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

Answer:

48 km

Step-by-step explanation:

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

We know that,

$\text{Speed}=\dfrac{\text{Distance}}{\text{Time}}$

or $\text{Time}=\dfrac{\text{Distance}}{\text{Speed}}$

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

$t_1=\dfrac{x}{12}$

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

$t_2=\dfrac{x}{8}$

Bob finishes 2 hours before Sue, so

$\Rightarrow t_2-t_1=2$

$\Rightarrow \dfrac{x}{8}-\dfrac{x}{12}=2$

$\Rightarrow \dfrac{3x-2x}{24}=2$

$\Rightarrow \dfrac{x}{24}=2$

$\Rightarrow x=2\times 24=48$ km