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Math Expert V
Joined: 02 Sep 2009
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Difficulty:   25% (medium)

Question Stats: 68% (01:26) correct 32% (01:58) wrong based on 75 sessions

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

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

(2) $$\frac{w^2}{v^2} \lt 1$$

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Math Expert V
Joined: 02 Sep 2009
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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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Retired Moderator G
Joined: 26 Nov 2012
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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 ?

Math Expert V
Joined: 02 Sep 2009
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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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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|

Math Expert V
Joined: 02 Sep 2009
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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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Intern  B
Joined: 09 Jan 2018
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What if the values are fractions? Then statement 2 becomes insufficient as well
Math Expert V
Joined: 02 Sep 2009
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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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I think this is a poor-quality question and I don't agree with the explanation. w and v may or may not be integer for fractions between 0 and 1 w^2<v^2 doesnt mean w < v
Math Expert V
Joined: 02 Sep 2009
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jayyash321 wrote:
I think this is a poor-quality question and I don't agree with the explanation. w and v may or may not be integer for fractions between 0 and 1 w^2<v^2 doesnt mean w < v

Before posting a reply I'd suggest:
1. To read a question carefully.
2. To read the solution carefully.
3. To read the discussion carefully.

For example, here the question does not ask whether w < v! It asks whether |w| < |v|. Moreover, your doubt is addressed in the post just above yours.
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Intern  B
Joined: 06 Jun 2018
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I think the question is whether we are able to answer...|w|<|v|?, it doen not ask is absolute value of w is less than absolute value of v.
Hence in both cases we are able to answer.... so alone is sufficient
Math Expert V
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jareen wrote:
I think the question is whether we are able to answer...|w|<|v|?, it doen not ask is absolute value of w is less than absolute value of v.
Hence in both cases we are able to answer.... so alone is sufficient

"|w| < |v|" and "absolute value of w is less than absolute value of v" is the same. The answer to the question is B, which is explained in detail above.
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Intern  B
Joined: 29 May 2019
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I don't agree with the explanation.
Manager  B
Joined: 27 Mar 2017
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Isn't question basically asking us to compare the values of W and V irrespective of signs ? Doesn't that mean that in order to check that condition we can cross multiply in (1) because we know that question is just asking for the value comparison irrespective of sign.

Had the question asked for just w<v (without modulus) then yes we could't cross multiply in (1) as the OE states. But since we don't care about the signs we can get the idea of values from (1) which makes it sufficient. Re: S96-10   [#permalink] 04 Nov 2019, 04:09
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# S96-10

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