This was an official explanation:
It is easier to break this motion up into different segments. Let's first consider the 40 minutes up until John stops to fix his flat.
40 minutes is 2/3 of an hour.
In 2/3 of an hour, John traveled 15 × 2/3 = 10 miles (rt = d)
In that same 2/3 of an hour, Jacob traveled 12 × 2/3 = 8 miles
John therefore had a two-mile lead when he stopped to fix his tire.
It took John 1 hour to fix his tire, during which time Jacob traveled 12 miles. Since John began this 1-hour period 2 miles ahead, at the end of the period he is 12 – 2 = 10 miles behind Jacob.
The question now becomes "how long does it take John to bridge the 10-mile gap between him and Jacob, plus whatever additional distance Jacob has covered, while traveling at 15 miles per hour while Jacob is traveling at 12 miles per hour?" We can set up an rt = d chart to solve this.
John's travel during this "catch-up period" can be represented as 15t = d + 10
Jacob's travel during this "catch-up period" can be represented as 12t = d
If we solve these two simultaneous equations, we get:
15t = 12t + 10
3t = 10
t = 3 1/3 hours
Another way to approach this question is to note that when John begins to ride again, Jacob is 10 miles ahead. So John must make up those first 10 miles plus whatever additional distance Jacob has covered while both are riding. Since Jacob's additional distance at any given moment is 12t (measuring from the moment when John begins riding again) we can represent the distance that John has to make up as 12t + 10. We can also represent John's distance at any given moment as 15t. Therefore, 15t = 12t + 10, when John catches up to Jacob. We can solve this question as outlined above.
The correct answer is B.
Last edited by
lilu on Fri Mar 13, 2009 10:43 am, edited 1 time in total.
