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

Answer:

account 1 investment: 18,000

account 2 investment: 8,500

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

$B_0=\frac{680}{0.08}$

$\boxed{B_0=8500}$

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.