MTH 201

Business Mathematics - I

TU BBA · Semester 1 · BBA curriculum effective from 2021

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Chapter-wise questions

63 past questions across 6 units

Open a chapter to study questions grouped by unit and syllabus topic, with step-by-step solutions.

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Unit 1: Straight lines and Functions

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  1. Asked on 2024 Exam[2 marks]2024 paper

    Find the equation of a straight line which passes through the points (2, 5) and (3, 1).

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    Step-by-Step Solution:

    Given two points:

    (x1,y1)=(2,5)and(x2,y2)=(3,1)(x_1, y_1) = (2, 5) \quad \text{and} \quad (x_2, y_2) = (3, 1)

    Step 1: Calculate the slope (mm)

    m=y2−y1x2−x1=1−53−2=−41=−4m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 5}{3 - 2} = \frac{-4}{1} = -4

    Step 2: Write the equation using point-slope form

    y−y1=m(x−x1)y - y_1 = m(x - x_1)
    y−5=−4(x−2)y - 5 = -4(x - 2)
    y−5=−4x+8y - 5 = -4x + 8

    Rearranging into standard form:

    4x+y−5−8=04x + y - 5 - 8 = 0
    4x+y−13=04x + y - 13 = 0

    Final Answer: The equation of the straight line is 4x+y−13=04x + y - 13 = 0 (or y=−4x+13y = -4x + 13).

  2. Asked on 2024 Exam[2 marks]2024 paper

    Calculate the price elasticity of demand in the demand function P = 40 - 0.2Q when P=20.

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    Step-by-Step Solution:

    Given demand function:

    P=40−0.2QP = 40 - 0.2Q

    Step 1: Find quantity (QQ) when P=20P = 2020=40−0.2Q20 = 40 - 0.2Q$

    0.2Q=40−20=200.2Q = 40 - 20 = 20
    Q=200.2=100 unitsQ = \frac{20}{0.2} = 100 \text{ units}

    Step 2: Differentiate with respect to PP

    From P=40−0.2QP = 40 - 0.2Q:

    dPdQ=−0.2\frac{dP}{dQ} = -0.2
    Taking the reciprocal:
    dQdP=1dPdQ=1−0.2=−5\frac{dQ}{dP} = \frac{1}{\frac{dP}{dQ}} = \frac{1}{-0.2} = -5

    Step 3: Compute point price elasticity of demand (ϵd\epsilon_d)

    ϵd=−PQ⋅dQdP\epsilon_d = -\frac{P}{Q} \cdot \frac{dQ}{dP}
    ϵd=−20100×(−5)=−0.2×(−5)=1.0\epsilon_d = -\frac{20}{100} \times (-5) = -0.2 \times (-5) = 1.0

    In absolute magnitude:

    ∣ϵd∣=1.0|\epsilon_d| = 1.0

    Final Answer: The price elasticity of demand is ∣ϵd∣=1.0|\epsilon_d| = 1.0 (unitary elastic demand).

  3. Asked on 2024 Exam[2 marks]2024 paper

    Form a quadratic equation whose roots are 555\sqrt{5} and −5-\sqrt{5}

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    Step-by-Step Solution:

    Let the given roots be:

    α=55andβ=−5\alpha = 5\sqrt{5} \quad \text{and} \quad \beta = -\sqrt{5}

    Step 1: Calculate the sum of the roots (SS)

    S=α+β=55+(−5)=45S = \alpha + \beta = 5\sqrt{5} + (-\sqrt{5}) = 4\sqrt{5}

    Step 2: Calculate the product of the roots (PP)

    P=α⋅β=(55)(−5)=−5×(5)2=−5×5=−25P = \alpha \cdot \beta = (5\sqrt{5})(-\sqrt{5}) = -5 \times (\sqrt{5})^2 = -5 \times 5 = -25

    Step 3: Form the quadratic equation

    A quadratic equation with sum of roots SS and product of roots PP is given by:

    x2−Sx+P=0x^2 - Sx + P = 0

    Substituting S=45S = 4\sqrt{5} and P=−25P = -25:

    x2−45x−25=0x^2 - 4\sqrt{5}x - 25 = 0

    Final Answer: The quadratic equation is x2−45x−25=0x^2 - 4\sqrt{5}x - 25 = 0.

