November 22, 2018 November 22, 2018 10:00 PM PST 11:00 PM PST Mark your calendars  All GMAT Club Tests are free and open November 22nd to celebrate Thanksgiving Day! Access will be available from 0:01 AM to 11:59 PM, Pacific Time (USA) November 23, 2018 November 23, 2018 10:00 PM PST 11:00 PM PST Practice the one most important Quant section  Integer properties, and rapidly improve your skills.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 50711

Question Stats:
47% (01:44) correct 53% (01:16) wrong based on 70 sessions
HideShow timer Statistics



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re M3120
[#permalink]
Show Tags
09 Jun 2015, 07:38
Official Solution: If \(x\) is a positive integer, and two sides of a certain triangle have lengths \(3x+2\) and \(4x+5\) respectively, which of the following could be the length of the third side of the triangle? I. \(6x + 7\) II. \(6x + 9\) III. \(8x + 1\)
A. I only B. II only C. I and II only D. II and III only E. I, II and III Notice that the question asks which of the following could be the length of the third side of the triangle. Next, the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Therefore: \((4x+5)  (3x+2) < (third \ side) < (3x+2) + (4x+5)\) \(x + 3 < (third \ side) < 7x + 7\). Let's check options: I. \(6x + 7\). For any positive value of \(x\), \(x + 3 < 6x + 7 < 7x + 7\). OK. II. \(6x + 9\). For \(x = 1\) or \(x = 2\), \(x + 3 < 6x + 9 < 7x + 7\) does not hold true but if \(x = 3\), then it does. OK. III. \(8x + 1\). Right away we can find that if \(x = 1\), then \(x + 3 < 8x + 1 < 7x + 7\) is true. OK. Answer: E
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 06 Nov 2014
Posts: 31

Re: M3120
[#permalink]
Show Tags
31 Aug 2015, 03:53
Bunuel wrote: Official Solution:
If \(x\) is a positive integer, and two sides of a certain triangle have lengths \(3x+2\) and \(4x+5\) respectively, which of the following could be the length of the third side of the triangle? II. \(6x + 7\) III. \(6x + 9\) IIII. \(8x + 1\)
A. I only B. II only C. I and II only D. II and III only E. I, II and III
Notice that the question asks which of the following could be the length of the third side of the triangle. Next, the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Therefore: \((4x+5)  (3x+2) < (third \ side) < (3x+2) + (4x+5)\) \(x + 3 < (third \ side) < 7x + 7\). Let's check options: I. \(6x + 7\). For any positive value of \(x\), \(x + 3 < 6x + 7 < 7x + 7\). OK. II. \(6x + 9\). For \(x = 1\) or \(x = 2\), \(x + 3 < 6x + 9 < 7x + 7\) does not hold true but if \(x = 3\), then it does. OK. III. \(8x + 1\). Right away we can find that if \(x = 1\), then \(x + 3 < 8x + 1 < 7x + 7\) is true. OK.
Answer: E Please correct me if I am wrong but option 2 seems incorrect. And therefore the answers choices too aren't right. for x=1 6(1)+9=15 7(1)+7=14



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M3120
[#permalink]
Show Tags
31 Aug 2015, 06:52
anurag356 wrote: Bunuel wrote: Official Solution:
If \(x\) is a positive integer, and two sides of a certain triangle have lengths \(3x+2\) and \(4x+5\) respectively, which of the following could be the length of the third side of the triangle? II. \(6x + 7\) III. \(6x + 9\) IIII. \(8x + 1\)
A. I only B. II only C. I and II only D. II and III only E. I, II and III
Notice that the question asks which of the following could be the length of the third side of the triangle. Next, the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Therefore: \((4x+5)  (3x+2) < (third \ side) < (3x+2) + (4x+5)\) \(x + 3 < (third \ side) < 7x + 7\). Let's check options: I. \(6x + 7\). For any positive value of \(x\), \(x + 3 < 6x + 7 < 7x + 7\). OK. II. \(6x + 9\). For \(x = 1\) or \(x = 2\), \(x + 3 < 6x + 9 < 7x + 7\) does not hold true but if \(x = 3\), then it does. OK. III. \(8x + 1\). Right away we can find that if \(x = 1\), then \(x + 3 < 8x + 1 < 7x + 7\) is true. OK.
Answer: E Please correct me if I am wrong but option 2 seems incorrect. And therefore the answers choices too aren't right. for x=1 6(1)+9=15 7(1)+7=14 Please reread what the explanation for the second option says: II. \(6x + 9\). For \(x = 1\) or \(x = 2\), \(x + 3 < 6x + 9 < 7x + 7\) does not hold true but if \(x = 3\), then it does. OK.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 06 Nov 2014
Posts: 31

Re: M3120
[#permalink]
Show Tags
31 Aug 2015, 07:08
Before writing to u I actually used the calculator to check 6+9=15. I seriously doubted that ur solution can be wrong .
I should have read your solution more carefully.
My bad Thanks



