Question

Renaldo catches the bus at 4:00 p.m. to ride 3.2 miles from his house to the dentists office. He arrives at 4:30 p.m., for a one hour appointment. Then he will ride a bus traveling at the same rate of speed for 4.8 miles to the soccer field. Will he be on time for 6:30 p.m. soccer practice?

Answer 1

Answer:

yes he will reach before his practice time at 6:15 p.m.

Step-by-step explanation:

Renaldo traveled 3.2 miles in 30 min

he will travel 1 mile in 30÷3.2 min

he will travel 4.8 miles in 30/3.2 × 4.8 = 45 min

his appointment with the doctor - 1hr

his travelling time to the field - 45min

4.30p.m.+1hr 45 min = 6:15p.m.

Answer 2

Renaldo will reach before his practice time at 6 : 15 p.m

Solution:

Renaldo catches the bus at 4:00 p.m. to ride 3.2 miles from his house to the dentists office

He arrives at 4:30 p.m

4 : 00 pm to 4 : 30 pm = 30 minutes = 0.5 hour

Thus, he has covered 3.2 miles in 30 minutes

Find speed

$Speed = \frac{distance}{time}\\\\Speed = \frac{3.2}{0.5} = 6.4\ miles\ per\ hour$

He arrives at 4:30 p.m., for a one hour appointment

i.e At 5:30 pm he took bus

Then he will ride a bus traveling at the same rate of speed for 4.8 miles to the soccer field

$Time = \frac{distance}{speed} \\\\Time = \frac{4.8}{6.4} = 0.75\ hour$

Now,

Time he will arrive = 5 : 30 pm + 0.75 hour = 5 : 30 pm + 45 minutes

Time he will arrive = 6 : 15 pm

Will he be on time for 6:30 p.m. soccer practice?

Yes, he will reach before his practice time at 6 : 15 p.m

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