Question

An investor decides to invest some cash in an account paying12%

annual interest, and to put the rest in a stock fund that ends up earning 8%over the course of a year. The investor puts $1500more in the first account than in the stock fund, and at the end of the year finds the total interest from the two investments was $1880. How much money was invested at each of the two rates? Round to the nearest integer.

Answer 1

Let A be the amount of money invested at a 12% annual interest rate and B be the amount of money invested at a 8% annual interest rate.

We know that A = B + 1500, and the total interest earned by the two investments is $1880.

The interest earned by the investment at a 12% annual interest rate is 0.12 * A = 0.12A

The interest earned by the investment at a 8% annual interest rate is 0.08 * B = 0.08B

Therefore, we can write the following equation to represent the situation:

0.12A + 0.08B = 1880

Since A = B + 1500, we can substitute this into the equation to get:

0.12(B + 1500) + 0.08B = 1880

Solving for B, we get:

0.12B + 180 + 0.08B = 1880

Combining like terms, we get:

0.2B + 180 = 1880

Subtracting 180 from both sides, we get:

0.2B = 1700

Dividing both sides by 0.2, we get:

B = 8500

Since A = B + 1500, we can substitute this value into the equation to find the value of A:

A = 8500 + 1500 = 10000

Therefore, the investor invested $8,500 at an 8% annual interest rate and $10,000 at a 12% annual interest rate.

Answer 2

Account A (12%)= $10,000

Account B (8% stock fund) = $8,500

Explanation:

Annual Interest Formula

`\large \text{$ \sf I=P\left(1+r\right)^(t) -P$}`

where:

• I = Interest.
• P = Principal amount.
• r = Interest rate (in decimal form).
• t = Time (in years).

Account A:

• P = x + 1500
• r = 12% = 0.12
• t = 1 year

`\implies \sf Interest=(x+1500) (1+0.12)^1-(x+1500)`

`\implies \sf Interest=(x+1500) (1+0.12)-(x+1500)`

`\implies \sf Interest=(x+1500)+0.12(x+1500)-(x+1500)`

`\implies \sf Interest=0.12(x+1500)`

`\implies \sf Interest=0.12x+180`

Account B (stock fund):

• P = x
• r = 8% = 0.08
• t = 1 year

`\implies \sf Interest=x (1+0.08)^1-x`

`\implies \sf Interest=x (1+0.08)-x`

`\implies \sf Interest=x+0.08x-x`

`\implies \sf Interest=0.08x`

If the total interest from the two investments was $1880:

`\implies \sf 0.12x+180+0.08x=1880`

`\implies \sf 0.2x+180=1880`

`\implies \sf 0.2x=1700`

`\implies \sf x=8500`

Therefore, the money invested in each of the two accounts is:

• Account A = $8,500 + $1,500 = $10,000
• Account B = $8,500