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Math Expert V
Joined: 02 Sep 2009
Posts: 59728

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Difficulty:   75% (hard)

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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59728

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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.

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Intern  Joined: 06 Nov 2014
Posts: 28

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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.

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 V
Joined: 02 Sep 2009
Posts: 59728

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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.

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

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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Intern  Joined: 06 Nov 2014
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1
Before writing to u I actually used the calculator to check 6+9=15. I seriously doubted that ur solution can be wrong .

Thanks
Intern  Joined: 29 Jun 2016
Posts: 11

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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 V
Joined: 02 Sep 2009
Posts: 59728

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subhajit1 wrote:
quality of the question is ok, but the quality of solution is not.
8x+1 cannot be the third side for x>8

Why should x be greater than 8?

If x = 1, then the sides are 3x + 2 = 5, 4x + 5 = 9 and 8x + 1 = 9. We CAN have a triangle withe these sides.
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Intern  Joined: 29 Jun 2016
Posts: 11

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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
Intern  Joined: 29 Jun 2016
Posts: 11

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In other words x>= 8; the third option doesn't satisfy
Math Expert V
Joined: 02 Sep 2009
Posts: 59728

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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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Intern  Joined: 29 Jun 2016
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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 V
Joined: 02 Sep 2009
Posts: 59728

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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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Thanks for the info
Senior Manager  Joined: 31 Mar 2016
Posts: 375
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34 GPA: 3.8
WE: Operations (Commercial Banking)

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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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Joined: 20 May 2014
Posts: 12
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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.

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
Current Student B
Joined: 23 Nov 2015
Posts: 15
GMAT 1: 690 Q45 V39 GPA: 3.47

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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.
Math Expert V
Joined: 02 Sep 2009
Posts: 59728

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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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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 V
Joined: 02 Sep 2009
Posts: 59728

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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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Intern  Joined: 18 Nov 2019
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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. Re: M31-20   [#permalink] 22 Nov 2019, 03:46

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# M31-20

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