Question

The annual interest on a $2000 investment exceeds the interest earned on a $1000 investment by $55.The $2000 is invested at a 0.5% higher rate of interest thanthe $1000. What is the interest rate of each investment?

Answer 1

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

`\begin{gathered} \text{Let} \\ P_1=\text{ \$1000 investment} \\ I_1\text{ = x} \\ R_1\text{ = r\%} \end{gathered}`

given the above information on $1000 investment above and the information above

`\begin{gathered} \text{let } \\ P_2\text{ = \$2000 investment} \\ I_2\text{ = x + 55} \\ R_2\text{ = (r + 0.5)\%} \end{gathered}`

applying the formula for Interest

`\text{Interest I = }\frac{P\text{ }* T\text{ }*\text{ R}}{100}`

for $1000 investment

`\begin{gathered} u\sin g\text{ the interest formula and inserting the appropriate values} \\ we\text{ get} \\ x\text{ = }\frac{1000\text{ }*\text{ 1 }*\text{ r}}{100} \\ x\text{ = 10r -------- (1)} \end{gathered}`

for $2000 investmet

`\begin{gathered} u\sin g\text{ the interest formula and applying the appropriate values} \\ we\text{ get} \\ x\text{ + 55 = }\frac{2000\text{ }*\text{ 1 }*\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}`

combine the two equations;

`\begin{gathered} x\text{ = 10r -------------(1)} \\ x\text{ - 20r = - 45 ---------(2)} \end{gathered}`

substitute x=10r into equation 2; we get

`\begin{gathered} 10r\text{ - 20r= -45} \\ -10r\text{ = - 45} \\ \text{divide both sides by -10} \\ (-10r)/(-10)\text{ = }(-45)/(-10) \\ r\text{ = 4.5} \end{gathered}`

Therefore,

the interest rate on $1000 investment = 4.5%

The interest rate on $2000 investment is (4.5 + 0.5)% = 5.0%