Question

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?

Answer 1

Answer:

$0.83\text{ miles per minute}$

Step-by-step explanation:

GIVEN: In a $10$ mile race, Janet covered the first $2$ miles at a constant rate. She then speed 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.

TO FIND: What was Janet's average speed (in miles per minute) for the whole race?

SOLUTION:

let the speed of Janet in first $2\text{ miles}$  $=x\text{ miles per minute}$

speed of Janet's bike in last $8\text{ miles}$  $=x\text{+0.5 miles per minute}$

total time taken by Janet,

$\text{Time}=\frac{\text{distance}}{\text{speed}}$

$\text{Time taken}=\frac{2}{x}+\frac{8}{x+0.5}\text{minute}$

Time taken if Janet rides whole race at faster pace $=\frac{10}{x+0.5}\text{minute}$

 

As, Janet's overall time would have been $2\text{ minutes}$ faster had she ridden her bike the whole race at the faster pace.

$\frac{2}{x}+\frac{8}{x+0.5}=\frac{10}{x+0.5}+2$

$\frac{2}{x}-\frac{2}{x+0.5}=2$

$\frac{1}{x}-\frac{1}{x+0.5}=1$

$x^{2} +0.5x-1=0$

on solving we get

$x=0.5\text{ miles per minute}$

$\text{Time taken}=\frac{2}{x}+\frac{8}{x+0.5}\text{minute}$

$\text{Time taken}=\frac{2}{0.5}+\frac{8}{1}\text{minute}$

$\text{Time taken}=\frac{2}{x}+\frac{8}{x+0.5}\text{minute}$

$\text{Time taken}=12\text{minute}$

Average speed $=\frac{\text{total distance}}{\text{time taken}}$

Average speed $=\frac{10}{12}=0.83\text{miles per minute}$

Therefore average speed of Janet was $0.83\text{ miles per minute}$