SOLUTION: A retailer planned to buy some computers form a wholesaler for a total of ksh 1,800,000, before the retailer could buy the computers the price per unit was reduced by ksh.4000. thi

Let n be the number of computers the retailer bought (the value under the question).


The price for each single computer then was  .


The planned price (before reducing) was  .


The difference in prices was 4000:


     -  = 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).


     -  = 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.


ANSWER.  The retailer bought 50 computers.