While on a family vacation in a national park, Carson, who is training

We see that Carson’s first two legs of running path are 6 miles and 8 miles and we can let his return path to his family RV be the third leg.

We can say that 8 and 6 represent two legs of a triangle. Thus, we can follow the triangle inequality rule in which the sum of 2 sides of the triangle must be greater than the 3rd side. So 12 is a possible distance of the 3rd leg.

However, if the first two legs are part of a straight-line path and his return path is also along this straight-line path, then his return path can be exactly 6 + 8 = 14 miles. Thus 14 is also a possible distance of the 3rd leg.

At this point, we already know the answer since there is only one answer choice containing both I and II, but we can also easily justify that III is indeed impossible by noticing 6 + 8 = 14 < 16; which by the triangle inequality cannot be the length of the third leg of a triangle when the first two legs measure 6 and 8 units..

Alternate Solution:

Let’s assume Carson has traveled a straight-line path for 6 miles to get to the ranger station, and then he proceeds an additional 8 miles along that straight-line path, resulting in his being 6 + 8 = 14 miles from the family RV. Statement II is correct. Note that 14 miles is the maximum distance that he can be located from the family RV. Thus, Statement III is impossible.

Note also that Carson’s minimum distance from the family RV at the end of his 14-mile run is 0; he could have run straight, or in circles, or in any direction during his run. We see that any answer between 0 and 14, inclusive, is possible for his distance from the family RV at the end of his run. Thus, Statement I is also correct.

Answer: C

Elaborating ScottTargettestprep alternate solution

Triangle ∆
If we open this triangle
\_/
___
Then it will become a straight line
Maximum distance here is 14
So any value less than equal to 14 will be and
So 12 and 14 are answers

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