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In a 10 mile race, Janet covered the first 2 miles at a constant rate. She then sped up and rode her bike the last 8 miles at a rate that was 0.5 miles per minute faster. Janet's overall time would have been 2 minutes faster had she ridden her bike the whole race at the faster pace. What was Janet's average speed (in miles per minute) for the whole race?

Apr 7, 2020

#1
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Let x be the speed for the first two miles and x+.5 be the speed for the last 8 miles. Since D=rt, $$2\over x$$+$$8\over x+.5$$is the time it took Janet to do the 10 mile race. If she took her bike the whole way, her time would be $$10\over x+.5$$. Since that was two minutes faster than her real time, $$2\over x$$+$$8\over x+.5$$-$$10\over x+.5$$=2. Solving for s you get s=.5. The question asks for the average speed for the whole race so, $$2\over .5$$+$$8\over .5+.5$$=$$10\over a$$ where a is the average speed. A= $$5\over 6$$ miles per minute.

Hope it helps!

Apr 7, 2020
#2
+1

2miles  /   f mile/m     +   8 miles  /  (f+.5) mile/min  =   2min     +   10 miles  /  (f+.5) mile /min

solve for 'f'

2/f  +  8(f + .5)  = 2 + 10/(f + .5)

2/f = 2 + 2/(f+.5)

2(f+.5)  = 2(f+.5)f + 2f

2f+1 = 2f^2 +f + 2f

2f^2 +f - 1 = 0                              f = .5          then faster speed =   1 mile /min

2miles / .5 mile/min     +     8 miles/ 1 mile/min    = 12 minutes for 10 miles

10 miles/12 min =   5/6 mile / min  average         (Note:  this is 50 miles per hour average....on a bike !   Was it a MOTORbike ??)

Apr 7, 2020