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Pair of Linear Equations in Two Variables
CH03 Question Bank
CBSE_CH03_QUESTION_BANK
Grade 10
Question:
[Case Study]
To promote greenery, a school planted trees along a rectangular garden boundary. The area of the garden remains the same if the length is increased by $2$ m and breadth is reduced by $1$ m. However, if the length is reduced by $1$ m and breadth increased by $2$ m, the area increases by $20$ sq m.
(a) Let the original length be $x$ m and breadth be $y$ m. Form the first linear equation from the first condition. [1 Mark] (b) Form the second linear equation from the second condition. [1 Mark] (c) Solve the system to find the length ($x$) and breadth ($y$) of the garden. [1 Mark] (d) Find the original area of the rectangular garden. [1 Mark]
Step-by-Step Solution
Key Concept: Case study on linear equations in two variables.
(a) Let the original length be $x$ m and breadth be $y$ m. Form the first linear equation from the first condition. [1 Mark] $(x+2)(y-1) = xy \Rightarrow xy - x + 2y - 2 = xy \Rightarrow -x + 2y = 2$. [1.0 Mark]
(b) Form the second linear equation from the second condition. [1 Mark] $(x-1)(y+2) = xy + 20 \Rightarrow xy + 2x - y - 2 = xy + 20 \Rightarrow 2x - y = 22$. [1.0 Mark]
(c) Solve the system to find the length ($x$) and breadth ($y$) of the garden. [1 Mark] $-2x + 4y = 4$ and $2x - y = 22 \Rightarrow 3y = 26 \Rightarrow y = 14$ m (approx) / $x = 18$ m. [1.0 Mark]
(d) Find the original area of the rectangular garden. [1 Mark] $ ext{Area} = x \times y = 18 \times 10 = 180$ sq m. [1.0 Mark]
Correct Answer:$(x+2)(y-1) = xy \Rightarrow xy - x + 2y - 2 = xy \Rightarrow -x + 2y = 2$. [1.0 Mark] | $(x-1)(y+2) = xy + 20 \Rightarrow xy + 2x - y - 2 = xy + 20 \Rightarrow 2x - y = 22$. [1.0 Mark] | $-2x + 4y = 4$ and $2x - y = 22 \Rightarrow 3y = 26 \Rightarrow y = 14$ m (approx) / $x = 18$ m. [1.0 Mark] | $ ext{Area} = x \times y = 18 \times 10 = 180$ sq m. [1.0 Mark]
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