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S96-10

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S96-10  [#permalink]

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New post 16 Sep 2014, 01:50
2
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A
B
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D
E

Difficulty:

  35% (medium)

Question Stats:

61% (01:08) correct 39% (01:27) wrong based on 79 sessions

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Re S96-10  [#permalink]

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New post 16 Sep 2014, 01:51
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 \(|w| \lt |v|\).


Answer: B
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Re: S96-10  [#permalink]

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New post 28 Jun 2016, 07:24
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 \(|w| \lt |v|\).


Answer: B


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 ?

Please let us know.
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Re: S96-10  [#permalink]

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New post 28 Jun 2016, 07:44
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 ?

Please let us know.


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).

Check below links for more:

Inequalities Made Easy!

Solving Quadratic Inequalities - Graphic Approach
Inequality tips

Hope it helps.
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Re: S96-10  [#permalink]

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New post 07 Oct 2016, 05:58
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|

Please clarify
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Re: S96-10  [#permalink]

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New post 07 Oct 2016, 06:22
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|

Please clarify


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|.

Please check the links provided above for more.
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Re: S96-10  [#permalink]

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New post 25 Jan 2018, 07:30
What if the values are fractions? Then statement 2 becomes insufficient as well
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Re: S96-10  [#permalink]

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New post 25 Jan 2018, 07:47
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.

Answer: B.
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Re S96-10  [#permalink]

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New post 27 Sep 2018, 11:07
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
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Re: S96-10  [#permalink]

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New post 27 Sep 2018, 20:47
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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Re: S96-10  [#permalink]

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New post 30 Sep 2018, 07:04
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
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Re: S96-10  [#permalink]

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New post 30 Sep 2018, 08:59
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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Re: S96-10 &nbs [#permalink] 30 Sep 2018, 08:59
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