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average speed of an unknown distance

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I drove x miles at 55 mph. I then drove x + 20 miles at 40 mph. I drove for a total of 100 minutes. How far did I drive?

Edited by BMAD
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Posted · Report post

27 miles @55mph, then 47 miles@40 mph

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Posted · Report post

t1 + t2 = 100 minutes = 100/60 hours


t1 = x/55
t2 = (x+20)/40

x/55 + (x+20)/40 = 100/60
x = 1540/57

Total distance = x + (x+20) = 4220/57 miles = 74 2/57 miles
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Posted · Report post

X = miles at 55 mph

X+20 = miles driven at 40 mph

and

55 mi/ hr = 0.917 mi/ min

40 mi/ hr = 0.667 mi/ min

D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph)

T = Total Time = T1 (at 55 mph) + D2 (at 40 mph)

D1 = T1 ( 0.917 mi/ min)

D2 = T2 ( 0.667 mi/ min)

Therefore

D = D1 + D2 = T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min)

We also know that D1 +20 = D2

These can be rearranged into 2 simultaneous equations in the form

20

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Posted · Report post

X = miles at 55 mph

X+20 = miles driven at 40 mph

and

55 mi/ hr = 0.917 mi/ min

40 mi/ hr = 0.667 mi/ min

D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph)

T = Total Time = T1 (at 55 mph) + D2 (at 40 mph)

D1 = T1 ( 0.917 mi/ min)

D2 = T2 ( 0.667 mi/ min)

Therefore

D = D1 + D2 = T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min)

We also know that D1 +20 = D2

These can be rearranged into 2 simultaneous equations in the form

20

I hit the wrong key

20 = - T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min) [Eqn 1]

100 = T1 +T2 [Eqn 2]

Multiplying all terms in Eqn 1 by {- 1/0.667} yeilds

-30 = 1.375 T1 - T2

Adding this to Eqn2 results in

70 = 2.375 T1 Therefore

T1 = 70 /2.375 = 29.47 min, so

D1 = 29.47 min * ( 0.917 mi/ min) = 27.02

D2 = D1 + 20 = 47.02

Therefore Total distance = 74.04 Miles

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