AIOU 1349 Past Paper Spring 2025

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ALLAMA IQBAL OPEN UNIVERSITY

Level: Intermediate

Semester: Spring 2025

Course Code: Introduction to Business Mathematics (1349)

Maximum Marks: 100

Time Allowed: 03 Hours

Pass Marks: 40

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Note: Attempt any Five Questions. Each Question Carries Equal Marks.

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Q1(a). Calculate the continued ratio \(a:b:c\) when \(a:b=5:6\) and \(b:c=6:7\). ▶

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Step 1: Write down the given ratios
We are given two ratios: \(a:b = 5:6\) and \(b:c = 6:7\).

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Step 2: Express the ratios in fraction form
We can write the ratios as fractions: \(\frac{a}{b} = \frac{5}{6}\) and \(\frac{b}{c} = \frac{6}{7}\).

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Step 3: Make the common term equal
The term \(b\) is common in both ratios. Its value is already 6 in both ratios, so no adjustment is needed.

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Step 4: Write the continued ratio
Now we can combine the ratios directly: \(a:b:c = 5:6:7\).

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Step 5: Verification
Check the ratios: \(\frac{a}{b} = \frac{5}{6}\) and \(\frac{b}{c} = \frac{6}{7}\). Both are satisfied.

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5:6:7

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Q1(b). A manager earns Rs. 200,000 per month and spends Rs. 190,000. An assistant earns Rs. 40,000 and saves Rs. 8,000. Who saves more regarding their income? ▶

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Step 1: Write down the given information
The manager earns Rs. 200,000 and spends Rs. 190,000. The assistant earns Rs. 40,000 and saves Rs. 8,000.

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Step 2: Calculate the manager's savings
Savings is income minus expenditure. For the manager: \(200,000 - 190,000 = 10,000\). So the manager saves Rs. 10,000.

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Step 3: Calculate the savings percentage for the manager
The savings percentage is given by \(\frac{\text{savings}}{\text{income}} \times 100\). For the manager: \(\frac{10,000}{200,000} \times 100 = 5\%\).

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Step 4: Calculate the assistant's savings percentage
The assistant earns Rs. 40,000 and saves Rs. 8,000. Savings percentage: \(\frac{8,000}{40,000} \times 100 = 20\%\).

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Step 5: Compare savings percentages
The manager saves 5% of income, while the assistant saves 20% of income. Therefore, regarding income, the assistant saves more.

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The assistant saves more regarding their income

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Q2(a). Define discount and explain its types and calculate the discounted price of an item originally priced at Rs. 20,000 with a discount of Rs. 800. ▶

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Step 1: Define discount
A discount is a reduction in the original price of a product or service offered by the seller to the buyer. It is used to encourage purchases or reward customers.

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Step 2: Explain the types of discount
There are several types of discounts: 1. Trade discount: Given by manufacturers or wholesalers to retailers when buying in bulk. 2. Cash discount: Offered to buyers for early payment of bills. 3. Seasonal discount: Offered during certain seasons or festivals to boost sales. 4. Promotional discount: Temporary reduction in price to attract customers.

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Step 3: Write down the given values
The original price of the item is Rs. 20,000 and the discount given is Rs. 800.

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Step 4: Calculate the discounted price
The discounted price is calculated as: \[ \text{Discounted Price} = \text{Original Price} - \text{Discount} \] \[ \text{Discounted Price} = 20,000 - 800 \] \[ \text{Discounted Price} = 19,200 \]

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The discounted price is Rs.19,200/-

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Q2(b). If a person buys a book rack for Rs. 18,000 with a discount of 12%, what was the original price? ▶

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Step 1: Write down the given information
The person buys a book rack for Rs. 18,000 after a discount of 12%.

