CBSE Class 10 Mathematics Pair Of Linear Equations In 2 Variables Worksheet Set 03

Pair of Linear Equations in Two Variables

(Key Points)

• An equation of the form ax + by + c = 0, where a, b, c are real nos. (a  0, b  0) i.e (a2+b2 ≠0) is called a linear equation in two variables x and y.

Ex : (i) x – 5y + 2 =0

(ii) 32 x – y =1

• The general form for a pair of linear equations in two variables x and y is

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

Where a1, b1, c1, a2, b2, c2 are all real nos and a1 0, b1 0, a2 0, b2 0.

Examples

• Graphical representation of a pair of linear equations in two variables:

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

(i) Will represent intersecting lines if

I.e. unique solution. And these types of equations are called consistent pair of linear equations.

Ex: x – 2y = 0

3x + 4y – 20 = 0

(ii) will represent overlapping or coincident lines if

i.e. Infinitely many solutions, consistent or dependent pair of linear equations

Ex: 2x + 3y – 9 = 0

4x + 6y – 18 = 0

(iii) will represent parallel lines if

i.e. no solution and called inconsistent pair of linear equations.

Ex: x + 2y – 4 = 0

2x + 4y – 12 = 0

• Algebraic methods of solving a pair of linear equations:

(i) Substitution method

(ii) Elimination Method

(iii) Cross multiplication method

Level - I

1. Find the value of ‘a’ so that the point(2,9) lies on the line represented by ax-3y=5

2. Find the value of k so that the lines 2x – 3y = 9 and kx-9y =18 will be parallel.

3. Find the value of k for which x + 2y =5, 3x+ky+15=0 is inconsistent

4. Check whether given pair of lines is consistent or not 5x – 1 = 2y, y = +

5. Determine the value of ‘a’ if the system of linear equations 3x+2y -4 =0 and ax – y – 3 = 0 will represent intersecting lines.

6. Write any one equation of the line which is parallel to 2x – 3y =5

7. Find the point of intersection of line -3x + 7y =3 with x-axis

8. For what value of k the following pair has infinite number of solutions.

(k-3)x + 3y = k

k(x+y)=12

9. Write the condition sothat a1x + b1y = c1 and a2x + b2y = c2 have unique solution.

Level - II

1. 5 pencils and 7pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.

2. Solve the equations:

3x – y = 3

7x + 2y = 20

3. Find the fraction which becomes to 2/3 when the numerator is increased by 2 and equal to 4/7 when the denominator is increased by 4

4. Solve the equation:

px + qy = p – q

qx – py = p + q

5. Solve the equation using the method of substitution:

6. Solve the equations:

Where, x

7. Solve the equations by using the method of cross multiplication:

5x + 12y =7

Level - III

1. Draw the graph of the equations

4x – y = 4

4x + y = 12

Determine the vertices of the triangle formed by the lines representing these equations and the x-axis. Shade the triangular region so formed

2. Solve Graphically

x – y = -1 and

3x + 2y = 12

Choose the correct answer from the given options:

Question. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, (in years) of the son and the father are, respectively
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24
Answer : C

Question. A purse contains 25 paise and 10 paise coins. The total amount in the purse is ₹ 8.25. If the number of 25 paise coins is one-third of the number of 10 paise coins in the purse, then the total number of coins in the purse is
(a) 60
(b) 40
(c) 80
(d) 72
Answer : A

Question. Aruna has only ₹ 1 and ₹ 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is ₹ 75, then the number of ₹ 1 and ₹ 2 coins are, respectively
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Answer : A

Question. A fraction becomes 3/1 when 2 is subtracted from the numerator and it becomes 2/1 when 1 is subtracted from the denominator. The fraction is
(a) 2/5
(b) 5/18
(c) 4/13
(d) 7/15
Answer : D

Question. The difference between two numbers is 26 and the larger number exceeds thrice of the smaller number by 4. The numbers are
(a) 39, 13
(b) 12, 38
(c) 37, 11
(d) None of these
Answer : C

