Q1. Divide Rs.80000 in the ratio 2:6
\[
\text{Sum} = \left( 2+6 \right) = 8
\]
\[
\text{First Share} = \left( \frac{2}{8} \right) \times 80,000 = 20,000
\]
\[
\text{Second Share} = \left( \frac{6}{8} \right) \times 80,000 = 60,000
\]
Q2. Write \( \frac{1}{2} : 200 \) in simplest form.
Step 1: Convert the fraction into a whole number by multiplying both terms by 2:
\( \left(\frac{1}{2} \times 2 \right) : (200 \times 2) \)
Step 2: Simplify the ratio:
\( 1 : 400 \)
So, the simplest form of \( \frac{1}{2} : 200 \) is 1:400.
Q3. Rs. 5,00,000 profit is earned from business by four partners. This profit is to be allocated to the four partners in the ratio 5: 8: 9: 3. Determine the amount of profit of each.
Profit Distribution Among Four Partners
Total Profit: Rs. 5,00,000
Given Ratio: 5:8:9:3
Steps to Allocate Profit:
Step 1: Add the ratio parts: 5 + 8 + 9 + 3 = 25
Step 2: Find the value of one part:
\[
\text{One part value} = \left( \frac{5,00,000}{25} \right) = 20,000
\]
Profit Distribution:
- First Partner = 5 × Rs. 20,000 = Rs. 1,00,000
- Second Partner = 8 × Rs. 20,000 = Rs. 1,60,000
- Third Partner = 9 × Rs. 20,000 = Rs. 1,80,000
- Fourth Partner = 3 × Rs. 20,000 = Rs. 60,000
Q4. Express 58.5% to a common fraction.
Step 1: Write as a fraction over 100:
\( 58.5\% = \frac{58.5}{100} \)
Step 2: Eliminate the decimal by multiplying numerator and denominator by 10:
\( \frac{58.5 \times 10}{100 \times 10} = \frac{585}{1000} \)
Step 3: Simplify by dividing both terms by the greatest common divisor (GCD = 5):
\( \frac{585 \div 5}{1000 \div 5} = \frac{117}{200} \)
Final Answer: 58.5% as a common fraction is \(\frac{117}{200}\)
Q5. The height of a building A is 30 feet and its shadow is 36 feet. If the height of a building B is 40 feet then how long a shadow would be?
Calculating the Shadow Length of Building B
Given Data:
- Building A Height = 30 feet
- Building A Shadow = 36 feet
- Building B Height = 40 feet
Step 1: Calculate the Ratio
\[
\frac{\text{Height of A}}{\text{Shadow of A}} = \frac{30}{36} = \frac{5}{6}
\]
Step 2: Apply Ratio to Building B
\[
\frac{40}{x} = \frac{5}{6}
\]
Step 3: Solve for x
\[
x = \frac{40 \times 6}{5} = \frac{240}{5} = 48 \text{ feet}
\]
Final Answer: The shadow of Building B will be 48 feet.
Q6. Solve for x in each case 4 ∶ 8 ∷ x ∶ 50 and 2 ∶ 40 ∷ 10 ∶ x
Problem 1
\( 4 : 8 :: x : 50 \)
\( \frac{4}{8} = \frac{x}{50} \)
\( \frac{1}{2} = \frac{x}{50} \)
\( 1 \times 50 = 2 \times x \Rightarrow 50 = 2x \)
\( x = \frac{50}{2} = 25 \)
Answer: x = 25
Problem 2
\( 2 : 40 :: 10 : x \)
\( \frac{2}{40} = \frac{10}{x} \)
\( \frac{1}{20} = \frac{10}{x} \)
\( 1 \times x = 20 \times 10 \Rightarrow x = 200 \)
Answer: x = 200
Q7. If Rs.3000 is the price of three suits of five meters in the amount of Rs.8000 how many suits, can be purchased? Also find the price of cloth per meter.
Suits and Cloth Price Calculation
Given:
- Rs. 3000 = price of 3 suits
- Each suit = 5 meters of cloth
- Available amount = Rs. 8000
Step 1: Price of One Suit
\( \frac{3000}{3} = 1000 \) Rs
Step 2: Number of Suits from Rs. 8000
\( \frac{8000}{1000} = 8 \) suits
Answer: 8 suits can be purchased
Step 3: Total Cloth in 3 Suits
\( 3 \times 5 = 15 \) meters
Step 4: Price of Cloth Per Meter
\( \frac{3000}{15} = 200 \) Rs
Answer: Rs. 200 per meter
Q8. A car runs 60 km in 5 liters of petrol. If tank is full of 20 liters, how far will it cover the distance?
Petrol Usage and Distance Calculation
Given:
- 60 km can be covered with 5 liters of petrol
- The car tank holds 20 liters
Step 1: Find Distance per Liter
\( \frac{60}{5} = 12 \, \text{km per liter} \)
Step 2: Distance Covered with Full Tank
\( 12 \times 20 = 240 \, \text{km} \)
Answer: The car will cover 240 km with a full 20-liter tank.
Q9. Convert 77.87% and 24 1/5% into decimal and common fraction.
1. Convert 77.87%
Decimal:
\( \frac{77.87}{100} = 0.7787 \)
Fraction:
\( \frac{7787}{10000} \)
2. Convert 24 \( \frac{1}{5} \)%
Convert to improper fraction:
\( 24 \frac{1}{5} = \frac{121}{5} \)
Decimal:
\( \frac{121}{5 \times 100} = \frac{121}{500} = 0.242 \)
Fraction:
\( \frac{121}{500} \)
Q10. A person spend 80% of his income and saves Rs.600. What is his total income?
Income and Savings Problem
Given:
- The person spends 80% of their income
- They save Rs. 600
Step: Find Total Income
Savings represent 20% of the total income:
\( \frac{20}{100} \times \text{Income} = 600 \)
Solving for Income:
\( \text{Income} = \frac{600 \times 100}{20} = 3000 \)
Answer: The total income is Rs. 3000
Q1. Price of a T.V set is Rs. 20000 is available for sale at discount of Rs. 800. Find the discount rate and purchaser price.
