M31-51

Question

The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If  d  is the distance, in meters, between points B and C, which of the following must be correct?

A.  0 < d < 24 
B.  1 < d < 25 
C.  2 < d < 26 
D.  3 < d < 27 
E.  4 < d < 28

Bunuel · Answer

Official Solution:

The distance from point A to point B is 15 meters, and the distance from point A to point C is 11 meters. If  d  is the distance, in meters, between points B and C, which of the following must be correct?

A.  0 < d < 24 
B.  1 < d < 25 
C.  2 < d < 26 
D.  3 < d < 27 
E.  4 < d < 28

If point C is on line AB, between points A and B, then the distance between B and C will be 15 - 11 = 4 meters (the least possible distance).
 
If point C is on line AB, to the left of point A, then the distance between B and C will be 15 + 11 = 26 meters (the greatest possible distance).
 
  https://gmatclub.com/gmat-practice-tests/resources/import/img/M31-51.png 
 
Therefore,  4 \le BC \le 26 . The correct option must cover this entire range. Only option D offers a range that includes all possible values of BC.
 
Note that this is not a Geometry question. While it uses basic knowledge of lines and figures, it is actually an Inequalities question. There are 8 questions within GMAT Prep Focus Edition that use similar principles.  Here is one example .

Answer: D

rbongartz · Answer

4≤AC≤26
D. 3<d<27

The question stem doesn't limit us to integers. Accordingly, a value like d=3.1 or d=26.9 is said to be acceptable by this answer but is outside the defined range taken from solving the problem.

Bunuel · Answer

You did not understand the question.

The question asks: which of the following must be correct for d.

4 \le d \le 26 . Whatever d is, integer or not, if it's between 4 and 26, inclusive, then it must also be between 3 and 27.

sasidharrs · Answer

Hi Banuel,
I pictured it as triangle.. you can ask me why..
We are talking A to B, A to C and B to C.. tht gives a triangle in mind.
But, why cant we start with triangle anyway.. or is the question trying to give a triangle picture as a trap?

Bunuel · Answer

The question asks about the range of distance, so you should consider the case which gives you the smallest and greatest distances. The smallest and the greatest distances occur if the points are on a straight line.

RenanBragion · Answer

I think this is a  high-quality  question and I agree with explanation.

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.

MBAToronto2024 · Answer

KarishmaB  Very difficult one. How'd you solve it?

KarishmaB · Answer

Basically we have lengths of AB = 15 and AC = 11 and we are asked about BC (like 3 sides of a triangle). The only difference here is that we don't need to make a triangle so the sides could overlap too. The difference between the length of the sides is 4 so the third side could be 4 or more. The sum of the sides is 26 so the third side could be 26 or less. Hence 4 <= BC <= 26 Only option (D) works. I have illustrated this concept with straws in this video: https://www.youtube.com/watch?v=y05XIAWgAT0 Watch it from 12:01 onwards.

agrasan · Answer

I like the solution - it’s helpful.

Adit_ · Answer

I don’t quite agree with the solution.  Bunuel 
The question stem says must be true and hence triangle should be one for the considerations. If it is indeed a triangle with ABC

AB=15m
AC=11m
Distance between BC asked.
Here BC cant be 4 because 11+4=15 and not > 15 for any triangle having 3 sides.

Hence 4 cannot I rejected 3<d<27 as a reason.

Kindly clarify!

Bunuel · Answer

Triangle inequality only applies if points A, B, C are not noncollinear The stem does not say that, so C can be on line AB, which allows d = 4. Therefore any “must be true” range must include 4. This is clearly shown to be the case with the diagram given in the solution:

https://gmatclub.com/gmat-practice-tests/resources/import/img/M31-51.png

Adit_ · Answer

However "must be true" cases are inclusive of all possiblities right? They havent mentioned the points are collinear. By chance, if it were triangle(which we aren't aware of), the situation changes.

So "must be true" cases can be solved by considering one particular case and working all possibilities of it alone?

Bunuel · Answer

You are missing a point. It’s the other way around. “Must be true” means it has to hold for every placement of C that matches the given distances AB = 15 and AC = 11.

Since the stem never says A, B, C are noncollinear, C is allowed to lie on line AB. In that case, BC can be as small as 4 (when C is between A and B) and as large as 26 (when C is on the extension past A). If A, B, and C were not collinear, then  4 ≤ d ≤ 26  would still be correct, and thus a must-be-true statement.

So the correct choice is the one whose interval contains every possible BC value from 4 through 26, which is why the wider open interval in D is the only one that works.

M31-51

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