SOLUTION: A motorist travels a distance of 120 miles. Had she increased her average speed by 10 mi/hr she would have cut 36 minutes off of her travel time. Find her average speed.

Question 319987: A motorist travels a distance of 120 miles. Had she increased her average speed by 10 mi/hr she would have cut 36 minutes off of her travel time. Find her average speed.
Found 2 solutions by mananth, MathTherapy:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let her speed be x
increased speed = x+10
distance traveled=120 miles
36/60 = 3/5 hours
..
time taken with increased speed= original time -3/5 hours
120/x+10= 120/x -3/5
120/x+10=600-3x /5x
5x*120=(x+10)(600-3x)
600x=600x-30x+6000-3x^2
3x^2+30x-6000=0
x^2+10x-2000=0
x^2+50x-40x-2000=0
x(x+50)-40(x+50)=0
(x+50)(x-40)=0
x=40 mph
Answer by MathTherapy(10551) (Show Source): You can put this solution on YOUR website!
A motorist travels a distance of 120 miles. Had she increased her average speed by 10 mi/hr she would have cut 36 minutes off of her travel time. Find her average speed.

Let her average speed be S

Then we'll have: ----->

Multiply by LCD, 5S(S + 10) to get: 120(5)(S + 10) - 3S(S + 10) = 120(5S)

----- Dividing by GCF, 3

----- (S + 50)(S - 40) = 0

Ignoring the negative value (- 50), we get S, or average speed as: mph

------------------
Check
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Initial speed: 40 mph

Time taken: 120/40 = 3 hours

Increased speed = 50 (40 + 10)mph

New time = 120/50 = 2.4 hours

Difference in travel times = .6 hour (3 - 2.4)

.6 hour = = 36 minutes
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