Discount Rate Formula:
\[\text{Discount Rate} = \left( \frac{\text{Discount Amount}}{\text{Original Price}} \right) \times 100\]
Substituting values:
\[\text{Discount Rate} = \left( \frac{800}{20000} \right) \times 100 = 4\%\]
Purchaser Price Calculation:
\[\text{Purchaser Price} = 20000 - 800 = 19200\]
So, the discount rate is 4%, and the final purchaser price is Rs. 19,200.
Q2. A person buy a book rack of Rs. 18000, on the discount of 12%. What was the original price of book rack?
Using the formula:
\[\text{Original Price} = \frac{\text{Discounted Price}}{(1 - \text{Discount Rate})}\]
Substituting values:
\[\text{Original Price} = \frac{18000}{(1 - 0.12)} = \frac{18000}{0.88} = 20454.55\]
Thus, the original price of the book rack was approximately Rs. 20,454.55.
Q3. 25 packets of papers purchased by a photo copier. Market price of each packet is Rs.300 and he got a discount of 18% and an additional discount @ 5% on the cash payment. Find the value of the deal.
Step 1: Calculate the total market price
\[\text{Total Market Price} = \text{Number of Packets} \times \text{Price per Packet}\]
\[= 25 \times 300 = 7500\]
Step 2: Apply the first discount of 18%
\[\text{Discount Amount} = 7500 \times \frac{18}{100} = 1350\]
\[\text{Price after first discount} = 7500 - 1350 = 6150\]
Step 3: Apply the additional cash payment discount of 5%
\[\text{Additional Discount Amount} = 6150 \times \frac{5}{100} = 307.50\]
\[\text{Final Price} = 6150 - 307.50 = 5842.50\]
Thus, the final value of the deal is Rs. 5,842.50.
Q4. A company offers buy three pens and get four. Convert this offer of quantity discount into equivalent discount rate.
Using the formula:
\[\text{Discount Rate} = \left( \frac{\text{Free Items}}{\text{Total Items Received}} \right) \times 100\]
Substituting values:
\[\text{Discount Rate} = \left( \frac{1}{4} \right) \times 100 = 25\%\]
Thus, the equivalent discount rate for this offer is 25%.
Q5. A property dealer sold a shop in Lahore for 65 lakh and commission that he received is 3%. Find the commission of the property dealer.
Using the formula:
\[\text{Commission} = \text{Selling Price} \times \frac{\text{Commission Rate}}{100}\]
Substituting values:
\[\text{Commission} = 65,00,000 \times \frac{3}{100}\]
\[= 65,00,000 \times 0.03 = 1,95,000\]
Thus, the property dealer's commission is Rs. 1,95,000.
Q6. A man bought a second hand scooter for Rs.1,65,000 and he sold it for Rs.1,70,000. Calculate the amount of profit or loss that he made by selling his second hand scotty.
\[\text{Profit} = \text{Selling Price} - \text{Cost Price}\]
Substituting values:
\[\text{Profit} = 1,70,000 - 1,65,000 = 5,000\]
Profit Percentage Calculation
Using the formula:
\[\text{Profit Percentage} = \left( \frac{\text{Profit}}{\text{Cost Price}} \right) \times 100\]
Substituting values:
\[\text{Profit Percentage} = \left( \frac{5000}{165000} \right) \times 100\]
\[= \left( 0.0303 \right) \times 100 = 3.03\%\]
Thus, the man made a profit of Rs. 5,000, which is 3.03% of the cost price.
Q7. A person bought a T.V set at cost of Rs. 5000 and sold it at a profit of 25%. Find the profit and selling price of T.V.
\[\text{Profit} = \text{Cost Price} \times \frac{\text{Profit Percentage}}{100}\]
Substituting values:
\[\text{Profit} = 5000 \times \frac{25}{100} = 5000 \times 0.25 = 1250\]
Thus, the profit earned is Rs. 1,250.
Selling Price Calculation
\[\text{Selling Price} = \text{Cost Price} + \text{Profit}\]
Substituting values:
\[\text{Selling Price} = 5000 + 1250 = 6250\]
Thus, the selling price of the TV set is Rs. 6,250.
Q8. Calculate the annual installment of the depreciation for a machine of having price Rs.5,00,000 at the end of 30 years its scrap value is Rs.60,000. Also find the rate of depreciation and value of the asset after 20 years.
Using the formula for annual depreciation:
\[\text{Annual Depreciation} = \frac{\text{Cost of Asset} - \text{Scrap Value}}{\text{Life of Asset}}\]
Substituting values:
\[\text{Annual Depreciation} = \frac{5,00,000 - 60,000}{30} = \frac{4,40,000}{30} = 14,666.67\]
Thus, the annual installment of depreciation is Rs. 14,667.
Depreciation Rate Calculation
Using the formula:
\[\text{Depreciation Rate} = \left( \frac{\text{Annual Depreciation}}{\text{Cost of Asset - Scrap Value}} \right) \times 100\]
Substituting values:
\[\text{Depreciation Rate} = \left( \frac{14,667}{5,00,000 - 60,000} \right) \times 100\]
\[= \left( 0.0333 \right) \times 100 = 3.33\%\]
Thus, the depreciation rate is 3.33% per year.
Asset Value After 20 Years
Using the formula:
\[\text{Asset Value} = \text{Cost of Asset} - (n \times \text{Annual Depreciation})\]
For \( n = 20 \) years:
\[\text{Asset Value} = 5,00,000 - (20 \times 14,667)\]
\[= 5,00,000 - 2,93,340 = 2,06,660\]
Thus, the value of the asset after 20 years is Rs. 2,06,660.
