Beth had planned to run an average of 6 miles per hour in a race. She had a very good race and ran at an average speed of 7 mile per hour, finishing 10 minutes sooner than she would if she had averaged 6 miles per hour. How long was the race?

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Answer 1 · 2018-02-19T18:35:49

#7# miles

#s = d/t#

speed = distance / time

#d/t = 6 mi//h#

here, #d# is the total distance of the race

and #t# is the time, in hours, that Beth would have taken, if she ran at #6 mi//h#.

#1h = 60m#
she ran #10# minutes sooner than planned, so the time taken for her to run was #t - 1/6#. (where #t# is in hours)

when she ran for #10# minutes less than planned, her speed was #7 mi//h#.

#d/(t-1/6) = 7 mi//h#

this gives the two equations, #d/t = 6# and #d/(t-1/6) = 7#

you can solve both by isolating distance #d:#

#d/t = 6#

#d = 6t#

#d/(t-1/6) = 7#

#d = 7(t - 1/6)#

#d = 6t, d = 7(t - 7/6)#

this means that #6t = 7(t - 1/6)#.

using this, you can solve for time #t:#

#6t = 7(t - 1/6)#

#6t = 7t - 7/6#

#t - 7/6 = 6t - 6t = 0#

#t = 7/6 h# (or #7h# #10m#)

then #t# can be substituted into the speed equation:

#d/t = 6#

#d = 6t#

#d = 6 * 7/6#

#= 7/1#

#= 7#

the distance of the race is #7# miles.