  4. Asked on 2024 Exam[2 marks]2024 paper

    Solve the equation: 2x−2×8x2^{x-2} \times 8^x = 16 + 5

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    Step-by-Step Solution:

    Given equation:

    2x−2×8x=16+52^{x-2} \times 8^x = 16 + 5

    Step 1: Simplify both sides

    Left-hand side:

    8x=(23)x=23x8^x = (2^3)^x = 2^{3x}
    2x−2×23x=2(x−2)+3x=24x−22^{x-2} \times 2^{3x} = 2^{(x-2) + 3x} = 2^{4x - 2}

    Right-hand side:

    16+5=2116 + 5 = 21

    Thus:

    24x−2=212^{4x - 2} = 21

    Step 2: Solve for xx using logarithms

    Take natural logarithms on both sides:

    ln⁡(24x−2)=ln⁡(21)\ln(2^{4x - 2}) = \ln(21)
    (4x−2)ln⁡(2)=ln⁡(21)(4x - 2) \ln(2) = \ln(21)
    4x−2=ln⁡(21)ln⁡(2)=log⁡2(21)4x - 2 = \frac{\ln(21)}{\ln(2)} = \log_2(21)

    Evaluating the numerical values:

    ln⁡(21)≈3.044522\ln(21) \approx 3.044522
    ln⁡(2)≈0.693147\ln(2) \approx 0.693147
    ln⁡(21)ln⁡(2)≈4.392317\frac{\ln(21)}{\ln(2)} \approx 4.392317

    Now solve for xx:

    4x=4.392317+2=6.3923174x = 4.392317 + 2 = 6.392317
    x=6.3923174≈1.598x = \frac{6.392317}{4} \approx 1.598

    (Note: If the right side was a misprint for 16=2416 = 2^4, then 4x−2=4  ⟹  x=1.54x - 2 = 4 \implies x = 1.5; if 16×2=32=2516 \times 2 = 32 = 2^5, 4x−2=5  ⟹  x=1.754x - 2 = 5 \implies x = 1.75. As printed, the exact mathematical solution is x=2+log⁡2(21)4x = \frac{2 + \log_2(21)}{4}).

    Final Answer: x=2+log⁡2(21)4≈1.598x = \frac{2 + \log_2(21)}{4} \approx 1.598.

  5. Asked on 2024 Exam[5 marks]2024 paper

    Solve the following system of linear equations: x+2y+3z=13; 2x + 4y + z = 11 ; 3x+2y+2z=14.

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    Step-by-Step Solution:

    Given system of linear equations:

    x+2y+3z=13— (1)x + 2y + 3z = 13 \quad \text{--- (1)}
    2x+4y+z=11— (2)2x + 4y + z = 11 \quad \text{--- (2)}
    3x+2y+2z=14— (3)3x + 2y + 2z = 14 \quad \text{--- (3)}

    Step 1: Eliminate xx and yy using equations (1) and (2)

    Multiply equation (1) by 2:

    2(x+2y+3z)=2(13)2(x + 2y + 3z) = 2(13)
    2x+4y+6z=26— (4)2x + 4y + 6z = 26 \quad \text{--- (4)}

    Now subtract equation (2) from equation (4):

    (2x+4y+6z)−(2x+4y+z)=26−11(2x + 4y + 6z) - (2x + 4y + z) = 26 - 11
    5z=155z = 15
    z=155=3z = \frac{15}{5} = 3

    Step 2: Substitute z=3z = 3 into the system

    From equation (2):

    2x+4y+3=11  ⟹  2x+4y=8  ⟹  x+2y=4  ⟹  x=4−2y— (5)2x + 4y + 3 = 11 \implies 2x + 4y = 8 \implies x + 2y = 4 \implies x = 4 - 2y \quad \text{--- (5)}

    From equation (3):

    3x+2y+2(3)=143x + 2y + 2(3) = 14
    3x+2y+6=14  ⟹  3x+2y=8— (6)3x + 2y + 6 = 14 \implies 3x + 2y = 8 \quad \text{--- (6)}

    Step 3: Solve for yy and xx

    Substitute equation (5) into equation (6):

    3(4−2y)+2y=83(4 - 2y) + 2y = 8
    12−6y+2y=812 - 6y + 2y = 8
    12−4y=812 - 4y = 8
    −4y=8−12=−4  ⟹  y=1-4y = 8 - 12 = -4 \implies y = 1

    Now find xx:

    x=4−2(1)=4−2=2x = 4 - 2(1) = 4 - 2 = 2

    Step 4: Verification

    • Eq (1): 2+2(1)+3(3)=2+2+9=132 + 2(1) + 3(3) = 2 + 2 + 9 = 13 (Matches)
    • Eq (2): 2(2)+4(1)+3=4+4+3=112(2) + 4(1) + 3 = 4 + 4 + 3 = 11 (Matches)
    • Eq (3): 3(2)+2(1)+2(3)=6+2+6=143(2) + 2(1) + 2(3) = 6 + 2 + 6 = 14 (Matches)

    Final Answer: The solution is x=2x = 2, y=1y = 1, z=3z = 3.

