Question

James invests a total of $26,500 in two accts. paying 10% and 2% annual interest, respectively. How much was invested in each account if after one year the total interest was $1970?

Answer 1

account 1 investment: 18,000

account 2 investment: 8,500

Step-by-step explanation:

Let us call A0 and B0 the principle amount in each account, then we know that

`A_0+B_0=26,500`

Furthermore, the simple interest earned on account A0, for example, is

`I=A_0(1+r_At)-A_0=A_0r_At`

where A0 ( 1+ rt) is the account balance after time t on simple interest. If we subtract the initial balance from the final, we would get the total interest earned. The expression above finds exactly just that ( the interest earned).

Now the interest earned on account B0 is

`I=B_0(1+rt)-B_0=B_0+B_0rt-B_0=B_0r_Bt`
`\Rightarrow I=B_0r_Bt`

Now we know that the total interest earned is $1970, therefore,

`B_0r_Bt+A_0r_At=1970`

putting in rA = 10% = 0.10 , rB = 2% = 0.02, and t = 1 gives us the system:

`\begin{gathered} A_0+B_0=26,500 \\ 0.10A_0+0.02B_0=1970 \end{gathered}`

Now, this is a system of equations with unknowns A0 and B0.

We multiply the first equation by 0.10 t0 get:

`\begin{gathered} 0.10A_0+0.10B_0=0.10\cdot26,500 \\ 0.10A_0+0.02B_0=1970 \end{gathered}`

subtracting the second equation from the first gives

`0.10A_0-0.10A_0+0.10B_0-0.02B_0=(0.10\cdot26,500)-1970`
`0.08B_0=(0.10\cdot26,500)-1970`

`0.08B_0=680`

finally, dividing both sides by 0.08 gives

`B_0=(680)/(0.08)`

`\boxed{B_0=8500}`

which is our answer!

Now that we have the value of B0, we now find the value of A0 from the following equation:

`A_0+B_0=26,500`

putting in b0 = 8500 gives

`A_0+8500=26,500`

finally, subtracting 8500 from both sides gives

`\boxed{A_0=18,000.}`

which is our answer!

Hence, the amount invested in the accounts was 8,500 and 18,000.