Wendy took a trip from one city to another, a distance of 290 mi. She traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train.

Question

Wendy took a trip from one city to another, a distance of 290 mi. She traveled part of the way by bus, which arrived at the train station just in time for Wendy to complete her journey by train. The bus averaged 40 mi/h, and the train averaged 110 mi/h. The entire trip took 5 1/2 h. How long did Wendy spend on the train?

Answer 1

Let DB be the distance traveled by bus,

DT be the distance traveled by train,

HB be the time in hours traveled by bus,

and HT be the time in hours traveled by train.

So from the basic rule of distance = rate * time, we know that

DB = 40 mi/hr * HB, and

DT = 110 mi/hr * HT

We also know that DB + DT = 290 mi, or

40*HB + 110 * HT = 290 Label this equation (A).

We also know that HB + HT = 5.5 hrs.

We observe that if we multiply both sides by 40, we can get the HB term to drop out:

40*HB + 40*HT = 220 Label this equation (B).

Then Equation (A) minus Equation (B) is:

40*HB - 40*HB + 110*HT - 40*HT = 290 - 220

70*HT = 70

HT = 1 hour

So the time spent on the train is 1 hour.

Answer 2

40 t + (5.5-t) 110= 290

40t + 605-110t= 290

-70t= -315

T=4.5

4.5*40 + 1*110=290

4.5 hours for 180 miles by hus

1 hour for 110 miles by train