Question 1138859
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<pre>
Let n be the number of computers the retailer bought (the value under the question).


The price for each single computer then was  {{{1800000/n}}}.


The planned price (before reducing) was  {{{1800000/(n-5)}}}.


The difference in prices was 4000:


    {{{1800000/(n-5)}}} - {{{1800000/n}}} = 4000.


To solve this equation and to find "n", first cancel the factor "1000" in both sides; then multiply both sides by n*(n-5).


    {{{1800/(n-5)}}} - {{{1800/n}}} = 4

     1800*n - 1800*(n-5) = 4n*(x-5)

     1800x + 1800n + 9000 = 4n^2 - 20n

      4n^2 - 20n + 9000 = 0

      n^2  -  5n + 2250 = 0

      (n+45)*(n-50) = 0


The two roots are n= -45  and  n= 50,  but only positive value x= 50 is a meaningful solution.


<U>ANSWER</U>.  The retailer bought 50 computers.
</pre>

Solved.


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