MathsCBSE Class 12 · Sample paper, session 2026-27 · 4 marks · Case-based

CBSE Class 12 Mathematics sample paper 2026-27: question 38

From the official CBSE sample question paper for 2026-27, section E, question 38, for 4 marks. Try it first; the hints and our step-by-step solution are below.

Q38 · 4 marks · Case-basedCompetency-basedMatrices

To celebrate their child's birthday, a family decided to distribute fruits among children in three orphanages. The fruit vendor packed three different types of baskets, according to the quantity required at each orphanage, details are given as follows:
Basket A contains 5 kg Apples, 10 kg Oranges and 5 kg Pomegranate
Basket B contains 10 kg Apples, 10 kg Oranges and 5kg Pomegranate
Basket C contains 8 kg Apples, 15 kg Oranges and 7kg Pomegranate
Cost of basket A, B and C are respectively ₹2750, ₹3500, and ₹4100.
Based on the given information, answer the following questions:

(i)
Write the matrix form of the equations involved in this situation and find the adjoint of the coefficient matrix involved.
2 marks
(ii)
Using inverse of the matrix, calculate the price of each fruit per kg.
2 marks

Figure in the paper: Photo of a family buying fruits from a fruit vendor with baskets of fruits on a cart. See it in the official paper.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

Each basket gives one equation: (kg of each fruit × price per kg) summed equals the basket cost.

Hint 2

The adjoint is the transpose of the matrix of cofactors.

Hint 3

Use $A^{- 1} = \frac{1}{∣ A ∣} adj A$ and multiply by $B$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

(i) Let the prices per kg of apples, oranges and pomegranates be ₹ $x$ , ₹ $y$ , ₹ $z$ . Then:
- $5 x + 10 y + 5 z = 2750$
- $10 x + 10 y + 5 z = 3500$
- $8 x + 15 y + 7 z = 4100$
In matrix form this is $A X = B$ , where $A = 5108101015557$ , $X = x y z$ , $B = 275035004100$ .
1 mark
(i) Cofactors of $A$ :
- $C_{11} = - 5, C_{12} = - 30, C_{13} = 70$
- $C_{21} = 5, C_{22} = - 5, C_{23} = 5$
- $C_{31} = 0, C_{32} = 25, C_{33} = - 50$
Taking the transpose, $adj A = - 5 - 30 70 5 - 5 5 025 - 50$ .
1 mark
(ii) $∣ A ∣ = 5 (70 - 75) - 10 (70 - 40) + 5 (150 - 80) = - 25 - 300 + 350 = 25 \neq = 0$ , so $A^{- 1}$ exists and the system has a unique solution. Then $X = A^{- 1} B = \frac{1}{25} - 5 - 30 70 5 - 5 5 025 - 50 275035004100$ .
1 mark
(ii) Multiplying out:
- $x = \frac{- 13750 + 17500}{25} = 150$
- $y = \frac{- 82500 - 17500 + 102500}{25} = 100$
- $z = \frac{192500 + 17500 - 205000}{25} = 200$
So apples cost ₹150/kg, oranges ₹100/kg and pomegranates ₹200/kg.
1 mark

Answer: (i) $A X = B$ with $A = 5108101015557$ , $adj A = - 5 - 30 70 5 - 5 5 025 - 50$ ; (ii) Apples ₹150/kg, Oranges ₹100/kg, Pomegranates ₹200/kg

Where the marks are

Forms the equations and writes them as $A X = B$ (1 mark)
Correct adjoint of $A$ (1 mark)
Finds $∣ A ∣ = 25 \neq = 0$ and writes $X = A^{- 1} B$ (1 mark)
Obtains $x = 150, y = 100, z = 200$ with the conclusion (1 mark)

Revise the chapter

Chapter: Matrices (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Matrices, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Mathematics sample paper 2026-27: question 38

Hints

Step-by-step solution

Revise the chapter

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