Amy drove the 200 miles to New Orleans at an average speed 1

Amy drove the 200 miles to New Orleans at an average speed 10 miles per hour faster than her usual average speed. If she completed the trip in 1 hour less than usual, what is her usual driving speed in miles per hour.

Solution

his makes the speed she traveled at equal to X + 10.

Problem states that she took 1 hour less.

This makes the time she took equal to T - 1.

She normally travels at X miles per hours and takes T hours.

Since rate * time = distance, we get:

X * T = 200

This time she traveled 10 miles per hour faster and took 1 hour less.

This time, we get:

(X+10) * (T-1) = 200

Since they are both equal to 200, then they are both equal to each other, so we get:

X*T = (X+10)*(T-1)

We simplify this by multiplying out all terms to get:

X*T = X*T - X + 10T - 10

We have one equation in two unknowns, but we can substitute for one of the unknowns and solve for the other.

We\'ll take the equation X*T = 200 and solve for T to get T = 200/X

We\'ll then replace T in our equation of X*T = X*T - X + 10T - 10 to get:

X*200/X = X*200/X - X + 10*200/X - 10.

We\'ll simplify this equation to get:

200 = 200 - X + 2000/X - 10.

We\'ll subtract 200 and we\'ll add X and we\'ll subtract 2000/X and we\'ll add 10 to both sides of this equation to get:

X - 2000/X + 10 = 0

We\'ll multiply both sides of this equation by X to get:

X^2 + 10X - 2000 = 0

This factors out to:

(X-40) * (X+50) = 0

X can be either 40 or -50.

since X can\'t be negative, we are left with X = 40.

Since X*T = 200, We get T = 5

Our answer is:

X = 40
T = 5

40*5 = 200 which is the normal time at the normal rate.

50*4 = 200 which is 1 hour less than the normal time at 10 miles per hour more than the normal rate.