Maths · Coordinate geometry · chapter 9 of 14
Class 11 Maths Chapter 9: Straight Lines case study question with answers
One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.
Case study · 4 marks
A home bakery in Indore keeps weekly accounts. The owner finds that both her total weekly cost C (in ₹) and her weekly sales revenue R (in ₹) are linear in n, the number of cakes made and sold in the week. Her notebook shows the figures below. Plotted with n on the horizontal axis, each set of figures lies on a straight line, and the two lines cross at the break-even point.
| Cakes in a week (n) | Total cost C (₹) | Revenue R (₹) |
|---|---|---|
| 50 | 8,000 | 5,000 |
| 150 | 12,000 | 15,000 |
Part 1 · 1 mark
The slope of the cost line, the extra cost of one more cake, is
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Answer: (d) ₹40
Slope = (12000 − 8000)/(150 − 50) = 4000/100 = ₹40 per cake.
Part 2 · 1 mark
The fixed weekly cost, the C-intercept of the cost line, is
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Answer: (a) ₹6,000
C − 8000 = 40(n − 50); at n = 0, C = 8000 − 2000 = ₹6,000.
Part 3 · 2 marks
Write the equations of the cost line and the revenue line, and find the number of cakes at the break-even point.
Answer and marking scheme
C = 40n + 6000. Revenue slope = (15000 − 5000)/(150 − 50) = 100, R − 5000 = 100(n − 50), so R = 100n. At break-even R = C: 100n = 40n + 6000, n = 100 cakes.
- C = 40n + 6000 and R = 100n (1)
- n = 100 (1)
Or
Using the two lines, find the number of cakes she must sell in a week to make a profit of ₹4,800.
Answer and marking scheme
C = 40n + 6000 and R = 100n, so profit R − C = 60n − 6000. 60n − 6000 = 4800 gives 60n = 10800, n = 180 cakes.
- profit = 60n − 6000 from the two lines (1)
- n = 180 (1)
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