Answer
The distance between to her house and the school is 1.575 miles.
Step by step explanation
Eugene walked all the way to school = 3mph
She ran back at the rate (r) = 7 mph
Let's "d" is distance between her house and the school.
time(t) = distance/rate
Time taken = 45 minutes
d/3 + d/7 = 45/60
Now we have to find the LCD of 3 and 7, that is 21
(7d + 3d)/21 = 45/60
10d/21 = 45/60
Cross multiply and find the value of "d"
10d * 60 = 45*21
600d = 945
d = 945/600
d = 1.575 miles
Answer:
1.575 miles
Step-by-step explanation:
We know that,
the rate (r) at which Eugene walked to school = 3 mph
the rate (r) at which Eugene ran back to home = 7 mph
Time taken (t) for her to get back to house after she left = 45 minutes
Using the given data we can write an equation following the formula [tex]time(t)=\frac{distance(d)}{speed(r)}[/tex]
[tex]\frac{d}{3} +\frac{d}{7} = \frac{45}{60}[/tex]
Taking the LCM:
[tex]\frac{7d+3d}{21} = \frac{45}{60} \\\\\frac{10d}{21} = \frac{45}{60}[/tex]
By cross multiplication:
[tex]10d*60=21*45\\\\600d=945\\\\d=1.575[/tex]
Therefore, Eugene's lives 1.575 miles away from her school.
