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Steve made a business trip of 200.5 miles. He averaged 51 mph for the first part of the trip and 62 mph for the second part. If the trip took 3.5 hours, how long did hetravel at each rate?

Sagot :

Let t = time traveled at 51 mph

The total time is given as 3.5 hours

So (3.5- t )= time traveled at 62 mph

We are going to use the distance formula:

distance = speed* time

51t + 62(3.5-t) = 200.5

51t + 62*3.5 - 62*t = 200.5

51t + 217 - 62t = 200.5

Solve the equal terms

51t - 62t = 200.5 - 217

-11t = -16.5

t = -16.5/-11

t = 1.5

Then he took 1.5 at 51mph

and (3.5- t ) = (3.5-1.5) = 2h at 62 mph

To confirm these results, find the actual speed of each speed:

speed* time = distance

51*1.5 = 76.5miles

62*2. = 124 miles

76.5miles + 124 miles = 200.5miles

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