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# S96-10

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Math Expert
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16 Sep 2014, 00:50
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35% (medium)

Question Stats:

58% (01:12) correct 42% (01:13) wrong based on 64 sessions

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If $$v \neq 0$$, is $$|w| \lt |v|$$?

(1) $$\frac{w}{v} \lt 1$$

(2) $$\frac{w^2}{v^2} \lt 1$$
[Reveal] Spoiler: OA

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16 Sep 2014, 00:51
Official Solution:

In words, we are asked whether the absolute value of $$w$$ is less than the absolute value of $$v$$. We could rephrase this geometrically: is $$w$$ closer to 0 on the number line than $$v$$ is? However, we should also retain the original phrasing, which uses concise algebra. (By the way, the condition that $$v$$ does not equal 0 is only there to prevent division by 0 in the statements.)

Statement (1): INSUFFICIENT. We should not cross-multiply this statement, since we don't know whether v is positive or negative. That is, we cannot claim that $$w \lt v$$ results from $$\frac{w}{v} \lt 1$$, since we can't multiply both sides of an inequality by a variable and preserve the inequality sign as is, unless we know that the variable is positive.

Let's instead plug a couple of quick pairs of numbers, trying for a "Yes" case and a "No" case. (The "Yes" case would be one in which $$|w|$$ is in fact less than $$|v|$$, and the "No" case would be one in which $$|w|$$ is not less than $$|v|$$.) Remember to try negative numbers!

Yes case: $$w = 2$$, $$v = 3$$. $$\frac{2}{3}$$ is indeed less than 1, satisfying the statement. As for the question, we get $$|2| \lt |3|$$, giving us a "Yes" answer to the question.

No case: $$w = -5$$, $$v = 3$$. $$-\frac{5}{3}$$ is indeed less than 1, satisfying the statement. But $$|-5|$$ is NOT less than $$|3|$$, so we would answer the question "No."

Since we have both a Yes case and a No case, this statement is insufficient.

Statement (2): SUFFICIENT. Here, we CAN cross-multiply, since $$v^2$$ is definitely positive. Thus, we can rephrase the statement as follows:
$$\frac{w^2}{v^2} \lt 1$$
$$w^2 \lt v^2$$

Now, since both sides are positive (or at least not negative), we can take the positive square root of both sides. The positive square root of a square is in fact the absolute value, so we get
$$|w| \lt |v|$$

If you like, test numbers that satisfy $$w^2 \lt v^2$$. In every case, you will find that $$|w| \lt |v|$$.

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28 Jun 2016, 06:24
Bunuel wrote:
Official Solution:

In words, we are asked whether the absolute value of $$w$$ is less than the absolute value of $$v$$. We could rephrase this geometrically: is $$w$$ closer to 0 on the number line than $$v$$ is? However, we should also retain the original phrasing, which uses concise algebra. (By the way, the condition that $$v$$ does not equal 0 is only there to prevent division by 0 in the statements.)

Statement (1): INSUFFICIENT. We should not cross-multiply this statement, since we don't know whether v is positive or negative. That is, we cannot claim that $$w \lt v$$ results from $$\frac{w}{v} \lt 1$$, since we can't multiply both sides of an inequality by a variable and preserve the inequality sign as is, unless we know that the variable is positive.

Let's instead plug a couple of quick pairs of numbers, trying for a "Yes" case and a "No" case. (The "Yes" case would be one in which $$|w|$$ is in fact less than $$|v|$$, and the "No" case would be one in which $$|w|$$ is not less than $$|v|$$.) Remember to try negative numbers!

Yes case: $$w = 2$$, $$v = 3$$. $$\frac{2}{3}$$ is indeed less than 1, satisfying the statement. As for the question, we get $$|2| \lt |3|$$, giving us a "Yes" answer to the question.

No case: $$w = -5$$, $$v = 3$$. $$-\frac{5}{3}$$ is indeed less than 1, satisfying the statement. But $$|-5|$$ is NOT less than $$|3|$$, so we would answer the question "No."

Since we have both a Yes case and a No case, this statement is insufficient.

Statement (2): SUFFICIENT. Here, we CAN cross-multiply, since $$v^2$$ is definitely positive. Thus, we can rephrase the statement as follows:
$$\frac{w^2}{v^2} \lt 1$$
$$w^2 \lt v^2$$

Now, since both sides are positive (or at least not negative), we can take the positive square root of both sides. The positive square root of a square is in fact the absolute value, so we get
$$|w| \lt |v|$$

If you like, test numbers that satisfy $$w^2 \lt v^2$$. In every case, you will find that $$|w| \lt |v|$$.

Hi Bunuel,

When |w| < |v| given, can't we square both the inequalities. ( as we do |w| = |v| , we can square both to get rid of the modulus sign).

Is there any way to get rid of modulus for < or > sign in between inequalities ?

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28 Jun 2016, 06:44
msk0657 wrote:

Hi Bunuel,

When |w| < |v| given, can't we square both the inequalities. ( as we do |w| = |v| , we can square both to get rid of the modulus sign).

Is there any way to get rid of modulus for < or > sign in between inequalities ?

Yes, you can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).

Solving Quadratic Inequalities - Graphic Approach
Inequality tips

Hope it helps.
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Joined: 12 Sep 2016
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07 Oct 2016, 04:58
Hi

It is not mentioned that w and v are integers, if w and v are real numbers then it is possible that w^2 < v^2 and |w| > |v|

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07 Oct 2016, 05:22
NLMurthy wrote:
Hi

It is not mentioned that w and v are integers, if w and v are real numbers then it is possible that w^2 < v^2 and |w| > |v|

Do you have examples of this?

When both sides of an inequality are non-negative we can safely take the square root: w^2 < v^2 --> |w| < |v|.

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25 Jan 2018, 06:30
What if the values are fractions? Then statement 2 becomes insufficient as well
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25 Jan 2018, 06:47
harshitasarwal wrote:
What if the values are fractions? Then statement 2 becomes insufficient as well

The second statement is sufficient for any values of v and w.

If v ≠ 0, is |w| < |v|?

(1) w/v < 1 --> if $$w=1$$ and $$v=2$$ the answer is YES but if $$w=-2$$ and $$v=1$$ the answer is NO. Not sufficient.

(2) w^2/v^2 < 1 --> since $$v^2>0$$ then we can safely cross multiply: $$w^2<v^2$$ --> $$|w|<|v|$$. Sufficient.

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Re: S96-10   [#permalink] 25 Jan 2018, 06:47
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# S96-10

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