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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
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Re: M3120
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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



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Re: M3120
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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.
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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.
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Re: M3120
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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



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Re: M3120
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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



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02 Jul 2016, 06:04



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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



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



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Re: M3120
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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.



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02 Jul 2016, 09:31



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02 Jul 2016, 21:33
Thanks for the info



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Re: M3120
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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



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Re: M3120
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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



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Re: M3120
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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.



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20 Jun 2017, 01:03



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Re M3120
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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?



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31 Oct 2018, 21:42










