Problema Solution
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 provided by our tutors
the quicher plane = Q+75 miles per hour
the slower plane = Q miles per hour
1000/Q = 3 + 1000/(Q+75) where 1000/Q is trip time for the slower plane and 1000/(Q+75) is trip time for the quicker plane
If you solve the equality, we have
1000(Q+75)=3Q(Q+75)+1000Q =======> Q^2+75Q-25000=0 =====> (Q+200)*(Q-125)=0
Then, Q=125.
the slower plane completes 1000/Q = 1000/125= 8 hours
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!PLEASE CAN YOU RATE MY SOLUTION!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!