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John and Brian leave Williston at the same time. John drives north and Brian drives east.
John's average speed is 10 miles per hour slower than Brian's. At the end of one hour
they are 50 miles apart. Find Brian's average speed.
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@mananth incorrectly read the problem, incorrectly interpreted it and,
as a consequence of it, incorrectly solved.
I came to make a job and to bring a correct solution.
Obviously, we have a right angled triangle.
One leg is x miles (the traveled distance for Brian, which is numerically equal to the Brian's average speed).
The other leg is (x-10) miles (the traveled distance for John, which is numerically equal to the John's average speed).
The hypotenuse is 50 miles, according to the condition.
Now recall (3,4,5)-right angled triangle, and the answer will momentarily come to your mind:
+------------------------------------------------------------------------+
| The Brian's average speed was 40 miles per hour |
| (because 40 miles is the length of the longer leg of this triangle). |
+------------------------------------------------------------------------+
Solved.
Alternatively, you can solve the quadratic equation
x^2 + (x-10)^2 = 50^2,
x^2 + x^2 - 20x + 100 = 2500
2x^2 - 20x - 2400 = 0,
x^2 - 10x - 1200 = 0
(x-40)*(x+30) = 0,
which leads you to the same answer.
Solved in two ways: mentally and via an equation, for your better understanding.