AIOU 1349 Chapter 3 - Mathematicsof Finance – I
Q1. 3 years loan of Rs.60,000 issued by a bank. It charges simple interest at the rate of 12 percent per year. At the end of the 3rd year principal plus interest is to be repaid. Calculate the interest for the 3 year period and what amount will be repaid at the end of the 3rd year?
\[
\text{Simple Interest} = P \times r \times t
\]
Where:
- \( P = 60000 \) (Principal Amount)
- \( r = 12\% = 0.12\) (Annual Interest Rate)
- \( t = 3 \) (Time in Years)
\[
\text{Interest} = 60,000 \times 0.12 \times 3
\]
\[
\text{Interest} = 60,000 \times 0.36
\]
\[
\text{Interest} = 21,600
\]
\[
\text{Total Amount} = \text{Principal} + \text{Interest}
\]
\[
\text{Total Amount} = 60,000 + 21,600
\]
\[
\text{Total Amount} = 81,600
\]
Q2. Suppose a person deposited Rs.9000 in a bank, which pays interest of 9 percent per year-compounded quarterly. He wants to determine the amount of money he will have on deposit at the end of the year if all interest amounts are retained in the account as reinvestment.
\[
A = P \left(1 + \frac{r}{n} \right)^{n t}
\]
Where:
- \( P = 9000 \) (Principal)
- \( r = 9\% = 0.09 \) (Annual Interest Rate)
- \( n = 4 \) (Quarterly Compounding)
- \( t = 1 \) (Time in Years)
Substituting the values:
\[
A = 9000 \left(1 + \frac{0.09}{4} \right)^{4 \times 1}
\]
\[
A = 9000 \left(1 + 0.0225 \right)^4
\]
\[
A = 9000 \times (1.0225)^4
\]
Approximating:
\[
(1.0225)^4 \approx 1.093
\]
\[
A \approx 9000 \times 1.093
\]
\[
A \approx 9837
\]
Total Amount at the End of the Year: Rs. 9837
Q3. Find the interest on Rs.2180 for one year at simple interest 4%?
\[
\text{Simple Interest} = P \times r \times t
\]
Where:
- \( P = 2180 \) (Principal Amount)
- \( r = 4\% = 0.04 \) (Annual Interest Rate)
- \( t = 1 \) (Time in Years)
Substituting the values:
\[
\text{Interest} = 2180 \times 0.04 \times 1
\]
\[
\text{Interest} = 2180 \times 0.04
\]
\[
\text{Interest} = 87.20
\]
Total Simple Interest: Rs. 87.20
Q4. Calculate the difference between compound interest and simple interest both are charged on Rs.3500 at the rate 25% annually for 6 years.
\[
SI = P \times r \times t
\]
Where:
- \( P = 3500 \) (Principal Amount)
- \( r = 25\% = 0.25 \) (Annual Interest Rate)
- \( t = 6 \) (Time in Years)
Substituting the values:
\[
SI = 3500 \times 0.25 \times 6
\]
\[
SI = 3500 \times 1.5
\]
\[
SI = 5250
\]
Compound Interest Formula:
\[
CI = P \left(1 + r \right)^t - P
\]
Substituting the values:
\[
CI = 3500 \left(1 + 0.25 \right)^6 - 3500
\]
\[
CI = 3500 \times (1.25)^6 - 3500
\]
Approximating:
\[
(1.25)^6 = 3.814697265625
\]
\[
CI = 3500 \times 3.814697265625 - 3500
\]
\[
CI = 13356.44 - 3500
\]
\[
CI = 9856.44
\]
Difference Between Compound Interest and Simple Interest:
\[
\text{Difference} = CI - SI
\]
\[
\text{Difference} = 9856.44 - 5250
\]
\[
\text{Difference} = 4606.44
\]
Total Difference: Rs. 4606.44
Q5. Calculate the principal that collects to Rs. 1500 in 3 years at tha rate 12% annualy.
\[
P = \frac{A}{(1 + r \times t)}
\]
Where:
- \( A = 1500 \) (Final Amount)
- \( r = 12\% = 0.12 \) (Annual Interest Rate)
- \( t = 3 \) (Time in Years)
- \( P \) (Principal to be found)
Substituting the values:
\[
P = \frac{1500}{(1 + 0.12 \times 3)}
\]
\[
P = \frac{1500}{(1 + 0.36)}
\]
\[
P = \frac{1500}{1.36}
\]
\[
P \approx 1102.94
\]
Final Principal Amount: Rs. 1102.94
AIOU 1349 Chapter 4 - Linear, Quadratic and Simultaneous Equations
Q1. Solve the given equation to find the value of x.
5x - 10 = 3x + 6
Given Equation:
$$5x - 10 = 3x + 6$$
$$5x - 3x = 6 + 10$$
$$2x = 16$$
$$x = \frac{16}{2}$$
$$x = 8$$
Q2. Find two consecutive odd integers whose sum are 40.
Let the two consecutive odd integers be: 2y + 1 and 2y + 3
According to the question, their sum is 40:
(2y + 1) + (2y + 3) = 40
Combine like terms:
4y + 4 = 40
4y = 40 - 4
4y = 36
$$y = \frac{36}{4} = 9$$
y = 9
Substitute back to find the integers:
2y + 1 => 2(9) + 1 = 19
2y + 3 => 2(9) + 3 = 21
Final Answer: The two consecutive odd integers are 19 and 21.
Q3. Length of a rectangle is 3 feets less than 2 times of its width and its parameter is 60 feets. Calculate the width and length of the rectangle.
