Question

Number 7. Not sure how to solve

Answer 1

let's say the amounts invested were "a" at 10% or 10/100 and "b" at 8.5% or 8.5/100

the total amount invested was 300,000, so, whatever "a" and "b" are, they add up to that much, thus

a + b = 300,000

10% of a is 10/100 * a or 0.10a and 8.5% of b is 8.5/100 * b or 0.085b

and we know their yield is 28,500 thus

0.10a + 0.085b = 28,500

thus
`\begin{array}{lcclll} &amount&\%&total\\ &-----&-----&-----\\ \textit{invested at 10\%}&a&0.10&0.10a\\ \textit{invested at 8.5\%}&b&0.085&0.085b\\ \end{array}\\\\ -----------------------------\\\\ \begin{cases} a+b=300,000\implies \boxed{b}=300,000-a\\\\ 0.10a+0.085b=28,500\\ ----------\\ 0.10a+0.085\left( \boxed{300,000-a} \right)=28,500 \end{cases}`

solve for "a", to see how much was invested at 10%

what about b? well, b = 300000 - a