$5400 is invested, part of it at 12% and part of it at 9%. For a certain year, the total yield is 576.00. How much was invested at each rate?

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Answer 1 · 2016-11-19T04:17:42

$$3000$ was invested at $12%$ and $$2400$ was invested at $9%$

Let the money invested at $12%$ be $$x$ . Interest on it for a year will be $12/100 xx x$ .

Then money invested at $9%$ would be $$(5400-x)$ and interest on it for a year will be $9/100 xx (5400-x)$ .

As total yield is $576$ , we have

$(12x)/100+(9(5400-x))/100=576$ - multiplying both sides by $100$

or $12x+9(5400-x)=57600$

or $12x+48600-9x=57600$

or $3x=57600-48600=9000$

i.e. $x=9000/3=3000$ and $5400-x=2400$

Hence, $$3000$ was invested at $12%$ and $$2400$ was invested at $9%$