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Question

# An inheritance of $24,000 is to be divided among three trusts with the second trust receiving twice as much as the first trust. The three trusts pay interest at the rates of 9%, 10%, and 6% annually, respectively, and return a total in interest of$2210 at the end of the first year. How much was invested in each trust?

Solution

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Let $x$ dollars be the amount invested in the first trust. Twice as much is invested in the second trust so $2x$ dollars is invested in the second trust. A total of $\24,000$ is invested in the three trusts so $24,000-x-2x=24,000-3x$ dollars is invested in the third trust. Since the trusts earn 9%, 10%, and 6% annual interest, respectively, then the amount of interest earned on the three trusts is $0.09x+0.1(2x)+0.06(24,000-3x)$. It is given that the amount of interest earned is $\2210$ so $0.09x+0.1(2x)+0.06(24,000-3x)=2210$. Solve this equation for $x$:

\begin{align*} 0.09x+0.1(2x)+0.06(24,000-3x)&=2210\\ 0.09x+0.2x+1440-0.18x&=2210&&\text{Simplify and distribute.}\\ 0.11x+1440&=2210&&\text{Combine like terms.}\\ 0.11x&=770&&\text{Subtract 1440 on both sides.}\\ x&=7000 \end{align*}

Therefore $x=\7,000$ is invested in the first trust, $2x=2(\7,000)=\14,000$ is invested in the second trust, and $\24,000-3(\7000)=\24,000-\21,000=\3,000$ was invested in the third trust.

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