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.
did you check the response you received this morning?
4000(x+5)=1800000
4000x+20000=1800000
4000x=1800000-20000
X=445
He was able to buy 450 computer
4000(x+5)=1800000
4000x+20000=1800000
4000x=1800000-20000
X=445
He was able to buy 450 computers
To determine the number of computers the retailer bought, we need to set up an equation.
Let's suppose the original price per unit of the computers was P, and the original number of computers the retailer planned to buy was N.
The total amount of money the retailer planned to spend on the computers was 1800000.
Therefore, the original equation is:
Original Total Cost = Original Price per Unit × Original Number of Computers
1800000 = P × N
After the reduction in price, the new price per unit became P - 4000.
The retailer was able to buy 5 more computers using the same amount of money as originally planned. This means the new number of computers bought is N + 5.
The new equation is:
New Total Cost = New Price per Unit × New Number of Computers
1800000 = (P - 4000) × (N + 5)
Now we have a system of two equations:
1800000 = P × N (Equation 1)
1800000 = (P - 4000) × (N + 5) (Equation 2)
Let's solve the system of equations to find the values of P and N.
From Equation 1, we can rearrange and isolate P:
P = 1800000 / N
Substitute this value of P into Equation 2:
1800000 = (1800000 / N - 4000) × (N + 5)
Multiply through by N to eliminate the denominator:
1800000N = (1800000 - 4000N) × (N + 5)
Expand and simplify:
1800000N = 1800000N + 9000000 - 4000N^2 - 20000N
Rearrange the equation to form a quadratic equation:
4000N^2 + 20000N + 9000000 = 0
Solve this quadratic equation using the quadratic formula or factoring methods.
Once you find the value of N, you can substitute it back into Equation 1 to find the value of P. The number of computers the retailer bought will be N + 5.
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