lheiannie07 wrote:Bob bikes to school every day at a steady rate of x miles per hour. On a particular day, Bob had a flat tire exactly halfway to school. He immediately started walking to school at a steady pace of y miles per hour. He arrived at school exactly t hours after leaving his home. How many miles is it from the school to Bob's home?
A. (x + y) / t
B. 2(x + t) / xy
C. 2xyt / (x + y)
D. 2(x + y + t) / xy
E. x(y + t) + y(x + t)
Let the distance to school = 20 miles.
Let x = 5 miles per hour.
Time spent biking = 10/5 = 2 hours.
Let y = 2 miles per hour.
Time spent walking = 10/2 = 5 hours.
Total time = t = 2+5 = 7.
The questions asks for the total distance: 20 miles. This is our target.
New we plug x=5, y=2, and t=7 into the answers to see which yields our target of 20.
Only answer choice
C works:
2xyt/(x+y) = (2*5*2*7)/(5+2) = 140/7 = 20.
The correct answer is
C.
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