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# If y > 0, is x/y < 3?

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Director
Joined: 11 Feb 2015
Posts: 621
If y > 0, is x/y < 3?  [#permalink]

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27 Jul 2018, 01:49
3
00:00

Difficulty:

75% (hard)

Question Stats:

52% (02:47) correct 48% (02:45) wrong based on 27 sessions

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If y > 0, is $$\frac{x}{y}$$ < 3?

1. x(x + y) − 4y(x + y) < 0
2. 5(y + 8) < 20y – 3x + 40

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Manish

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Director
Joined: 11 Feb 2015
Posts: 621
Re: If y > 0, is x/y < 3?  [#permalink]

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27 Jul 2018, 02:00

If y > 0, is $$\frac{x}{y}$$ < 3?

Step 1: Analyze the Question Stem
Since y is positive, you can multiply both sides of the inequality $$\frac{x}{y}$$ < 3 by the positive number y to obtain the equivalent inequality x < 3y. So the question is equivalent to “If y > 0, is x < 3y?”

Step 2: Evaluate the Statements
Statement (1): Factor x + y out of the left side of the inequality to yield x(x + y) − 4y(x + y) = (x + y)(x − 4y). So statement (1) is equivalent to (x + y)(x − 4y) < 0. When the product of two quantities is negative, one of the quantities must be negative and the other quantity must be positive. So either (i) x + y < 0 and x − 4y > 0, or (ii) x + y > 0 and x − 4y < 0. (Keep in mind that y is positive.) In case (i), if x + y < 0, then x < −y. If x − 4y > 0, then x > 4y. So in case (i), x < −y and x
> 4y. Since y is positive, x < −y means that x is less than −y, where −y is a negative number. Since y is positive, x > 4y means that x is greater than 4y, where 4y is a positive number. So (i) requires that x be less than the negative number −y and also requires that x be greater than the positive number 4y. This is impossible. So case (i), which is x + y < 0 and x − 4y > 0, cannot happen. On to case (ii), which is x + y > 0 and x − 4y < 0. (Again, keep in mind that y > 0.) If x + y > 0, then x > −y. If x − 4y < 0, then x < 4y. So this time, x > −y and x < 4y. Since y is positive, x > −y means that x is greater than the negative number −y. Since y is positive, x < 4y means that x is less than the positive number 4y. So case (ii), x + y > 0 and x − 4y < 0, is possible. In case (ii), you can conclude that −y < x < 4y. Because case (i) is impossible, Statement (1) is equivalent to −y < x < 4y. Since y is positive, 3y < 4y. You know that x < 4y. However, you do not know whether or not x < 3y.
More than one answer to the question is possible, so Statement (1) is insufficient. Eliminate choices (A) and (D).

Statement (2): Simplify the inequality 5(y + 8) < 20y − 3x + 40.
Statement (2) is equivalent to x < 5y. Since y is positive, 3y < 5y. You know that x < 5y. However, you do not know whether or not x < 3y. Statement (2) is insufficient. Eliminate choice (B).

The statements taken together: Statement (1) is equivalent to −y < x < 4y and Statement (2) is equivalent to x < 5y. Because the question requires that y > 0, 4y < 5y and the range in Statement (1) is a subset of the range in Statement (2). Therefore, the statements taken together require that −y < x < 4y, which you have already determined to be insufficient.

Choice (E) is correct.
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Re: If y > 0, is x/y < 3? &nbs [#permalink] 27 Jul 2018, 02:00
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