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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 :

sihanamritsingh

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.

semsee45

Answer:

Account A (12%)= $10,000

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

Step-by-step explanation:

Annual Interest Formula

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

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

[tex]\implies \sf Interest=(x+1500) (1+0.12)^1-(x+1500)[/tex]

[tex]\implies \sf Interest=(x+1500) (1+0.12)-(x+1500)[/tex]

[tex]\implies \sf Interest=(x+1500)+0.12(x+1500)-(x+1500)[/tex]

[tex]\implies \sf Interest=0.12(x+1500)[/tex]

[tex]\implies \sf Interest=0.12x+180[/tex]

Account B (stock fund):

  • P = x
  • r = 8% = 0.08
  • t  = 1 year

[tex]\implies \sf Interest=x (1+0.08)^1-x[/tex]

[tex]\implies \sf Interest=x (1+0.08)-x[/tex]

[tex]\implies \sf Interest=x+0.08x-x[/tex]

[tex]\implies \sf Interest=0.08x[/tex]

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

[tex]\implies \sf 0.12x+180+0.08x=1880[/tex]

[tex]\implies \sf 0.2x+180=1880[/tex]

[tex]\implies \sf 0.2x=1700[/tex]

[tex]\implies \sf x=8500[/tex]

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

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

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