<PRE><B>Question 1162960</B>
&lt;br&gt;
The other tutor demonstrates one good algebraic method for solving the problem: setting the two time (=distance/rate) expressions equal to each other and solving the equation.&lt;br&gt;
{{{260/(r+3) = 200/(r-3)}}}&lt;br&gt;
If you are good with mental arithmetic, the problem can be solved without algebra by setting up the problem differently.&lt;br&gt;
Since the times are the same, the ratio of distances is equal to the ratio of the speeds:&lt;br&gt;
{{{260/200 = (r+3)/(r-3)}}}&lt;br&gt;
The difference between numerator and denominator on the right is 6 -- so write the fraction on the left as an equivalent fraction in which the difference between numerator and denominator is also 6:&lt;br&gt;
{{{26/20 = (r+3)/(r-3)}}}&lt;br&gt;
r+3 = 26 and r-3 = 20...&lt;br&gt;
ANSWER: r = 23&lt;br&gt;
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