A man invests some money at 5%, and $49,000 more than three times the amount at 11%. The total annual interest earned from the investment is $51,370. How much did he invest at each amount?

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Answer 1 · 2017-12-20T06:27:13

$$121,000$ was invested at $5%$

$$412,000$ was invested at $11%$

There is a given relationship between the two amounts, so you can define them using one variable.

Let the amount at $5%$ be $x$ .
Therefore the larger amount at $11%$ is $3x+49,000$

Now form an equation - you know the total amount of interest is $51,370$

'Interest at $5%$ + Interest at $11% = 51,370$

$5/100xx x + 11/100xx(3x+49000) = 51370" "larr xx 100$

$5x +11(3x+49,000) = 51,370$

$5x +33x+539,000 = 5,137,000$

$38x = 5,137,000-539,000$

$38x = 4,598,000$

$x=121,000$

$121xx3+49000 = 412,000$