Oct 22 08:00 AM PDT  09:00 AM PDT Join to learn strategies for tackling the longest, wordiest examples of Counting, Sets, & Series GMAT questions Oct 22 09:00 AM PDT  10:00 AM PDT Watch & learn the Do's and Don’ts for your upcoming interview Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions) Oct 23 08:00 AM PDT  09:00 AM PDT Join an exclusive interview with the people behind the test. If you're taking the GMAT, this is a webinar you cannot afford to miss! Oct 23 12:00 PM EDT  01:00 PM EDT Are you curious about life at Tuck? Join current secondyear students as they share an insiderview of life as a Tuckie. Oct 26 07:00 AM PDT  09:00 AM PDT Want to score 90 percentile or higher on GMAT CR? Attend this free webinar to learn how to prethink assumptions and solve the most challenging questions in less than 2 minutes. Oct 27 07:00 AM EDT  09:00 AM PDT Exclusive offer! Get 400+ Practice Questions, 25 Video lessons and 6+ Webinars for FREE.
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 04 Mar 2019
Posts: 4

if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 02:50
Question Stats:
30% (02:21) correct 70% (02:33) wrong based on 175 sessions
HideShow timer Statistics
if (10x)/3 < 2x, which of the following must be true? I) 2 < x II) x5 >= 7 III) (x1/x) > 1 A. I Only B. II Only C. III Only D. II and III E. I, II, and III
Official Answer and Stats are available only to registered users. Register/ Login.




Director
Status: Manager
Joined: 27 Oct 2018
Posts: 685
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)

if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
Updated on: 11 May 2019, 06:18
Hi eabhgoyin the "must be true questions", you have to deal with question the other way around. the inequality : (10x)/3 < 2x , is the TRUE ground that you will start from. by simplifying it, it will be x< 2. so the question says that: if the value of x is less than 2 ... (2.5, 3, ........) these values are covered by which of the following I, II , III? I) 2 < x > for sure false, because it is not covering any of the values of x< 2 II) x5 >= 7 : so the range is x>=12 and x<= 2 > it covers ALL the values of "x<2" and MORE (ALL is a must, MORE doesn't affect the judgement because if you try ANY value of "x<2", it will fit into "x5 >= 7" III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE however, if in the choices, there is (for example): x<4 > this will be false because it is not covering ALL values of "x<2" (if x = 3 for instance)
_________________
Originally posted by MahmoudFawzy on 31 Mar 2019, 08:42.
Last edited by MahmoudFawzy on 11 May 2019, 06:18, edited 1 time in total.




Senior Manager
Joined: 25 Feb 2019
Posts: 336

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 03:20
IMO D
after simplifying the given equation we get
x+2<0
x<2
now only 2nd and 3rd hold true for this
award kudos if helpful
award kudos if helpful
Posted from my mobile device



Manager
Joined: 12 Apr 2011
Posts: 149
Location: United Arab Emirates
Concentration: Strategy, Marketing
GMAT 1: 670 Q50 V31 GMAT 2: 720 Q50 V37
WE: Marketing (Telecommunications)

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 04:52
chakrads wrote: if (10x)/3 < 2x, which of the following must be true?
I) 2 < x II) x5 >= 7 III) (x1/x) > 1
A. I Only B. II Only C. III Only D. II and III E. I, II, and III If X=13, then x5>=7 is true but (10x)/3<2x is not true, then how is the answer D?
_________________
Hit Kudos if you like what you see!



Director
Joined: 27 May 2012
Posts: 903

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 11:27
chakrads wrote: if (10x)/3 < 2x, which of the following must be true?
I) 2 < x II) x5 >= 7 III) (x1/x) > 1
A. I Only B. II Only C. III Only D. II and III E. I, II, and III Given \(\frac{10x}{3}< 2x\) 10x<6x 5x<10 x<2 Take x=2.1 I)) 2<2.1 No II) \(2.15\geq 7\)or \(7.1\geq7\) Yes III)\(\frac{2.11}{2.1}\) >1 \(\Rightarrow\) \(\frac{3.1}{2.1}\)>1 Yes Hence only II and III must be true . Hope this helps.
_________________



