# One morning John drove 5 hour before stopping To eat. After lunch he increased his speed by 10 mph if he completed a 430-mile in 8 hours of driving time how fast did he drive in the morning?

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The total distance traveled is given as 430 miles with a total time of 8 hours. The rate for the first 5 hours is 10mph less than the rate for the last three hours.

We are asked to find the speed in the first 5 hours.

Use the formula d=rt where d is the distance (here in miles), r is the rate (here in miles per hour or mph) and t is time (here in hours.)

The total distance is the sum of the distances traveled in the morning and after lunch.

Let x represent the unknown speed in the morning. Then the distance traveled in the morning is d(m)=5x since the time driven was 5 hours.

Then the distance traveled in the afternoon is d(a)=3(x+10). The time traveled was 3 hours and the rate is increased by 10.

So the total distance, 430 miles, is found by:

430=5x+3(x+10)

430=5x+3x+30 Using the distributive property.

430=8x+30 Collect like terms.

400=8x Subtract 30 from both sides of the equation.

x=50.

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The speed in the morning was 50mph

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Check: If the speed in the morning was 50mph then in 5 hours John covers 5(50)=250 miles. In the afternoon John would have been traveling at 60 mph so he would have gone 60(3)=180 miles further for a total of 430 miles as required.

** Before break After Break**

Speed of travel s kph s + 10 kph

Time taken 5 hours `8hrs - 5 hrs = 3 hrs`

` `

Distance traveled(= Speed * time taken) 5s 3(s + 10)

Now, Total distance traveled = 430 km

`:.` `5s + 3(s +10)` = 430

`5s + 3s + 30 = 430`

`8s = 430 - 30`

`8s = 400`

s = 50

So Jon drove at a speed of 50 kph that morning

5s + 3(s +10) = 430

5s + 3s + 30 = 430

8s = 430 - 30

8s = 400

s = 50

So Jon drove at a speed of 50 kph that morning.

5s + 3(s +10) = 430

5s + 3s + 30 = 430

8s = 430 - 30

8s = 400

s = 50