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xy < 0 [#permalink]
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Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx



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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Re: xy < 0 [#permalink]
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CarlosSG wrote:
Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


First, use a/x < b/y

(a/x) - (b/y) < 0
(ay-bx)/xy < 0

but xy<0, so ay-bx>0
Finally ay>bx

ANSWER (A)


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

Cheers,
Brent
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Re: xy < 0 [#permalink]
I just cross multiplied the second equation and got:
ay < bx.
Therefore B>A. Is that wrong?
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Re: xy < 0 [#permalink]
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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
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Re: xy < 0 [#permalink]
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bellavarghese wrote:
I just cross multiplied the second equation and got:
ay < bx.
Therefore B>A. Is that wrong?


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
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Re: xy < 0 [#permalink]
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Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx



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
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Re: xy < 0 [#permalink]
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Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


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

Thank you! :)
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Re: xy < 0 [#permalink]
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Thank you :)

The book still reports A as far as I could see.
Mah :roll:
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Re: xy < 0 [#permalink]
should we not consider the sign of a & b?
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Re: xy < 0 [#permalink]
good question
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Re: xy < 0 [#permalink]
GreenlightTestPrep wrote:
Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx



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


Shouldn’t we consider the signs of a and b?
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Re: xy < 0 [#permalink]
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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
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Re: xy < 0 [#permalink]
CarlosSG wrote:
Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx


A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.


First, use a/x < b/y

(a/x) - (b/y) < 0
(ay-bx)/xy < 0

but xy<0, so ay-bx>0
Finally ay>bx

ANSWER (A)

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Re: xy < 0 [#permalink]
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Re: xy < 0 [#permalink]
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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!
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Re: xy < 0 [#permalink]
Beautifully simple! What an elegant solution.

GreenlightTestPrep wrote:
Carcass wrote:
\(xy < 0\)

\(\frac{a}{x} < \frac{b}{y}\)


Quantity A
Quantity B
ay
bx



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
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Re: xy < 0 [#permalink]
Carcass wrote:
Thank you :)

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



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?
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