Answer :
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:
[tex]\text { Simple interest }=\frac{\text {amount} \times \text {rate} \times \text {time}}{100}[/tex]
Then, for 10% account:
[tex]\begin{array}{l}{\text { S. } \mathrm{l}=\frac{(7000-n) \times 10 \times 1}{100}} \\\\ {\text { S. } 1=\frac{(7000-n)}{10}}\end{array}[/tex]
And for 11% account:
[tex]\begin{array}{l}{S .1=\frac{n \times 11 \times 1}{100}} \\\\ {S .1=\frac{11 n}{100}}\end{array}[/tex]
Now, we know that, total simple interest = S.I from 10% account + S.I from 11% account
[tex]800=\frac{7000-n}{10}+\frac{11 n}{100}[/tex]
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