Maharashtra Board Class 11 Maths Part 2 Chapter 9 Commercial Mathematics 9.4 Solutions

11th Commerce Maths 2 Chapter 9 Exercise 9.4 Answers Maharashtra Board

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.4 Answers Maharashtra Board

Exercise 9.4 Solutions Commerce Maths

Question 1. Kanchan purchased a Maruti car for Rs. 2,45,000/- and the rate of depreciation is \(14\frac{2}{7}\%\) per annum. Find the value of the car after two years?
Answer:
Solution:
Given, purchase price of the car = V = Rs. 2,45,000
Rate of depreciation per annum = r
\( = 14\frac{2}{7}\% \)
\( = \frac{100}{7}\% \)
\[ \therefore \text{Value of the car after two years} = V\left(1 - \frac{r}{100}\right)^n \]
\[ = 2,45,000 \left(1 - \frac{100/7}{100}\right)^2 \]
\[ = 2,45,000 \left(1 - \frac{1}{7}\right)^2 \]
\[ = 2,45,000 \left(\frac{6}{7}\right)^2 \]
\[ = 2,45,000 \times \frac{36}{49} \]
\( = 5000 \times 36 \)
\( = 1,80,000 \)
∴ The value of the car after two years is Rs. 1,80,000.
In simple words: The problem asks to calculate the depreciated value of a car over two years using a given purchase price and an annual depreciation rate. The formula for depreciated value \( V(1 - \frac{r}{100})^n \) is applied by substituting the given values.

🎯 Exam Tip: Remember to convert mixed percentage rates to a proper fraction or decimal before applying them in the depreciation formula. Show all calculation steps clearly.

Question 2. The value of a machine depreciates from Rs. 32,768 to Rs. 21,952/- in three years. What is the rate of depreciation?
Answer:
Solution:
Given, initial value of machine = V = Rs. 32,768/-
Depreciated value of the machine = D.V. = Rs. 21,952/-
Number of years = n = 3
\[ \text{Using D.V.} = V \left(1-\frac{r}{100}\right)^n \]
\[ 21,952 = 32,768 \left(1-\frac{r}{100}\right)^3 \]
\[ \left(1-\frac{r}{100}\right)^3 = \frac{21,952}{32,768} \]
\[ \left(1-\frac{r}{100}\right)^3 = \frac{2^6 \times 7^3}{2^{15}} \]
\[ 1-\frac{r}{100} = \sqrt[3]{\frac{2^6 \times 7^3}{2^{15}}} \]
\[ 1-\frac{r}{100} = \frac{2^2 \times 7}{2^5} \]
\[ 1-\frac{r}{100} = \frac{4 \times 7}{32} \]
\[ 1-\frac{r}{100} = \frac{28}{32} = \frac{7}{8} \]
\[ \therefore \frac{100-r}{100} = \frac{7}{8} \]
\( \implies 8(100-r) = 700 \)
\( \implies 800-8r = 700 \)
\( \implies 8r = 100 \)
\( \implies r = \frac{100}{8} \)
\( \implies r = 12.5\% \)
∴ Rate of depreciation is 12.5% per annum.
In simple words: This problem requires finding the annual rate of depreciation given the initial value, final depreciated value, and the number of years. By rearranging the depreciation formula, we solve for 'r' after taking the cube root of the ratio of depreciated value to initial value.

🎯 Exam Tip: When dealing with cubic roots, try to simplify fractions to their lowest terms or recognize perfect cubes to make calculations easier. Double-check your algebraic manipulations.

Question 3. The value of a machine depreciates at the rate of 10% every year. It was purchased 3 years ago. Its present value is Rs. 2,18,700/-. What was the purchase price of the machine?
Answer:
Solution:
Given, the rate of depreciation per annum = r = 10%
Number of years = n = 3
Present value of the machine = P.V. = Rs. 2,18,700/-
∴ Purchase price of the machine
\[ = P.V \div \left(1-\frac{r}{100}\right)^n \]
\[ = 2,18,700 \div \left(1-\frac{10}{100}\right)^3 \]
\[ = 2,18,700 \div \left(\frac{9}{10}\right)^3 \]
\[ = \frac{2,18,700 \times 1,000}{729} \]
\( = 300 \times 1,000 \)
\( = 3,00,000 \)
∴ The purchase price of the machine is Rs. 3,00,000.
In simple words: To find the original purchase price of a depreciated asset, you need to reverse the depreciation calculation. Divide the present depreciated value by the depreciation factor \((1 - \frac{r}{100})^n\) to get the initial value.

🎯 Exam Tip: Pay attention to whether the question asks for future value or past value. For past value, you generally divide by the depreciation factor; for future value, you multiply.

Question 4. Mr. Manish purchased a motorcycle at Rs. 70,000/-. After some years he sold his motorcycle at its exact depreciated value of it that is Rs. 51,030/-. The rate of depreciation was taken as 10%. Find out how many years he sold his motorcycle.
Answer:
Solution:
Given, purchase price of the motorcycle = V = Rs. 70,000/-
Depreciated value of the motorcycle = D.V. = Rs. 51,030/-
∴ Rate of depreciation = r = 10%
\[ \text{Using, D.V.} = V \left(1-\frac{r}{100}\right)^n \]
\[ 51,030 = 70,000 \left(1-\frac{10}{100}\right)^n \]
\[ \left(1-\frac{10}{100}\right)^n = \frac{51,030}{70,000} \]
\[ \left(\frac{9}{10}\right)^n = \frac{5103}{7000} \]
\[ \left(\frac{9}{10}\right)^n = \frac{729}{1000} \]
\[ \left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^3 \]
∴ n = 3
∴ Manish sold his motorcycle after 3 years.
In simple words: This problem involves determining the number of years for depreciation. By setting up the depreciation formula with the given initial, final, and rate values, you can solve for 'n' by comparing the base and exponents.

