Question 1 (a)
Sales relation S = a + bX + cY by determinants (Cramer's rule)
6 marks
- Time to write
- About 12 minutes
Question
Question 1 (a) · 6 marksA firm sells two products X and Y. Its total sales (S), in thousands of rupees, obey the relation S = a + bX + cY. Data for the first three months are given below:
| Month | Total Sales | X | Y |
|---|---|---|---|
| 1 | 17 | 1 | 3 |
| 2 | 12 | 2 | 1 |
| 3 | 29 | 5 | 3 |
Using the determinant method, find the sales in the month when it sells 12 units of X and 15 units of Y.
Reworded from the university paper: names and figures differ, the task is the same. The original paper belongs to the University of Delhi. See DU's question bank.
Before you write
Read off the question and its marks
Read from the question
- determinantFormula first, then the working, then the answer with its unit (₹, %, units).
Write it in this order
- Read the data; note exactly what is asked 1 min
- 1 Equations 1 mark 2 min
- 2 Δ, Δₐ, Δ_b, Δ_c 3 marks 5 min
- 3 A, b, c with check 1 mark 2 min
- 4 Sales for X = 12, Y = 15 1 mark 1 min
- Check: arithmetic, units (₹, %), final answer underlined 1 min
Short on time? Step 2 carries 3 of the 6 marks: guard its minutes.
Try it first. About 12 minutes for 6 marks. Write it out, then open the solution.
Solution
Hover or tap a marked figure, like 72,000, to see how it is made and where else it appears.
Equations from the three months (S in thousands of rupees):
a + b + 3c = 17 a + 2b + c = 12 a + 5b + 3c = 29
Coefficient matrix A = [[1, 1, 3], [1, 2, 1], [1, 5, 3]] and constants [17, 12, 29]ᵀ.
Determinants (Cramer's rule, one of the two methods the syllabus names)
Δ = |A| = 1(2 × 3 − 1 × 5) − 1(1 × 3 − 1 × 1) + 3(1 × 5 − 2 × 1) = 1 − 2 + 9 = 8 ≠ 0, so the system has a unique solution.
Δₐ (first column replaced by the constants) = |[[17, 1, 3], [12, 2, 1], [29, 5, 3]]| = 17(1) − 1(7) + 3(2) = 16 Δ_b (second column replaced) = |[[1, 17, 3], [1, 12, 1], [1, 29, 3]]| = 1(7) − 17(2) + 3(17) = 24 Δ_c (third column replaced) = |[[1, 1, 17], [1, 2, 12], [1, 5, 29]]| = 1(−2) − 1(17) + 17(3) = 32
a = Δₐ/Δ = 16/8 = 2, b = Δ_b/Δ = 24/8 = 3, c = Δ_c/Δ = 32/8 = 4.
So S = 2 + 3X + 4Y (₹ thousand).
Check by substitution: month 1: 2 + 3 + 12 = 17; month 2: 2 + 6 + 4 = 12; month 3: 2 + 15 + 12 = 29.
| Term | Value (₹ thousand) |
|---|---|
| a | 2 |
| bX = 3 × 12 | 36 |
| cY = 4 × 15 | 60 |
| S | 98 |
Sales in that month = ₹98 thousand (₹98,000).
Where the 6 marks sit
| # | Component | Marks |
|---|---|---|
| 1 | Equations | 1 |
| 2 | Δ, Δₐ, Δ_b, Δ_c | 3 |
| 3 | a, b, c with check | 1 |
| 4 | Sales for X = 12, Y = 15 | 1 |
| Total | 6 |
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