CH03 Question Bank
Pair of Linear Equations in Two Variables • 50 Questions
Question 1
Hint available
For the pair of linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ to have a unique solution, the condition is:
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$
Key Idea
This is the standard algebraic condition for a unique solution (intersecting lines).
Step-by-Step Solution
A unique solution occurs exactly when $\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$. [1.0 Mark]
eq\dfrac{b_1}{b_2}$. [1.0 Mark]
Question 2
Hint available
For the pair of linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ to have no solution, the condition is:
$\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$
$\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}$
$\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$
$\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}$
Key Idea
This is the standard algebraic condition for no solution (parallel, non-coincident lines).
Step-by-Step Solution
No solution occurs exactly when $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$. [1.0 Mark]
eq\dfrac{c_1}{c_2}$. [1.0 Mark]
Question 3
Hint available
For the pair of linear equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ to have infinitely many solutions, the condition is:
$\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
$a_1=a_2, b_1=b_2$
$\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$
$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$
$a_1=a_2, b_1=b_2$
Key Idea
This is the standard algebraic condition for infinitely many solutions (coincident lines).
Step-by-Step Solution
Infinitely many solutions occur exactly when $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$. [1.0 Mark]
Question 4
Hint available
Which of the following methods is NOT part of the current CBSE Class 10 syllabus for solving a pair of linear equations?
Graphical method
Substitution method
Elimination method
Cross-multiplication method
Graphical method
Substitution method
Elimination method
Cross-multiplication method
Key Idea
The cross-multiplication method has been removed from the current rationalised CBSE syllabus.
Step-by-Step Solution
Only the graphical, substitution, and elimination methods are part of the current syllabus; cross-multiplication has been deleted. [1.0 Mark]
Question 5
Hint available
For what value of $k$ do the equations $2x+3y=7$ and $4x+6y=k$ have infinitely many solutions?
$7$
$10$
$14$
$21$
$7$
$10$
$14$
$21$
Key Idea
Apply the condition $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$.
Step-by-Step Solution
$\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{1}{2}$. For infinitely many solutions, $\dfrac{7}{k}$ must also equal $\dfrac{1}{2}$. [0.5 Mark]
$k=14$. [0.5 Mark]
$k=14$. [0.5 Mark]
Question 6
Hint available
For what value of $k$ does the pair of equations $2x+ky=1$ and $3x-5y=7$ have a unique solution?
$k=-\dfrac{10}{3}$
$k
eq-\dfrac{10}{3}$
$k=0$
$k=1$
$k=-\dfrac{10}{3}$
$k
eq-\dfrac{10}{3}$
$k=0$
$k=1$
Key Idea
Apply the condition $\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$.
eq\dfrac{b_1}{b_2}$.
Step-by-Step Solution
Unique solution requires $\dfrac{2}{3}
eq\dfrac{k}{-5}$. [0.5 Mark]
$\dfrac{k}{-5}=\dfrac23 \Rightarrow k=-\dfrac{10}{3}$ is the excluded value, so the condition is $k
eq-\dfrac{10}{3}$. [0.5 Mark]
eq\dfrac{k}{-5}$. [0.5 Mark]
$\dfrac{k}{-5}=\dfrac23 \Rightarrow k=-\dfrac{10}{3}$ is the excluded value, so the condition is $k
eq-\dfrac{10}{3}$. [0.5 Mark]
Question 7
Hint available
The graph of a pair of linear equations is shown below. What can you conclude about the system?
It has no solution (inconsistent)
It has a unique solution (consistent)
It has infinitely many solutions (dependent)
It cannot be determined from a graph
It has no solution (inconsistent)
It has a unique solution (consistent)
It has infinitely many solutions (dependent)
It cannot be determined from a graph
Key Idea
Two lines intersecting at exactly one point represent a consistent system with a unique solution.
Step-by-Step Solution
The two lines intersect at exactly one point, $(3,2)$, so the system is consistent with a unique solution. [1.0 Mark]
Question 8
Hint available
The graph of a pair of linear equations is shown below. What can you conclude about the system?
It has a unique solution
It has no solution (inconsistent)
It has infinitely many solutions
The lines are perpendicular
It has a unique solution
It has no solution (inconsistent)
It has infinitely many solutions
The lines are perpendicular
Key Idea
Two distinct parallel lines (never meeting) represent an inconsistent system with no solution.
Step-by-Step Solution
The two lines are parallel and never intersect, so the system has no solution — it is inconsistent. [1.0 Mark]
Question 9
Hint available
The graph of a pair of linear equations is shown below. What can you conclude about the system?
It has no solution
It has a unique solution
It has infinitely many solutions (the lines coincide)
The lines are perpendicular
It has no solution
It has a unique solution
It has infinitely many solutions (the lines coincide)
The lines are perpendicular
Key Idea
Two lines that lie exactly on top of each other (coincident) represent a dependent, consistent system with infinitely many solutions.
Step-by-Step Solution
The two lines coincide completely (every point on one line is also on the other), so the system has infinitely many solutions. [1.0 Mark]
Question 10
Hint available
Using substitution, if $x+y=5$ and $x-y=1$, the value of $x$ is:
$1$
$2$
$3$
$4$
$1$
$2$
$3$
$4$
Key Idea
Add the two equations to eliminate $y$, or substitute directly.
Step-by-Step Solution
Adding the equations: $2x=6\Rightarrow x=3$. [1.0 Mark]
Question 11
Hint available
By the elimination method, if $2x+3y=12$ and $2x-3y=0$, the value of $x$ is:
$1$
$2$
$3$
$6$
$1$
$2$
$3$
$6$
Key Idea
Add the two equations to eliminate $y$.
Step-by-Step Solution
Adding: $4x=12\Rightarrow x=3$. [1.0 Mark]
Question 12
Hint available
The sum of two numbers is $10$ and their difference is $4$. The two numbers are:
$7$ and $3$
$8$ and $2$
$6$ and $4$
$9$ and $1$
$7$ and $3$
$8$ and $2$
$6$ and $4$
$9$ and $1$
Key Idea
Let the numbers be $x,y$; solve $x+y=10$ and $x-y=4$ by elimination.
Step-by-Step Solution
Adding: $2x=14\Rightarrow x=7$; then $y=10-7=3$. [1.0 Mark]
Question 13
Hint available
If a pair of linear equations has infinitely many solutions, the lines representing them are:
Intersecting
Parallel (distinct)
Coincident
Perpendicular
Intersecting
Parallel (distinct)
Coincident
Perpendicular
Key Idea
Infinitely many solutions correspond geometrically to the two lines being identical (coincident).
Step-by-Step Solution
Infinitely many solutions means every point on one line also satisfies the other equation — the lines coincide. [1.0 Mark]
Question 14
Hint available
If $a_1b_2=a_2b_1$ but $a_1c_2
eq a_2c_1$, the pair of linear equations $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ has:
A unique solution
No solution
Infinitely many solutions
Exactly two solutions
eq a_2c_1$, the pair of linear equations $a_1x+b_1y+c_1=0$, $a_2x+b_2y+c_2=0$ has:
A unique solution
No solution
Infinitely many solutions
Exactly two solutions
Key Idea
$a_1b_2=a_2b_1$ is equivalent to $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$, and $a_1c_2
eq a_2c_1$ is equivalent to $\dfrac{a_1}{a_2}
eq\dfrac{c_1}{c_2}$ — the no-solution condition.
eq a_2c_1$ is equivalent to $\dfrac{a_1}{a_2}
eq\dfrac{c_1}{c_2}$ — the no-solution condition.
Step-by-Step Solution
$a_1b_2=a_2b_1 \Rightarrow \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$, and $a_1c_2
eq a_2c_1 \Rightarrow \dfrac{a_1}{a_2}
eq\dfrac{c_1}{c_2}$. [0.5 Mark]
This is exactly the no-solution condition, so the system is inconsistent. [0.5 Mark]
eq a_2c_1 \Rightarrow \dfrac{a_1}{a_2}
eq\dfrac{c_1}{c_2}$. [0.5 Mark]
This is exactly the no-solution condition, so the system is inconsistent. [0.5 Mark]
Question 15
Hint available
The general form of a linear equation in two variables $x$ and $y$ is:
$ax^2+by+c=0$
$ax+by+c=0\ (a,b \text{ not both zero})$
$ax+by^2=c$
$ax+b=0$
$ax^2+by+c=0$
$ax+by+c=0\ (a,b \text{ not both zero})$
$ax+by^2=c$
$ax+b=0$
Key Idea
Standard definition of a linear equation in two variables.
Step-by-Step Solution
The general form is $ax+by+c=0$, where $a$ and $b$ are not both zero. [1.0 Mark]
Question 16
Hint available
Assertion (A): The pair of equations $x+2y-4=0$ and $2x+4y-6=0$ has no solution.
Reason (R): For this pair, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac12$, but $\dfrac{c_1}{c_2}=\dfrac{-4}{-6}=\dfrac23$, which is not equal to $\dfrac12$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): For this pair, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac12$, but $\dfrac{c_1}{c_2}=\dfrac{-4}{-6}=\dfrac23$, which is not equal to $\dfrac12$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Direct application of the no-solution condition.
Step-by-Step Solution
Checking: $\dfrac{1}{2}=\dfrac{2}{4}=\dfrac12$, and $\dfrac{-4}{-6}=\dfrac23
eq\dfrac12$ — this matches the no-solution condition, so A is true. [0.5 Mark]
R correctly computes and compares these exact ratios, which is precisely the justification for A, so R correctly explains A. [0.5 Mark]
eq\dfrac12$ — this matches the no-solution condition, so A is true. [0.5 Mark]
R correctly computes and compares these exact ratios, which is precisely the justification for A, so R correctly explains A. [0.5 Mark]
Question 17
Hint available
Assertion (A): The pair of equations $2x+3y=5$ and $4x+6y=10$ has infinitely many solutions.
Reason (R): For this pair, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=\dfrac12$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): For this pair, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}=\dfrac12$.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Direct application of the infinite-solutions condition.
Step-by-Step Solution
$\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{5}{10}=\dfrac12$ — all three ratios are equal, so A is true. [0.5 Mark]
R correctly states this exact equality of ratios, which is precisely the condition for infinitely many solutions, so R correctly explains A. [0.5 Mark]
R correctly states this exact equality of ratios, which is precisely the condition for infinitely many solutions, so R correctly explains A. [0.5 Mark]
Question 18
Hint available
Assertion (A): The cross-multiplication method is used to solve a pair of linear equations in the current CBSE Class 10 syllabus.
Reason (R): The cross-multiplication method provides direct formulas for $x$ and $y$ in terms of the coefficients of the two equations.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): The cross-multiplication method provides direct formulas for $x$ and $y$ in terms of the coefficients of the two equations.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
The cross-multiplication method has been removed from the current syllabus, even though the method itself (as a mathematical technique) does what R describes.
Step-by-Step Solution
The cross-multiplication method has been deleted from the current CBSE syllabus — only graphical, substitution, and elimination methods are prescribed — so A is false. [0.5 Mark]
R is a factually correct general description of what the cross-multiplication method does (it is simply no longer part of the syllabus), so R is true. [0.5 Mark]
R is a factually correct general description of what the cross-multiplication method does (it is simply no longer part of the syllabus), so R is true. [0.5 Mark]
Question 19
Hint available
Assertion (A): The graphical method can be used to determine whether a pair of linear equations is consistent or inconsistent.
