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Is x/y < 0?

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by Max@Math Revolution » Fri Jan 04, 2019 12:02 am

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[Math Revolution GMAT math practice question]

Is x/y < 0?

1) x^4y^5 < 0
2) x^5y^3 < 0
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Source: — Data Sufficiency |

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by Brent@GMATPrepNow » Fri Jan 04, 2019 6:20 am
Max@Math Revolution wrote:Is x/y < 0?

1) (x^4)(y^5) < 0
2) (x^5)(y^3) < 0
Target question: Is x/y < 0?

Statement 1: (x^4)(y^5) < 0
There are several values of x and y that satisfy statement 1. Here are two:
Case a: x = 1 and y = -1. Notice that (x^4)(y^5) = (1^4)(-1^5) = (1)(-1) = -1, which is less than 0. In this case, x/y = 1/(-1) = -1. So, the answer to the target question is YES, x/y IS less than 0
Case b: x = -1 and y = -1. Notice that (x^4)(y^5) = (-1^4)(-1^5) = (1)(-1) = -1, which is less than 0. In this case, x/y = -1/(-1) = 1. So, the answer to the target question is NO, x/y is NOT less than 0
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: (x^5)(y^3) < 0
Since x^4 is POSITIVE, we can safely divide both sides of the inequality by x^4 to get: (x)(y^3) < 0
Since y^4 is POSITIVE, we can safely divide both sides of the inequality by y^4 to get: x/y < 0
So, the answer to the target question is YES, x/y IS less than 0
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by Manasa3190 » Fri Jan 04, 2019 6:25 am
Is x/y < 0?

1) x^4*y^5 < 0
As x^4 is always positive, it implies y < 0
if x > 0
then x/y < 0

if x < 0, then x/y > 0
Insufficient

2) x^5*y^3 < 0
As X,Y are both having the odd power, they retain their sign, and
Since the product is positive , both the variables have same sign

Positive * positive = positive
negative*negative = positive

So, x/y will be greater than zero as sign cancels out
Sufficient

Option B is correct
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by fskilnik@GMATH » Fri Jan 04, 2019 10:34 am
Max@Math Revolution wrote:[Math Revolution GMAT math practice question]

Is x/y < 0?

1) x^4y^5 < 0
2) x^5y^3 < 0
$$\frac{x}{y}\,\,\mathop < \limits^? \,\,0$$
$$\left( 1 \right)\,\,{x^4}{y^5} < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,x \ne 0\,\,\,{\rm{and}}\,\,\,y < 0\,\,\,\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( {1, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {x,y} \right) = \left( { - 1, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,{x^5}{y^3} < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \matrix{
\,x > 0\,\,\,{\rm{and}}\,\,\,y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr
\,{\rm{or}} \hfill \cr
\,x < 0\,\,\,{\rm{and}}\,\,\,y > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle $$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br
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by Max@Math Revolution » Sun Jan 06, 2019 5:05 pm
=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The question is equivalent to asking if xy < 0. This can be seen by multiplying both sides of the inequality by y^2.

Since we can ignore even exponents in inequalities like x^4y^5 < 0, condition 1) is equivalent to the statement y < 0 and condition 2) is equivalent to the statement xy < 0.

Therefore, the answer is B.
Answer: B
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