Mike decides he wants to replace his movie collection with digital versions.  He has 600 movies.  A third of the movies are in various series and he knows he can get those for only $6 of the cost of a normal movie by just buying the series together.  40% of the remaining movies are older movies which are $5.  How much does replacing the movies cost if a normal movie costs $10?
There are 600/3=
<gadget id="calculator">600/3</gadget>
<output>200</output>
200 movies in series.
So he pays 200*6=$
<gadget id="calculator">200*6</gadget>
<output>1_200</output>
1200.
He has 600-200=
<gadget id="calculator">600-200</gadget>
<output>400</output>
400 movies not in series
400*.4=
<gadget id="calculator">400*.4</gadget>
<output>160</output>
160 movies are old.
So those cost 160*5=$
<gadget id="calculator">160*5</gadget>
<output>800</output>
800
400-160=
<gadget id="calculator">400-160</gadget>
<output>240</output>
240 movies are neither old nor in series.
They cost 240*10=$
<gadget id="calculator">240*10</gadget>
<output>2_400</output>
2400.
So the total cost is 1200+800+2400=$
<gadget id="calculator">1200+800+2400</gadget>
<output>4_400</output>
4400.

<result>4_400</result>