Vardaan Learning Institute

Principal $P = \text{₹}10,800$, $R = 12\frac{1}{2}\% = 12.5\%$, $n = 3$.
Amount $A = P(1 + \frac{R}{100})^n$
$= 10800(1 + \frac{12.5}{100})^3$
$= 10800(1 + \frac{1}{8})^3$ $= 10800(\frac{9}{8})^3$
$= 10800 \times \frac{729}{512}$ $= \text{₹}15,377.34$ (approx)
CI $= A - P$ $= 15377.34 - 10800$ $= \text{₹}4,577.34$

$P = \text{₹}4,096$. Time = 18 months = 3 half-years ($n=3$).
Rate = $12\frac{1}{2}\%$ p.a. = $6.25\%$ per half-year = $\frac{1}{16}$.
Amount $A = P(1 + \frac{R}{100})^n$ $= 4096(1 + \frac{1}{16})^3$
$= 4096(\frac{17}{16})^3$ $= 4096 \times \frac{4913}{4096}$ $= \text{₹}4,913$.

(i) Population in 2001 ($P_{2001}$):
$P_{2003} = P_{2001}(1 + \frac{R}{100})^n$ $\Rightarrow$ $54000 = P_{2001}(1 + \frac{5}{100})^2$
$54000 = P_{2001}(\frac{21}{20})^2$ $\Rightarrow$ $P_{2001} = \frac{54000 \times 400}{441}$ $\approx 48,980$.

(ii) Population in 2005:
$A = 54000(1 + \frac{5}{100})^2$ $= 54000(\frac{441}{400})$ $= 135 \times 441$ $= 59,535$.

$P = \text{₹}42,000$, Rate = $8\%$ depreciation.
Value after 1 year $= P(1 - \frac{R}{100})$
$= 42000(1 - \frac{8}{100})$ $= 42000(\frac{92}{100})$
$= 420 \times 92$ $= \text{₹}38,640$.

$P = 10000, R = 8\%, T = 1$ year.
Compounded Half-yearly: $R=4\%, n=2$.
$A = 10000(1.04)^2$ $= 10816$ $\Rightarrow$ $CI = \text{₹}816$.
Compounded Annually: $R=8\%, n=1$.
$A = 10000(1.08)^1$ $= 10800$ $\Rightarrow$ $CI = \text{₹}800$.
Difference $= 816 - 800$ $= \text{₹}16$.

Difference for 2 years $= P(\frac{R}{100})^2$.
Given Diff = ₹500, R = 10%.
$500 = P(\frac{10}{100})^2$ $\Rightarrow$ $500 = P(\frac{1}{10})^2$
$500 = P(\frac{1}{100})$ $\Rightarrow$ $P = 500 \times 100$ $= \text{₹}50,000$.

Amount in 1 year (2 half-years) = 13,230.
Amount in 1.5 years (3 half-years) = 13,891.50.
Interest for 1 half-year on 13,230 is $13891.50 - 13230 = \text{₹}661.50$.
Rate per half-year $= \frac{661.50}{13230} \times 100$ $= 5\%$.
Rate per annum = 10%.
For Sum: $13230 = P(1 + \frac{5}{100})^2$ $= P(\frac{21}{20})^2$
$P = 13230 \times \frac{400}{441}$ $= 30 \times 400$ $= \text{₹}12,000$.

$A = P(1 + \frac{R}{100})^n$ $\Rightarrow$ $6615 = 6000(1 + \frac{R}{100})^2$
$\frac{6615}{6000} = (1 + \frac{R}{100})^2$ $\Rightarrow$ $1.1025 = (1 + \frac{R}{100})^2$
Taking square root: $1.05 = 1 + \frac{R}{100}$ $\Rightarrow$ $0.05 = \frac{R}{100}$
$R = 5\%$.

$P = 50000$. Rates: 5%, 4%, 3%.
$A = 50000(1 + \frac{5}{100})(1 + \frac{4}{100})(1 + \frac{3}{100})$
$= 50000(\frac{105}{100})(\frac{104}{100})(\frac{103}{100})$
$= 5 \times 105 \times 1.04 \times 103 \div 10$
$= 50000 \times 1.05 \times 1.04 \times 1.03$ $\approx 56,238$.

$P = 5,06,000$, Rate = $2.5\%$ per hour, Time = 2 hours.
$A = 506000(1 + \frac{2.5}{100})^2$ $= 506000(1.025)^2$
$= 506000 \times 1.050625$ $\approx 5,31,616.25$.
Rounding to nearest whole number.

Let Sum be $P$. It becomes $2P$ in 5 years.
$2P = P(1+R)^5$ $\Rightarrow$ $(1+R)^5 = 2$.
We want amount to be $8P$.
$8P = P(1+R)^n$ $\Rightarrow$ $(1+R)^n = 8$.
Since $8 = 2^3$, then $(1+R)^n = ((1+R)^5)^3$ $= (1+R)^{15}$.
$n = 15$.

$P = 1,60,000, T = 2$ yrs, $R = 10\%$ p.a.
Semi-annual: $R=5\%, n=4$.
$A = 160000(1.05)^4$ $= 160000(1.2155...)$ $= \text{₹}1,94,481$.
$CI = 194481 - 160000$ $= \text{₹}34,481$.
Annual: $R=10\%, n=2$.
$A = 160000(1.1)^2$ $= 160000(1.21)$ $= \text{₹}1,93,600$.
$CI = \text{₹}33,600$.
Difference $= 34481 - 33600$ $= \text{₹}881$.