<PRE><B>Question 537557</B>
Call the rate of speed of train B = B.&lt;P&gt;
Call the rate of speed of train A = B-20&lt;P&gt;
Do that because the problem said A&#39;s rate is 20 mph less than B&#39;s.&lt;P&gt;
D=rt for both, and the t is the same.&lt;P&gt;
Train A:  {{{190=(B-20)t}}}&lt;P&gt;
Divide both sides by B-20, so&lt;P&gt;
Train A:  {{{190/(B-20) = t}}}&lt;P&gt;
Train B:  {{{290=Bt}}}&lt;P&gt;
Divide both sides by B.&lt;P&gt;
Train B:  {{{290/B=t}}}&lt;P&gt;
t is the same in both cases, so set the equations equal to each other.&lt;P&gt;
{{{190/(B-20) = 290/B}}}&lt;P&gt;
Cross multiply.&lt;P&gt;
{{{190B = 290(B-20) = 290B-5800}}}&lt;P&gt;
Subtract 290B from both sides.&lt;P&gt;
{{{-100B=-5800}}}&lt;P&gt;
Divide both sides by -100.&lt;P&gt;
{{{B = 58}}}&lt;P&gt;
Train B&#39;s rate = B = 58 mph.&lt;P&gt;
Train A&#39;s rate = B-20 = 58-20 = 38mph
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