You can put this solution on YOUR website!
The ratio of the two speeds is .
Since the distances are the same, the ratio of times spent at the two speeds is .
The solution by tutor @MathLover1 is absolutely not right.
If equal TIMES are spent at two different speeds, then the average speed is the average of the two speeds.
But if equal DISTANCES are covered at two different speeds, more TIME will be spent at the lower speed, so the average speed for the round trip will be LESS THAN the average of the two speeds.
There are many correct ways to solve the problem; my preferred method is probably not a very common one. Here it is.
The ratio of the two speeds is 100:80 = 5:4.
Since the distances are the same, the ratio of times spent at the two speeds is 4:5.
Then the average speed is
There is a general formula for the average speed of the forth and back trip
= ,
which you can learn from this lesson, also.
By substituting the given data into the formula, you get the answer momentarily
= = = = km/h.
consistent with the other tutors.
I do not convince you to memorize this formula.
But every student, who is competent in Math or who wants to be considered as competent in Math,
must know that such formula does exist and should be able to deduce and to apply it (!)
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How @MathLover1 treats this problem, it is a classic example on how this problem SHOULD NOT be treated.
Every textbook and every teacher explains it to their students with all needed warnings (!) (!) (!)
I've always questioned what some people are doing in this forum, since they seem not to know certain basic math facts!
How can this woman claim to be a TUTOR when she doesn't even know that
In this case, let distance (one way) be D
This means that outgoing and return times are: , respectively
We then get: