Alice requires $10,000 in two years. If the interest rate is 6% and compounded
quarterly, how much must Alice place in the account to have the required amount? (a) $ 6,274.10 (b) $ 8,877.10 (c)$ 8,573.40 (d) $ 5,000.00 (e) $ 10,000.00
Pt = Po(1+r)^n.
r = (6% / 4) / 100% = 0.015 = QPR expressed as a decimal.
n = 4comp/yr * 2yrs = 8 Compounding periods.
$10,000 = Po(1.015)^8,
Po = 10,000 / (1.015)^8 = $8877.11.
How much compound interest will an account of $12,000 bear after 6 years at 3.2% APR, compounded annually? (Hint: Use the formula for annual compound interest.)
$2,496.38
$14,496.38
$1,896.38
$496.38
Question 6
After a golf ball struck Charl on the head he was awarded an amount from the Three Iron Fund as compensation
for his injuries. He chose to receive R18 900 per month indefinitely. If money is worth 9,95% per year,
compounded monthly, then the amount awarded is approximately
[1] R189 950.
[2] R2 279 397.
[3] R6 565 554.
[4] R7 252 333.
[5] none of the above.
To find out how much Alice must place in the account to have $10,000 in two years with a 6% interest rate compounded quarterly, we can use the formula for compound interest:
A = P(1 + r/n)^(nt)
Where:
A is the future amount
P is the principal amount (initial investment)
r is the annual interest rate (written as a decimal)
n is the number of times that interest is compounded per year
t is the number of years
In this case, Alice wants to have $10,000, so A = $10,000, the interest rate is 6% or 0.06 as a decimal, it is compounded quarterly, so n = 4, and t = 2 years. We need to find the value of P.
Substituting the values into the formula, we have:
$10,000 = P(1 + 0.06/4)^(4*2)
Simplifying further:
$10,000 = P(1.015)^8
Dividing both sides of the equation by (1.015)^8:
P = $10,000 / (1.015)^8
Using a calculator to compute (1.015)^8, we find (1.015)^8 ≈ 1.126775.
P ≈ $10,000 / 1.126775
P ≈ $8,877.104
Hence, Alice must place approximately $8,877.10 in the account to have the required amount of $10,000 in two years.
Therefore, the correct option is (b) $8,877.10.