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

`0.83\text{ miles per minute}`

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}=(2)/(x)+(8)/(x+0.5)\text{minute}`

Time taken if Janet rides whole race at faster pace
`=(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.

`(2)/(x)+(8)/(x+0.5)=(10)/(x+0.5)+2`

`(2)/(x)-(2)/(x+0.5)=2`

`(1)/(x)-(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}=(2)/(x)+(8)/(x+0.5)\text{minute}`

`\text{Time taken}=(2)/(0.5)+(8)/(1)\text{minute}`

`\text{Time taken}=(2)/(x)+(8)/(x+0.5)\text{minute}`

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

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

Average speed
`=(10)/(12)=0.83\text{miles per minute}`

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