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when a bus travels a certain route at an average speed of 40 km/h it arrives one hour late at its destination,
and when it averages 48 km/h it arrives one hour early.
Q1;What is the length of the journey?
Q2: How fast should the bus travel in order to arrive on the time at its destination?
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1. Let D be the length of the journey.
Then the time of traveling at the speed of 40 km/h is hours,
and the time of traveling at the speed of 48 km/h is hours.
According to the condition, the difference is 2 hours.
It gives an equation
= 2.
To solve it, multiply both sides by 40*48. You will get
48D - 40D = 2*40*48 ---> 8D = 2*40*48 ---> D = 2*40*6 = 840.
So, the distance is 840 kilometers, and the first question is answered.
2. Next, the travel time at the speed of 40 km/h is = 21 hours.
Hence, the regular/normal/scheduled time is 21 - 1 = 20 hours, and so
the regular/normal/scheduled speed is 840/20 = 44 km/h.
Solved.
Similar problem was solved in the lesson
- How far do you live from school?
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this textbook under the section "Word problems", the topic "Travel and Distance problems".