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

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 7120
GMAT 1: 760 Q51 V42
GPA: 3.82

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13 Dec 2017, 00:39
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Difficulty:

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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MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Senior Manager Joined: 15 Jan 2017 Posts: 352 Re: Is |s+t| < |s|+|t|? [#permalink] ### Show Tags 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! Manager Joined: 05 Dec 2016 Posts: 241 Concentration: Strategy, Finance GMAT 1: 620 Q46 V29 Re: Is |s+t| < |s|+|t|? [#permalink] ### Show Tags 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 DS Forum Moderator Joined: 22 Aug 2013 Posts: 1443 Location: India Is |s+t| < |s|+|t|? [#permalink] ### Show Tags 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. Manhattan Prep Instructor Joined: 04 Dec 2015 Posts: 709 GMAT 1: 790 Q51 V49 GRE 1: Q170 V170 Re: Is |s+t| < |s|+|t|? [#permalink] ### Show Tags 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. _________________ Chelsey Cooley | Manhattan Prep | Seattle and Online My latest GMAT blog posts | Suggestions for blog articles are always welcome! Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 7120 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: Is |s+t| < |s|+|t|? [#permalink] ### Show Tags 15 Dec 2017, 08:21 1 => 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 _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$149 for 3 month Online Course"
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Re: Is |s+t| < |s|+|t|?   [#permalink] 15 Dec 2017, 08:21
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