Question

) A bank customer invests a total of $7000in two savings accounts. One account yields 10% simple interest and the other 11% simple interest. The customer earned a total of $800interest for the year. How much was invested in the 11% account?

Answer 1

The amount of money invested in 11% account is $10000

Solution:

Given that, A bank customer invests a total of $7000in two savings accounts.

One account yields 10% simple interest and the other 11% simple interest.

The customer earned a total of $800interest for the year.

Now, let the amount invested in 11% account be $n, then amount in other account will be 7000 – n

The simple interest is given as:

`\text { Simple interest }=\frac{\text {amount} * \text {rate} * \text {time}}{100}`

Then, for 10% account:

`\begin{array}{l}{\text { S. } \mathrm{l}=((7000-n) * 10 * 1)/(100)} \\\\ {\text { S. } 1=((7000-n))/(10)}\end{array}`

And for 11% account:

`\begin{array}{l}{S .1=(n * 11 * 1)/(100)} \\\\ {S .1=(11 n)/(100)}\end{array}`

Now, we know that, total simple interest = S.I from 10% account + S.I from 11% account

`800=(7000-n)/(10)+(11 n)/(100)`

80000 = 10(7000 - n) + 11n

80000 = 70000 – 10n + 11n

11n – 10n = 80000 – 70000

n = 10000

Hence, the amount of money invested in 11% account is $10000