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M31-20

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M31-20  [#permalink]

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New post 09 Jun 2015, 08:38
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Question Stats:

46% (02:08) correct 54% (02:14) wrong based on 54 sessions

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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

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Re M31-20  [#permalink]

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New post 09 Jun 2015, 08:38
1
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: M31-20  [#permalink]

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New post 31 Aug 2015, 04:53
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: M31-20  [#permalink]

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New post 31 Aug 2015, 07:52
1
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 re-read 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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Re: M31-20  [#permalink]

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New post 31 Aug 2015, 08:08
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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
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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 06:52
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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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 07:04
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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 07:18
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 2-Third side=6(9)+9=63(ok)< 70
Option 3- Third side=8(8)+1=65>63 (not ok) Option 3-Third side=8(9)+1=73(not ok)>70
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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 07:19
In other words x>= 8; the third option doesn't satisfy
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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 07:25
subhajit1 wrote:
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 2-Third side=6(9)+9=63(ok)< 70
Option 3- Third side=8(8)+1=65>63 (not ok) Option 3-Third side=8(9)+1=73(not ok)>70


You are missing the point completely. The questions asks which of the following could be the length of the third side of the triangle? For x = 1, third option is possible. It's absolutely irrelevant whether it's not possible for x > 8.
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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 07:48
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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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 10:31
subhajit1 wrote:
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.


If a question asks which of the options MUST be true, an option to be considered true, MUST be true for ANY value.

If a question asks which of the options COULD be true, an option to be considered true, should be true for at least one value.

Here are must or could be true questions to practice: search.php?search_id=tag&tag_id=193
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Re: M31-20  [#permalink]

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New post 02 Jul 2016, 22:33
Thanks for the info
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Re: M31-20  [#permalink]

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New post 02 Aug 2016, 03:30
I think this is a high-quality 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: M31-20  [#permalink]

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New post 11 Sep 2016, 01:30
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: M31-20  [#permalink]

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New post 19 Jun 2017, 15:06
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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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Re: M31-20  [#permalink]

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New post 20 Jun 2017, 02:03
MikeMighty wrote:
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.


Thank you for noticing. Corrected. +1.
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Re M31-20  [#permalink]

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New post 31 Oct 2018, 15:50
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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Re: M31-20  [#permalink]

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New post 31 Oct 2018, 22:42
bks170882 wrote:
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?


I think it's explained above. The question asks: which of the following could be the length of the third side of the triangle? If x = 3, then 6x + 9 COULD be the length of the third side.
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Re: M31-20  [#permalink]

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New post 22 Nov 2019, 03:46
The question is confusing. I actually googled "certain triangle" and it is a legit term for triangles with 2 sides of the same length. That's why I only got "III" for the answer, nevertheless, still a question worth reviewing.
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Re: M31-20   [#permalink] 22 Nov 2019, 03:46

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