Length of the rectangle is 2x − 3
Parameter of the rectangle = Sum of all sides of the rectangle
x + (2x-3) + x + (2x-3) = 60
2x + 2x + 2x - 3 - 3 = 60
6x - 6 = 60
6x = 60 + 6
6x = 66
$$x = \frac{66}{6} = 11$$
x = 11
Width of the rectangle is 11 feets
Length of the rectangle is 2x - 3
Length of the rectangle is 2(11) - 3
Length of the rectangle is 22 - 3
Length of the rectangle is 19
Length of the rectangle is 19 feets
Quadratic Formula
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Q4. Solve the given equation using quadratic formula \(x^2 + 2x - 3 = 0\)
Solving the Equation: \(x^2 + 2x - 3 = 0\)
Using the quadratic formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Substitute
\(a = 1\),
\(b = 2\),
\(c = -3\)
$$x = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)}$$
$$x = \frac{-2 \pm \sqrt{4 + 12}}{2}$$
$$x = \frac{-2 \pm \sqrt{16}}{2}$$
$$x = \frac{-2 \pm 4}{2}$$
Solutions:
$$x = \frac{-2 + 4}{2} = 1$$
$$x = \frac{-2 - 4}{2} = -3$$
Final Answer: \(x = 1\) or \(x = -3\)
Q5. Solve the given equation using quadratic formula \(y^2 + 6y = -9\)
Solving the Equation: \(y^2 + 6y + 9 = 0\)
Here, \(a = 1\), \(b = 6\), and \(c = 9\)
Use the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
\[
y = \frac{-6 \pm \sqrt{6^2 - 4(1)(9)}}{2(1)}
\]
\[
y = \frac{-6 \pm \sqrt{36 - 36}}{2(1)}
\]
\[
y = \frac{-6 \pm \sqrt{0}}{2(1)}
\]
\[
y = \frac{-6 \pm 0}{2}
\]
\[
y = \frac{-6}{2}
\]
\[
y = -3
\]
Final Answer: \(y = -3\)
Solve using factorization
Q8. Solve the given equation using factorization \(y^2 + 6y = -9\)
Solving the Equation: y² + 6y = -9
y² + 6y + 9 = 0
y² + 3y + 3y + 9 = 0
y (y + 3) + 3(y + 3) = 0
(y + 3)(y + 3) = 0
y + 3 = 0 or y + 3 = 0
y = -3 or y = -3
Q9. Solve the given equation using factorization \(x^2 + 2x - 3 = 0\)
Solving the Equation: x² + 2x - 3 = 0
x² + 3x - x - 3 = 0
x(x + 3) - 1(x + 3) = 0
(x + 3)(x - 1) = 0
x + 3 = 0 or x - 1 = 0
x = -3 or x = 1
Solve using completing square method
(a + b)² = a² + 2ab + b²
(a - b)² = a² - 2ab + b²
Q10. Solve the given equation completing square method \(x^2 + 2x - 3 = 0\)
Solving the Equation: x² + 2x - 3 = 0
x² + 2x = 3
x² + 2x + 1 = 3 + 1
x² + 2x + 1 = 4
(a + b)² = a² + 2ab + b²
x² + 2x + 1 = 4
(x + 1)² = 4
\( \sqrt{(x + 1)^2} = \pm \sqrt{4} \)
x + 1 =\( \pm 2 \)
x = 2 - 1 or x = - 2 - 1
x = 1 or x = -3
Q11. Solve the given equation completing square method \(y^2 + 6y = -9\)
Solving the Equation: y² + 6y = -9
y² + 6y + 9 = 0
(a + b)² = a² + 2ab + b²
y² + 6y + 9 = 0
(y + 3)² = 0
\( \sqrt{(y + 3)^2} = \pm \sqrt{0} \)
y + 3 =\( \pm 0 \)
y = -3
Q12. Find the transpose of matrix A
\[ A =
\begin{bmatrix}
1 & 2 \\
3 & 4\\
5 & 6
\end{bmatrix}
\]
\[ A^t =
\begin{bmatrix}
1 & 3 & 5 \\
2 & 4 & 6
\end{bmatrix}
\]
Q13. Find the transpose of matrix A
\[ A =
\begin{bmatrix}
7 & 3 & 5\\
9 & 2 & 8\\
\end{bmatrix}
\]
\[ A^t =
\begin{bmatrix}
7 & 9 \\
3 & 2\\
5 & 8
\end{bmatrix}
\]
Q14. What is symmetric matrix?
Symmetric matrix is a square matrix. A matrix is said to be symmetric if the transpose of a given matrix is equal to the matrix itself that is \[ A^t = A\]
\[ A =
\begin{bmatrix}
-1 & 3 \\
3 & -1
\end{bmatrix}
\]
\[ A^t =
\begin{bmatrix}
-1 & 3 \\
3 & -1
\end{bmatrix}
\]
Q15. Determine the following matrices are singular or non-singular.
\[ A =
\begin{bmatrix}
2 & 6 \\
1 & 8
\end{bmatrix}
\]
\[ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\]
det(A) = ad − bc
\[ A =
\begin{bmatrix}
2 & 6 \\
1 & 8
\end{bmatrix}
\]
det(A) = (2 × 8) − (6 × 1) = 16 − 6 = 10
Conclusion: Since the determinant is 10, which is not equal to zero, matrix A is non-singular.
Q16. Determine the following matrices are singular or non-singular.
\[ A =
\begin{bmatrix}
6 & 12 \\
4 & 4
\end{bmatrix}
\]
det(A) = (6 × 4) − (12 × 2) = 24 − 24 = 0
Conclusion: Since the determinant is 0, matrix A is singular.
Q17. Find the value of y. Where A is a singular matrix.
\[ A =
\begin{bmatrix}
5 & 2 \\
y & 8
\end{bmatrix}
\]
det(A) = (5 × 8) − (2 × y)
det(A) = 40 - 2y
|A| = 40 - 2y
As A is singular matrix, so |A| = 0
40 - 2y = 0
2y = 40
y = 40/2
y = 20
Q18. Determine the inverse of matrix A.
\[ A =
\begin{bmatrix}
1 & 2 \\
3 & 4
\end{bmatrix}
\]
Compute the determinant of matrix A.
\[
\det(A) = (1)(4) - (2)(3)\] \[\det(A) = 4 - 6\] \[\det(A) = -2
\]
Formula for the inverse of a matrix.
If
\[
A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
\]
then
\[
A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
OR
\[
A^{-1} = \frac{1}{det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
\[
A^{-1} = \frac{1}{-2} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}
\]
\[
A^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}
\]
\[
A^{-1} = \begin{bmatrix} -2 & 1 \\ 1.5 & -0.5 \end{bmatrix}
\]
Q19. Let A and B are two matrix. Find A-1, B-1 and the products of A-1B-1 and B-1A-1
Is it true that (AB)-1 = A-1B-1 ?