  6. Asked on 2024 Exam[10 marks]2024 paper

    The following table shows the yearly income of a company:

    Year 2017 2018 2019 2020 2021 2022 2023
    Income (Rs millions) 52 54 61 59 62 60 65

    Obtain the equation of line by least squares method. Also, estimate the income of the company for the years 2024 and 2026.

    question_19: | In an economy which engages in foreign trade, it is assumed that Y = C + I + G + X - M, where C=C0+0.8YdC = C_0 + 0.8Y_d, C0=90mC_0 = 90\text{m}, I=500mI = 500\text{m}, G=700mG = 700\text{m}, X=250mX = 250\text{m}, M=0.3YdM = 0.3Y_d, t=0.4t = 0.4, Yd=Y−TY_d = Y - T, T = tY.

    Find the expenditure equation and hence find the equilibrium level of national income and consumption. Also, calculate the total tax.

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    Step-by-Step Solution:

    This question contains two comprehensive analytical components:


    Part 1: Least Squares Method for Company Yearly Income

    Given Data:

    Number of years (NN) = 77 (odd). Let the middle year 20202020 be taken as the origin, defining transformed variable u=Year−2020u = \text{Year} - 2020.

    Year u=Year−2020u = \text{Year} - 2020 Income yy (Rs million) u2u^2 uyuy
    2017 -3 52 9 -156
    2018 -2 54 4 -108
    2019 -1 61 1 -61
    2020 0 59 0 0
    2021 1 62 1 62
    2022 2 60 4 120
    2023 3 65 9 195
    Total ∑u=0\sum u = 0 ∑y=413\sum y = 413 ∑u2=28\sum u^2 = 28 ∑uy=52\sum uy = 52

    Step 1: Determine Trend Line Coefficients

    Since ∑u=0\sum u = 0:

    a=∑yN=4137=59a = \frac{\sum y}{N} = \frac{413}{7} = 59
    b=∑uy∑u2=5228=137≈1.85714b = \frac{\sum uy}{\sum u^2} = \frac{52}{28} = \frac{13}{7} \approx 1.85714

    The trend equation is:

    y=59+1.8571u(where u=Year−2020)y = 59 + 1.8571 u \quad (\text{where } u = \text{Year} - 2020)

    Step 2: Estimate Income for 2024 and 2026

    1. For Year 2024: u=2024−2020=4u = 2024 - 2020 = 4y2024=59+(137)(4)=59+527=59+7.4286=66.43 million Rsy_{2024} = 59 + \left(\frac{13}{7}\right)(4) = 59 + \frac{52}{7} = 59 + 7.4286 = \mathbf{66.43\text{ million Rs}}$

    2. For Year 2026: u=2026−2020=6u = 2026 - 2020 = 6y2026=59+(137)(6)=59+787=59+11.1429=70.14 million Rsy_{2026} = 59 + \left(\frac{13}{7}\right)(6) = 59 + \frac{78}{7} = 59 + 11.1429 = \mathbf{70.14\text{ million Rs}}$


    Part 2: Open-Economy Keynesian Macroeconomic Model

    Given:

    • National Income Identity: Y=C+I+G+X−MY = C + I + G + X - M
    • Consumption function: C=C0+0.8Yd=90+0.8YdC = C_0 + 0.8Y_d = 90 + 0.8Y_d
    • Investment: I=500mI = 500\text{m}
    • Government expenditure: G=700mG = 700\text{m}
    • Exports: X=250mX = 250\text{m}
    • Import function: M=0.3YdM = 0.3Y_d
    • Tax rate: t=0.4t = 0.4, with T=tY=0.4YT = tY = 0.4Y
    • Disposable Income: Yd=Y−T=Y−0.4Y=0.6YY_d = Y - T = Y - 0.4Y = 0.6Y

    Step 1: Derive the Aggregate Expenditure Equation (EE)

    Substitute Yd=0.6YY_d = 0.6Y into CC and MM:

