Practice Questions
Find compound interest on ₹6000 at 10% for 2 years.
$A = 6000(1.1)^2 = 6000 \times 1.21 = 7260$. $CI = 7260 - 6000 = 1260$.
At what rate will ₹5000 amount to ₹6050 in 2 years compounded annually?
$6050/5000 = (1 + R/100)^2 \Rightarrow 1.21 = (1 + R/100)^2 \Rightarrow 1.1 = 1 + R/100 \Rightarrow R = 10\%$.
Find amount on ₹10000 at 12% for 1 year compounded half-yearly.
$R = 12/2 = 6\%, T = 2$. $10000(1.06)^2 = 10000 \times 1.1236 = 11236$.
Find difference between CI and SI for 3 years at 8% on ₹8000.
Diff = $P(R/100)^2 \times (300+R)/100 = 8000(0.0064) \times 3.08 = 51.2 \times 3.08 = 157.696 \approx 157.44$ (Check: $SI = 1920$. $CI = 8000(1.08^3 - 1) = 8000(0.259712) = 2077.696$. Diff = 157.696).
A sum becomes ₹9261 in 3 years at CI if rate is 5% per annum. Find principal.
$9261 = P(1.05)^3 \Rightarrow 9261 = P(1.157625) \Rightarrow P = 8000$.
If ₹4000 becomes ₹5324 in 3 years compounded annually, find rate.
$5324/4000 = (1+R/100)^3 \Rightarrow 1.331 = (1+R/100)^3 \Rightarrow 1.1 = 1+R/100 \Rightarrow R=10\%$.
Find principal if CI for 2 years at 10% per annum is ₹1260.
$1260 = P(1.1^2 - 1) = P(1.21 - 1) = P × 0.21 \Rightarrow P = 1260 / 0.21 = 6000$.
A sum of ₹16000 invested at 10% CI for 3 years. Find the amount.
$A = 16000(1.1)^3 = 16000 × 1.331 = 21296$.
Find the compound interest on ₹20000 at 15% for 2 years.
$A = 20000(1.15)^2 = 20000 × 1.3225 = 26450$. $CI = 26450 - 20000 = 6450$.
The CI on a sum for 2 years is ₹820 and SI is ₹800 for the same period. Find rate.
Difference = $820 - 800 = 20$. $20 = P(R/100)^2$. Also $SI: 800 = P × R × 2 / 100 \Rightarrow PR = 40000$. Substituting: $20 = 40000 × R / 10000 \Rightarrow 20 = 4R \Rightarrow R = 5\%$.
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