Is it true that (AB)-1 = B-1A-1 ?
\[
A = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix}, B = \begin{bmatrix} 5 & -1 \\ 9 & -2 \end{bmatrix}
\]
\[
A^{-1} = \frac{1}{det(A)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
Compute the determinant of matrix A.
\[
\det(A) = (2)(2) - (3)(1)\] \[\det(A) = 4 - 3\] \[\det(A) = 1
\]
\[
A^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}
\]
\[
A^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}
\]
\[
B^{-1} = \frac{1}{det(B)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}
\]
Compute the determinant of matrix B.
\[
\det(B) = (5)(-2) - (9)(-1)\] \[\det(B) = -10 + 9\] \[\det(B) = -1
\]
\[
B^{-1} = \frac{1}{-1} \begin{bmatrix} -2 & -(-1) \\ -9 & 5 \end{bmatrix}
\]
\[
B^{-1} = \frac{1}{-1} \begin{bmatrix} -2 & 1 \\ -9 & 5 \end{bmatrix}
\]
\[
B^{-1} = \begin{bmatrix} 2 & -1 \\ 9 & -5 \end{bmatrix}
\]
\[
A^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}, B^{-1} = \begin{bmatrix} 2 & -1 \\ 9 & -5 \end{bmatrix}
\]
Find A-1B-1
\[
A^{-1}B^{-1} = \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ 9 & -5 \end{bmatrix}
\]
\[
A^{-1}B^{-1} = \begin{bmatrix} (2)(2)+(−1)(9) & (2)(−1)+(−1)(-5) \\ (-3)(2)+(2)(9) & (-3)(−1)+(2)(-5) \end{bmatrix}
\]
\[
A^{-1}B^{-1} = \begin{bmatrix} 4-9 & -2+5 \\ -6+18 & 3-10 \end{bmatrix}
\]
\[
A^{-1}B^{-1} = \begin{bmatrix} -5 & 3 \\ 12 & -7 \end{bmatrix}
\]
Find B-1A-1
\[B^{-1}A^{-1} = \begin{bmatrix} 2 & -1 \\ 9 & -5 \end{bmatrix} \begin{bmatrix} 2 & -1 \\ -3 & 2 \end{bmatrix}\]
\[
B^{-1}A^{-1} = \begin{bmatrix} (2)(2)+(−1)(−3) & (2)(−1)+(−1)(2) \\ (9)(2)+(−5)(−3) & (9)(−1)+(−5)(2) \end{bmatrix}
\]
\[
B^{-1}A^{-1} = \begin{bmatrix} 4+3 & -2-2 \\ 18+15 & -9-10 \end{bmatrix}
\]
\[
B^{-1}A^{-1} = \begin{bmatrix} 7 & -4 \\ 33 & -19 \end{bmatrix}
\]
1) Is it true that (AB)-1 = A-1B-1 ?
2) Is it true that (AB)-1 = B-1A-1 ?
Find (AB)-1
\[
AB = \begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} \begin{bmatrix} 5 & -1 \\ 9 & -2 \end{bmatrix}
\]
\[
AB = \begin{bmatrix} (2)(5)+(1)(9) & (2)(−1)+(1)(−2) \\ (3)(5)+(2)(9) & (3)(−1)+(2)(−2) \end{bmatrix}
\]
\[
AB = \begin{bmatrix} 10+9 & -2-2 \\ 15+18 & -3-4 \end{bmatrix}
\]
\[AB = \begin{bmatrix} 19 & -4 \\ 33 & -7 \end{bmatrix}\]
det(AB) = (19)(−7)−(−4)(33)
det(AB) = −133−(−132)
det(AB) = −133+132
det(AB) = −1
\[(AB)^{-1} = \frac{1}{-1} \begin{bmatrix} -7 & 4 \\ -33 & 19 \end{bmatrix}
\]
\[(AB)^{-1} = \begin{bmatrix} 7 & -4 \\ 33 & -19 \end{bmatrix}
\]
1) Is it true that (AB)-1 = A-1B-1
\[(AB)^{-1} = \begin{bmatrix} 7 & -4 \\ 33 & -19 \end{bmatrix}\]
From calculations it is not true (AB)-1 = A-1B-1
(AB)-1 ≠ A-1B-1
\[
A^{-1}B^{-1} = \begin{bmatrix} -5 & 3 \\ 12 & -7 \end{bmatrix}
\]
2) Is it true that (AB)-1 = B-1A-1
\[(AB)^{-1} = \begin{bmatrix} 7 & -4 \\ 33 & -19 \end{bmatrix}\]
\[
B^{-1}A^{-1} = \begin{bmatrix} 7 & -4 \\ 33 & -19 \end{bmatrix}
\]
From calculations it is true (AB)-1 = B-1A-1
(AB)-1 = B-1A-1
Q20. Find determinant of the following matrix.
\[A = \begin{bmatrix} 3 & 2 & 1 \\ 1 & 0 & 3 \\ 4 & 1 & 5 \end{bmatrix}\]
det(A) = 3(0x5−3x1) − 2(1x5−3x4) + 1(1x1−0x4)
det(A) = 3(-3) - 2(-7) + 1(1)
det(A) = -9 + 14 + 1
det(A) = 6
Q21. There are two matrix A and B find A + B.
\[A = \begin{bmatrix} 2 & 4 \\ -1 & 5 \end{bmatrix}, B = \begin{bmatrix} 7 & 0 \\ 4 & -3 \end{bmatrix}\]
\[A + B = \begin{bmatrix} 2+7 & 4+0 \\ -1+4 & 5-3 \end{bmatrix}\]
\[A + B = \begin{bmatrix} 9 & 4 \\ 3 & 2 \end{bmatrix}\]
Q22. The quarterly sales of wheat, cotton and corn for the year 2015 and 2016 are represented below in the form of matrix A and B.
Find the total quarterly sales of wheat, cotton and corn for these two years.
\[A = \begin{bmatrix} 22 & 15 & 26 & 10 \\ 5 & 20 & 17 & 15 \\ 32 & 24 & 18 & 9 \end{bmatrix},
B = \begin{bmatrix} 35 & 12 & 25 & 31 \\ 20 & 14 & 10 & 15 \\ 22 & 33 & 27 & 10 \end{bmatrix}\]
\[A + B = \begin{bmatrix} 22+35 & 15+12 & 26+25 & 10+31 \\ 5+20 & 20+14 & 17+10 & 15+15 \\ 32+22 & 24+33 & 18+27 & 9+10 \end{bmatrix}\]
\[A + B = \begin{bmatrix} 57 & 27 & 51 & 41 \\ 25 & 34 & 27 & 30 \\ 54 & 57 & 45 & 19 \end{bmatrix}\]
Q23. There are two matrix A and B find A - B.