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Step 2: Let the original price be \(P\)
If the discount is 12%, the selling price is 88% of the original price. So we can write: \[ \text{Selling Price} = P \times \frac{100 - 12}{100} = P \times 0.88 \]

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Step 3: Substitute the selling price
The selling price is Rs. 18,000, so: \[ 18,000 = P \times 0.88 \]

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Step 4: Solve for \(P\)
\[ P = \frac{18,000}{0.88} = 20,454.55 \]

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The original price Rs.20,454.55

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Q2(b). If a person buys a book rack for Rs. 18,000 with a discount of 12%, what was the original price. ▶

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Find the Original Price
A person buys a book rack for Rs. 18000 after receiving a discount of 12 percent. We need to find the original price of the rack.

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Let the Original Price be x
Assume the original price of the rack is Rs. \( x \). Since a 12 percent discount is given, the person pays 88 percent of the original price.

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Equation for Discounted Price
The discounted price is equal to 88 percent of the original price. Therefore, we can write it as
\[ 0.88x = 18000 \]

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Solving for x
To find the original price, divide both sides by 0.88
\[ x = \frac{18000}{0.88} \]

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Calculating the Value
Performing the division gives
\[ x = 20454.545\ldots \]

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The original price of the book rack is approximately Rs. 20454.55 or rounded to Rs. 20455.

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Q3(a). If a person invests Rs. 5,000 at a simple interest rate of 10% per annum, how much will he have after 5 years? ▶

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Step 1: Write down the given information
Principal \(P = 5,000\), Rate \(R = 10\%\) per annum, and Time \(T = 5\) years.

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Step 2: Recall the formula for simple interest
The formula is \[ I = \frac{P \times R \times T}{100} \]

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Step 3: Substitute the given values
\[ I = \frac{5,000 \times 10 \times 5}{100} = 2,500 \]

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Step 4: Calculate the total amount
The total amount after 5 years is \[ A = P + I = 5,000 + 2,500 = 7,500 \]

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The total amount after 5 years Rs.7,500

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Q3(b). If a principal amount of Rs. 10,000 is borrowed at 5% p.a. for 3 years, what is the total interest paid? ▶

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Step 1: Write down the given information
Principal \(P = 10,000\), Rate \(R = 5\%\) per annum, and Time \(T = 3\) years.

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Step 2: Recall the formula for simple interest
The formula is \[ I = \frac{P \times R \times T}{100} \]

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Step 3: Substitute the given values
\[ I = \frac{10,000 \times 5 \times 3}{100} = 1,500 \]

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The total interest paid Rs.1,500

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Q4(a). Calculate the future value of an ordinary annuity of Rs. 1,500 payable at the end of each year for 12 years at an interest rate of 6%. ▶

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Step 1: Write down the given information
Payment per year \(R = 1,500\), Number of years \(n = 12\), and Interest rate \(i = 6\% = 0.06\).

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Step 2: Recall the formula for the future value of an ordinary annuity
\[ FV = R \times \frac{(1 + i)^{n} - 1}{i} \]

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Step 3: Substitute the given values
\[ FV = 1,500 \times \frac{(1 + 0.06)^{12} - 1}{0.06} \]

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Step 4: Simplify the expression
\[ FV = 1,500 \times \frac{(1.06)^{12} - 1}{0.06} \] \[ (1.06)^{12} = 2.012196 \] \[ FV = 1,500 \times \frac{2.012196 - 1}{0.06} = 1,500 \times 16.8699 = 25,304.85 \]

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The future value Rs.25,304.85

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Q4(b). What is the present value of an annuity due of Rs. 2,000 per year for 5 years at an interest rate of 5%? ▶

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Step 1: Write down the given information
Payment per year \(R = 2,000\), Number of years \(n = 5\), and Interest rate \(i = 5\% = 0.05\).