    C=90+0.8(0.6Y)=90+0.48YC = 90 + 0.8(0.6Y) = 90 + 0.48Y
    M=0.3(0.6Y)=0.18YM = 0.3(0.6Y) = 0.18Y

    Aggregate Expenditure:

    E=C+I+G+X−ME = C + I + G + X - M
    E=(90+0.48Y)+500+700+250−0.18YE = (90 + 0.48Y) + 500 + 700 + 250 - 0.18Y
    E=(90+500+700+250)+(0.48−0.18)YE = (90 + 500 + 700 + 250) + (0.48 - 0.18)Y
    E=1540+0.30Y\mathbf{E = 1540 + 0.30Y}

    Step 2: Determine Equilibrium National Income (Y∗Y^*)

    Set Y=EY = E:

    Y=1540+0.30YY = 1540 + 0.30Y
    Y−0.30Y=1540Y - 0.30Y = 1540
    0.70Y=15400.70Y = 1540
    Y∗=15400.70=2,200 million RsY^* = \frac{1540}{0.70} = \mathbf{2,200\text{ million Rs}}

    Step 3: Determine Equilibrium Consumption (C∗C^*) and Total Tax (T∗T^*)

    • Consumption (C∗C^*):
      C∗=90+0.48(2,200)=90+1,056=1,146 million RsC^* = 90 + 0.48(2,200) = 90 + 1,056 = \mathbf{1,146\text{ million Rs}}
    • Total Tax (T∗T^*):
      T∗=0.4(2,200)=880 million RsT^* = 0.4(2,200) = \mathbf{880\text{ million Rs}}

    Final Answer:

    • Least Squares Trend Line: y=59+1.8571(Year−2020)y = 59 + 1.8571(\text{Year} - 2020)
      • 2024 Income: Rs 66.43 million
      • 2026 Income: Rs 70.14 million
    • Expenditure Equation: E=1540+0.30YE = 1540 + 0.30Y
    • Equilibrium Income (Y∗Y^*): Rs 2,200 million
    • Equilibrium Consumption (C∗C^*): Rs 1,146 million
    • Total Tax (T∗T^*): Rs 880 million
  7. Asked on 2025 Exam[2 marks]2025 paper

    Find the equation of straight line which passes through the points (3, 6) and (-5, 2).

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    Step-by-Step Solution:

    Given two points:

    (x1,y1)=(3,6)and(x2,y2)=(−5,2)(x_1, y_1) = (3, 6) \quad \text{and} \quad (x_2, y_2) = (-5, 2)

    Step 1: Compute the slope (mm)

    m=y2−y1x2−x1=2−6−5−3=−4−8=12m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 6}{-5 - 3} = \frac{-4}{-8} = \frac{1}{2}

    Step 2: Use point-slope form

    y−y1=m(x−x1)y - y_1 = m(x - x_1)

    Substitute (x1,y1)=(3,6)(x_1, y_1) = (3, 6) and m=12m = \frac{1}{2}:

    y−6=12(x−3)y - 6 = \frac{1}{2}(x - 3)
    2(y−6)=x−32(y - 6) = x - 3
    2y−12=x−32y - 12 = x - 3

    Rearrange into standard form Ax+By+C=0Ax + By + C = 0:

    x−2y−3+12=0x - 2y - 3 + 12 = 0
    x−2y+9=0x - 2y + 9 = 0

    Final Answer: The equation of the straight line is x−2y+9=0x - 2y + 9 = 0 (or y=12x+92y = \frac{1}{2}x + \frac{9}{2}).

  8. Asked on 2025 Exam[2 marks]2025 paper

    A manufacturing company has an investment of Rs 90,000 to purchase two raw materials X and Y whose prices are Rs 1,500 and Rs 1,800 each respectively. Write down the slope of budget constraint.

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    Step-by-Step Solution:

    Given:

    • Total investment / budget (MM) = Rs 90,000\text{Rs } 90,000
    • Unit price of material XX (PXP_X) = Rs 1,500\text{Rs } 1,500
    • Unit price of material YY (PYP_Y) = Rs 1,800\text{Rs } 1,800

    Step 1: Formulate the Budget Constraint Equation

    PX⋅X+PY⋅Y=MP_X \cdot X + P_Y \cdot Y = M
    1,500X+1,800Y=90,0001,500X + 1,800Y = 90,000

    Step 2: Determine the Slope

    Expressing YY as a function of XX (YY on the vertical axis, XX on the horizontal axis):

    1,800Y=90,000−1,500X1,800Y = 90,000 - 1,500X
    Y=90,0001,800−1,5001,800XY = \frac{90,000}{1,800} - \frac{1,500}{1,800}X
    Y=50−56XY = 50 - \frac{5}{6}X

    The slope of the budget constraint line is:

    Slope=−PXPY=−1,5001,800=−56≈−0.8333\text{Slope} = -\frac{P_X}{P_Y} = -\frac{1,500}{1,800} = -\frac{5}{6} \approx -0.8333

    Final Answer: The slope of the budget constraint is −56-\frac{5}{6} (or approximately −0.833-0.833).

