<PRE><B>Question 1187293</B>
&lt;br&gt;
Let x be the original number of long rulers
Then 540-x is the original number of short rulers&lt;br&gt;
At the end, he had 1/3 of the original number of long rulers left, so he sold 2/3 of the original number.&lt;br&gt;
At the end, he had 1/6 of the original number of short rulers left, so he sold 5/6 of the original number.&lt;br&gt;
The numbers of long and short rulers he sold were the same:&lt;br&gt;
{{{(2/3)x = (5/6)(540-x)}}}&lt;br&gt;
Solve using basic algebra; I leave that to you.&lt;br&gt;
Perhaps start by multiplying both sides of the equation by 6 to clear fractions:&lt;br&gt;
{{{4x = 5(540-x)}}}
....&lt;br&gt;
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