Problem 22 asks for the annual interest rate required to triple an investment in 29 years. Problem 23 states that three numbers are in the ratio 2:3:4 and the sum of their squares is 116. We need to find the numbers.

1. Problem Description

Problem 22 asks for the annual interest rate required to triple an investment in 29 years.

Problem 23 states that three numbers are in the ratio 2:3:4 and the sum of their squares is

1

1

6. We need to find the numbers.

2. Solution Steps

Problem 22:

The formula for compound interest is:

$A = P(1 + r)^n$

Where:

$A$ = final amount

$P$ = principal amount

$r$ = annual interest rate

$n$ = number of years

We want the investment to triple, so $A = 3P$. The number of years is $n = 29$. Substitute these values into the formula:

$3P = P(1 + r)^{29}$

Divide both sides by $P$:

$3 = (1 + r)^{29}$

Take the 29th root of both sides:

$3^{\frac{1}{29}} = 1 + r$

$r = 3^{\frac{1}{29}} - 1$

$r \approx 1.0387 - 1$

$r \approx 0.0387$

The interest rate is approximately $3.87\%$.

Problem 23:

Let the three numbers be $2x$, $3x$, and $4x$. The sum of their squares is 116:

$(2x)^2 + (3x)^2 + (4x)^2 = 116$

$4x^2 + 9x^2 + 16x^2 = 116$

$29x^2 = 116$

$x^2 = \frac{116}{29}$

$x^2 = 4$

$x = \pm 2$

If $x = 2$, the numbers are:

$2(2) = 4$

$3(2) = 6$

$4(2) = 8$

If $x = -2$, the numbers are:

$2(-2) = -4$

$3(-2) = -6$

$4(-2) = -8$

3. Final Answer

Problem 22: The annual interest rate is approximately 3.87%.

Problem 23: The numbers are 4, 6, and 8, or -4, -6, and -

8.