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

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 · Accepted Answer

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.

askhere · Answer

Let the third side of triangle be AB
Triangle rule states that:
 Difference of other 2 sides < any side < sum of other two sides

So 4x+5 - (3x+2) < AB < 4x+5 + (3x+2)
i.e. x + 3 < AB < 7x +7
Let's take x =5 :
 So 8 < AB < 42
Check if the options lies between 8 & 42.
I: 6x+7 = 37 ; could be a possible value.
II: 6x+9 =39 ; could be a possible value.
III: 8x+1 = 41; could be a possible value.
I, II and III could be a possible value.

Answer :  E 
Ambarish

Irtim2307 · Answer

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.

Irtim2307 · Answer

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.

askhere · Answer

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.

RudeyboyZ · Answer

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.

Irtim2307 · Answer

Hi Rudraksh,

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

RudeyboyZ · Answer

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.

Irtim2307 · Answer

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.

Jackal · Answer

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.

UJs · Answer

third side need to fall between  < 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)

BuggerinOn · Answer

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.

stonecold · Answer

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

UJs · Answer

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

stonecold · Answer

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 ??/

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

Excellent stuff..!!!
Thanks

Kinshook · Answer

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?

For a triangle sum of any 2 sides should be greater than the third side.

I. 6x + 7; 3x+2 + 4x+5 > 6x + 7; x > 0; 3x+2 + 6x + 7 > 4x+5; 5x> -4; 4x+5 + 6x + 7 > 3x+2; 7x>-10; Feasible
II. 6x + 9; 3x+2 + 4x+5 > 6x + 9; x > 2; 3x+2 + 6x + 9 > 4x+5; 5x> -6; 4x+5 + 6x + 9 > 3x+2; 7x>-12; Could be true if x>2
III. 8x + 1; 3x+2 + 4x+5 > 8x + 1; x < 6; 3x+2 + 8x + 1 > 4x+5; 7x> 2; 4x+5 + 8x + 1 > 3x+2; 9x>-4; Could be true if 0<x<6

A. I only
B. II only
C. I and II only
D. II and III only
E. I, II and III

IMO E

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

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