Asked by Austin

A Retailer planned to buy some computers from a total of sh 1800000. before the retailer could buy the computers the price per unit was reduced by sh 4000. this reduction in price enabled the retailer to buy 5 more computers using the same amount of money as originally planned.
determine the number of computers the retailer bought.
Answered by Reiny
let n be the number of computers he could buy initially.
cost price of computer initially = 1800000/n
new cost price = 1800000/n - 4000

(n+5)(1800000/n - 4000) = 1800000
1800000 - 4000n + 9000000/n - 20000 = 1800000
9000000/n - 4000n - 20000 = 0
9000/n- 4n - 20 = 0
9000 - 4n^2 - 20n = 0
carry on



Answered by Steve
did you check the response you received this morning?
Answered by Léticia
4000(x+5)=1800000
4000x+20000=1800000
4000x=1800000-20000
X=445
He was able to buy 450 computer
Answered by Léticia
4000(x+5)=1800000
4000x+20000=1800000
4000x=1800000-20000
X=445
He was able to buy 450 computers