Question

Help please!!! Thank you

Answer 1

Answer:

Option (G)

Step-by-step explanation:

Let the length of the race = a miles

Since, Speed = $\frac{\text{Distance}}{\text{Time}}$

Time taken to cover 'a' miles with the speed = 12 mph,

Time taken '$t_1$' = $\frac{a}{12}$

Time taken to cover 'a' miles with the speed = 11 mph,

Time taken '$t_2$' = $\frac{a}{11}$

Since the time taken by David to cover 'a' miles was 10 minutes Or $\frac{1}{6}$ hours more than the time he expected.

So, $t_2=t_1+\frac{1}{6}$

$\frac{a}{11}=\frac{a}{12}+\frac{1}{6}$

$\frac{a}{11}-\frac{a}{12}=\frac{1}{6}$

$\frac{12a-11a}{132}=\frac{1}{6}$

a = 22 mi

Therefore, distance of the race = 22 mi

Option (G) is the correct option.