Intern
Joined: 29 Jun 2016
Posts: 11

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 05:52
quality of the question is ok, but the quality of solution is not. 8x+1 cannot be the third side for x>8



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 06:04



Intern
Joined: 29 Jun 2016
Posts: 11

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 06:18
but we cannot have a triangle with x=8; First side+second side= 7(8) +7= 63 X=9 First side+second side= 7(9) +7= 70
Testing for x=8 or higher Testing for x=9 Option 1 Third side = 6(8)+7=55 (ok) Option 1 Third side = 6(9)+7=61 (ok) < 70 Option 2 Third side= 6(8)+9=57(ok) Option 2Third side=6(9)+9=63(ok)< 70 Option 3 Third side=8(8)+1=65>63 (not ok) Option 3Third side=8(9)+1=73(not ok)>70



Intern
Joined: 29 Jun 2016
Posts: 11

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 06:19
In other words x>= 8; the third option doesn't satisfy



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 06:25



Intern
Joined: 29 Jun 2016
Posts: 11

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 06:48
Are you telling that since, the word could be is used so option 3 can be considered?
please explain me as I am getting confused.



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 09:31



Intern
Joined: 29 Jun 2016
Posts: 11

Re: M3120
[#permalink]
Show Tags
02 Jul 2016, 21:33
Thanks for the info



Senior Manager
Joined: 31 Mar 2016
Posts: 386
Location: India
Concentration: Operations, Finance
GPA: 3.8
WE: Operations (Commercial Banking)

Re: M3120
[#permalink]
Show Tags
02 Aug 2016, 02:30
I think this is a highquality question and I agree with explanation. The "Could" part is the real trap i failed to consider option 3 and eliminated it when it was greater than the sum of the other 2 sides when x=10. Great question



Intern
Status: student
Affiliations: delhi university
Joined: 20 May 2014
Posts: 12
Concentration: Finance, International Business
WE: Accounting (Accounting)

Re: M3120
[#permalink]
Show Tags
11 Sep 2016, 00:30
Bunuel wrote: Official Solution:
If \(x\) is a positive integer, and two sides of a certain triangle have lengths \(3x+2\) and \(4x+5\) respectively, which of the following could be the length of the third side of the triangle? II. \(6x + 7\) III. \(6x + 9\) IIII. \(8x + 1\)
A. I only B. II only C. I and II only D. II and III only E. I, II and III
Notice that the question asks which of the following could be the length of the third side of the triangle. Next, the length of any side of a triangle must be larger than the positive difference of the other two sides, but smaller than the sum of the other two sides. Therefore: \((4x+5)  (3x+2) < (third \ side) < (3x+2) + (4x+5)\) \(x + 3 < (third \ side) < 7x + 7\). Let's check options: I. \(6x + 7\). For any positive value of \(x\), \(x + 3 < 6x + 7 < 7x + 7\). OK. II. \(6x + 9\). For \(x = 1\) or \(x = 2\), \(x + 3 < 6x + 9 < 7x + 7\) does not hold true but if \(x = 3\), then it does. OK. III. \(8x + 1\). Right away we can find that if \(x = 1\), then \(x + 3 < 8x + 1 < 7x + 7\) is true. OK.
Answer: E Hi, very good explanation. ! for option 2  i checked for values when x=1 and 2 and they both didnt satisfy so i marked it incorrect. So, i started looking for choice which had 1 and 3 as correct. Finally went with option A. Up To which value would you suggest we must verify our answers to get comfort in the question ? thanks



Intern
Joined: 23 Nov 2015
Posts: 15
GPA: 3.47

Re: M3120
[#permalink]
Show Tags
19 Jun 2017, 14:06
Bunuel wrote: If \(x\) is a positive integer, and two sides of a certain triangle have lengths \(3x+2\) and \(4x+5\) respectively, which of the following could be the length of the third side of the triangle? II. \(6x + 7\) III. \(6x + 9\) IIII. \(8x + 1\)
A. I only B. II only C. I and II only D. II and III only E. I, II and III Bunuel, please check the labeling of possible options, since the Roman numbers do not match with the answer choices. The typos have confused me during the test. Thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M3120
[#permalink]
Show Tags
20 Jun 2017, 01:03



Intern
Joined: 09 Oct 2018
Posts: 3
Location: India
GPA: 3.9

Re M3120
[#permalink]
Show Tags
31 Oct 2018, 14:50
I think this the explanation isn't clear enough, please elaborate. Can you please clear my doubt on II being the third side. If for x=1,2 it cannot be the third side then it should be good reason to exclude it> Isn't it?



Math Expert
Joined: 02 Sep 2009
Posts: 50711

Re: M3120
[#permalink]
Show Tags
31 Oct 2018, 21:42










