Answer:
Option (G)
Step-by-step explanation:
Let the length of the race = a miles
Since, Speed = [tex]\frac{\text{Distance}}{\text{Time}}[/tex]
Time taken to cover 'a' miles with the speed = 12 mph,
Time taken '[tex]t_1[/tex]' = [tex]\frac{a}{12}[/tex]
Time taken to cover 'a' miles with the speed = 11 mph,
Time taken '[tex]t_2[/tex]' = [tex]\frac{a}{11}[/tex]
Since the time taken by David to cover 'a' miles was 10 minutes Or [tex]\frac{1}{6}[/tex] hours more than the time he expected.
So, [tex]t_2=t_1+\frac{1}{6}[/tex]
[tex]\frac{a}{11}=\frac{a}{12}+\frac{1}{6}[/tex]
[tex]\frac{a}{11}-\frac{a}{12}=\frac{1}{6}[/tex]
[tex]\frac{12a-11a}{132}=\frac{1}{6}[/tex]
a = 22 mi
Therefore, distance of the race = 22 mi
Option (G) is the correct option.