Reason (R): The point of intersection of the two lines gives the solution of the pair of equations, whenever the lines intersect at a unique point.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): The point of intersection of the two lines gives the solution of the pair of equations, whenever the lines intersect at a unique point.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
R correctly describes only the intersecting-lines (unique solution) case, but A is a broader statement covering all three cases (intersecting, parallel, coincident) — so R is true but incomplete as an explanation of A.
Step-by-Step Solution
A is true: by plotting the graphs, one can see whether the lines intersect, are parallel, or coincide, and hence classify the system. [0.5 Mark]
R is also true, but it only explains the unique-solution case; it says nothing about how the graphical method identifies the no-solution (parallel) or infinite-solutions (coincident) cases, so R does not fully explain A. [0.5 Mark]
R is also true, but it only explains the unique-solution case; it says nothing about how the graphical method identifies the no-solution (parallel) or infinite-solutions (coincident) cases, so R does not fully explain A. [0.5 Mark]
Question 20
Hint available
Assertion (A): Substituting $y=5-x$ (obtained from $x+y=5$) into $2x+3y=12$ gives a valid method to solve the pair of equations.
Reason (R): This is an application of the substitution method, where one variable is expressed in terms of the other from one equation and substituted into the second equation.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Reason (R): This is an application of the substitution method, where one variable is expressed in terms of the other from one equation and substituted into the second equation.
Both Assertion (A) and Reason (R) are true, and R is the correct explanation of A.
Both Assertion (A) and Reason (R) are true, but R is NOT the correct explanation of A.
Assertion (A) is true, but Reason (R) is false.
Assertion (A) is false, but Reason (R) is true.
Key Idea
Direct description of the substitution method as prescribed in the syllabus.
Step-by-Step Solution
Expressing $y$ in terms of $x$ from the first equation and substituting into the second is indeed valid, so A is true. [0.5 Mark]
R correctly names and describes this exact procedure as the substitution method, so R correctly explains A. [0.5 Mark]
R correctly names and describes this exact procedure as the substitution method, so R correctly explains A. [0.5 Mark]
Question 21
Hint available
Solve the following pair of linear equations by the substitution method: $x+y=7$, $x-y=1$.
Key Idea
Express one variable in terms of the other from one equation, then substitute into the second.
Step-by-Step Solution
From the first equation, $x=7-y$. Substituting into the second: $(7-y)-y=1\Rightarrow 7-2y=1\Rightarrow y=3$. [1.0 Mark]
$x=7-3=4$. So $x=4,\ y=3$. [1.0 Mark]
$x=7-3=4$. So $x=4,\ y=3$. [1.0 Mark]
Question 22
Hint available
Solve the following pair of linear equations by the elimination method: $2x+y=7$, $x-y=2$.
Key Idea
Add the two equations directly to eliminate $y$.
Step-by-Step Solution
Adding the equations: $3x=9\Rightarrow x=3$. [1.0 Mark]
Substituting into $x-y=2$: $3-y=2\Rightarrow y=1$. So $x=3,\ y=1$. [1.0 Mark]
Substituting into $x-y=2$: $3-y=2\Rightarrow y=1$. So $x=3,\ y=1$. [1.0 Mark]
Question 23
Hint available
Without solving, determine whether the pair of equations $3x+2y=5$ and $6x+4y=10$ is consistent or inconsistent.
Key Idea
Compare the ratios of coefficients.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{3}{6}=\dfrac12,\ \dfrac{b_1}{b_2}=\dfrac{2}{4}=\dfrac12,\ \dfrac{c_1}{c_2}=\dfrac{5}{10}=\dfrac12$. [1.0 Mark]
Since all three ratios are equal, the pair is consistent (in fact, dependent, with infinitely many solutions). [1.0 Mark]
Since all three ratios are equal, the pair is consistent (in fact, dependent, with infinitely many solutions). [1.0 Mark]
Question 24
Hint available
Without solving, determine whether the pair of equations $x+2y=3$ and $2x+4y=7$ is consistent or inconsistent.
Key Idea
Compare the ratios of coefficients.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{1}{2},\ \dfrac{b_1}{b_2}=\dfrac{2}{4}=\dfrac12,\ \dfrac{c_1}{c_2}=\dfrac{3}{7}$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$, the pair is inconsistent (no solution). [1.0 Mark]
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$, the pair is inconsistent (no solution). [1.0 Mark]
Question 25
Hint available
Without solving, determine the nature of solutions of the pair $x-y=2$ and $2x+y=7$.
Key Idea
Compare the ratios of coefficients.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{1}{2}$ and $\dfrac{b_1}{b_2}=\dfrac{-1}{1}=-1$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$, the pair has a unique solution. [1.0 Mark]
Since $\dfrac{a_1}{a_2}
eq\dfrac{b_1}{b_2}$, the pair has a unique solution. [1.0 Mark]
Question 26
Hint available
For what value of $k$ will the pair of equations $2x+3y=5$ and $4x+ky=10$ have infinitely many solutions?
Key Idea
Apply the condition for infinitely many solutions to all three ratios.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac12$ and $\dfrac{c_1}{c_2}=\dfrac{5}{10}=\dfrac12$; for infinitely many solutions we also need $\dfrac{b_1}{b_2}=\dfrac12$. [1.0 Mark]
$\dfrac{3}{k}=\dfrac12 \Rightarrow k=6$. [1.0 Mark]
$\dfrac{3}{k}=\dfrac12 \Rightarrow k=6$. [1.0 Mark]
Question 27
Hint available
Without drawing a graph, state whether the lines represented by $4x-2y=6$ and $2x-y=3$ intersect at a point, are parallel, or coincide.
