Question

A driver of a car took a day trip around the coastline driving at two different speeds. He drove 7070 miles at a slower speed and 200200​ miles at a speed 3030 miles per hour faster. If the time spent driving at the faster speed was twicetwice that spent driving at the slower​ speed, find the two speeds during the trip.

Answer 1

Answer:

Slower speed = 70 mph

Faster speed = 100 mph

Explanation:

Let the slower speed be x miles per hour

Then, the faster speed = (x+30) miles per hour

Let the time spent driving at the slower speed (that is, at x mph) = t

Then, time spent driving at the faster speed (that is, at (x+30) mph) = 2t

speed = (distance)/(time)

Distance = speed × time

Distance covered during the slower speed = x × t = xt = 70 (given in the question)

xt = 70 (eqn 1)

At the faster speed

Distance covered = (x+30)(2t) = 2t(x+30)

Distance covered during faster speed = 200 miles

2t(x+30) = 200

2xt + 60t = 200

Recall (eqn 1)

xt = 70

2(70) + 60t = 200

60t = 60

t = 1 hour.

xt = 70

Slower speed = x = 70 mph

Faster speed = (x+30) = 100 mph

Answer 2

complete question:

A driver of a car took a day trip around the coastline driving at two different speeds. He drove 70 miles at a slower speed and 200​ miles at a speed 30 miles per hour faster. If the time spent driving at the faster speed was twice that spent driving at the slower​ speed, find the two speeds during the trip.

Answer:

slower speed

speed   = 70 miles/hr

Faster speed

speed = 100 miles/hr

Explanation:

The driver took a day trip at two different speed. The first speed was slower while the second was faster.

let the speed be divided into 2

Slower speed

speed = a

distance = 70

speed = distance/time

time = distance/speed

time = 70/a

Faster speed

speed = a + 30

distance = 200

speed = distance/time

time = distance/speed

time = 200/a + 30

Since the faster speed time is twice the slower speed time it can be represented as follows:

2 × 70/a = 200/a + 30

140/a = 200/ a + 30

cross multiply

140a + 140(30) = 200a

4200 = 200a - 140a

4200 = 60a

divide both sides by 60

4200/60 = a

a = 70

Inserting the value of a in the time of the faster speed formula

time = 200/a + 30

time = 200/100

time = 2 hr

slower speed

speed = distance/time

speed = 70/1

speed   = 70 miles/hr

Faster speed

speed = distance/time

speed = 200/2

speed = 100 miles/hr