

Take: $\frac{a}{x} < \frac{b}{y}$

Multiply both sides by y to get: $\frac{ay}{x} < b$ [since y is POSITIVE, the inequality symbol REMAINS facing the same direction]

Now multiply both sides by x to get: $ay > bx$ [since x is NEGATIVE, the direction of the inequality symbol is REVERSED]

In this case, we get $ay > bx$, which means Quantity A is greater


In BOTH possible cases, we concluded that Quantity A is greater
So, the correct answer is A

Cheers,
Brent

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Beautiful solution, Carlos!!
Kudos for you!!!!

Cheers,
Brent

I just cross multiplied the second equation and got:
ay < bx.
Therefore B>A. Is that wrong?

Is wrong because the stem says that xy<0

Which means that you cannot cross multiply because you do not know which is positive or negative.

Could x or y but not both.

Hope now is more clear.

Regards

IMPORTANT CONCEPT: If we multiply both sides of an inequality by a NEGATIVE value, we must REVERSE the direction of the inequality sign.
Your solution assumes that x and y are both positive.

Cheers,
Brent

Here's another (faster) approach that one of my students pointed out.


Take: $\frac{a}{x} < \frac{b}{y}$

Multiply both sides by $xy$ to get: $ay > bx$ [since we multiplied both sides by a NEGATIVE value, we REVERSED the direction of the inequality symbol]

Answer: A

Cheers,
Brent

Hi Carcass could you please update the solution? (there is still B as answer)

Thank you!

Thank you

The book still reports A as far as I could see.
Mah

should we not consider the sign of a & b?

good question

Shouldn’t we consider the signs of a and b?

The second condition which is (a/x) < (b/y) is what's governing the signal alternation between a & b.

In other words, if x was -ve then a should be +ve to fulfill the condition above, and y should be positive to fulfill the first condition xy<0.

For example:

Take x = -1 , y = 2

a = 3 , b = -4 (or 4)

it will give us a/x < b/y

-3 < -2 (2) => ay = 6, bx = 4 (-4). And both ways ay > bx

The second condition makes the question solvable i believe

BEST APPROACH

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Hello!

Solution:

Firstly as xy<0

x=+ve; y=-ve
x=-ve; y=+ve

Now, let work on the other given info, using the above
a/x<b/y
This means, x and a carry different signs because if the have the same signs it can be greater than the RHS and y and b carry the same signs.

So, if x=+ve; a--> -ve & y=-ve and b-->-ve
Thus, Qty A: -ve x -ve= +ve
Qty B: +ve x -ve= -ve

Similarly, x=-ve; y=+ve then,
Qty A: +ve x +ve= +ve
Qty B: +ve x -ve= -ve

Thus, in both of the above cases A>B

IMO A

Hope this helps!

Beautifully simple! What an elegant solution.

Can I tell that y is -ve, b is -ve so b/y is +ve and x is +ve a/x is +ve
So, both a/x and b/y are +ve but magnitude of b/y is greater
In this case how will you proceed?