\[A = \begin{bmatrix} 18 & 12 \\ 10 & 9 \end{bmatrix}, B = \begin{bmatrix} -5 & 8 \\ 9 & 16 \end{bmatrix}\]
\[A - B = \begin{bmatrix} 18-(-5) & 12-8 \\ 10-9 & 9-16 \end{bmatrix}\]
\[A - B = \begin{bmatrix} 18+5 & 4 \\ 1 & -7 \end{bmatrix}\]
\[A - B = \begin{bmatrix} 23 & 4 \\ 1 & -7 \end{bmatrix}\]
Q24. The number of books, journals and CDs available in Library A and Library B are represented by two matrices. How many of each of these items is more in Library A?
\[A = \begin{bmatrix} 3081 \\ 70 \\ 251 \end{bmatrix}, B = \begin{bmatrix} 2948 \\ 33 \\ 146 \end{bmatrix}\]
\[A - B = \begin{bmatrix} 3081-2948 \\ 70-33 \\ 251-146 \end{bmatrix}\]
\[A - B = \begin{bmatrix} 133 \\ 37 \\ 105 \end{bmatrix}\]
Q25. Amina, Fatima and Sahar bought cookies of different brands G, H and I. Amina bought 12 packets of G, 8 packets of H and 5 packets of I. Fatima bought 5 packets of G, 9 packets of H and 11 packets of I. Sahar bought 5 packets of G, 8 packets of H and 9 packets of I. If brand G costs Rs 5, H costs Rs 6 and I costs Rs 7 each, then using matrix operation, find the total amount of money spent by these three persons individually.
Let Q represents the matrix for the quantity of each brand of biscuit bought by G, H and I and let C is the matrix representing the cost of each brand of biscuit.
\[Q = \begin{bmatrix} 12 & 8 & 5 \\ 5 & 9 & 11 \\ 5 & 8 & 9 \end{bmatrix}, C = \begin{bmatrix} 5 \\ 6 \\ 7 \end{bmatrix}\]
\[QC = \begin{bmatrix} (12)(5) + (8)(6) + (5)(7) \\ (5)(5) + (9)(6) + (11)(7) \\ (5)(5) + (8)(6) + (9)(7) \end{bmatrix}\]
\[QC = \begin{bmatrix} 60 + 48 + 35 \\ 25 + 54 + 77 \\ 25 + 48 + 63 \end{bmatrix}\]
\[QC = \begin{bmatrix} 143 \\ 156 \\ 136 \end{bmatrix}\]
Q26. Use Cramer’s rule to solve the system.
2x + 2y + z = 3
3x - 2y - 2z = 1
5x + y - 3z = 2
Solution
\[\begin{bmatrix} 2 & 2 & 1 \\ 3 & -2 & -2 \\ 5 & 1 & -3 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}\]
Let
\[A = \begin{bmatrix} 2 & 2 & 1 \\ 3 & -2 & -2 \\ 5 & 1 & -3 \end{bmatrix}, X =\begin{bmatrix} x \\ y \\ z \end{bmatrix}, B = \begin{bmatrix} 3 \\ 1 \\ 2 \end{bmatrix}\]
Find |A|
|A| = 2((-2)(-3) - (1)(-2)) -2((3)(-3) - (-2)(5)) + 1((3)(1) - (-2)(5))
|A| = 2(6 + 2) − 2(−9 + 10) + 1(3 + 10)
|A| = 2(8) -2(1) +1(13)
|A| = 16 -2 + 13
|A| = 27
For x
\[x = \frac{1}{27} \begin{bmatrix} 3 & 2 & 1 \\ 1 & -2 & -2 \\ 2 & 1 & -3 \end{bmatrix}\]
\[x = \frac{1}{27} (3((-2 )( -3) - (-2 )( 1)) − 2((1 )( -3) - (-2 )( 2)) + 1((1 )( 1) - (-2 )( 2))) \]
\[x = \frac{1}{27} (3((6) - (-2)) − 2((-3) - (-4)) + 1((1) - (-4))) \]
\[x = \frac{1}{27} (3(6 + 2) − 2(−3 + 4) + 1(1 + 4)) \]
\[x = \frac{1}{27} (3(8) − 2(1) + 1(5)) \]
\[x = \frac{1}{27} (24 − 2 + 5) \]
\[x = \frac{1}{27} (27) \]
\[x = \frac{27}{27} \]
\[x = 1 \]
For y
\[y = \frac{1}{27} \begin{bmatrix} 2 & 3 & 1 \\ 3 & 1 & -2 \\ 5 & 2 & -3 \end{bmatrix}\]
\[y = \frac{1}{27} (2((1)(-3) - (-2)(2)) − 3((3)(-3) - (-2)(5)) + 1((3)(2) − (1)(5))) \]
\[y = \frac{1}{27} (2((-3) - (-4)) − 3((-9) - (-10)) + 1((6) − (5))) \]
\[y = \frac{1}{27} (2(-3 + 4) − 3(-9 + 10) + 1(6 − 5)) \]
\[y = \frac{1}{27} (2(1) − 3(1) + 1(1)) \]
\[y = \frac{1}{27} (2 − 3 + 1) \]
\[y = \frac{1}{27} (-1 + 1) \]
\[y = \frac{1}{27} (0) \]
\[y = \frac{0}{27} \]
\[y = 0 \]
For z
\[z = \frac{1}{27} \begin{bmatrix} 2 & 2 & 3 \\ 3 & -2 & 1 \\ 5 & 1 & 2 \end{bmatrix}\]
\[z = \frac{1}{27} (2((-2)(2) - (1)(1)) − 2((3)(2) - (1)(5)) + 3((3)(1) − (-2)(5))) \]
\[z = \frac{1}{27} (2(-4 - 1) − 2(6 - 5) + 3(3 − (-10)) \]
\[z = \frac{1}{27} (2(-4 - 1) − 2(6 - 5) + 3(3 + 10)) \]
\[z = \frac{1}{27} (2(-5) − 2(1) + 3(13)) \]
\[z = \frac{1}{27} (-10 − 2 + 39) \]
\[z = \frac{1}{27} (-12 + 39) \]
\[z = \frac{1}{27} (27) \]
\[z = \frac{27}{27} \]
\[z = 1 \]
Hence,
x = 1
y = 0
z = 1
Q27. The sum of three numbers is 26. The third number is twice the second and1 less than three times the first. What are the three numbers?