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Step 2: Recall the formula for the present value of an annuity due
\[ PV = R \times \frac{1 - (1 + i)^{-n}}{i} \times (1 + i) \]

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Step 3: Substitute the given values
\[ PV = 2,000 \times \frac{1 - (1 + 0.05)^{-5}}{0.05} \times (1 + 0.05) \]

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Step 4: Simplify the expression
\[ (1 + 0.05)^{-5} = \frac{1}{(1.05)^5} = 0.783526 \] \[ \frac{1 - 0.783526}{0.05} = 4.32948 \] \[ PV = 2,000 \times 4.32948 \times 1.05 = 2,000 \times 4.545954 = 9,091.91 \]

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The present value Rs.9,091.91

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Q5(a). Matrix Types: Differentiate between a scalar matrix and an identity matrix. Provide examples. ▶

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Definition of a Scalar Matrix
A scalar matrix is a square matrix in which all diagonal elements are equal and all non-diagonal elements are zero. It can be represented as \(A = kI\), where \(k\) is a scalar and \(I\) is the identity matrix.

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Example of a Scalar Matrix
\[ A = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix} \] Here, each diagonal element is 3 and all other elements are zero.

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Definition of an Identity Matrix
An identity matrix is a special type of square matrix where all diagonal elements are 1 and all non-diagonal elements are zero. It is denoted by \(I\).

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Example of an Identity Matrix
\[ I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]

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Difference Between Scalar and Identity Matrices
In a scalar matrix, all diagonal elements are equal to any constant \(k\), while in an identity matrix, the diagonal elements are always equal to 1. Every identity matrix is a scalar matrix with \(k = 1\), but not every scalar matrix is an identity matrix.

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Identity matrix is a special case of scalar matrix when k = 1

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Q5(b). What is the determinant of a \(2×2\) matrix? Find the determinant of \(\begin{bmatrix}2 & 3 \\ 1 & -5\end{bmatrix}\). Is this matrix singular? ▶

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Definition of Determinant of a 2×2 Matrix
For a 2×2 matrix \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the determinant is calculated as \[ \text{det}(A) = ad - bc \]

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Given Matrix
\[ A = \begin{bmatrix} 2 & 3 \\ 1 & -5 \end{bmatrix} \]

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Calculation of Determinant
\[ \text{det}(A) = (2 \times -5) - (3 \times 1) \] \[ \text{det}(A) = -10 - 3 = -13 \]

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Checking if the Matrix is Singular
A matrix is singular if its determinant equals zero. Since \(\text{det}(A) = -13 \neq 0\), the matrix is not singular.

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The determinant is -13 and the matrix is not singular

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Q6(a). Inverse Matrix: Find the inverse of \(\begin{bmatrix}5 & 9 \\ 4 & 1\end{bmatrix}\). ▶

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\[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] the inverse is given by \[ A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \] provided that \(ad - bc \neq 0\).

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Given Matrix
\[ A = \begin{bmatrix} 5 & 9 \\ 4 & 1 \end{bmatrix} \]

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Step 1: Calculate the determinant
\[ \text{det}(A) = (5 \times 1) - (9 \times 4) = 5 - 36 = -31 \]

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Step 2: Apply the formula for the inverse
\[ A^{-1} = \frac{1}{-31} \begin{bmatrix} 1 & -9 \\ -4 & 5 \end{bmatrix} \]

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Step 3: Simplify the result
\[ A^{-1} = \begin{bmatrix} -\frac{1}{31} & \frac{9}{31} \\ \frac{4}{31} & -\frac{5}{31} \end{bmatrix} \]

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The inverse matrix is
\[ A^{-1} = \begin{bmatrix} -\frac{1}{31} & \frac{9}{31} \\ \frac{4}{31} & -\frac{5}{31} \end{bmatrix} \]

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Q6(b). Conversion: Convert (101101)₂ to decimal. ▶

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Definition of Binary to Decimal Conversion
To convert a binary number to decimal, multiply each bit by \(2\) raised to the power of its position (counting from right to left, starting at 0) and then add all the results.

