Question

Eugene walked all the way to school at 3 mph then realized she forgot her math book. She ran back at 7 mph. If she got back to her house 45 minutes after she left, how far does she live from school? Please answer using d, r, t

Answer 1

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 2

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 $time(t)=\frac{distance(d)}{speed(r)}$

$\frac{d}{3} +\frac{d}{7} = \frac{45}{60}$

Taking the LCM:

$\frac{7d+3d}{21} = \frac{45}{60} \\\\\frac{10d}{21} = \frac{45}{60}$

By cross multiplication:

$10d*60=21*45\\\\600d=945\\\\d=1.575$

Therefore, Eugene's lives 1.575 miles away from her school.