Question. A man has some hens and cows. If the number of heads be 48 and the number of feet equals 140, the number of hens will be
(a) 18
(b) 26
(c) 32
(d) 40
Answer : B

Question. If 3 chairs and 1 table costs ₹ 1500 and 6 chairs and 1 table costs ₹ 2400, the pair of linear equations to represent this situation is
(a) 6x + y = 1500, 3x + y = 2400
(b) x/3 + y = 1500, x/6 + y = 2400
(c) 3x + y = 1500, 6x + y = 2400
(d) None of these
Answer : C

Question. Meena went to a bank to withdraw ₹ 2,000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. How many notes of ₹ 50 and ₹ 100 she received?
(a) ₹ 50 : 10, ₹ 100 : 15
(b) ₹ 50 : 12, ₹ 100 : 10
(c) ₹ 50 : 15, ₹ 100 : 10
(d) None of these
Answer : A

Question. A two-digit number is obtained by either multiplying the sum of digits by 8 and then subtracting 5 or by multiplying the difference of digits by 16 and adding 3. The number is
(a) 23
(b) 34
(c) 83
(d) 119
Answer : C

Question. The sum of a two-digit number and the number obtained by interchanging the digits is 132. If the two digits differ by 2, the number is
(a) 45
(b) 75
(c) 85
(d) 115
Answer : B

Question. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the son. The present age of father and son, respectively are
(a) 40 years, 12 years
(b) 30 years, 6 years
(c) 32 years, 8 years
(d) 42 years, 10 years
Answer : D

Question. A motor boat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. The speed of the stream is
(a) 60 km/hr
(b) 6 km/min
(c) 6 km/s
(d) 6 km/hr
Answer : B

Question. The two consecutive odd positive integers, sum of whose squares is 290 are
(a) 5, 13
(b) 11, 13
(c) 13, 17
(d) None of these
Answer : B

Question. Does the point (1, – 2) lie on the line whose equation is 3x – y – 5 = 0?
Answer: yes

Question. How many solutions of the equation 5x – 4y + 11 = 0 are possible?
Answer: Infinite solution

Question. Write a linear equation in two variables which is consistent to equation 5(x – y) = 3.
Answer: 2x + 3y = 7 or more others

Question. Which axis is the graph of equation x = 0?
Answer: y-axis

Question. What is the solution of √2 x – √5 y = 0 and 2. √3 x – √7 y = 0.
Answer: x = 0, y = 0

Question. Find the co – ordinates of the point where the line 2x – 3y = 6 meets x – axis and y – axis.
Answer: (3,0),(0,-2)

Question. What is the value of ‘K’ for which the graph of the equations 2x – 3y = 9 and kx – 9y = 18 are parallel lines?
Answer: k=6

Question. For what value of ‘K’ the pair of equations x – ky + 4 = 0 and 2x – 4y – 8 = 0 is inconsistent?
Answer: k=2

Questions of 2 Marks

Question. Aftab tells his daughter, “Seven years ago I was 7 times as old as you were then, also 3 years from now I shall be 3 times as old as you will be. Represent the situation algebraically.
Answer: x – 7y + 42 = 0
x – 3y – 6 = 0

Question. For what value of ’K’ will the following equations has no solution.
3x + y = 1, (2K – 1)x + (K – 1)y = 2K + 1.
Answer: k = 2

Question. Solve for x and y:
103x + 51y = 617,
97x + 49y = 583.
Answer: x = 5, y = 2

Question. The sum of two numbers is 35 and their difference is 13 find the Numbers.
Answer: x = 24, y = 11

Linear Equation in Two Variables

An equation of the form \( ax + by + c = 0 \), where \( a, b, c \) are real numbers such that \( a \neq 0 \) and \( b \neq 0 \) (which can also be written as \( a^2 + b^2 \neq 0 \)), is called a linear equation in two variables \( x \) and \( y \).

• Example (i): \( x - 5y + 2 = 0 \)
• Example (ii): \( \frac{3}{2}x - y = 1 \)

General Form of a Pair of Linear Equations

The general algebraic form for a pair of linear equations in two variables \( x \) and \( y \) is:

\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)

where \( a_1, b_1, c_1, a_2, b_2, c_2 \) are real numbers and \( a_1, b_1, a_2, b_2 \neq 0 \).