Key Idea
Compare the ratios of coefficients.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{4}{2}=2,\ \dfrac{b_1}{b_2}=\dfrac{-2}{-1}=2,\ \dfrac{c_1}{c_2}=\dfrac{6}{3}=2$. [1.0 Mark]
Since all three ratios are equal, the lines coincide. [1.0 Mark]
Since all three ratios are equal, the lines coincide. [1.0 Mark]
Question 28
Hint available
The sum of two numbers is $8$ and their difference is $2$. Find the two numbers.
Key Idea
Form a pair of linear equations and solve by elimination.
Step-by-Step Solution
Let the numbers be $x,y$ with $x+y=8$ and $x-y=2$. Adding: $2x=10\Rightarrow x=5$. [1.0 Mark]
$y=8-5=3$. So the numbers are $5$ and $3$. [1.0 Mark]
$y=8-5=3$. So the numbers are $5$ and $3$. [1.0 Mark]
Question 29
Hint available
Solve by substitution: $2x+3y=16$ and $x-y=3$.
Key Idea
Express $x$ in terms of $y$ from the second equation, then substitute.
Step-by-Step Solution
From the second equation, $x=y+3$. Substituting into the first: $2(y+3)+3y=16\Rightarrow 5y+6=16\Rightarrow y=2$. [1.0 Mark]
$x=2+3=5$. So $x=5,\ y=2$. [1.0 Mark]
$x=2+3=5$. So $x=5,\ y=2$. [1.0 Mark]
Question 30
Hint available
Solve by elimination: $3x-5y=4$ and $9x-10y=2$.
Key Idea
Scale the first equation to match a coefficient, then subtract.
Step-by-Step Solution
Multiply the first equation by $2$: $6x-10y=8$. Subtracting the second equation ($9x-10y=2$) from this: $6x-10y-(9x-10y)=8-2\Rightarrow -3x=6\Rightarrow x=-2$. [1.0 Mark]
Substituting into $3x-5y=4$: $-6-5y=4\Rightarrow y=-2$. So $x=-2,\ y=-2$. [1.0 Mark]
Substituting into $3x-5y=4$: $-6-5y=4\Rightarrow y=-2$. So $x=-2,\ y=-2$. [1.0 Mark]
Question 31
Hint available
Determine whether the pair of equations $x+3y=6$ and $2x+6y=8$ is consistent.
Key Idea
Compare the ratios of coefficients.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac12,\ \dfrac{b_1}{b_2}=\dfrac{3}{6}=\dfrac12,\ \dfrac{c_1}{c_2}=\dfrac{6}{8}=\dfrac34$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$, the pair is inconsistent (no solution). [1.0 Mark]
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$, the pair is inconsistent (no solution). [1.0 Mark]
Question 32
Hint available
The cost of $2$ pens and $3$ pencils is Rs $90$. The cost of one pen is Rs $10$ more than the cost of one pencil. Represent this situation algebraically as a pair of linear equations (you do NOT need to solve it).
Key Idea
Translate the given statements directly into two linear equations using two variables.
Step-by-Step Solution
Let the cost of one pen be Rs $x$ and one pencil be Rs $y$. [0.5 Mark]
\"Cost of 2 pens and 3 pencils is Rs 90\" gives: $2x+3y=90$. [0.75 Mark]
\"Cost of a pen is Rs 10 more than a pencil\" gives: $x=y+10$, i.e. $x-y=10$. [0.75 Mark]
\"Cost of 2 pens and 3 pencils is Rs 90\" gives: $2x+3y=90$. [0.75 Mark]
\"Cost of a pen is Rs 10 more than a pencil\" gives: $x=y+10$, i.e. $x-y=10$. [0.75 Mark]
Question 33
Hint available
Solve the following pair of linear equations by the elimination method: $2x+3y=13$ and $3x-2y=0$.
Key Idea
Scale both equations so that the coefficient of one variable matches, then eliminate.
Step-by-Step Solution
Multiply the first equation by $2$ and the second by $3$: $4x+6y=26$ and $9x-6y=0$. [1.0 Mark]
Adding these: $13x=26\Rightarrow x=2$. [1.0 Mark]
Substituting into $3x-2y=0$: $6-2y=0\Rightarrow y=3$. So $x=2,\ y=3$. [1.0 Mark]
Adding these: $13x=26\Rightarrow x=2$. [1.0 Mark]
Substituting into $3x-2y=0$: $6-2y=0\Rightarrow y=3$. So $x=2,\ y=3$. [1.0 Mark]
Question 34
Hint available
Solve the following pair of linear equations by the substitution method: $2x+y=6$ and $3x-2y=2$.
Key Idea
Express $y$ in terms of $x$ from the first equation, then substitute into the second.
Step-by-Step Solution
From the first equation, $y=6-2x$. [1.0 Mark]
Substituting into the second: $3x-2(6-2x)=2\Rightarrow 3x-12+4x=2\Rightarrow 7x=14\Rightarrow x=2$. [1.0 Mark]
$y=6-2(2)=2$. So $x=2,\ y=2$. [1.0 Mark]
Substituting into the second: $3x-2(6-2x)=2\Rightarrow 3x-12+4x=2\Rightarrow 7x=14\Rightarrow x=2$. [1.0 Mark]
$y=6-2(2)=2$. So $x=2,\ y=2$. [1.0 Mark]
Question 35
Hint available
The sum of a two-digit number and the number obtained by reversing its digits is $66$. The digits differ by $2$. Find the number(s).
Key Idea
Let the digits be $x$ (tens) and $y$ (units); form and solve a pair of linear equations.
