Practice Questions
Find the simple interest on ₹8000 at 12% for 3 years.
$SI = (8000 \times 12 \times 3) / 100 = 80 \times 36 = 2880$.
A sum amounts to ₹7200 in 4 years at 10% SI. Find principal.
$7200 = P(1 + 0.4) \Rightarrow P = 7200 / 1.4 = 5142.86$.
In how many years will ₹5000 amount to ₹6500 at 8% SI?
$SI = 6500 - 5000 = 1500$. $1500 = (5000 \times 8 \times T) / 100 \Rightarrow 1500 = 400T \Rightarrow T = 15/4 = 3.75$ years.
Find the rate if ₹4000 amounts to ₹5200 in 5 years at SI.
$SI = 1200$. $1200 = (4000 \times R \times 5) / 100 \Rightarrow 1200 = 200R \Rightarrow R = 6\%$.
A sum doubles in 8 years at SI. Find the rate of interest.
$R = (2-1) \times 100 / 8 = 100 / 8 = 12.5\%$.
Find difference between SI and CI for 2 years on ₹10000 at 10%.
$SI = (10000 \times 10 \times 2) / 100 = 2000$. $CI = 10000(1.1^2 - 1) = 10000(0.21) = 2100$. Difference = 100. (Short formula: $P(R/100)^2 = 10000 \times (0.1)^2 = 10000 \times 0.01 = 100$).
A sum of money triples in 10 years at SI. Find the rate of interest.
$A = 3P \Rightarrow SI = 2P$. $2P = (P × R × 10) / 100 \Rightarrow R = 20\%$.
What sum will amount to ₹7440 in 3 years at 8% SI?
$7440 = P(1 + 0.08 × 3) = P(1.24) \Rightarrow P = 7440 / 1.24 = 6000$.
A person invests ₹5000 partly at 6% and partly at 8% SI. Total annual interest is ₹350. Find the amount invested at 8%.
Let x be at 8%. $(5000-x)×0.06 + x×0.08 = 350 \Rightarrow 300 - 0.06x + 0.08x = 350 \Rightarrow 0.02x = 50 \Rightarrow x = 2500$.
The SI on a sum for 3 years at 10% is ₹1500. Find the CI for the same sum at same rate and time.
$SI = 1500 \Rightarrow P = 1500 × 100 / (10 × 3) = 5000$. $CI = 5000[(1.1)^3 - 1] = 5000(1.331 - 1) = 5000 × 0.331 = 1655$.
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