If x is a positive integer, and two sides of a certain triangle have

If you use the method of substituting x for a real integer, someone may substitutes the number which is NOT in the range of 2 < x < 6 (see my above solution), they will choose A or C (wrong answer) instead of E.

For example: if x = 1, substitute to II we have 6x + 9 = 15
Your inequality becomes 4 < 15 <14 -> wrong

or if x = 10, substitute to III, 8x + 1 = 81
Your inequality becomes 13 < 81 < 77 -> wrong

You luckily substituted the right number, so you get the right answer. However, if someone is not lucky like you, they will choose wrong answer or they will waste time trying some different numbers until get I, II and III are possible length of the 3rd side.

I DO NOT say your method is the incorrect one, but in my perspective a method which is based on luck is not a good method.

third side need to fall between [ (4x+5)-(3x+2) < Side 3 < (4x+5)+(3x+2) ]

(x+3) < Side 3 <(7x+7)

now as x is positive integer (lets try some +ve integers say : 1 ,2, 3)

for x = 1 ;
I. 6x + 7 = 13 ; satisfy cond. (x+3) < Side 3 < (7x+7)
II. 6x + 9 = 15 ; does not satisfy
III. 8x + 1 = 9 ; satisfy cond. (x+3) < Side 3 < (7x+7)

for x = 2 ; same as above

for x = 3 ;
II. 6x + 9 = 15 ; satisfy cond. (x+3) < Side 3 < (7x+7)

hence all three are possible values of 3rd side

Answer : E. (I, II and III)

Let the lenth of the third side is Y. The sum of the 2 other sides will be 7x + 7.
Use the triangle rule: the sum of any 2 sides will be always larger than the third side.
-> 7x + 7 > Y

Look at I, simply because X is a positive number so that 7x + 7 definitely larger than 6x + 7. Hence, I is right.
Therefore, we can eliminate answer B and D.

Now we can easily solve the 2 inequalities:

* 7x + 7 > 6x + 9 -> x > 2

* 7x + 7 > 8x + 1 -> x <6

Given the condition 2 < x < 6 (x > 0), II and II COULD BE the length of the 3rd side as the thread asks us.

Hence, I, II and III could be the length of the third side.
The right answer is E.

Good point. It wasn't by luck I chose 5. I knew that there are numbers which give you wrong answers as well. But since this is a 'could be' question one should choose a number such that you get to the answer in one shot. Looking at the upper limits (7x+7 and 8x+1 ) one can deduce that they cannot work with integers greater than 5. For example when x =6; 7x+7 = 8x+1 which is out of the range.This is why I chose 5.

This questions asks us to keep in mind when finding the values as "could be"

As per formula the sum of 2 sides of a triangle are more than the third side and the difference of the 2 sides is less than than the third side.

I and II can be seen to fit into the equation, though if we look at the III option, 8x+1, We can plug in values to check, if x=1, then the 2 sides given are 9 and 5, the III option is 9 as well, considering the question is asking "could be". all three options are correct. E is the right answer.

Warm Regards,

Rudraksh.

Dear Irtim2307,

I read your solution and from what I understand, it is a more correct approach than mine, as rightly said, one may come to a wrong conclusion.

That said, I paid heed to word "could" knowing that 7x+7 is (4x+5 summed 3x+2) will have possibilities where it might and might not work when tried against 6x+9.

I just needed to know that it could, as the question stem says.

Hope I am able to assist you.

Warm Regards,

Rudraksh.

Hello guys my attempt:

Concept used: Triangle inequality theorem which states that sum of two sides of a triangle is always equal to or greater than the third side.

Point to keep in mind that x is a positive integer and sum of the given two sides is 7x+7

1. 6x+7
This is always greater than the other two numbers and 7x+7>6x+7. Thus possible.

2. 6x+9
This is always greater than the other two numbers and 7x+7=6x+9 is possible for x=2. Thus possible.

3. 8x+1
This number can be equal to 4x+5 at x=1 where two sides of the isosceles triangle will be 9 each and the third side will be 5. Thus possible.

All possible. Answer E.

Note : solving for If questions had asked for must be true instead of could be (then we have to plug values of x)

Now, as x is a positive integer, it can have values = {0, 1 , 2 , 3.... 10..}

(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.
\(x + 3 < (third Side = 6x + 7) < 7x + 7\)
should hold for any positive value of x


II. 6x + 9. For x = 1 or x = 2, x + 3 < 6x + 9 < 7x + 7 does not hold true but if x = 2, No.
\(x + 3 < (third Side = 6x + 9) < 7x + 7\)
\(2 + 3 < (third Side = 6.2 + 9) < 7.2 + 7\)
\(5 < 21 < 21\)
\(Nope\)

III. 8x + 1. Right away we can find that if x = 1, then x + 3 < 8x + 1 < 7x + 7 is true. OK, But what about bigger value of x say x=10 does not hold true but if x = 10, No.
\(x + 3 < (third Side = 8x + 1) < 7x + 7\)
\(10 + 3 < (third Side = 8.10 + 1) < 7.10 + 7\)
\(13 < 81 < 77\)
\(Nope\)

so in case of must be true only (i) is true

in case of could be true all 3 true

Hi Rudraksh,

Why do you say "II can be seen to fit into the equation"?

Thank you for answering me Rudraksh.

But I just don't understand why just looking II makes you say "II can be seen to fit into the equation". May be you have something I can learn from.

Regard.

I plug in x =1 and didn't think to check for additional values of x. It led me to answer choice A. Your explanations helped! +1 UJs, thanks.

What if the Question Had Asked Must be instead of could be => Then 1 and 3 will be true i am guessing ??/

Notice that the question asks which of the following could be the length of the third side of the triangle.


What if the Question Had Asked Must be instead of could be => Then 1 and 3 will be true i am guessing ??/[/quote][/quote]

Note : solving for If questions had asked for must be true instead of could be (then we have to plug values of x)

Now, as x is a positive integer, it can have values = {0, 1 , 2 , 3.... 10..}

(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.
\(x + 3 < (third Side = 6x + 7) < 7x + 7\)
should hold for any positive value of x


II. 6x + 9. For x = 1 or x = 2, x + 3 < 6x + 9 < 7x + 7 does not hold true but if x = 2, No.
\(x + 3 < (third Side = 6x + 9) < 7x + 7\)
\(2 + 3 < (third Side = 6.2 + 9) < 7.2 + 7\)
\(5 < 21 < 21\)
\(Nope\)

III. 8x + 1. Right away we can find that if x = 1, then x + 3 < 8x + 1 < 7x + 7 is true. OK, But what about bigger value of x say x=10 does not hold true but if x = 10, No.
\(x + 3 < (third Side = 8x + 1) < 7x + 7\)
\(10 + 3 < (third Side = 8.10 + 1) < 7.10 + 7\)
\(13 < 81 < 77\)
\(Nope\)

so in case of must be true only (i) is true

in case of could be true all 3 true[/quote]



Excellent stuff..!!!
Thanks