Step-by-Step Solution
Let the tens digit be $x$ and units digit be $y$. Number $=10x+y$, reversed number $=10y+x$. Given: $(10x+y)+(10y+x)=66\Rightarrow 11x+11y=66\Rightarrow x+y=6$. [1.0 Mark]
Also given the digits differ by $2$: $x-y=2$ (taking $x>y$). [1.0 Mark]
Adding: $2x=8\Rightarrow x=4,\ y=2$. So the number is $42$ (or, taking $y-x=2$ instead, the number would be $24$). [1.0 Mark]
Also given the digits differ by $2$: $x-y=2$ (taking $x>y$). [1.0 Mark]
Adding: $2x=8\Rightarrow x=4,\ y=2$. So the number is $42$ (or, taking $y-x=2$ instead, the number would be $24$). [1.0 Mark]
Question 36
Hint available
Determine, without solving completely, whether the pair of equations $2x-3y=8$ and $4x-6y=9$ is consistent or inconsistent, and describe how the lines would appear if graphed.
Key Idea
Compare all three ratios of coefficients.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac12,\ \dfrac{b_1}{b_2}=\dfrac{-3}{-6}=\dfrac12,\ \dfrac{c_1}{c_2}=\dfrac{8}{9}$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$, the pair is inconsistent (has no solution). [1.0 Mark]
Graphically, this means the two lines are parallel to each other but do not coincide — they never meet. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}
eq\dfrac{c_1}{c_2}$, the pair is inconsistent (has no solution). [1.0 Mark]
Graphically, this means the two lines are parallel to each other but do not coincide — they never meet. [1.0 Mark]
Question 37
Hint available
$5$ pens and $6$ pencils together cost Rs $9$, and $3$ pens and $2$ pencils cost Rs $5$. Find the cost of one pen and one pencil.
Key Idea
Form a pair of linear equations and solve by elimination.
Step-by-Step Solution
Let cost of a pen be Rs $x$ and a pencil be Rs $y$: $5x+6y=9$ — (1); $3x+2y=5$ — (2). [1.0 Mark]
Multiply (2) by $3$: $9x+6y=15$. Subtracting (1): $9x+6y-(5x+6y)=15-9\Rightarrow 4x=6\Rightarrow x=1.5$. [1.0 Mark]
Substituting into (2): $4.5+2y=5\Rightarrow y=0.25$. So a pen costs Rs $1.50$ and a pencil costs Rs $0.25$. [1.0 Mark]
Multiply (2) by $3$: $9x+6y=15$. Subtracting (1): $9x+6y-(5x+6y)=15-9\Rightarrow 4x=6\Rightarrow x=1.5$. [1.0 Mark]
Substituting into (2): $4.5+2y=5\Rightarrow y=0.25$. So a pen costs Rs $1.50$ and a pencil costs Rs $0.25$. [1.0 Mark]
Question 38
Hint available
For what value of $k$ does the system $kx+3y=k-3$ and $12x+ky=k$ have infinitely many solutions?
Key Idea
Set up the equal-ratios condition and solve, then verify the extraneous root.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac{k}{12}$ and $\dfrac{b_1}{b_2}=\dfrac{3}{k}$. Setting these equal: $\dfrac{k}{12}=\dfrac{3}{k}\Rightarrow k^2=36\Rightarrow k=\pm6$. [1.0 Mark]
Check $k=6$: $\dfrac{c_1}{c_2}=\dfrac{6-3}{6}=\dfrac12$, and $\dfrac{a_1}{a_2}=\dfrac{6}{12}=\dfrac12$ — matches. [1.0 Mark]
Check $k=-6$: $\dfrac{c_1}{c_2}=\dfrac{-6-3}{-6}=\dfrac32$, but $\dfrac{a_1}{a_2}=\dfrac{-6}{12}=-\dfrac12$ — does NOT match, so $k=-6$ is rejected. Hence $k=6$. [1.0 Mark]
Check $k=6$: $\dfrac{c_1}{c_2}=\dfrac{6-3}{6}=\dfrac12$, and $\dfrac{a_1}{a_2}=\dfrac{6}{12}=\dfrac12$ — matches. [1.0 Mark]
Check $k=-6$: $\dfrac{c_1}{c_2}=\dfrac{-6-3}{-6}=\dfrac32$, but $\dfrac{a_1}{a_2}=\dfrac{-6}{12}=-\dfrac12$ — does NOT match, so $k=-6$ is rejected. Hence $k=6$. [1.0 Mark]
Question 39
Hint available
For what value of $k$ does the system $3x+y=1$ and $(2k-1)x+(k-1)y=2k+1$ have no solution?
Key Idea
Apply the no-solution condition: equal first two ratios, unequal third.
Step-by-Step Solution
For no solution: $\dfrac{3}{2k-1}=\dfrac{1}{k-1}$. Cross-multiplying: $3(k-1)=2k-1\Rightarrow 3k-3=2k-1\Rightarrow k=2$. [1.0 Mark]
Check at $k=2$: $\dfrac{a_1}{a_2}=\dfrac{3}{3}=1$, $\dfrac{c_1}{c_2}=\dfrac{1}{5}$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=1
eq\dfrac15=\dfrac{c_1}{c_2}$, the no-solution condition is genuinely satisfied at $k=2$. [1.0 Mark]
Check at $k=2$: $\dfrac{a_1}{a_2}=\dfrac{3}{3}=1$, $\dfrac{c_1}{c_2}=\dfrac{1}{5}$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=1
eq\dfrac15=\dfrac{c_1}{c_2}$, the no-solution condition is genuinely satisfied at $k=2$. [1.0 Mark]
Question 40
Hint available
Points $A$ and $B$ are $100$ km apart. Two cars start simultaneously from $A$ and $B$, travelling towards each other, and meet after $5$ hours. If instead they travel in the same direction (starting from the same points), the faster car catches up to the slower one after $25$ hours. Find the speed of each car.
