Financial Mathematics

Given: Principal (P) = Rs 10,000
Rates: $R_1 = 5\%$, $R_2 = 6\%$, $R_3 = 8\%$ p.a.
Time = 3 years

Formula: $A = P\left(1+\frac{R_1}{100}\right)\left(1+\frac{R_2}{100}\right)\left(1+\frac{R_3}{100}\right)$

Substitution: $A = 10000\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right)\left(1+\frac{8}{100}\right)$ $A = 10000 \times 1.05 \times 1.06 \times 1.08$ $A = 12020.40$

Compound Interest: $CI = A - P = 12020.40 - 10000 = 2020.40$

Therefore, the compound interest is Rs 2,020.40.

For an annuity due, payment is made at the beginning of each year.

$PV_{\text{due}}=P\left[1+\frac{1-(1+i)^{-(n-1)}}{i}\right]$

Given:
$P=5000$, $n=10$, $i=9\%=0.09$

$PV=5000\left[1+\frac{1-(1.09)^{-9}}{0.09}\right]$

$(1.09)^{-9}=0.4604278$

$PV=5000\left[1+\frac{1-0.4604278}{0.09}\right]$

$PV=5000[1+5.995247]$

$PV=5000\times 6.995247=34976.235$

$\boxed{PV \approx \text{Rs. }34,976.24}$

So, the present value of the annuity due is Rs. 34,976.24.

$AR = \dfrac{R}{Q}$

Average revenue is total revenue divided by output (quantity sold).

The formula for elasticity of supply is:

$e_s = \frac{\%\ \text{change in quantity supplied}}{\%\ \text{change in price}}$

Equivalently, using the ratio method:

$e_s = \frac{\Delta q}{\Delta p} \times \frac{p}{q}$

where $q$ is the original quantity supplied, $p$ is the original price, and $\Delta q$, $\Delta p$ are the respective changes.

At break-even, total revenue equals total cost, i.e. $R = C$ (equivalently, profit $= R - C = 0$).

Given: A sum doubles in 5 years at compound interest.

To find: Time to become four times the sum.

Solution:

Let the principal be $P$ and the rate be $R\%$ per annum.

Step 1 — Use the doubling condition.

$A = P\left(1 + \frac{R}{100}\right)^n$

When the sum doubles, $A = 2P$ and $n = 5$:

$2P = P\left(1 + \frac{R}{100}\right)^5$

$2 = \left(1 + \frac{R}{100}\right)^5 \quad \text{... (i)}$

Step 2 — Apply the condition for four times the sum.

When the sum becomes four times, $A = 4P$ and let the time be $n$ years:

$4P = P\left(1 + \frac{R}{100}\right)^n$

$4 = \left(1 + \frac{R}{100}\right)^n$

$2^2 = \left(1 + \frac{R}{100}\right)^n \quad \text{... (ii)}$

Step 3 — Compare (i) and (ii).

From (i), $2 = \left(1 + \dfrac{R}{100}\right)^5$. Squaring both sides:

$2^2 = \left(1 + \frac{R}{100}\right)^{10}$

So (ii) becomes:

$\left(1 + \frac{R}{100}\right)^{10} = \left(1 + \frac{R}{100}\right)^n$

$\therefore n = 10$

Result: The sum becomes four times itself in 10 years.

(Note: Four times the sum means doubling twice, so the time is $2 \times 5 = 10$ years.)

Given: Loan amount, $P = \text{Rs. } 2,00,000$
Number of installments, $n = 10$ years
Rate of interest, $i = 15\% = 0.15$

Using the present value of an ordinary annuity formula:

where $R$ is the yearly installment.

Therefore, each yearly installment is approximately:

So, the loan can be repaid by paying about Rs. 39,850.41 at the end of each year for 10 years.

(a) $P = \dfrac{P_T}{(1+i)^T}$

The present value of a future return is found by discounting the future amount $P_T$ back to the present at the rate $i$ for $T$ years, i.e. $P = \dfrac{P_T}{(1+i)^T}$.

(d) Rs. 32000

Value after 2 years $= 50000\left(1-\frac{20}{100}\right)^2 = 50000(0.8)^2 = 50000 \times 0.64 = \text{Rs. } 32000$.

The scholarship is paid at the end of each year, so the required deposit is the present value of an ordinary annuity.

Given: $R=\text{Rs }50{,}000$, $n=5$, $i=8\%=0.08$.

$PV=R\cdot\frac{1-(1+i)^{-n}}{i}=50000\cdot\frac{1-(1.08)^{-5}}{0.08}$

$(1.08)^5=1.469328\ \Rightarrow\ (1.08)^{-5}=0.680583$

$PV=50000\times\frac{1-0.680583}{0.08}=50000\times 3.99271=199{,}635.6$

Hence the donor should deposit about Rs 1,99,636.