Question

. Leah invested some money at 4.5% annual interest and $1000 more than four times that amount at 6%. Her total annual income from these two investments was $801. How much did she invest at each rate

Answer 1

Answer:

The amount invested are $2600 and $11400 respectively

Step-by-step explanation:

Let the first amount be x

Given:

(First Investment)

Principal (P1) = x

Rate (R1) = 4.5%

Time (T) = 1 year

(Second Investment)

Principal (P2) = 4x + 1000

Rate (R2) = 6%

Time (T) = 1 year

Income = $801

Calculating the income from the first investment;

$I_1 = \frac{P_1R_1T}{100}$

Substitute values for P1, R1 and T

$I_1 = \frac{x * 4.5 * 1}{100}$

$I_1 = \frac{4.5x}{100}$

Calculating the income from the second investment;

$I_2 = \frac{P_2R_2T}{100}$

Substitute values for P2, R2 and T

$I_2 = \frac{(4x + 1000) * 6 * 1}{100}$

$I_2 = \frac{6(4x + 1000)}{100}$

$I_1 + I_2 = Annual\ Income$

So:

$\frac{4.5x}{100} + \frac{6(4x + 1000)}{100} = 801$

Multiply through by 100

$100 * \frac{4.5x}{100} +100 * \frac{6(4x + 1000)}{100} = 801 * 100$

$4.5x +6(4x + 1000) = 801 * 100$

$4.5x +24x + 6000 = 80100$

Collect Like Terms

$4.5x +24x = 80100 - 6000$

$28.5x = 74100$

Divide through by 28.5

$x = \frac{74100}{28.5}$

$x = \ 2600$

Recall that; the second invest

Amount Invested = 4x + 1000

This gives

$Amount = 4 * \ 2600 + 1000$

$Amount = \ 11400$

Hence;

The amount invested are $2600 and $11400 respectively