Senior Manager
Joined: 12 Sep 2017
Posts: 302

if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 13:14
Mahmoudfawzy83 wrote: Hi eabhgoyin the "must be true questions", you have to deal with question the other way around. the inequality : (10x)/3 < 2x , is the TRUE ground that you will start from. by simplifying it, it will be x< 2. so the question says that: if the value of x is less than 2 ... (2.5, 3, ........) these values are covered by which of the following I, II , III? I) 2 < x > for sure false, because it is not covering any of the values of x< 2 II) x5 >= 7 : so the range is x>=12 and x<= 2 > it covers ALL the values of "x<2" and MORE (ALL is a must, MORE doesn't affect the judgement because if you try ANY value of "x<2", it will fit into "x5 >= 7"III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE however, if in the choices, there is (for example): x<4 > this will be false because it is not covering ALL values of "x<2" (if x = 3 for instance) Hello Mahmoudfawzy83 ! Could you please explain to me the red statement? If x must be less than 2, how come could II holds true? What happens if x is 2 as the II statement says that x could be equal or less than 2 if it's equal to 2 then it can't be true. Kind regards!



Director
Status: Manager
Joined: 27 Oct 2018
Posts: 685
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 13:58
Hi jfranciscocuencagfirst, you must know that it was confusing for me at the beginning. I will try explaining through a simplified question: if x = 1, which of the following must be true 1) x >1 2) x >1 3) \(x\geq{1}\) what the question is really asking about is: is 1 > 1 ? > no ..... why? .... because 1 is not in the range of >1 is 1 > 1? > yes ...... why? ..... because 1 is in the range of >1 is \(1\geq{1}\)? > yes ..... why? ..... because 1 is in the range of \(\geq{1}\) I may haven't answered your question yet, but do you agree with the simplified question so far?
_________________



Senior Manager
Joined: 12 Sep 2017
Posts: 302

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 14:32
Mahmoudfawzy83 wrote: Hi jfranciscocuencagfirst, you must know that it was confusing for me at the beginning. I will try explaining through a simplified question: if x = 1, which of the following must be true 1) x >1 2) x >1 3) \(x\geq{1}\) what the question is really asking about is: is 1 > 1 ? > no ..... why? .... because 1 is not in the range of >1 is 1 > 1? > yes ...... why? ..... because 1 is in the range of >1 is \(1\geq{1}\)? > yes ..... why? ..... because 1 is in the range of \(\geq{1}\) I may haven't answered your question yet, but do you agree with the simplified question so far? Mahmoudfawzy83, yes, it actually helped. But now, for example in statement two we got two values when x is equal or more than 12 and equal or less than 2. So, in this kind of questions, is it ok if the statement just satisfies one of the two inequalities? Kind regards!



Director
Status: Manager
Joined: 27 Oct 2018
Posts: 685
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 15:01
Mahmoudfawzy83 wrote: is 1 > 1? > yes ...... why? ..... because 1 is in the range of >1
Now I am going to attack the answer I just said myself: What happens if x is 0 as the II statement says that x could be any value more than 1? if it's equal to 0 then it can't be truethe counter argument (in red) I used is irrelevant, because we am not testing whether x = 1 or not in the first place (it is just a fact I am using to test the other statements) we am testing the fitting of x in the ranges stated in I, II, and III it is similar to the question you just wrote: jfranciscocuencag wrote: If x must be less than 2, how come could II holds true? What happens if x is 2 as the II statement says that x could be equal or less than 2 if it's equal to 2 then it can't be true. Kind regards! In summary, "must be true" questions are different, and has a special mindset to deal with it. as long as ALL the values are covered by the proposed range (even though there are extra irrelevant values), then the statement is true, but if the proposed range is not covering ALL the values, then false (as in the example in green below) Mahmoudfawzy83 wrote: Hi eabhgoyin the "must be true questions", you have to deal with question the other way around. the inequality : (10x)/3 < 2x , is the TRUE ground that you will start from. by simplifying it, it will be x< 2. so the question says that: if the value of x is less than 2 ... (2.5, 3, ........) these values are covered by which of the following I, II , III? I) 2 < x > for sure false, because it is not covering any of the values of x< 2 II) x5 >= 7 : so the range is x>=12 and x<= 2 > it covers ALL the values of "x<2" and MORE (ALL is a must, MORE doesn't affect the judgement because if you try ANY value of "x<2", it will fit into "x5 >= 7" III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE however, if in the choices, there is (for example): x<4 > this will be false because it is not covering ALL values of "x<2" (if x = 3 for instance)Try practicing on the "Must or Could be True Questions" under the problem solving part of the question bank: https://gmatclub.com/forum/search.php?view=search_tagsstart with the easy level to get familiar with this type of questions.
_________________



