Given:
Total amount invested = $3000
Rate of annul interest on investments are 5% and 8%.
Total interest earned for the year = $210
To find:
The amount invested at each rate.
Solution:
Let the amount invested at 5% be [tex]\$x[/tex] and the amount invested at 8% be [tex]\$(3000-x)[/tex].
Total interest earned for the year = $210
It means. 5% of [tex]x[/tex] + 8% of [tex](3000-x)[/tex] = 210
[tex]x\times \dfrac{5}{100}+(3000-x)\times \dfrac{8}{100}=210[/tex]
Multiply both sides by 100.
[tex]5x+(3000-x)8=21000[/tex]
[tex]5x+24000-8x=21000[/tex]
[tex]-3x=21000-24000[/tex]
[tex]-3x=-3000[/tex]
Divide both sides by -3.
[tex]x=1000[/tex]
Now,
[tex]3000-x=3000-1000[/tex]
[tex]3000-x=2000[/tex]
Therefore, the amount invested at the rate of 5% is $1000 and the amount invested at the rate of 8% is $2000.