Key Idea
Form equations from 'towards each other' (relative speed = sum) and 'same direction' (relative speed = difference) scenarios.
Step-by-Step Solution
Let the speeds be $x$ km/h and $y$ km/h ($x>y$). Moving towards each other: $5(x+y)=100\Rightarrow x+y=20$. [1.0 Mark]
Moving in the same direction, the gap closes at rate $(x-y)$: $25(x-y)=100\Rightarrow x-y=4$. [1.0 Mark]
Adding the two equations: $2x=24\Rightarrow x=12$; then $y=20-12=8$. So the speeds are $12$ km/h and $8$ km/h. [1.0 Mark]
Moving in the same direction, the gap closes at rate $(x-y)$: $25(x-y)=100\Rightarrow x-y=4$. [1.0 Mark]
Adding the two equations: $2x=24\Rightarrow x=12$; then $y=20-12=8$. So the speeds are $12$ km/h and $8$ km/h. [1.0 Mark]
Question 41
Hint available
A fraction becomes $\dfrac{9}{11}$ if $2$ is added to both its numerator and denominator. It becomes $\dfrac56$ if $3$ is added to both. Find the fraction.
Key Idea
Let the fraction be $x/y$; translate both conditions into linear equations and solve.
Step-by-Step Solution
Let the fraction be $\dfrac{x}{y}$. From the first condition: $\dfrac{x+2}{y+2}=\dfrac{9}{11}\Rightarrow 11x+22=9y+18\Rightarrow 11x-9y=-4$ — (1). [1.0 Mark]
From the second condition: $\dfrac{x+3}{y+3}=\dfrac{5}{6}\Rightarrow 6x+18=5y+15\Rightarrow 6x-5y=-3$ — (2). [1.0 Mark]
Multiply (1) by $5$ and (2) by $9$: $55x-45y=-20$ and $54x-45y=-27$. Subtracting: $x=7$. Substituting into (2): $42-5y=-3\Rightarrow y=9$. The fraction is $\dfrac{7}{9}$. [1.0 Mark]
From the second condition: $\dfrac{x+3}{y+3}=\dfrac{5}{6}\Rightarrow 6x+18=5y+15\Rightarrow 6x-5y=-3$ — (2). [1.0 Mark]
Multiply (1) by $5$ and (2) by $9$: $55x-45y=-20$ and $54x-45y=-27$. Subtracting: $x=7$. Substituting into (2): $42-5y=-3\Rightarrow y=9$. The fraction is $\dfrac{7}{9}$. [1.0 Mark]
Question 42
Hint available
Check whether the pair of equations $x-2y=0$ and $3x+4y=20$ is consistent; if so, solve it.
Key Idea
Compare ratios to check consistency, then solve by substitution.
Step-by-Step Solution
$\dfrac{a_1}{a_2}=\dfrac13,\ \dfrac{b_1}{b_2}=\dfrac{-2}{4}=-\dfrac12$; since these are unequal, the pair is consistent with a unique solution. [1.0 Mark]
From $x-2y=0$, $x=2y$. Substituting into $3x+4y=20$: $3(2y)+4y=20\Rightarrow 10y=20\Rightarrow y=2$. [1.0 Mark]
$x=2(2)=4$. So $x=4,\ y=2$. [1.0 Mark]
From $x-2y=0$, $x=2y$. Substituting into $3x+4y=20$: $3(2y)+4y=20\Rightarrow 10y=20\Rightarrow y=2$. [1.0 Mark]
$x=2(2)=4$. So $x=4,\ y=2$. [1.0 Mark]
Question 43
Hint available
Solve the pair of equations $x-y+1=0$ and $3x+2y-12=0$ graphically. Also find the coordinates of the vertices of the triangle formed by these two lines and the $x$-axis, and hence find the area of this triangle.
Key Idea
Plot both lines using a table of values, read off the intersection point as the solution, then find where each line meets the $x$-axis to identify the triangle's vertices.
Step-by-Step Solution
For $x-y=-1$: when $x=-1,y=0$; when $x=2,y=3$. For $3x+2y=12$: when $x=4,y=0$; when $x=2,y=3$. Plotting both lines on the same graph. [1.0 Mark]
The two lines intersect at $(2,3)$, so the solution is $x=2,\ y=3$. [1.0 Mark]
Line $x-y=-1$ meets the $x$-axis at $(-1,0)$ (setting $y=0$); line $3x+2y=12$ meets the $x$-axis at $(4,0)$ (setting $y=0$). [1.0 Mark]
So the triangle has vertices $(2,3),\ (-1,0),\ (4,0)$. Its base lies along the $x$-axis, from $(-1,0)$ to $(4,0)$, so base $=4-(-1)=5$ units; the height is the $y$-coordinate of the third vertex $=3$ units. [1.0 Mark]
Area $=\dfrac12\times\text{base}\times\text{height}=\dfrac12\times5\times3=7.5$ square units. [1.0 Mark]
The two lines intersect at $(2,3)$, so the solution is $x=2,\ y=3$. [1.0 Mark]
Line $x-y=-1$ meets the $x$-axis at $(-1,0)$ (setting $y=0$); line $3x+2y=12$ meets the $x$-axis at $(4,0)$ (setting $y=0$). [1.0 Mark]
So the triangle has vertices $(2,3),\ (-1,0),\ (4,0)$. Its base lies along the $x$-axis, from $(-1,0)$ to $(4,0)$, so base $=4-(-1)=5$ units; the height is the $y$-coordinate of the third vertex $=3$ units. [1.0 Mark]
Area $=\dfrac12\times\text{base}\times\text{height}=\dfrac12\times5\times3=7.5$ square units. [1.0 Mark]
Question 44
Hint available
Ten years ago, a father was $12$ times as old as his son. Ten years from now, the father will be twice as old as the son will be then. Find their present ages.
