a) Machinery costing Rs 4,50,000 has an estimated life of 15 years. If its cost is reduced by Rs.3,00,000 in 15 years, find the rate of depreciation. [3] b) Manoj borrows Rs.4368 which he promises…

a) Rate of Depreciation

Given:

• Initial Cost of Machinery ($P$) = Rs. 4,50,000
• Estimated Life ($n$) = 15 years
• Reduction in Cost (Total Depreciation) = Rs. 3,00,000

Assumption: Since the method of depreciation is not specified, we assume the Straight Line Method (SLM), which is the standard default for such problems in NEB Business Mathematics unless "Reducing Balance" is explicitly mentioned.

Formula: $$\text{Annual Depreciation (D)}=\frac{\text{Cost}-\text{Scrap Value}}{\text{Useful Life}}$$

First, calculate the Scrap Value (Salvage Value): $$\text{Scrap Value}=\text{Cost}-\text{Total Depreciation}$$ $$\text{Scrap Value}=4,50,000-3,00,000=\text{Rs. }1,50,000$$

Now, calculate the Annual Depreciation: $$D=\frac{4,50,000-1,50,000}{15}$$ $$D=\frac{3,00,000}{15}=\text{Rs. }20,000$$

The Rate of Depreciation ($r$) is calculated as a percentage of the initial cost: $$r=\left(\frac{D}{P}\right)\times 100\%$$ $$r=\left(\frac{20,000}{4,50,000}\right)\times 100\%$$ $$r=\frac{2}{45}\times 100\%$$ $$r=4.44\%$$

Answer: The rate of depreciation is 4.44%.

---

b) Amount of the First Instalment

Given:

• Total Amount to be paid ($S$) = Rs. 4,368
• Number of instalments ($n$) = 6
• Each instalment is treble (3 times) the preceding one. This forms a Geometric Progression (G.P.).
• Common Ratio ($r$) = 3

Let the first instalment be $a$. The instalments are: $a,3a,9a,27a,81a,243a$.

Formula for Sum of G.P.: $$S_n=\frac{a(r^n-1)}{r-1}$$

Substituting the given values: $$4,368=\frac{a(3^6-1)}{3-1}$$

Calculate $3^6$: $$3^6=729$$

Substitute back into the equation: $$4,368=\frac{a(729-1)}{2}$$ $$4,368=\frac{a(728)}{2}$$ $$4,368=a\times 364$$

Solving for $a$: $$a=\frac{4,368}{364}$$ $$a=12$$

Answer: The amount of the first instalment is Rs. 12.

---

c) Sum and Rate of Interest

Given:

• Compound Interest (C.I.) for 2 years = Rs. 204
• Simple Interest (S.I.) for 3 years = Rs. 300
• Let Principal be $P$ and Rate of interest be $R\%$.

Step 1: Find the Rate ($R$)

We know that: $$\text{C.I. for 2 years}=\text{S.I. for 2 years}+\text{Interest on Interest}$$ Alternatively, using the direct formula for C.I. difference: $${\text{C.I.}}_2-{\text{S.I.}}_2=P{\left(\frac{R}{100}\right)}^2$$

First, we need to find the Simple Interest for 2 years. Since S.I. for 3 years is Rs. 300, the S.I. for 1 year is: $${\text{S.I.}}_1=\frac{300}{3}=\text{Rs. }100$$

Therefore, S.I. for 2 years is: $${\text{S.I.}}_2=100\times 2=\text{Rs. }200$$

Now, calculate the difference between C.I. and S.I. for 2 years: $$\text{Difference}={\text{C.I.}}_2-{\text{S.I.}}_2=204-200=\text{Rs. }4$$

Using the formula: $$4=P{\left(\frac{R}{100}\right)}^2\text{— (Equation 1)}$$

We also know that S.I. for 2 years is Rs. 200: $$\text{S.I.}=\frac{P\times R\times T}{100}$$ $$200=\frac{P\times R\times 2}{100}$$ $$200=\frac{P\times R}{50}$$ $$P\times R=10,000\text{— (Equation 2)}$$

From Equation 2, we can express $P$ as: $$P=\frac{10,000}{R}$$

Substitute this value of $P$ into Equation 1: $$4=\left(\frac{10,000}{R}\right){\left(\frac{R}{100}\right)}^2$$ $$4=\frac{10,000}{R}\times \frac{R^2}{10,000}$$ $$4=\frac{10,000\times R^2}{10,000\times R}$$ $$4=R$$

So, the rate of interest $R=4\%$.

Step 2: Find the Principal ($P$)

Substitute $R=4$ into Equation 2: $$P\times 4=10,000$$ $$P=\frac{10,000}{4}$$ $$P=2,500$$

Answer: The Sum (Principal) is Rs. 2,500 and the Rate of Interest is 4%.