Solution
Let
First number is x
Second number is y
Third number is z
x + y + z = 26
-2y + z = 0
-3x + z = -1
Let
\[A = \begin{bmatrix} 1 & 1 & 1 \\ 0 & -2 & 1 \\ -3 & 0 & 1 \end{bmatrix}, x = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\ B = \begin{bmatrix} 26 \\ 0 \\ -1 \end{bmatrix}\]
Find |A|
\[|A| = 1(-2-0)-1(0+3)+1(0-6)\]
\[|A| = -2-3-6\]
\[|A| = -11\]
For x
\[A = \begin{bmatrix} 26 & 1 & 1 \\ 0 & -2 & 1 \\ -1 & 0 & 1 \end{bmatrix}\]
\[x = \frac{1}{-11} (26(-2-0)-1(0+1)+1(0-2))\]
\[x = \frac{1}{-11} (-52-1-2)\]
\[x = \frac{1}{-11} (-55)\]
\[x = \frac{-55}{-11}\]
\[x = 5\]
For y
\[A = \begin{bmatrix} 1 & 26 & 1 \\ 0 & 0 & 1 \\ -3 & -1 & 1 \end{bmatrix}\]
\[y = \frac{1}{-11} (1(0+1)-26(0+3)+1(0-0))\]
\[y = \frac{1}{-11} (1-78+0)\]
\[y = \frac{1}{-11} (-77)\]
\[y = \frac{-77}{-11}\]
\[y = 7\]
For z
\[A = \begin{bmatrix} 1 & 1 & 26 \\ 0 & -2 & 0 \\ -3 & 0 & -1 \end{bmatrix}\]
\[z = \frac{1}{-11} (1(2-0)-1(0+0)+26(0-6))\]
\[z = \frac{1}{-11} (2-0-156)\]
\[z = \frac{1}{-11} (-154)\]
\[z = \frac{-154}{-11}\]
\[z = 14\]
AIOU 1349 Binary Number System and Its Operations
Binary Number System
Binary number system consists of only two digits 0 and 1.
Binary numbers: 010001, 100001, 000000, 111111, 1, 0, 101010, 010101
Representation of Binary Numbers
100001
1000012
(100001)2
Decimal Number System
Decimal number system consists of ten digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Decimal numbers: 049, 55, 99, 5745, 7785, 0, 1, 9.
Representation of Decimal Numbers
55
5510
(55)10
Q28. Convert the following binary number into equivalent decimal number.
100101
(1 x 2)5 + (0 x 2)4 + (0 x 2)3 + (1 x 2)2 + (0 x 2)1 + (1 x 2)0
(2)5 + (0)4 + (0)3 + (2)2 + (0)1 + (2)0
32 + 0 + 0 + 4 + 0 + 1
(37)10
(111111)2 = (?)10
(1 x 2)5 + (1 x 2)4 + (1 x 2)3 + (1 x 2)2 + (1 x 2)1 + (1 x 2)0
(2)5 + (2)4 + (2)3 + (2)2 + (2)1 + (2)0
32 + 16 + 8 + 4 + 2 + 1
(63)10
11101112 = (?)10
(1 x 2)6 + (1 x 2)5 + (1 x 2)4 + (0 x 2)3 + (1 x 2)2 + (1 x 2)1 + (1 x 2)0
(2)6 + (2)5 + (2)4 + (0)3 + (2)2 + (2)1 + (2)0
64 + 32 + 16 + 0 + 4 + 2 + 1
11910
Q29. Convert the following binary number into equivalent decimal number.
0.01101
(0 x 2)-1 + (1 x 2)-2 + (1 x 2)-3 + (0 x 2)-4 + (1 x 2)-5
(0)-1 + (2)-2 + (2)-3 + (0)-4 + (2)-5
0 + + + 0 +
0 + 0.25 + 0.125 + 0 + 0.03125
(0.40625)10
(0.111111)2 = (?)10
(1 x 2)-1 + (1 x 2)-2 + (1 x 2)-3 + (1 x 2)-4 + (1 x 2)-5 + (1 x 2)-6
(2)-1 + (2)-2 + (2)-3 + (2)-4 + (2)-5 + (2)-6
+ + + + +
0.5 + 0.25 + 0.125 + 0.0625 + 0.03125 + 0.015625
(0.984375)10
(0.010010)2 = (?)10
(0 x 2)-1 + (1 x 2)-2 + (0 x 2)-3 + (0 x 2)-4 + (1 x 2)-5 + (0 x 2)-6
(0)-1 + (2)-2 + (0)-3 + (0)-4 + (2)-5 + (0)-6
0 + + 0 + 0 + + 0
0 + 0.25 + 0 + 0 + 0.03125 + 0
(0.28125)10
(1.1)2 to decimal.
(1 x 2)0 + (1 x 2)-1
1 +
1 + 0.5
(1.5)10
(11101.1011)2 = (?)10
(1 x 2)4 + (1 x 2)3 + (1 x 2)2 + (0 x 2)1 + (1 x 2)0 + (1 x 2)-1 + (0 x 2)-2 + (1 x 2)-3 + (1 x 2)-4
16 + 8 + 4 + 0 + 1 + + 0 + +
29 + 0.5 + 0 + 0.125 + 0.0625
29 + 0.6875
(29.6875)10
(11101)2 = (?)10
(1 x 2)4 + (1 x 2)3 + (1 x 2)2 + (0 x 2)1 + (1 x 2)0
OR
24 + 23 + 22 + 01 + 20
16 + 8 + 4 + 0 + 1
29
(29)10
Q30. Convert to Binary 26.
| Division |
Quotient |
Remainder |
| 26 ÷ 2 |
13 |
0 |
| 13 ÷ 2 |
6 |
1 |
| 6 ÷ 2 |
3 |
0 |
| 3 ÷ 2 |
1 |
1 |
| 1 ÷ 2 |
0 |
1 |
(26)10 = (11010)2
Convert 261 into binary number.
| Division |
Quotient |
Remainder |
| 261 ÷ 2 |
130 |
1 |
| 130 ÷ 2 |
65 |
0 |
| 65 ÷ 2 |
32 |
1 |
| 32 ÷ 2 |
16 |
0 |
| 16 ÷ 2 |
8 |
0 |
| 8 ÷ 2 |
4 |
0 |
| 4 ÷ 2 |
2 |
0 |
| 2 ÷ 2 |
1 |
0 |
| 1 ÷ 2 |
0 |
1 |
(261)10 = (100000101)2
Convert 2250 into binary number.
