One year, Shannon made $288 from two investments. $1100 was invested at one yearly rate and 1800 at a rate that was 1.5% higher. Find the two rates of interest
lower rate ---- r
higher rate ---- r+.015
1100r + 1800(r+.015) = 288
1100r + 1800r + 27 = 288
2900r = 261
r = .09
the one rate was 9% , the other 10.5%
check:
1100(.09) + 1800(.105) = 288
To find the two rates of interest, let's assign variables to the rates. Let's say the rate of the first investment is 'x' and the rate of the second investment is 'x + 1.5%'. Now, let's use the given information to set up the equation.
The total amount earned from the two investments is $288:
$1100 * x + $1800 * (x + 1.5%) = $288
First, let's convert 1.5% to decimal form by dividing it by 100:
1.5% = 1.5/100 = 0.015
Now, we can rewrite the equation:
$1100 * x + $1800 * (x + 0.015) = $288
Next, let's simplify the equation:
1100x + 1800(x + 0.015) = 288
Distribute 1800 to both terms in the parentheses:
1100x + 1800x + 27 = 288
Combine like terms:
2900x + 27 = 288
Move the constant term to the other side of the equation:
2900x = 288 - 27
2900x = 261
Divide both sides of the equation by 2900:
x = 261/2900
x ≈ 0.09
So, the rate of the first investment is approximately 0.09 or 9%.
To find the rate of the second investment, we can substitute the value of x back into our equation:
x + 0.015 ≈ 0.09 + 0.015
x + 0.015 ≈ 0.105
So, the rate of the second investment is approximately 0.105 or 10.5%.
Therefore, the two rates of interest are approximately 9% and 10.5%.