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Mathematics · 2026 · 5 marks
CBSE 2026 · Region 5 · Set 2 · Q32
Find cost (per kg) of each fertilizer A, B and C that the farmer needs to buy, such that 1 kg each of fertilizer A and C added to 2 kg of B costs him…
Find cost (per kg) of each fertilizer A, B and C that the farmer needs to buy, such that $\displaystyle 1$ kg each of fertilizer A and C added to $\displaystyle 2$ kg of B costs him ₹ 400. Also, cost of each kg of fertilizer B and C added together is equal to cost of $\displaystyle 1$ kg of fertilizer A. However, cost of $\displaystyle 3$ kg of fertilizer B added to ₹ $\displaystyle 200$ is the same as cost of $\displaystyle 1$ kg of fertilizer A and C together. Use matrix method to find the solution.
Marking-scheme solution
Let cost of one kg of fertilizers A, B and C be ₹x, ₹y and ₹z respectively
$\displaystyle x+2 y+z=400$; $\displaystyle -x+y+z=0$; $\displaystyle x-3 y+z=200$
Let $\displaystyle A=\begin{bmatrix} 1 & 2 & 1 \\ -1 & 1 & 1 \\ 1 & -3 & 1 \end{bmatrix}, X=\begin{bmatrix} x \\ y \\ z \end{bmatrix}, B=\begin{bmatrix} 400 \\ 0 \\ 200 \end{bmatrix}$
$\displaystyle |A|=10 \neq 0 \Rightarrow A^{-1}$ exists
System becomes $\displaystyle A X=B$. So, $\displaystyle X=A^{-1} B$
$\displaystyle \operatorname{adj} A=\begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix}$
$\displaystyle A^{-1}=\dfrac{1}{10}\begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix}$
$\displaystyle X=\dfrac{1}{10}\begin{bmatrix} 4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3 \end{bmatrix}\begin{bmatrix} 400 \\ 0 \\ 200 \end{bmatrix}$
$\displaystyle X=\begin{bmatrix} 180 \\ 40 \\ 140 \end{bmatrix}$
$\displaystyle \therefore$ cost of one kg of fertilizers A, B and C is ₹$\displaystyle 180$, ₹$\displaystyle 40$ and ₹$\displaystyle 140$ 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.