| Division |
Quotient |
Remainder |
| 2250 ÷ 2 |
1125 |
0 |
| 1125 ÷ 2 |
562 |
1 |
| 562 ÷ 2 |
281 |
0 |
| 281 ÷ 2 |
140 |
1 |
| 140 ÷ 2 |
70 |
0 |
| 70 ÷ 2 |
35 |
0 |
| 35 ÷ 2 |
17 |
1 |
| 17 ÷ 2 |
8 |
1 |
| 8 ÷ 2 |
4 |
0 |
| 4 ÷ 2 |
2 |
0 |
| 2 ÷ 2 |
1 |
0 |
| 1 ÷ 2 |
0 |
1 |
(2250)10 = (100011001010)2
Simplify the following in binary system:
{(10001101)2 x (235)10} - (27)10
First convert all binary numbers to decimal number system.
27 + 06 + 05 + 04 + 23 + 22 + 01 + 20
128+0+0+0+8+4+0+1
(141)10
(10001101)2 = (141)10
{(141)10 x (235)10} - (27)10
141 x 235 - 27
33135 - 27
33108
(33108)10
Convert (33108)10 to binary number.
| Division |
Quotient |
Remainder |
| 33108 ÷ 2 |
16554 |
0 |
| 16554 ÷ 2 |
8277 |
0 |
| 8277 ÷ 2 |
4138 |
1 |
| 4138 ÷ 2 |
2069 |
0 |
| 2069 ÷ 2 |
1034 |
1 |
| 1034 ÷ 2 |
517 |
0 |
| 517 ÷ 2 |
258 |
1 |
| 258 ÷ 2 |
129 |
0 |
| 129 ÷ 2 |
64 |
1 |
| 64 ÷ 2 |
32 |
0 |
| 32 ÷ 2 |
16 |
0 |
| 16 ÷ 2 |
8 |
0 |
| 8 ÷ 2 |
4 |
0 |
| 4 ÷ 2 |
2 |
0 |
| 2 ÷ 2 |
1 |
0 |
| 1 ÷ 2 |
0 |
1 |
(33108)10 = (1000000101010100)2
{(10001101)2 x (235)10} - (27)10 = (1000000101010100)2
Simplify into binary: {(681)10 ÷ (22)10} + (10111011)2
First convert all binary numbers to decimal number system.
27 + 06 + 25 + 24 + 23 + 02 + 21 + 20
128+0+32+16+8+0+2+1
(187)10
(10111011)2 = (187)10
{(681)10 ÷ (22)10} + (187)10
681 ÷ 22 + 187
(218)10
Convert (218)10 to binary number.
| Division |
Quotient |
Remainder |
| 218 ÷ 2 |
109 |
0 |
| 109 ÷ 2 |
54 |
1 |
| 54 ÷ 2 |
27 |
0 |
| 27 ÷ 2 |
13 |
1 |
| 13 ÷ 2 |
6 |
1 |
| 6 ÷ 2 |
3 |
0 |
| 3 ÷ 2 |
1 |
1 |
| 1 ÷ 2 |
0 |
1 |
(218)10 = (11011010)2
{(681)10 ÷ (22)10} + (10111011)2 = (11011010)2
Simplify this into binary number system. {(10111101)2 ÷ (133)10} x (10111)2
First convert binary to decimal.
(10111101)2 = (?)10
27 + 06 + 25 + 24 + 23 + 22 + 01 + 20
128+0+32+16+8+4+0+1
(189)10
(10111101)2 = (189)10
(10111)2 = (?)10
24 + 03 + 22 + 21 + 20
16+0+4+2+1
(23)10
(10111)2 = (23)10
{(189)10 ÷ (133)10} x (23)10
189 ÷ 133 x 23
Adding Binary Numbers Rule
Basic Binary Addition Rules
| Operation |
Sum |
Carry |
| 0 + 0 |
0 |
0 |
| 0 + 1 |
1 |
0 |
| 1 + 0 |
1 |
0 |
| 1 + 1 |
0 |
1 |
| 1 + 1 + 1 |
1 |
1 |
Q31. Add the binary numbers 11101 and 11011
1 1 1 0 1
+ 1 1 0 1 1
1 1 1 0 0 0
Add the binary numbers 00000 and 00000
0 0 0 0 0
+ 0 0 0 0 0
0 0 0 0 0
Add the binary numbers 100000 and 100000
1 0 0 0 0 0
+ 1 0 0 0 0 0
1 0 0 0 0 0 0
Add the binary numbers 1 and 0
Add the binary numbers 101 and 1101
1 0 1
+ 1 1 0 1
1 0 0 1 0
Add the binary numbers 101 and 1101
(101)2 = 22 + 01 + 20
(101)2 = 4 + 0 + 1
(101)2 = (5)10
(1101)2 = 23 + 22 + 01 + 20
(101)2 = 8 + 4 + 0 + 1
(101)2 = (13)10
Q31.The annual installment of the depreciation for a machine of value Rs. 7,00,000 with its scrap value Rs. 60,000 at the end of 20 years. Also calculate rate of depreciation and value of the asset after 10 years.
Depreciation Calculation for a Machine
Given:
Cost of machine \( C = 7,00,000 \)
Scrap value \( S = 60,000 \)
Useful life \( n = 20 \) years
Step 1: Annual Depreciation (Straight-Line Method)
\[
\text{Annual Depreciation} = \frac{C - S}{n} = \frac{700,000 - 60,000}{20} = 32,000
\]
Step 2: Rate of Depreciation
\[
\text{Rate of Depreciation} = \frac{\text{Annual Depreciation}}{C} \times 100 = \frac{32,000}{700,000} \times 100 \approx 4.57\% \text{ per year}
\]
Step 3: Value of Asset After 10 Years
\[
\text{Value after 10 years} = C - (\text{Annual Depreciation} \times 10) = 700,000 - (32,000 \times 10) = 380,000
\]
- Annual Depreciation: Rs. 32,000
- Rate of Depreciation: 4.57% per year
- Value of Asset after 10 years: Rs. 380,000