  9. Asked on 2025 Exam[10 marks]2025 paper

    The following table shows the yearly profit of a company [in million Rs].

    Year 2008 2009 2010 2011 2012 2013 2014
    Profit 150 180 170 190 160 170 210

    Obtain the equation of trend line by least squares method. Also, estimate the profit of the company for the years 2015.

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    Step-by-Step Solution:

    Given Data:

    Number of years (NN) = 77 (odd). Let the origin be the middle year 20112011, defining transformed time variable u=Year−2011u = \text{Year} - 2011.

    Year u=Year−2011u = \text{Year} - 2011 Profit yy (million Rs) u2u^2 uyuy
    2008 -3 150 9 -450
    2009 -2 180 4 -360
    2010 -1 170 1 -170
    2011 0 190 0 0
    2012 1 160 1 160
    2013 2 170 4 340
    2014 3 210 9 630
    Total ∑u=0\sum u = 0 ∑y=1,230\sum y = 1,230 ∑u2=28\sum u^2 = 28 ∑uy=150\sum uy = 150

    Step 1: Compute Trend Line Coefficients

    The trend equation is y=a+buy = a + bu. Since ∑u=0\sum u = 0:

    a=∑yN=1,2307≈175.7143a = \frac{\sum y}{N} = \frac{1,230}{7} \approx 175.7143
    b=∑uy∑u2=15028=7514≈5.3571b = \frac{\sum uy}{\sum u^2} = \frac{150}{28} = \frac{75}{14} \approx 5.3571

    The least squares trend line is:

    y=175.7143+5.3571u(where u=Year−2011)y = 175.7143 + 5.3571 u \quad (\text{where } u = \text{Year} - 2011)


    Step 2: Estimate Profit for the Year 2015

    For Year 2015:

    u=2015−2011=4u = 2015 - 2011 = 4
    y2015=1,2307+(7514)(4)=1,2307+1507=1,3807≈197.14y_{2015} = \frac{1,230}{7} + \left(\frac{75}{14}\right)(4) = \frac{1,230}{7} + \frac{150}{7} = \frac{1,380}{7} \approx 197.14

    Final Answer:

    • Equation of trend line: y=175.7143+5.3571(Year−2011)y = 175.7143 + 5.3571(\text{Year} - 2011)
    • Estimated profit for 2015: Rs 197.14 million
  10. Asked on 2023 Exam[2 marks]2023 paper

    Find the equation of straight line passing through the points (8, -3) and (-7, 2).

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    Step-by-Step Solution:

    Given the two points:

    (x1,y1)=(8,−3)and(x2,y2)=(−7,2)(x_1, y_1) = (8, -3) \quad \text{and} \quad (x_2, y_2) = (-7, 2)

    Step 1: Calculate the slope (mm) of the line

    Using the two-point slope formula:

    m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}
    m=2−(−3)−7−8=2+3−15=5−15=−13m = \frac{2 - (-3)}{-7 - 8} = \frac{2 + 3}{-15} = \frac{5}{-15} = -\frac{1}{3}

    Step 2: Use the point-slope form to find the equation

    y−y1=m(x−x1)y - y_1 = m(x - x_1)

    Substitute (x1,y1)=(8,−3)(x_1, y_1) = (8, -3) and m=−13m = -\frac{1}{3}:

    y−(−3)=−13(x−8)y - (-3) = -\frac{1}{3}(x - 8)
    y+3=−13(x−8)y + 3 = -\frac{1}{3}(x - 8)

    Multiply both sides by 33:

    3(y+3)=−(x−8)3(y + 3) = -(x - 8)
    3y+9=−x+83y + 9 = -x + 8

    Rearranging into the standard linear equation form Ax+By+C=0Ax + By + C = 0:

    x+3y+9−8=0x + 3y + 9 - 8 = 0
    x+3y+1=0x + 3y + 1 = 0

    Final Answer: The equation of the straight line is x+3y+1=0x + 3y + 1 = 0 (or in slope-intercept form, y=−13x−13y = -\frac{1}{3}x - \frac{1}{3}).

