Question 231453
distance = 200 miles
Her usual speed is X miles per hour since we don't know it.
Her usual time is T hours because we don't know that either.


Problem states that she traveled 10 miles per hour faster than her usual speed.


This 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.