Question

Bob and Sue enter a race. Bob runs an average of 12 kilometers per hour, and Sue runs an average of 8 kilometers per hour. If Bob finishes 2 hours before Sue, how long is the race?

Answer 1

Bob = 12 x T
Sue= 8 (T +2)

12T = 8T + 16
12T-8T=8T-8T +16
4T = 16
4T/4 = 16/4
T= 4 HOURS

PLUG T BACK INTO
12(4) = 8(4) + 16
48=48

DISTANCE = 48 kilometers

Answer 2

48 km

Explanation:

Let us assume that, the distance of the track where they raced is x km.

We know that,

`\text{Speed}=\frac{\text{Distance}}{\text{Time}}`

or
`\text{Time}=\frac{\text{Distance}}{\text{Speed}}`

Bob runs an average of 12 kilometers per hour. So time taken by Bob is,

`t_1=(x)/(12)`

Sue runs an average of 8 kilometers per hour. So time taken by Sue is,

`t_2=(x)/(8)`

Bob finishes 2 hours before Sue, so

`\Rightarrow t_2-t_1=2`

`\Rightarrow (x)/(8)-(x)/(12)=2`

`\Rightarrow (3x-2x)/(24)=2`

`\Rightarrow (x)/(24)=2`

`\Rightarrow x=2* 24=48` km