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Given Binary Number
\( (101101)_2 \)

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Step 1: Write each bit with its positional value
\[ (101101)_2 = (1 \times 2^5) + (0 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) \]

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Step 2: Simplify each term
\[ (101101)_2 = (32) + (0) + (8) + (4) + (0) + (1) \]

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Step 3: Add them together
\[ 32 + 0 + 8 + 4 + 0 + 1 = 45 \]

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The decimal equivalent is 45

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Q7(a). Find the transpose of the matrix \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \). ▶

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The transpose of a matrix is obtained by interchanging its rows and columns. If \[ A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \] then its transpose is \[ A^{T} = \begin{bmatrix} a & c \\ b & d \end{bmatrix} \]

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Given Matrix
\[ A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \]

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Transpose of A
Interchanging rows and columns gives \[ A^{T} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \]

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The transpose of matrix A is
\[ A^{T} = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \]

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Q7(b). Calculate the cofactor of the element in the first row and first column of the matrix B= |5 7; 8 6|. ▶

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The cofactor of an element in a matrix is found by multiplying the determinant of its minor by \((-1)^{i+j}\), where \(i\) and \(j\) represent the row and column positions of the element.

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Given Matrix
\[ B = \begin{bmatrix} 5 & 7 \\ 8 & 6 \end{bmatrix} \]

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Step 1: Identify the element
The element in the first row and first column is \(5\) (that is \(b_{11}\)).

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Step 2: Find the minor of the element
Remove the first row and first column. The remaining element is \(6\). So, the minor \(M_{11} = 6\).

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Step 3: Apply the cofactor formula
\[ C_{11} = (-1)^{1+1} \times M_{11} = (+1) \times 6 = 6 \]

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The cofactor of the element in the first row and first column is 6

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Q8(a). Use the quadratic formula to solve: \( 2x^2 - 4x + 1 = 0 \) ▶

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For a quadratic equation of the form \(ax^2 + bx + c = 0\), the quadratic formula is \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

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Given Equation
\[ 2x^2 - 4x + 1 = 0 \] Here, \(a = 2\), \(b = -4\), and \(c = 1\).

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Step 1: Substitute the values into the formula
\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(1)}}{2(2)} \]

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Step 2: Simplify the expression
\[ x = \frac{4 \pm \sqrt{16 - 8}}{4} \] \[ x = \frac{4 \pm \sqrt{8}}{4} \]

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Step 3: Simplify the square root
\[ \sqrt{8} = 2\sqrt{2} \] \[ x = \frac{4 \pm 2\sqrt{2}}{4} \]

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Step 4: Simplify the fraction
\[ x = 1 \pm \frac{\sqrt{2}}{2} \]

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The roots are \(x = 1 + \frac{\sqrt{2}}{2}\) and \(x = 1 - \frac{\sqrt{2}}{2}\)

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Q8(b). Solve the simultaneous equations:
\( 3x + 4y = 24 \)
\( 2x - 3y = 15 \) ▶

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Given Equations
\[ 3x + 4y = 24 \] \[ 2x - 3y = 15 \]

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Step 1: Eliminate one variable
To eliminate \(x\), multiply the first equation by 2 and the second by 3 to make the coefficients of \(x\) equal. \[ (3x + 4y) \times 2 \Rightarrow 6x + 8y = 48 \] \[ (2x - 3y) \times 3 \Rightarrow 6x - 9y = 45 \]

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Step 2: Subtract the second equation from the first
\[ (6x + 8y) - (6x - 9y) = 48 - 45 \] \[ 6x + 8y - 6x + 9y = 3 \] \[ 17y = 3 \] \[ y = \frac{3}{17} \]

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Step 3: Substitute the value of \(y\) into the first equation
\[ 3x + 4\left(\frac{3}{17}\right) = 24 \] \[ 3x + \frac{12}{17} = 24 \] \[ 3x = 24 - \frac{12}{17} = \frac{408 - 12}{17} = \frac{396}{17} \] \[ x = \frac{396}{51} = \frac{132}{17} \]

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The solution is \(x = \frac{132}{17}\) and \(y = \frac{3}{17}\)

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