Question

Michelle has money in two savings accounts. One rate is 6% and the other is 14 %. If she has $900 more in the 14 % account and the total interest is $209. how muchis invested in each savings account?

Answer 1

The Solution:

Let the amount in the savings account with 6% interest rate be represented with $x

So, the interest in this account is given as below:

$\begin{gathered} I_1=\frac{\text{PRT}}{100}=\frac{x\times6\times1}{100}=\frac{6x}{100} \\ \text{where} \\ I_1=\text{interest}=\text{?,P}=\text{amount saved=x, T=time=1 (assumed)} \end{gathered}$

Similarly, the amount in the savings account with 14% interest rate will be $(x+900)

So, the interest in this account is given as below:

$\begin{gathered} I_2=\frac{\text{PRT}}{100}=\frac{(x+900)\times14\times1}{100}=\frac{14x+12600}{100} \\ \text{where} \\ I_2=\text{interest}=\text{?,P}=\text{amount saved=x+900, T=time=1 (assumed)} \end{gathered}$

So, the total interest from the two savings accounts is $209. That is,

$I_1+I_2=\frac{6x}{100}+\frac{14x+12600}{100}=209$

This becomes

$\begin{gathered} \frac{6x+14x+12600}{100}=209 \\ \\ \frac{20x+12600}{100}=209 \end{gathered}$

Cross multiplying, we get

$\begin{gathered} 20x+12600=100\times209 \\ 20x+12600=20900 \end{gathered}$

Collecting the like terms, we get

$\begin{gathered} 20x=20900-12600 \\ 20x=8300 \end{gathered}$

Dividing both sides by 20, we get

$\begin{gathered} \frac{20x}{20}=\frac{8300}{20} \\ \\ x=\text{ \ 415} \end{gathered}$

So, the amount invested in the savings account that yields 6% is $415.00, while

the amount invested in the savings account that yields 14% is $1315.00 ( that is, 415+900)