# Dave Horn invested half of his money at 5%,one third of his money at 4%, and the rest of his money at 3.5%.If his total annual investment income was $530,how much had he invested?

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It is first necessary to make an equation out of the values.

Let x= the amount invested

`therefore 1/2 x` is earned at 5% (`5/100` )

`1/3 x ` is earned at 4% (4/100)

and the rest which is represented by the total minus what we have so far

`x-1/2 x - 1/3 x` is earned at 3.5% (`35/ 1000)` .

Simplify this part of the equation:

`x- 1/2x - 1/3x `

`= x-3/6 x - 2/6 x `

`=1/6 x`

Now you are ready to form the equation because we know that the total = 530 and we can split the interest up as follows:

`therefore 1/2 x times 5/100` + `1/3 x times 4/100` + `1/6 x times 35/1000` = 530

`5/200 x + 4/300 x + 35/6000 x` = 530

` (150x + 80x +35x)/ 6000` = 530

` 265/6000 x = 530`

`therefore x = 530 divide 265/6000`

`x=$12000`

There will be a slight discrepancy due to using fractions / or decimals and an amount of $12004 will also result.

However on checking your answer :

$6000 @ 5% = 300

$4000 @ 4%= 160

$2000 @ 3,5%= 70

**therefore the original investment was $12 000**

Let Dave invested $ x.

Acording to problem , he invested x/2 money in 5%

x/3 money in 4% and (x-x/2-x/3) in 3.5% i.e x/6 in 3.5%.

His return /incomes are 5x/200 ,4x/300 and 3.5x/600.

so Total income will be `5x/200+4x/300+3.5x/600=26.5x/600`

But we have `26.5x/600=530`

`x=530 . 600/26.5= 12000 `

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