Case Study – 1 Rohini wants to give a rectangular plot — CBSE Class 12 Applied Mathematics 2024 PYQ

Question

Q. Case Study - 1 Rohini wants to give a rectangular plot of land for a school in her village. When she was asked to mention the dimensions of the plot, she told that if its length is decreased by 50 m and breadth is increased by 50 m, then its area does not alter, but if its length is decreased by 10 m and breadth is decreased by 20 m, then its area will decrease by 5300 sq m. Based on the above information, answer the following questions :

(i) Assuming

x

m and

y

m as the length and breadth of the plot respectively, write the system of linear equations in

x

and

y

. [1]

(ii) Write the system of linear equations obtained in

(i) in the matrix equation

AX = B

. [1]

(iii)

(a) Determine

A^{-1}

. [2]

(OR)

(b) Find the area of the plot. [2]

CBSECBSE Class XII Board 2024Subjective· 4mImportance★★★★★

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Concept understanding — Solving Linear Systems

Solving Linear Systems: From Intuition to Precision

Imagine you're at a market. You buy 2 apples and 3 bananas for ₹17. Your friend buys 1 apple and 2 bananas for ₹10. How much does one apple cost? One banana?

You already know how to solve this with a single variable — but here, two things are unknown at once. The price of an apple and the price of a banana are linked together by two different conditions. That's a linear system: a set of equations that must all be true at the same time.

The Intuition: Two Lines, One Meeting Point

Each equation in a system of two variables (say $x$ and $y$ ) can be drawn as a straight line on a graph. The equation $2x + 3y = 17$ is one line; $x + 2y = 10$ is another.

Solving the system means finding the $(x, y)$ pair that lies on both lines at once. That's the point where the two lines cross — their intersection. If you think about it, that's the only place where both conditions are satisfied simultaneously.

Tip

Visualise it: two straight lines on a plane. They either cross at exactly one point (one solution), run parallel and never meet (no solution), or lie exactly on top of each other (infinitely many solutions). The first case is what you'll see most often.

The Precise Statement

A linear system (in two variables) is a collection of equations of the form:

\begin{cases} a_1 x + b_1 y = c_1 \\ a_2 x + b_2 y = c_2 \end{cases}

where $a_1, b_1, c_1, a_2, b_2, c_2$ are known numbers (constants), and $x$ and $y$ are the unknowns we need to find. The word "linear" means each variable appears only to the first power — no $x^2$ , no $xy$ , no $\frac{1}{x}$ .

A solution is an ordered pair $(x, y)$ that makes both equations true when substituted in.

a_1 x + b_1 y = c_1 \quad \text{and} \quad a_2 x + b_2 y = c_2

The Three Possibilities

Case	Graph	Number of Solutions
Lines intersect at one point	Two crossing lines	Exactly one
Lines are parallel (same slope, different intercept)	Two parallel lines	None
Lines are identical (same slope and intercept)	One line on top of another	Infinitely many

How to Solve: Two Core Methods

1. Substitution Method

Solve one equation for one variable, then plug that into the other equation.

From the apple-banana problem:

$\begin{cases} 2x + 3y = 17 \\ x + 2y = 10 \end{cases}$ …

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The conditions give $x-y=50$ and $2x+y=550$ ; as $AX=B$ with $A=\begin{bmatrix}1&-1\\2&1\end{bmatrix}$ , we get $A^{-1}=\tfrac13\begin{bmatrix}1&1\\-2&1\end{bmatrix}$ , $x=200,\ y=150$ and area $=30{,}000$ sq m.

=xy

(i) Form the equations

Length decreased by 50, breadth increased by 50, area unchanged: $(x-50)(y+50)=xy.$ Expanding: $50x-50y-2500=0\Rightarrow x-y=50.$
Length decreased by 10, breadth decreased by 20, area falls by 5300: $(x-10)(y-20)=xy-5300.$ Expanding: $-20x-10y+200=-5300\Rightarrow 2x+y=550.$

(ii) Matrix equation

3. $\begin{bmatrix}1&-1\\2&1\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}50\\550\end{bmatrix}$ , i.e. $A=\begin{bmatrix}1&-1\\2&1\end{bmatrix},\ X=\begin{bmatrix}x\\y\end{bmatrix},\ B=\begin{bmatrix}50\\550\end{bmatrix}.$

(iii)(a) Determine $A^{-1}$

4. $\det A=(1)(1)-(-1)(2)=1+2=3.$

5. $\operatorname{adj}A=\begin{bmatrix}1&1\\-2&1\end{bmatrix}$ (swap leading diagonal, negate off-diagonal). …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set 465/S/WXYZ/41 markMCQ
$2x + ky = 7$
›Reveal solutionSolution
The system fails (parallel lines) only at $k=\tfrac{4}{3}$ ; hence it is consistent for every $k \neq \tfrac{4}{3}$ .

$a_1x+b_1y=c_1$

Identify coefficients: $a_1=2,\ b_1=k,\ c_1=7$ and $a_2=3,\ b_2=2,\ c_2=7$ .

Unique solution (consistent) when $\dfrac{2}{3} \ne \dfrac{k}{2}$ , i.e. $k \ne \dfrac{4}{3}$ . …
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CBSE 2023Set 465/EF1GH/41 markMCQ
$\begin{bmatrix} x+y & x+2 \\ 2x-y & 16 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ 1 & 3y+1 \end{bmatrix}$
(a) $x = 3, y = 5$
(b) $x = 5, y = 3$
(c) $x = 2, y = 7$
(d) $x = 7, y = 2$
›Reveal solutionSolution
Equal matrices have equal corresponding entries; solving the simplest entries gives $x=3,\ y=5$ .

$a_{ij}=b_{ij}$

From the $(1,2)$ entries: $x+2=5 \Rightarrow x=3$ . …
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