Interest Problem

Question

Danny invests $4400 in two different accounts. The first account paid 4 %, the second account paid 3 % in interest. At the end of the first year he had earned $158 in interest. How much was in each account?

Answer 1

If we invest $x at 4% then we invest $(4400-x) at 3%

Also, I will assume the interest is compounded annually.

 

A = p(1+r/n)^nt

A = future amount = p +I = 4400+158 = 4558

p = principal investment = $x and $(4400-x)

r = 4% and 3% = .04 and .03

n = number of times compounded per year = 1

t = time in years = 1

 

4558 = x(1+.04/1)¹⁽¹⁾ + (4400-x)(1+.03)¹⁽¹⁾

4558 = 1.04x + (4400-x)(1.03)

4558 = 1.04x + 4532 - 1.03x

4558 - 4532 = 0.01x

 

x = 26/.01 = $2600 invested at 4%

4400-x = $1800 invested at 3%

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I corrected a simple math error above since 4558-4532 = 26, not 36.

Let's do this again the same way I did your other post since

this is still 1 year of interest compounded annually.

158 = 0.04x + 0.03(4400-x)

158 = 0.04x + 132 - 0.03x

26 = 0.01x

x = $2600 invested at 4%

4400-x = $1800 invested at 3%