[tex] \large \bold{PROBLEM:} [/tex]
A total of [tex]\$ 50,\!000[/tex] is invested in three funds paying [tex]6\%, 8\%,[/tex] and [tex]10\%[/tex] simple interests. The yearly is [tex]\$ 3,\!700.[/tex] Twice as much money is invested at [tex]6\%[/tex] as invested at [tex]10\%.[/tex]How much is invested in each of the funds?
[tex] \large \bold{SOLUTION:} [/tex]
Let [tex]x, y,[/tex] and [tex]z[/tex] be the amount invested in funds with simple interests [tex]6\%, 8\%,[/tex] and [tex]10\%,[/tex] respectively.
- [tex] 6\% = 0.06 [/tex]
- [tex] 8\% = 0.08 [/tex]
- [tex] 10\% = 0.10 [/tex]
A total of [tex]\$ 50,\!000[/tex] is invested in three funds.
[tex](1) \quad x + y + z = 50,\!000 [/tex]
The yearly is [tex]\$ 3,\!700[/tex].
[tex](2) \quad 0.06x + 0.08y + 0.10z = 3,\!700 [/tex]
Twice as much money is invested at [tex]6\%[/tex] as invested at [tex]10\%[/tex].
[tex](3) \quad x = 2z [/tex]
By elimination, multiply [tex](1)[/tex] by [tex]0.08[/tex] then subtract the resulting equivalent equation from [tex](2)[/tex] to eliminate [tex]y[/tex].
[tex] \begin{array}{l} 0.08(x + y + z) = 0.08(50,\!000) \\ \longrightarrow 0.08x + 0.08y + 0.08z = 4,\!000 \end{array} [/tex]
[tex] \begin{aligned} 0.06x + 0.08y + 0.10z &= 3,\!700 \\ -\quad 0.08x + 0.08y + 0.08z &= 4,\!000 \\ \hline -0.02x \: \quad \quad \quad + 0.02z &= -300 \end{aligned} [/tex]
Substituting [tex](3)[/tex] to the resulting equation,
[tex] \begin{aligned} -0.02(2z) + 0.02z &= -300 \\ -0.04z + 0.02z &= -300 \\ -0.02z &= -300 \\ \frac{\cancel{-0.02}z}{\cancel{-0.02}} &= \frac{-300}{-0.02} \\ z &= \boxed{\$ 15,\!000} \\ \\ x &= 2z \\ x &= 2(15,\!000) \\ x &= \boxed{\$ 30,\!000} \\ \\ y &= 50,\!000 - (x + z) \\ y &= 50,\!000 - 45,\!000 \\ y &= \boxed{\$ 5,\!000} \end{aligned} [/tex]
Therefore, the amounts invested in three funds paying [tex]6\%, 8\%,[/tex] and [tex]10\%[/tex] simple interests are [tex]\$ 30,\!000,[/tex] [tex]\$ 5,\!000,[/tex] and [tex]\$ 15,\!000,[/tex] respectively.
[tex] \blue{\mathfrak{\#CarryOnLearning}} [/tex]
