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Pair of Linear Equations in Two Variables
CH03 Question Bank
CBSE_CH03_QUESTION_BANK
Grade 10
Question:
[Case Study]
A taxi service charges a fixed charge for the first few kilometres plus an additional charge for every kilometre travelled thereafter. For a journey of $10$ km, the total fare is Rs $105$; for a journey of $15$ km, the total fare is Rs $155$.
(a) Taking the fixed charge as Rs $x$ and the charge per km as Rs $y$, form a pair of linear equations representing the given information. [1 Mark] (b) Find the value of $y$, the charge per km. [1 Mark] (c) Find the value of $x$, the fixed charge. [1 Mark] (d) Using these values, find the fare for a journey of $20$ km. [1 Mark]
Step-by-Step Solution
Key Concept: Case study on linear equations in two variables.
(a) Taking the fixed charge as Rs $x$ and the charge per km as Rs $y$, form a pair of linear equations representing the given information. [1 Mark] $x+10y=105$ and $x+15y=155$. [1.0 Mark]
(b) Find the value of $y$, the charge per km. [1 Mark] Subtracting: $(x+15y)-(x+10y)=155-105\Rightarrow 5y=50\Rightarrow y=10$. [1.0 Mark]
(c) Find the value of $x$, the fixed charge. [1 Mark] $x+10(10)=105\Rightarrow x=5$. [1.0 Mark]
(d) Using these values, find the fare for a journey of $20$ km. [1 Mark] Fare $=x+20y=5+20(10)=205$. So the fare for $20$ km is Rs $205$. [1.0 Mark]
Correct Answer:$x+10y=105$ and $x+15y=155$. [1.0 Mark] | Subtracting: $(x+15y)-(x+10y)=155-105\Rightarrow 5y=50\Rightarrow y=10$. [1.0 Mark] | $x+10(10)=105\Rightarrow x=5$. [1.0 Mark] | Fare $=x+20y=5+20(10)=205$. So the fare for $20$ km is Rs $205$. [1.0 Mark]
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