Respuesta :
from the question;
The annual interest on $2000 ivestment exceeds the interest earned on $1000 investment by $55 and the $2000 is invested at a 0.5% higher rate of interest than the $1000
let P = principal
I = interest
R = interest rate
first
[tex]\begin{gathered} \text{Let} \\ P_1=\text{ \$1000 investment} \\ I_1\text{ = x} \\ R_1\text{ = r\%} \end{gathered}[/tex]given the above information on $1000 investment above and the information above
[tex]\begin{gathered} \text{let } \\ P_2\text{ = \$2000 investment} \\ I_2\text{ = x + 55} \\ R_2\text{ = (r + 0.5)\%} \end{gathered}[/tex]applying the formula for Interest
[tex]\text{Interest I = }\frac{P\text{ }\times T\text{ }\times\text{ R}}{100}[/tex]for $1000 investment
[tex]\begin{gathered} u\sin g\text{ the interest formula and inserting the appropriate values} \\ we\text{ get} \\ x\text{ = }\frac{1000\text{ }\times\text{ 1 }\times\text{ r}}{100} \\ x\text{ = 10r -------- (1)} \end{gathered}[/tex]for $2000 investmet
[tex]\begin{gathered} u\sin g\text{ the interest formula and applying the appropriate values} \\ we\text{ get} \\ x\text{ + 55 = }\frac{2000\text{ }\times\text{ 1 }\times\text{ (r + 0.5)}}{100} \\ x\text{ + 55 = 20(r + 0.5)} \\ x\text{ + 55 = 20r + 10} \\ x\text{ - 20r = 10 - 55} \\ x\text{ - 20r = -45 ----------(2)} \end{gathered}[/tex]combine the two equations;
[tex]\begin{gathered} x\text{ = 10r -------------(1)} \\ x\text{ - 20r = - 45 ---------(2)} \end{gathered}[/tex]substitute x=10r into equation 2; we get
[tex]\begin{gathered} 10r\text{ - 20r= -45} \\ -10r\text{ = - 45} \\ \text{divide both sides by -10} \\ \frac{-10r}{-10}\text{ = }\frac{-45}{-10} \\ r\text{ = 4.5} \end{gathered}[/tex]Therefore,
the interest rate on $1000 investment = 4.5%
The interest rate on $2000 investment is (4.5 + 0.5)% = 5.0%
