Question

Due Sun 06/09/2019 11:59 pm Skip Navigation Questions correct Q 1 [1/1] correct Q 2 [1/1] correct Q 3 [1/1] untried Q 4 (0/1) untried Q 5 (0/1) untried Q 6 (0/1) untried Q 7 (0/1) untried Q 8 (0/1) untried Q 9 (0/1) untried Q 10 (0/1) Grade: 3/10 Print Version Start of Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

Answer 1

Due Sun 06/09/2019 11:59 pm (you're already more than a week late) Skip Navigation Questions, correct Q 1 [1/1], correct Q 2 [1/1], correct Q 3 [1/1], untried Q 4 (0/1), untried Q 5 (0/1), untried Q 6 (0/1), untried Q 7 (0/1), untried Q 8 (0/1), untried Q 9 (0/1), untried Q 10 (0/1), Grade: 3/10, Print Version, Start of Questions:

Questions Two cyclists, 42 miles apart, start riding toward each other at the same time. One cycles 2 times as fast as the other. If they meet 2 hours later, what is the speed (in mi/h) of the faster cyclist?

Start of Answer:

-- When they started, they were 42 miles apart.  When they met, 2 hours later, they were no miles apart.  The distance between them shrank at the rate of 21 miles per hour.  Since they rode directly toward each other, the sum of their individual speeds must have been 21 miles per hour.

-- One biker was two times the speed of the other.  Slower biker: 1 time.  Faster biker: 2 times.  Sum of their speeds:  3 times = 21 miles per hour.  Each 'time' = 7 miles per hour.

-- Slower cycler = 1 time = 7 mi/hr

Faster cycler = 2 times = 14 mi/hr