  11. Asked on 2023 Exam[2 marks]2023 paper

    If f(x)=7x2−3x+9thenfindf(3)f(x) = 7x^2 - 3x + 9 then find f(3)

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    Step-by-Step Solution:

    Given function:

    f(x)=7x2−3x+9f(x) = 7x^2 - 3x + 9

    To find the value of f(3)f(3), substitute x=3x = 3 into the function:

    f(3)=7(3)2−3(3)+9f(3) = 7(3)^2 - 3(3) + 9
    f(3)=7(9)−9+9f(3) = 7(9) - 9 + 9
    f(3)=63−9+9f(3) = 63 - 9 + 9
    f(3)=63f(3) = 63

    Final Answer: f(3)=63f(3) = 63.

  12. Asked on 2023 Exam[10 marks]2023 paper

    The following table shows the year of service and monthly income (thousand Rs) of the workers in a factory:

    year 5 8 7 9 11 10 12
    Income (000 Rs) 6 9 8 10 12 11 14

    Obtain the equation of straight line by least square method. Also estimate the income of the workers who have served for 13 years and 15 years.

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    Step-by-Step Solution:

    Given data:

    • Independent variable (xx) = Year of service
    • Dependent variable (yy) = Monthly income (in thousands Rs)
    • Number of observations (NN) = 77
    Worker Year of Service (xx) Income (yy) x2x^2 xyxy
    1 5 6 25 30
    2 8 9 64 72
    3 7 8 49 56
    4 9 10 81 90
    5 11 12 121 132
    6 10 11 100 110
    7 12 14 144 168
    Total ∑x=62\sum x = 62 ∑y=70\sum y = 70 ∑x2=584\sum x^2 = 584 ∑xy=658\sum xy = 658

    Step 1: Compute the means xˉ\bar{x} and yˉ\bar{y}xˉ=∑xN=627≈8.8571\bar{x} = \frac{\sum x}{N} = \frac{62}{7} \approx 8.8571$

    yˉ=∑yN=707=10\bar{y} = \frac{\sum y}{N} = \frac{70}{7} = 10

    Step 2: Compute regression slope (bb) and intercept (aa)

    The least squares regression line is y=a+bxy = a + bx.

    b=N∑xy−(∑x)(∑y)N∑x2−(∑x)2b = \frac{N \sum xy - (\sum x)(\sum y)}{N \sum x^2 - (\sum x)^2}
    b=7(658)−(62)(70)7(584)−(62)2=4,606−4,3404,088−3,844=266244=133122≈1.09016b = \frac{7(658) - (62)(70)}{7(584) - (62)^2} = \frac{4,606 - 4,340}{4,088 - 3,844} = \frac{266}{244} = \frac{133}{122} \approx 1.09016

    Now find intercept aa:

    a=yˉ−bxˉ=10−(133122)(627)=10−1,178122=1,220−1,178122=42122=2161≈0.34426a = \bar{y} - b\bar{x} = 10 - \left(\frac{133}{122}\right)\left(\frac{62}{7}\right) = 10 - \frac{1,178}{122} = \frac{1,220 - 1,178}{122} = \frac{42}{122} = \frac{21}{61} \approx 0.34426

    Thus, the least squares trend line is:

    y=0.3443+1.0902xy = 0.3443 + 1.0902 x


    Step 3: Estimate monthly income for 13 years and 15 years

    1. For x=13x = 13 years:

      y13=0.34426+1.09016(13)=0.34426+14.17208=14.51634 (thousand Rs)y_{13} = 0.34426 + 1.09016(13) = 0.34426 + 14.17208 = 14.51634 \text{ (thousand Rs)}
      Estimated Income≈Rs 14,516.34\text{Estimated Income} \approx \text{Rs } 14,516.34

    2. For x=15x = 15 years:

      y15=0.34426+1.09016(15)=0.34426+16.3524=16.69666 (thousand Rs)y_{15} = 0.34426 + 1.09016(15) = 0.34426 + 16.3524 = 16.69666 \text{ (thousand Rs)}
      Estimated Income≈Rs 16,696.66\text{Estimated Income} \approx \text{Rs } 16,696.66

    Final Answer:

    • Equation of straight line: y=0.3443+1.0902xy = 0.3443 + 1.0902 x
    • Estimated income for 13 years: Rs 14,516.34
    • Estimated income for 15 years: Rs 16,696.66