Manager
Joined: 12 Apr 2011
Posts: 149
Location: United Arab Emirates
Concentration: Strategy, Marketing
GMAT 1: 670 Q50 V31 GMAT 2: 720 Q50 V37
WE: Marketing (Telecommunications)

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
31 Mar 2019, 21:24
Mahmoudfawzy83 wrote: Hi eabhgoyin the "must be true questions", you have to deal with question the other way around. the inequality : (10x)/3 < 2x , is the TRUE ground that you will start from. by simplifying it, it will be x< 2. so the question says that: if the value of x is less than 2 ... (2.5, 3, ........) these values are covered by which of the following I, II , III? I) 2 < x > for sure false, because it is not covering any of the values of x< 2 II) x5 >= 7 : so the range is x>=12 and x<= 2 > it covers ALL the values of "x<2" and MORE (ALL is a must, MORE doesn't affect the judgement because if you try ANY value of "x<2", it will fit into "x5 >= 7" III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE however, if in the choices, there is (for example): x<4 > this will be false because it is not covering ALL values of "x<2" (if x = 3 for instance) Thanks buddy, makes sense!
_________________
Hit Kudos if you like what you see!



Intern
Joined: 20 Aug 2018
Posts: 9

if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
11 May 2019, 01:02
III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE
Hi Mahmoudfawzy83, Could you please help me with the steps to solve the 3rd inequality? I'm lost.



Director
Joined: 19 Oct 2018
Posts: 989
Location: India

if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
Updated on: 11 May 2019, 02:18
We know that x<2 Assume that 3rd statement is true x1/x > 1 Or x1>x (We can crossmultiply here, because x is always positive under given constraint) or x1  x >0 If x is less than 2, Then x1 is negative Hence x1= (x1) and x= x (x1)  (x)>0 x+1+x>0 1>0 Hence our assumption was true, 3rd statement will always true under given constraint 99ramanmehta wrote: III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE
Hi Mahmoudfawzy83, Could you please help me with the steps to solve the 3rd inequality? I'm lost.
Originally posted by nick1816 on 11 May 2019, 01:36.
Last edited by nick1816 on 11 May 2019, 02:18, edited 2 times in total.



Intern
Joined: 20 Aug 2018
Posts: 9

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
11 May 2019, 02:00
nick1816 wrote: We know that x<2 which means x is always negative Assume that 3rd statement is true x1/x > 1 Or x1>x (We can crossmultiply here, because x is always positive under given constraint) or x1  x >0 If x is negative Then x1 is also negative Hence x1= (x1) and x= x (x1)  (x)>0 x+1+x>0 1>0 Hence our assumption was true, 3rd statement will always true under given constraint 99ramanmehta wrote: III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE
Hi Mahmoudfawzy83, Could you please help me with the steps to solve the 3rd inequality? I'm lost. Thanks for the explanation Nick. I, too, was able reach to the conclusion that x1  x >0 and hence 1>0. What I want to understand is that how did Mahmoudfawzy83 solve this equation to find the range is x <1/2. Can you please help on this?



Director
Status: Manager
Joined: 27 Oct 2018
Posts: 685
Location: Egypt
Concentration: Strategy, International Business
GPA: 3.67
WE: Pharmaceuticals (Health Care)

