Respuesta :
$ 1850 was invested in 9% account
Solution:
Given that
Total amount invested by man in two saving accounts = $9493
Simple interest on one account =9%
Simple interest on second account = 10%
Total interest earned = $930.80
Need to determine amount invested in 9 % account.
Let assume amount invested in account where Simple Interest is 9% = x
And assume amount invested in account where Simple Interest is 10% = y
As total amount invested in two accounts is $9493
=> x + y = 9493
=> y = 9493 - x ------(1)
[tex]\text { Simple Interest }=\frac{\text { Amount Invested } \times \text {rate of interest } \times \text {time}}{100}[/tex]
[tex]\begin{array}{l}{\text { Simple interest when rate of interest is } 9 \%=\frac{x \times 9 \times 1}{100}=\frac{9 x}{100}} \\\\ {\text { Simple interest when rate of interest is } 10 \%=\frac{y \times 10 \times 1}{100}=\frac{10 y}{100}}\end{array}[/tex]
As total interest earned = $930.80
[tex]\begin{array}{l}{\Rightarrow \frac{9 x}{100}+\frac{10 y}{100}=930.80} \\\\ {\Rightarrow 9 x+10 y=930.80 \times 100} \\\\ {\Rightarrow 9 x+10 y=93080}\end{array}[/tex]
On substituting value of y from equation(1) in above equation , we get
9x + 10 (9493 – x) = 93080
=> 9x + 94930 – 10x = 93080
=> -x = 93080 – 94930
=> -x = -1850
=> x = 1850
Amount invested in account where Simple Interest is 9% = x = $1850
Hence $1850 was invested in 9% account.
