SOLUTION: Mr. Blackstone drove from his city hoe to his country house, stopping for lunch on the way. Before lunch, he traveled 80 miles farther than he did after lunch. Before lunch, he a

Algebra ->  Finance -> SOLUTION: Mr. Blackstone drove from his city hoe to his country house, stopping for lunch on the way. Before lunch, he traveled 80 miles farther than he did after lunch. Before lunch, he a      Log On


   

Question 1155126: Mr. Blackstone drove from his city hoe to his country house, stopping for lunch on the way. Before lunch, he traveled 80 miles farther than he did after lunch. Before lunch, he averaged 40 mph, after lunch he averaged 50 mph. His traveling time, not including the time spent for lunch, was 6 1/2 hours. Find the number of miles he drove before lunch and also after lunch.
Answer by ikleyn(52756) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let d be the distance traveled before lunch, in miles.


Then the distance traveled after lunch is d-80 miles.


He spent d%2F40 hours for the first part  and %28d-80%29%2F50 hours for the second part.


The total time equation is


    d%2F40 + %28d-80%29%2F50 = 61%2F2 hours = 13%2F2 hours.


To solve it, multiply both sides by 200. You will get

    5d + 4*(d-80) = 1300.


Simplify and find d.


    9d = 1620.

     d = 1620%2F9 miles = 180 miles.


Thus the partial distances are 180 miles and (180-80) = 100 miles, 

   and the total distance  is  180 + 100 = 280 miles.    ANSWER

Solved.