Given :
Roberto invested a total of $10,000, part at 6% and part at 9%.
To Find :
How much did he invest in the higher interest account if the total interest earned in one year was $810 .
Solution :
Let , x amount is invested in 6% interest .
Therefore , 10000-x amount is invested in 9% interest .
Now , total interest is $810 .
We know , simple interest is given by :
[tex]I=\dfrac{PRt}{100}[/tex]
So ,
[tex]\dfrac{x(1)(6)}{100}+\dfrac{(10000-x)(1)(9)}{100}=810\\[/tex]
Solving above equation , we get :
[tex]x=\$3000[/tex] .
Therefore , investment on higher i.e 9 % is 10000-3000 = $7000 .
Hence , this is the required solution .
