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Is |s+t| < |s|+|t|?

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Is |s+t| < |s|+|t|?  [#permalink]

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New post 13 Dec 2017, 00:39
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A
B
C
D
E

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  25% (medium)

Question Stats:

77% (01:07) correct 23% (01:39) wrong based on 113 sessions

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[GMAT math practice question]

Is \(|s+t| < |s|+|t|\)?

1) \(s>t\)
2) \(st<0\)

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Re: Is |s+t| < |s|+|t|?  [#permalink]

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New post 13 Dec 2017, 01:43
Im not sure how B is suff.

|-1 + 3| < |-1| + |3| --> 2 < 4 suff

|-1 + 3| < 1 - 3 (opening 3 as negative, -1 remains one) --> 2 not less than -2

We get two cases.

Please explain B. Thank you!
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Re: Is |s+t| < |s|+|t|?  [#permalink]

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New post 13 Dec 2017, 01:49
This inequality holds true if signs are different and none of the variables is Zero

(1) no information about the signs and actual values, so multiples outcomes are possible:
s=5 t=1
|5+1|=|5|+|1| answer is NO because we have an equality

s=5 t=-1
|5-1|<|5|+|1| answer is YES, signs are different

s=-5 t=0
again we have an equality so again the answer is NO

(2) now we get information that the signs of t & s, and that none if the variables is equal to Zero hence YES sufficicent

Answer B
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Is |s+t| < |s|+|t|?  [#permalink]

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New post 13 Dec 2017, 11:07
Madhavi1990 wrote:
Im not sure how B is suff.

|-1 + 3| < |-1| + |3| --> 2 < 4 suff

|-1 + 3| < 1 - 3 (opening 3 as negative, -1 remains one) --> 2 not less than -2

We get two cases.

Please explain B. Thank you!


Hi

I understand that you have plugged in certain values of s and t as -1 and 3 respectively. Your first case is correct, and that is how we will solve this. I did not understand 'opening 3 as negative', absolute value of 3 or |3| will be 3 only, not -3. And absolute value of |-1| will be 1, so RHS is 1+3 = 4.
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Re: Is |s+t| < |s|+|t|?  [#permalink]

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New post 13 Dec 2017, 14:28
Madhavi1990 wrote:
Im not sure how B is suff.

|-1 + 3| < |-1| + |3| --> 2 < 4 suff

|-1 + 3| < 1 - 3 (opening 3 as negative, -1 remains one) --> 2 not less than -2

We get two cases.

Please explain B. Thank you!


I think you've gotten turned around a little in how you're looking at absolute values.

If a problem says something like |x| = 3, then x can equal 3 or -3. So, if a problem tells you what |x| is, and asks you about the value of x, you do what you described here.

However, if you're putting a number (or any known value) inside of absolute value signs, then it will always become positive, no matter what. Since the problem talks about the value of |s| and the value of |t|, those two values, |s| and |t|, will always be positive numbers. Interpreting it as '1-3' is incorrect - it has to be 1 + 3 no matter what.
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Re: Is |s+t| < |s|+|t|?  [#permalink]

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New post 15 Dec 2017, 08:21
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=>

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 first step of the VA (Variable Approach) method is to modify the original condition and the question, and then recheck the question.

Modifying the question:
\(|s+t| < |s|+|t|\)
\(⇔ |s+t|^2 < (|s|+|t|)^2\)
\(⇔ (s+t)^2 < (|s|+|t|)^2\)
\(⇔ s^2 + 2st + t^2 < |s|^2 + 2|s||t|+ |t|^2\)
\(⇔ s^2 + 2st + t^2 < s^2 + 2|s||t|+ t^2\)
\(⇔ 2st < 2|s||t|\)
\(⇔ st < |st|\)
\(⇔ st < 0\)

This is exactly condition 2).

Therefore, the answer is B.

Answer: B
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Re: Is |s+t| < |s|+|t|?   [#permalink] 15 Dec 2017, 08:21
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