SOLUTION: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.       Log On


   

Question 48347: Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
Found 2 solutions by venugopalramana, AnlytcPhil:
Answer by venugopalramana(3286) About Me  (Show Source):
You can put this solution on YOUR website!
Steve traveled 200 miles at a certain speed. Had he gone 10mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
let his speed be =x mph
speed at 10 mph faster=x+10 mph
time to go 200 miles
1.at normal speed =200/x
at faster speed =200/(x+10)
difference = 200[(1/x)-1/(x+10)]=200(x+10-x)/x(x+10)=1
2000=x^2+10x
x^2+10x-2000=0
(x+50)(x-40)=0
x=40 mph
x=

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Steve traveled 200 miles at a certain speed. Had he gone 10mph faster,
the trip would have taken 1 hour less. Find the speed of his vehicle.



Make a chart like this


                   DISTANCE   RATE       TIME
Actual trip                                     
Hypothetical trip                                


Let the speed of his vehicle be x

Then if he had gone 10 mph faster, his speed would have been x+10
So we put x for the actual rate, and x+10 for the hypothetical rate

                   DISTANCE   RATE       TIME
Actual trip                     x            
Hypothetical trip             x+10                

We are told that his distance was 200, so we put 200 for
the distances in both cases:

                   DISTANCE   RATE       TIME
Actual trip           200       x            
Hypothetical trip     200     x+10              

Now we use TIME = DISTANCE/RATE to fill in the two rates:

                   DISTANCE   RATE       TIME
Actual trip           200       x       200/x
Hypothetical trip     200     x+10    200/(x+10)

We are told that the time for the hypothetical trip was 1
hour less, so 

HYPOTHETICAL TIME = ACTUAL TIME - 1

      200/(x+10) = 200/x - 1

Can you solve that?  Hint: Multiply thru by LCD = x(x+10)

Answer: x = 40 mph

Edwin