Mrs.Krauser invested a part of her 6,000 $i n h e r i t a n c e a t$ 9 $p e r c e n t s i m p l e a n n u a l i n t e r e s t a n d t h e r e s t a t$ 12 $p e r c e n t s i m p l e a n n u a l i n t e r e s t . I f t h e t o t a l i n t e r e s t e a r n e d i n o n e y e a r w a s$ $660$ , how much did she invest at $12$ percent?

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Solution

Let the amount invested in $9$ percent interest be $a$ , then the amount invested in $12$ percent anual interest is $6000 - a$ .
Given that $660 = a (\frac{9}{100}) + (6000 - a) (\frac{12}{100}) = 720 - \frac{3 a}{100}$
We get $\frac{3 a}{100} = 720 - 660 = 60$ $\Rightarrow a = 2000$
Therefore $6000 - a = 6000 - 2000 = 4000$
She invested $4000$ at $12$ percent.

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