Practice Questions
Find the difference between CI and SI for 2 years on ₹6000 at 10%.
Diff = $6000 \times (10/100)^2 = 6000 \times 0.01 = 60$.
A sum doubles in 5 years at SI. Find the rate %.
$R = (2-1) \times 100 / 5 = 20\%$.
In how many years will a sum become four times at 10% CI?
$(1.1)^n = 4$. Using $\log$: $n \log 1.1 = \log 4 \Rightarrow n \times 0.0414 = 0.6020 \Rightarrow n \approx 14.5$ years.
Compare SI and CI for 3 years on ₹10000 at 12%.
Diff = $10000 \times (0.12)^2 \times (3.12) = 10000 \times 0.0144 \times 3.12 = 144 \times 3.12 = 449.28$. CI is higher.
Find principal if amount is ₹14520 after 2 years at 10% CI.
$14520 = P(1.1)^2 \Rightarrow 14520 = 1.21P \Rightarrow P = 14520 / 1.21 = 12000$.
A sum amounts to ₹9680 in 2 years and ₹10648 in 3 years at CI. Find rate %.
Interest for middle year = $10648 - 9680 = 968$. Rate = $(968 / 9680) \times 100 = 10\%$.
Find the compound interest on ₹5000 at 8% for 2 years.
$A = 5000(1.08)^2 = 5000\times1.1664 = 5832$. CI $= 5832 - 5000 = ₹832$.
At what rate of simple interest will a sum triple in 10 years?
$P\times R\times10/100 = 2P$ (interest = 2 times principal). $R = 200/10 = 20\%$.
The difference between CI and SI for 2 years on a sum at 15% is ₹45. Find the sum.
Diff $= P(r/100)^2 \Rightarrow 45 = P(0.15)^2 = P\times0.0225 \Rightarrow P = 45/0.0225 = ₹2000$.
A sum of ₹24000 becomes ₹27783 in 2 years at compound interest. Find the rate.
$24000(1+r)^2 = 27783 \Rightarrow (1+r)^2 = 1.157625 \Rightarrow 1+r = \sqrt{1.157625} \approx 1.076 \Rightarrow r \approx 7.6\%$. Closest given option is $7.5\%$.
Premium Content
Unlock SI & CI Quiz and all premium lessons with a subscription.
From ₹199.99/year — See plans