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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 5 · Set 1 · Q32
A man goes to buy fruits from the market. The shopkeeper informs him that 4 apples, 3 oranges and 2 bananas cost ₹ 60; 2 apples, 4 oranges and 6…
A man goes to buy fruits from the market. The shopkeeper informs him that $\displaystyle 4$ apples, $\displaystyle 3$ oranges and $\displaystyle 2$ bananas cost ₹ $\displaystyle 60$ ; $\displaystyle 2$ apples, $\displaystyle 4$ oranges and $\displaystyle 6$ bananas cost ₹ $\displaystyle 90$; whereas $\displaystyle 6$ apples, $\displaystyle 2$ oranges and $\displaystyle 3$ bananas cost ₹ 70. Using matrix method, find the cost of one fruit of each kind.
Marking-scheme solution
Let cost of one apple, one orange, one banana be ₹ x, ₹ y and ₹ z respectively. According to the question,
$\displaystyle 4 x+3 y+2 z=60$; $\displaystyle 2 x+4 y+6 z=90$; $\displaystyle 6 x+2 y+3 z=70$
Let $\displaystyle A=\begin{bmatrix} 4 & 3 & 2 \\ 2 & 4 & 6 \\ 6 & 2 & 3 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=\begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$
$\displaystyle |A|=50 \neq 0 \Rightarrow A^{-1}$ exists
System becomes $\displaystyle A X=B$. So, $\displaystyle X=A^{-1} B$
$\displaystyle \operatorname{adj} A=\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$
$\displaystyle A^{-1}=\dfrac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}$
$\displaystyle X=\dfrac{1}{50}\begin{bmatrix} 0 & -5 & 10 \\ 30 & 0 & -20 \\ -20 & 10 & 10 \end{bmatrix}\begin{bmatrix} 60 \\ 90 \\ 70 \end{bmatrix}$
$\displaystyle X=\begin{bmatrix} 5 \\ 8 \\ 8 \end{bmatrix}$
$\displaystyle \therefore$ cost of one apple, one orange, one banana is ₹ $\displaystyle 5$, ₹ $\displaystyle 8$ and ₹ $\displaystyle 8$ respectively
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CBSE Class 12 Mathematics past-paper question from the 2026 board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.