if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
11 May 2019, 02:01
99ramanmehta wrote: III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE
Hi Mahmoudfawzy83, Could you please help me with the steps to solve the 3rd inequality? I'm lost. Hi 99ramanmehta\(\frac{x1}{x} > 1\) because x is positive (we are sure of the sign), multiplying, both sides by x is valid, so: \(x1 > x\) and because both sides are not negative , squaring both sides is valid (so we can get rid of the "") \(x^2  2x + 1 > x^2\) \(1 > 2x\) \(\frac{1}{2}> x\) an important step is to check the outcome by trying it in the original equation > choose \(\frac{1}{4}\) for example: \(\frac{\frac{1}{4}1}{\frac{1}{4}}\) = \(3 = 3 > 1\) > so confirmed (note that the range is all values less than \(\frac{1}{2}\), except 0 because \(x\) is a denominator) Thanks nick1816 for sharing your idea with me and 99ramanmehta , I would like to comment on your approach nick1816 wrote: If x is negative Then x1 is also negative
..... But x1 can be negative \(\frac{1}{2}\), while x is \(\frac{1}{2}\) (positive) , so your assumption is not perfect. to confirm, try \(x = \frac{3}{4}\) (which lies in your range), and you will find it invalid.
_________________



Director
Joined: 19 Oct 2018
Posts: 989
Location: India

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
11 May 2019, 02:17
I've corrected it now. Thanks!!! I would like to comment on your approach nick1816 wrote: If x is negative Then x1 is also negative
..... But x1 can be negative \(\frac{1}{2}\), while x is \(\frac{1}{2}\) (positive) , so your assumption is not perfect. to confirm, try \(x = \frac{3}{4}\) (which lies in your range), and you will find it invalid.[/quote]



Intern
Joined: 20 Aug 2018
Posts: 9

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
11 May 2019, 06:15
Mahmoudfawzy83 wrote: 99ramanmehta wrote: III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE
Hi Mahmoudfawzy83, Could you please help me with the steps to solve the 3rd inequality? I'm lost. Hi 99ramanmehta\(\frac{x1}{x} > 1\) because x is positive (we are sure of the sign), multiplying, both sides by x is valid, so: \(x1 > x\) and because both sides are not negative , squaring both sides is valid (so we can get rid of the "") \(x^2  2x + 1 > x^2\) \(1 > 2x\) \(\frac{1}{2}> x\) an important step is to check the outcome by trying it in the original equation > choose \(\frac{1}{4}\) for example: \(\frac{\frac{1}{4}1}{\frac{1}{4}}\) = \(3 = 3 > 1\) > so confirmed (note that the range is all values less than \(\frac{1}{2}\), except 0 because \(x\) is a denominator) Thanks nick1816 for sharing your idea with me and 99ramanmehta , I would like to comment on your approach nick1816 wrote: If x is negative Then x1 is also negative
..... But x1 can be negative \(\frac{1}{2}\), while x is \(\frac{1}{2}\) (positive) , so your assumption is not perfect. to confirm, try \(x = \frac{3}{4}\) (which lies in your range), and you will find it invalid. Thank you so much Mahmoudfawzy83 for the clarification. I followed the below mentioned approach: Critical points 0,1 1. For x<0; 1>0 => true 2. For x>=1; 1>0 => false 3. For 0<x<1; (x1)>x x+1>x 1>2x x<1/2in your earlier answer, you'd specified the range as x<1/2 instead of x<1/2, hence the confusion.



Manager
Joined: 07 Apr 2018
Posts: 101
Location: United States
Concentration: General Management, Marketing
GPA: 3.8

Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
Show Tags
13 May 2019, 12:12
Mahmoudfawzy83 wrote: Hi eabhgoyin the "must be true questions", you have to deal with question the other way around. the inequality : (10x)/3 < 2x , is the TRUE ground that you will start from. by simplifying it, it will be x< 2. so the question says that: if the value of x is less than 2 ... (2.5, 3, ........) these values are covered by which of the following I, II , III? I) 2 < x > for sure false, because it is not covering any of the values of x< 2 II) x5 >= 7 : so the range is x>=12 and x<= 2 > it covers ALL the values of "x<2" and MORE (ALL is a must, MORE doesn't affect the judgement because if you try ANY value of "x<2", it will fit into "x5 >= 7" III) (x1/x) > 1 : so the range is x <1/2 > the same as II, it covers ALL values of "x<2" and MORE Can you please explain how x <1/2 is same as x<2? What if x is greater than 2 and less than 1/2. however, if in the choices, there is (for example): x<4 > this will be false because it is not covering ALL values of "x<2" (if x = 3 for instance)




Re: if (10x)/3 <2x, which of the following must be true?
[#permalink]
13 May 2019, 12:12