Key Idea
Let the present ages be $F$ and $S$; translate both time-shifted conditions into linear equations and solve by substitution.
Step-by-Step Solution
Let the father's present age be $F$ years and the son's present age be $S$ years. [0.5 Mark]
\"Ten years ago, father was 12 times as old as son\": $F-10=12(S-10)\Rightarrow F=12S-110$ — (1). [1.5 Marks]
\"Ten years hence, father will be twice as old as son\": $F+10=2(S+10)\Rightarrow F=2S+10$ — (2). [1.5 Marks]
Equating (1) and (2): $12S-110=2S+10\Rightarrow 10S=120\Rightarrow S=12$. [1.0 Mark]
$F=2(12)+10=34$. So the father is $34$ years old and the son is $12$ years old. [0.5 Mark]
\"Ten years ago, father was 12 times as old as son\": $F-10=12(S-10)\Rightarrow F=12S-110$ — (1). [1.5 Marks]
\"Ten years hence, father will be twice as old as son\": $F+10=2(S+10)\Rightarrow F=2S+10$ — (2). [1.5 Marks]
Equating (1) and (2): $12S-110=2S+10\Rightarrow 10S=120\Rightarrow S=12$. [1.0 Mark]
$F=2(12)+10=34$. So the father is $34$ years old and the son is $12$ years old. [0.5 Mark]
Question 45
Hint available
Find the values of $a$ and $b$ for which the pair of equations $2x+3y=7$ and $(a-b)x+(a+b)y=3a+b-2$ has infinitely many solutions.
Key Idea
Set up the equal-ratios condition across all three coefficient pairs, and solve the resulting simultaneous equations in $a,b$.
Step-by-Step Solution
For infinitely many solutions: $\dfrac{2}{a-b}=\dfrac{3}{a+b}=\dfrac{7}{3a+b-2}$. [1.0 Mark]
From the first two: $2(a+b)=3(a-b)\Rightarrow 2a+2b=3a-3b\Rightarrow a=5b$ — (1). [1.0 Mark]
From the first and third: $2(3a+b-2)=7(a-b)\Rightarrow 6a+2b-4=7a-7b\Rightarrow a=9b-4$ — (2). [1.5 Marks]
Equating (1) and (2): $5b=9b-4\Rightarrow 4b=4\Rightarrow b=1$; then $a=5(1)=5$. [1.0 Mark]
Verification: $a-b=4,\ a+b=6,\ 3a+b-2=14$; ratios $\dfrac24=\dfrac36=\dfrac{7}{14}=\dfrac12$ — all equal, confirming $a=5,\ b=1$. [0.5 Mark]
From the first two: $2(a+b)=3(a-b)\Rightarrow 2a+2b=3a-3b\Rightarrow a=5b$ — (1). [1.0 Mark]
From the first and third: $2(3a+b-2)=7(a-b)\Rightarrow 6a+2b-4=7a-7b\Rightarrow a=9b-4$ — (2). [1.5 Marks]
Equating (1) and (2): $5b=9b-4\Rightarrow 4b=4\Rightarrow b=1$; then $a=5(1)=5$. [1.0 Mark]
Verification: $a-b=4,\ a+b=6,\ 3a+b-2=14$; ratios $\dfrac24=\dfrac36=\dfrac{7}{14}=\dfrac12$ — all equal, confirming $a=5,\ b=1$. [0.5 Mark]
Question 46
Hint available
Ritu can row downstream $20$ km in $2$ hours, and upstream $4$ km in $2$ hours. Find her speed of rowing in still water and the speed of the current.
Key Idea
Compute the downstream and upstream speeds directly from distance/time, then use (speed in still water + current speed) for downstream and (speed in still water − current speed) for upstream.
Step-by-Step Solution
Downstream speed $=\dfrac{20}{2}=10$ km/h; upstream speed $=\dfrac{4}{2}=2$ km/h. [1.0 Mark]
Let $x=$ speed in still water, $y=$ speed of current. Downstream: $x+y=10$ — (1). Upstream: $x-y=2$ — (2). [1.5 Marks]
Adding (1) and (2): $2x=12\Rightarrow x=6$. [1.5 Marks]
Substituting into (1): $6+y=10\Rightarrow y=4$. So Ritu's speed in still water is $6$ km/h and the speed of the current is $4$ km/h. [1.0 Mark]
Let $x=$ speed in still water, $y=$ speed of current. Downstream: $x+y=10$ — (1). Upstream: $x-y=2$ — (2). [1.5 Marks]
Adding (1) and (2): $2x=12\Rightarrow x=6$. [1.5 Marks]
Substituting into (1): $6+y=10\Rightarrow y=4$. So Ritu's speed in still water is $6$ km/h and the speed of the current is $4$ km/h. [1.0 Mark]
Question 47
Hint available
[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]
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]
Key Idea
Case study on linear equations in two variables.
Step-by-Step Solution
(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]
$x+10y=105$ 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]
Question 48
Hint available
[Case Study]
A boat goes $30$ km upstream and $44$ km downstream in $10$ hours. In $13$ hours, it can go $40$ km upstream and $55$ km downstream. Let the speed of the boat in still water be $x$ km/h and the speed of the stream be $y$ km/h ($x > y$).
