Word Problem w/ Distance

Question

Two planes make a 1000 mile flight, one flying 75 miles per hour faster than the other. The quicker plane makes the trip 3 hours faster. How long did it take the slower plane to complete the flight?

Answer 1

Distance = Rate times Time or D = R*T 

 

Let X = Rate of slower plane 

Then X + 75 = Rate of faster plane 

 

Let Y = Time of slower plane

Then Y - 3 = Time of faster plane 

 

So 1000 = X*Y for the slower plane  ==>  X = 1000/Y 

and 1000 = (X + 75)*(Y - 3) for the faster plane  

 

Multiply out the 2nd equation 

 

X*Y + 75*Y - 3*X - 225 = 1000 

 

Substitute X = 1000/Y into this equation 

 

(1000/Y)*Y + 75*Y - 3*(1000/Y) -225 = 1000

1000 + 75*Y - 3000/Y = 1225

75*Y - 3000/Y = 225

75*Y - 3000/Y - 225 = 0 

 

Multiply both sides by Y 

 

75*Y^2 - 225*Y - 3000 = 0

Y^2 - 3*Y - 40 = 0

(Y + 5)*(Y - 8) = 0 

 

Y = -5  or Y  = 8 

 

Time cannot be negative so pick Y = 8  

 

It took the slower plane 8 hours to complete the flight 

 

Check:  X = 1000/Y = 1000/8 = 125 mph for the slower plane 

           X + 75 = 200 mph for the faster plane

           Y - 3 = 5 hours for the faster plane 

           D = (5 hours)*(200 mph) = 1000 miles for the faster plane