🎯 Exam Tip: When the exponent 'n' is unknown, simplify the ratio of values to a fractional base. Then, express both sides of the equation with the same base to easily find 'n'.

Question 5. Mr. Chetan purchased furniture for his home at Rs. 5,12,000/-. Considering the rate of depreciation as 12.5%, what will be the value of furniture after 3 years.
Answer:
Solution:
Given, purchase price of furniture = V = Rs. 5,12,000/-
Rate of depreciation = r = 12.5%
Number of years = n = 3 years
\[ \therefore \text{Value of furniture after 3 years} = V\left(1 - \frac{r}{100}\right)^n \]
\[ = 5,12,000 \left(1 - \frac{12.5}{100}\right)^3 \]
\[ = 5,12,000 \left(1 - \frac{1}{8}\right)^3 \]
\[ = 5,12,000 \left(\frac{7}{8}\right)^3 \]
\[ = 5,12,000 \times \frac{343}{512} \]
\( = 1000 \times 343 \)
\( = 3,43,000 \)
∴ The value of furniture will be Rs. 3,43,000/-
In simple words: This problem calculates the future depreciated value of an asset. Apply the compound depreciation formula using the initial value, the annual depreciation rate, and the number of years to find the final value.

🎯 Exam Tip: Convert percentage rates to fractions (e.g., 12.5% to 1/8) to simplify calculations, especially when dealing with powers. Recognize common powers (like \(8^3 = 512\)) to speed up mental math.

Question 6. Grace Fashion Boutique purchased a sewing machine at Rs. 25,000/-. After 3 years machine was sold at depreciated value of Rs. 18,225/-. Find the rate of depreciation.
Answer:
Solution:
Given, purchase price of sewing machine = V = Rs. 25,000/-
Selling price of machine = D.V. = Rs. 18,225/-
Number of years = n = 3 years
\[ \text{By using, D.V.} = V \left(1-\frac{r}{100}\right)^n \]
\[ 18,225 = 25,000 \left(1 - \frac{r}{100}\right)^3 \]
\[ \left(1-\frac{r}{100}\right)^3 = \frac{18,225}{25,000} \]
\[ \left(1-\frac{r}{100}\right)^3 = \frac{729}{1000} \]
\[ \left(1-\frac{r}{100}\right)^3 = \left(\frac{9}{10}\right)^3 \]
\[ \therefore 1-\frac{r}{100} = \frac{9}{10} \]
\[ \therefore \frac{100-r}{100} = \frac{9}{10} \]
\( \implies 100-r = 90 \)
\( \implies r = 100-90 \)
\( \implies r = 10\% \)
∴ Rate of depreciation is 10% per annum.
In simple words: This problem determines the depreciation rate when the initial value, final value, and number of years are known. Set up the depreciation formula, solve for the term containing 'r', then isolate 'r' by taking the appropriate root and performing algebraic operations.

🎯 Exam Tip: Recognize perfect cubes (e.g., 729 is \(9^3\), 1000 is \(10^3\)) to simplify fractions quickly, which helps in solving for the unknown rate 'r' efficiently.

Question 7. Mr. Pritesh reduced the value of his assets by 5% each year, which were purchased for Rs. 50,00,000/-. Find the value of assets after 2 years.
Answer:
Solution:
Given, initial value of assets = V = Rs. 50,00,000/-
Rate of depreciation per annum = r = 5%
Number of years = n = 2 years
\[ \therefore \text{Value of assets after two years} = V \left(1-\frac{r}{100}\right)^n \]
\[ = 50,00,000 \left(1 - \frac{5}{100}\right)^2 \]
\[ = 50,00,000 \left(\frac{95}{100}\right)^2 \]
\[ = 50,00,000 \left(\frac{19}{20}\right)^2 \]
\[ = 50,00,000 \times \frac{361}{400} \]
\( = 12,500 \times 361 \)
\( = 45,12,500 \)
∴ The value of assets after two years is Rs. 45,12,500/-.
In simple words: This question calculates the depreciated value of assets over a specified period. Apply the standard depreciation formula using the initial value, the annual depreciation rate, and the number of years.

🎯 Exam Tip: Simplify the fraction `(1 - r/100)` before squaring or cubing to avoid large numbers and reduce potential calculation errors.

Question 8. A manufacturing company is allowed to charge 10% depreciation on its stock. The initial value of the stock was Rs. 60,000/-. After how many years value of the stock will be Rs. 39366?
Answer:
Solution:
Given, rate of depreciation = r = 10%
Initial value of stock = V = Rs. 60,000
Depreciated value of stock = D.V. = Rs. 39,366/-
\[ \text{By using, D.V.} = V \left(1-\frac{r}{100}\right)^n \]
\[ 39,366 = 60,000 \left(1-\frac{10}{100}\right)^n \]
\[ \left(1-\frac{10}{100}\right)^n = \frac{39,366}{60,000} \]
\[ \left(\frac{9}{10}\right)^n = \frac{6,561}{10,000} \]
\[ \left(\frac{9}{10}\right)^n = \left(\frac{9}{10}\right)^4 \]
∴ n = 4
∴ The value of the stock will be Rs. 39,366/- after 4 years.
In simple words: This problem asks for the number of years required for an asset to depreciate to a specific value. Use the depreciation formula, simplify the ratio of values, and then find the exponent 'n' by expressing both sides with the same base.

🎯 Exam Tip: When solving for 'n', it's crucial to simplify the fraction on the right-hand side and identify if it can be expressed as a power of the base on the left-hand side (e.g., \( (9/10)^n = (9/10)^4 \)).

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