Answer is 15 how do you get there

Question

At a cross-country track meet, Alicia ran 8 mph for the first part of the race, then increased her speed to 12 mph for the second part. If the race was 21 miles long and Alicia finished in 2 hours, how far did she run at the faster pace?

Answer 1

Our unknowns here are the distances of the two "parts" of the race.  We must assign them to variables - Alicia ran x miles for the first part of the race, and then y miles for the second part.

 

Since the race is 21 miles in total, x and y together must add up to 21.

x + y = 21

 

Next, we're told the speeds at which she ran and the total time involved; we can relate this to the distances by the speed and distance equation d = st, or t = d/s.  Since she finished in two hours, the time spent running the first part and time spent running the second part must add up to 2 hours, or:

 

x/8 + y/12 = 2

 

Two equations are sufficient to solve for two unknowns.  I'll multiply the second equation by the LCM, 24

 

3x + 2y = 48

 

And rearrange the first to get y = 21 - x, which I can plug into the above

 

3x + 2(21 - x) = 48

3x + 42 - 2x = 48

x = 6

 

Use y = 21 - x again:

y = 21 - (6) = 15.

 

The question asks how far she ran at the faster pace - this would be the second part of the race, which we have called y, so it would be 15 miles.  But let's check that our answer works, first.

 

Certainly, 15 + 6 = 21, so the total length works.

As for time, 6 miles at 8 mph would take 6/8 = .75 hours = 45 minutes, and 15 miles at 12 mph would take 15/12 = 1.25 hours = 75 minutes.  The total time to finish would then be .75 + 1.25 = 2 hours, as the question states.  So this answer is indeed valid, and Alicia ran 15 miles at the faster pace.