Compound Interest Practice Test Set 5

Treat each annual installment as a payment made at the end of its year, and express the loan value at the appropriate time.

Apply the formula and calculate:
End of Year 1: Amount = 10,000 * 1.1 = 11,000. Balance after payment = 11,000 - 4,000 = 7,000. End of Year 2: Amount due = 7,000 * 1.1 = Rs 7,700.

Write down the given values and identify the compound-interest relationship needed to find the required quantity.

Apply the formula and calculate:
Quarterly rate = 3%. Time = 3 quarters. Amount = 50,000 * (1.03)^3 = 50,000 * 1.092727 = 54,636.35. CI = Rs 4,636.35.

Identify the given principal, rate and time, and determine whether the question asks for SI, CI, their difference, or an unknown value.

Apply the formula and calculate:
Diff = P * (R/100)^2 => 40 = 16,000 * (R/100)^2 => (R/100)^2 = 40 / 16,000 = 1 / 400 => R/100 = 1/20 => R = 5%.

Use the principal and amount (or the given interest information) to determine the annual growth factor and hence the rate.

Apply the formula and calculate:
$A/P = 24,200 / 20,000 = 1.21 = (1.1)^2$. Rate R = 10%.

Compare the required amount with the principal using the compound-growth factor to determine the number of compounding periods.

Apply the formula and calculate:
$A/P = 3,645 / 3,125 = 729 / 625 = (27/25)^2$. Since rate is 8% (27/25 per year), time = 2 years.