(a) Write the equations for upstream speed and downstream speed in terms of $x$ and $y$. [1 Mark]
(b) Form a pair of linear equations representing the two journeys using $u = \dfrac{1}{x-y}$ and $v = \dfrac{1}{x+y}$. [1 Mark]
(c) Solve for $u$ and $v$. [1 Mark]
(d) Find the speed of the boat in still water ($x$) and speed of the stream ($y$). [1 Mark]
A boat goes $30$ km upstream and $44$ km downstream in $10$ hours. In $13$ hours, it can go $40$ km upstream and $55$ km downstream. Let the speed of the boat in still water be $x$ km/h and the speed of the stream be $y$ km/h ($x > y$).
(a) Write the equations for upstream speed and downstream speed in terms of $x$ and $y$. [1 Mark]
(b) Form a pair of linear equations representing the two journeys using $u = \dfrac{1}{x-y}$ and $v = \dfrac{1}{x+y}$. [1 Mark]
(c) Solve for $u$ and $v$. [1 Mark]
(d) Find the speed of the boat in still water ($x$) and speed of the stream ($y$). [1 Mark]
Key Idea
Case study on linear equations in two variables.
Step-by-Step Solution
(a) Write the equations for upstream speed and downstream speed in terms of $x$ and $y$. [1 Mark]
Upstream speed $= x - y$ km/h; Downstream speed $= x + y$ km/h. [1.0 Mark]
(b) Form a pair of linear equations representing the two journeys using $u = \dfrac{1}{x-y}$ and $v = \dfrac{1}{x+y}$. [1 Mark]
$30u + 44v = 10$ and $40u + 55v = 13$. [1.0 Mark]
(c) Solve for $u$ and $v$. [1 Mark]
$120u + 176v = 40$ and $120u + 165v = 39 \Rightarrow 11v = 1 \Rightarrow v = 1/11$. $u = 1/5$. [1.0 Mark]
(d) Find the speed of the boat in still water ($x$) and speed of the stream ($y$). [1 Mark]
$x - y = 5, x + y = 11 \Rightarrow 2x = 16 \Rightarrow x = 8$ km/h, $y = 3$ km/h. [1.0 Mark]
Upstream speed $= x - y$ km/h; Downstream speed $= x + y$ km/h. [1.0 Mark]
(b) Form a pair of linear equations representing the two journeys using $u = \dfrac{1}{x-y}$ and $v = \dfrac{1}{x+y}$. [1 Mark]
$30u + 44v = 10$ and $40u + 55v = 13$. [1.0 Mark]
(c) Solve for $u$ and $v$. [1 Mark]
$120u + 176v = 40$ and $120u + 165v = 39 \Rightarrow 11v = 1 \Rightarrow v = 1/11$. $u = 1/5$. [1.0 Mark]
(d) Find the speed of the boat in still water ($x$) and speed of the stream ($y$). [1 Mark]
$x - y = 5, x + y = 11 \Rightarrow 2x = 16 \Rightarrow x = 8$ km/h, $y = 3$ km/h. [1.0 Mark]
Question 49
Hint available
[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]
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]
Key Idea
Case study on linear equations in two variables.
Step-by-Step Solution
(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]
$(x+2)(y-1) = xy \Rightarrow xy - x + 2y - 2 = xy \Rightarrow -x + 2y = 2$. [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]
Question 50
Hint available
[Case Study]
A library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹$27$ for a book kept for seven days, while Susy paid ₹$21$ for the book she kept for five days.
(a) Let the fixed charge for 3 days be ₹$x$ and additional charge per day be ₹$y$. Form linear equations for Saritha and Susy. [1 Mark]
(b) Find the additional charge per day ($y$). [1 Mark]
(c) Find the fixed charge for the first 3 days ($x$). [1 Mark]
(d) Find the total amount paid by a student who keeps the book for 6 days. [1 Mark]
A library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹$27$ for a book kept for seven days, while Susy paid ₹$21$ for the book she kept for five days.
(a) Let the fixed charge for 3 days be ₹$x$ and additional charge per day be ₹$y$. Form linear equations for Saritha and Susy. [1 Mark]
(b) Find the additional charge per day ($y$). [1 Mark]
(c) Find the fixed charge for the first 3 days ($x$). [1 Mark]
(d) Find the total amount paid by a student who keeps the book for 6 days. [1 Mark]
Key Idea
Case study on linear equations in two variables.
Step-by-Step Solution
(a) Let the fixed charge for 3 days be ₹$x$ and additional charge per day be ₹$y$. Form linear equations for Saritha and Susy. [1 Mark]
$x + 4y = 27$ and $x + 2y = 21$. [1.0 Mark]
(b) Find the additional charge per day ($y$). [1 Mark]
$(x+4y) - (x+2y) = 27 - 21 \Rightarrow 2y = 6 \Rightarrow y = 3$. Additional charge = ₹$3$/day. [1.0 Mark]
(c) Find the fixed charge for the first 3 days ($x$). [1 Mark]
$x + 6 = 21 \Rightarrow x = 15$. Fixed charge = ₹$15$. [1.0 Mark]
(d) Find the total amount paid by a student who keeps the book for 6 days. [1 Mark]
$ ext{Total} = 15 + 3(3) = 15 + 9 = ₹24$. [1.0 Mark]
$x + 4y = 27$ and $x + 2y = 21$. [1.0 Mark]
(b) Find the additional charge per day ($y$). [1 Mark]
$(x+4y) - (x+2y) = 27 - 21 \Rightarrow 2y = 6 \Rightarrow y = 3$. Additional charge = ₹$3$/day. [1.0 Mark]
(c) Find the fixed charge for the first 3 days ($x$). [1 Mark]
$x + 6 = 21 \Rightarrow x = 15$. Fixed charge = ₹$15$. [1.0 Mark]
(d) Find the total amount paid by a student who keeps the book for 6 days. [1 Mark]
$ ext{Total} = 15 + 3(3) = 15 + 9 = ₹24$. [1.0 Mark]