M31-20

Question

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

Bunuel · Answer

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

Note that the question asks which of the following options  could  be the length of the third side of the triangle.
 
Remember that  the length of any side of a triangle must be greater than the positive difference of the other two sides and less than the sum of the other two sides . Therefore, we have:
 
 (4x+5) - (3x+2) < (third \ side) < (3x+2) + (4x+5) 
 
 x + 3 < (third \ side) < 7x + 7 .
 
Let's examine the options:
 
I.  6x + 7 . For any positive value of  x ,  x + 3 < 6x + 7 < 7x + 7 . This option works.
 
II.  6x + 9 . For  x = 1  or  x = 2 , the inequality  x + 3 < 6x + 9 < 7x + 7  does not hold true, but for  x = 3 , it does. This option works as well.
 
III.  8x + 1 . Immediately, we can see that for  x = 1 , the inequality  x + 3 < 8x + 1 < 7x + 7  is true. This option also works.

Answer: E

subhajit1 · Answer

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.

Bunuel · Answer

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

Senthil7 · Answer

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

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

arpitk17 · Answer

I agree with logic and that's the same I used to solve. But II doesn't make sense. If you substitute x with 1 then 7x + 7 = 14 and 6x+9 =15 hence it wouldn't fit the inequality.

How can option II. be correct?

Bunuel · Answer

The question asks for the lengths that  could  potentially be the length of the third side of the triangle, not the lengths that  must  be true for all values of x. Therefore, even if the inequality does not hold for certain values of x, as long as there exists at least one positive integer value of x for which the inequality does hold, the option is considered valid.

For example, while II does not hold true for x=1 or x=2, it does hold true for x=3. When x=3, the two given sides of the triangle are 3x+2 = 3(3)+2 = 11 and 4x+5 = 4(3)+5 = 17. For II, the third side would be 6x+9 = 6(3)+9 = 27. Since 27 is greater than 17-11=6 and less than 17+11=28, it " could " be the length of the third side of the triangle, making it a valid option.

Thus, the official solution is correct, and the answer is indeed E, I, II, and III.

M31-20

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