Question

A young equestrian loved to ride her horse all day long, dreaming of the day

she would ride in the Kentucky Derby. One day in early April, she left home at

10:00 am and rode her stallion at a nice, steady 10 miles per hour away from

her house. At noon a call came in from the Derby saying she was in! Her

father immediately jumped on his scooter and rides at a breezy 15 miles per

hour to try and track her down and tell her the good news.

On what scale would you measure the number of minutes the father must

ride his scooter before catching up to his daughter?

O

A. Ten thousands

O

B. Thousands

O

c. Tens

O

D. Hundreds

Answer 1

Answer:

Hundreds

Step-by-step explanation:

Given that :

Daughter rode out at 10:00 am

Speed = 10 miles per hour

Father rode out at 12:00 noon

Speed = 15 miles per hour

Distance = speed * time

Daughter's distance before father rode out = (10 * (12:00 - 10:00))

= 10 * 2

= 20 miles

Difference in speed (father - daughter) :

15mph - 10mph = 5mph

Time which father will catch up with her :

Time = distance / speed

Time = 20 miles / 5 mph

Time = 4 hours

In minutes ; 4 * 60 = 240 minutes

Highest place value of minutes = hundreds