IB Mathematics SL 1.8 Systems of linear equations AI HL Paper 2- Exam Style Questions- New Syllabus

Question

The following table shows the number of bicycles, \( x \), produced daily by a factory and their total production cost, \( y \), in US dollars (USD). The table shows data recorded over seven days.
\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Day} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline x \text{ (bicycles)} & 12 & 15 & 14 & 17 & 20 & 18 & 21 \\ \hline y \text{ (USD)} & 3900 & 4600 & 4100 & 5300 & 6000 & 5400 & 6000 \\ \hline \end{array} \]
a. (i) Write down the Pearson’s product–moment correlation coefficient, \( r \), for these data.
    (ii) Hence comment on the result [4].
b. Write down the equation of the regression line \( y \) on \( x \) for these data, in the form \( y = ax + b \) [2].
c. Estimate the total cost, to the nearest USD, of producing 13 bicycles on a particular day [3].
d. All the bicycles that are produced are sold. The bicycles are sold for 304 USD each. Explain why the factory does not make a profit when producing 13 bicycles on a particular day [2].
e. All the bicycles that are produced are sold. The bicycles are sold for 304 USD each.
    (i) Write down an expression for the total selling price of \( x \) bicycles.
    (ii) Write down an expression for the profit the factory makes when producing \( x \) bicycles on a particular day.
    (iii) Find the least number of bicycles that the factory should produce, on a particular day, in order to make a profit [5].

▶️ Answer/Explanation

Markscheme

a(i)
Method 1: Direct Calculation
Compute: \( \sum x = 117 \), \( \sum y = 34900 \), \( \sum xy = 606500 \), \( \sum x^2 = 2019 \), \( \sum y^2 = 182430000 \)
\( r = \frac{7 \times 606500 – 117 \times 34900}{\sqrt{[7 \times 2019 – 117^2][7 \times 182430000 – 34900^2]}} \approx \frac{162200}{\sqrt{444 \times 5900000}} \approx 0.985 \)
Method 2: Statistical Tool
Input data confirms \( r \approx 0.985 \)
Result: \( r = 0.985 \) [2].

a(ii)
\( r = 0.985 \) is close to 1, indicating a strong positive linear relationship
As number of bicycles increases, cost increases almost linearly
Result: Strong, positive correlation [2].

b
Slope: \( a = \frac{7 \times 606500 – 117 \times 34900}{7 \times 2019 – 117^2} \approx 259.909 \)
Intercept: \( b = \frac{34900 – 259.909 \times 117}{7} \approx 698.664 \)
Rounded: \( y = 260x + 699 \)
Result: \( y = 260x + 699 \) [2].

c
Method 1: Use Regression Line
Unrounded: \( y = 259.909 \times 13 + 698.664 \approx 4077.481 \approx 4077 \)
Rounded: \( y = 260 \times 13 + 699 = 4079 \)
Prefer unrounded: 4077
Method 2: Interpolation
Between \( (12, 3900) \), \( (14, 4100) \): \( y \approx 3900 + \frac{4100 – 3900}{14 – 12} \times (13 – 12) = 4000 \)
Regression more precise: 4077
Result: 4077 USD [3].

d
Method 1: Profit Calculation
Revenue: \( 13 \times 304 = 3952 \)
Cost: 4077
Profit: \( 3952 – 4077 = -125 \)
Negative profit indicates cost exceeds revenue
Method 2: Per Bicycle Cost
Cost per bicycle: \( \frac{4077}{13} \approx 313.62 \)
Selling price: 304
\( 313.62 > 304 \), cost exceeds selling price
Result: No profit because cost (4077 USD) exceeds revenue (3952 USD) [2].

e(i)
Selling price: \( 304 \times x \)
Result: \( 304x \) [1].

e(ii)
Profit: Revenue – Cost = \( 304x – (260x + 699) \)
Result: \( 304x – (260x + 699) \) [2].

e(iii)
Method 1: Solve Inequality
\( 304x – (260x + 699) > 0 \)
\( 44x – 699 > 0 \implies x > \frac{699}{44} \approx 15.886 \)
Least integer: \( x = 16 \)
Method 2: Test Values
For \( x = 15 \): Profit = \( 304 \times 15 – (260 \times 15 + 699) = -39 \)
For \( x = 16 \): Profit = \( 304 \times 16 – (260 \times 16 + 699) = 5 \)
Result: 16 bicycles [2].

IB Mathematics SL 1.8 Systems of linear equations AI HL Paper 2- Exam Style Questions- New Syllabus

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