• Example:
  \( x + 3y - 6 = 0 \)
  \( 2x - 3y - 12 = 0 \)

Graphical Representation and Consistency

When plotted on a graph, the behavior of the two lines determines the consistency of the system:

• (i) Intersecting Lines: This occurs if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). The system has a unique (single) solution, and the pair of equations is called consistent.
  Example:
  \( x - 2y = 0 \)
  \( 3x + 4y - 20 = 0 \)
• 
• The coordinates of the point of intersection give the unique solution to the system.
• (ii) Overlapping or Coincident Lines: This occurs if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). The system has infinitely many solutions, and the pair of equations is called consistent or dependent.
  Example:
  \( 2x + 3y - 9 = 0 \)
  \( 4x + 6y - 18 = 0 \)
• 
• (iii) Parallel Lines: This occurs if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). The system has no solution, and the pair of equations is called inconsistent.
  Example:
  \( x + 2y - 4 = 0 \)
  \( 2x + 4y - 12 = 0 \)
• 

Algebraic Methods of Solving Systems of Linear Equations

• Substitution method
• Elimination method
• Cross-multiplication method

Level - I

Question 1. Find the value of ‘a’ so that the point(2,9) lies on the line represented by ax-3y=5
Answer: Since the point \( (2, 9) \) lies on the line \( ax - 3y = 5 \), substituting \( x = 2 \) and \( y = 9 \) must satisfy the equation:
\( a(2) - 3(9) = 5 \)
\( \implies 2a - 27 = 5 \)
\( \implies 2a = 5 + 27 \)
\( \implies 2a = 32 \)
\( \implies a = 16 \)
Therefore, the value of \( a \) is 16.
In simple words: Substitute 2 for x and 9 for y in the equation, then solve for a. This gives us a value of 16.

Exam Tip: Whenever a point is stated to lie on a line, always plug its coordinates directly into the equation to find the unknown coefficient.

Question 2. Find the value of k so that the lines 2x – 3y = 9 and kx-9y =18 will be parallel.
Answer: For two linear equations to represent parallel lines, the coefficients must satisfy the condition:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Given equations are \( 2x - 3y = 9 \) and \( kx - 9y = 18 \). Comparing the coefficients:
\( a_1 = 2, b_1 = -3, c_1 = 9 \)
\( a_2 = k, b_2 = -9, c_2 = 18 \)
Using the parallel line condition:
\( \frac{2}{k} = \frac{-3}{-9} \neq \frac{9}{18} \)
\( \implies \frac{2}{k} = \frac{1}{3} \)
\( \implies k = 6 \)
Since \( \frac{2}{6} = \frac{1}{3} \) is not equal to \( \frac{9}{18} = \frac{1}{2} \), \( k = 6 \) is the correct value.
In simple words: For the lines to be parallel, the ratio of the x-coefficients must match the ratio of the y-coefficients but not the constants. Setting up this ratio gives us \( k = 6 \).

Exam Tip: Always check the constant ratio \( \frac{c_1}{c_2} \) to ensure the lines are not coincident instead of parallel.

Question 3. Find the value of k for which x + 2y =5, 3x+ky+15=0 is inconsistent
Answer: For a system to be inconsistent, the lines must be parallel, satisfying:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Given \( x + 2y - 5 = 0 \) and \( 3x + ky + 15 = 0 \):
\( a_1 = 1, b_1 = 2, c_1 = -5 \)
\( a_2 = 3, b_2 = k, c_2 = 15 \)
Substituting these values:
\( \frac{1}{3} = \frac{2}{k} \neq \frac{-5}{15} \)
\( \implies \frac{1}{3} = \frac{2}{k} \)
\( \implies k = 6 \)
Since \( \frac{1}{3} \neq -\frac{1}{3} \), the system is inconsistent for \( k = 6 \).
In simple words: An inconsistent system has no solutions because the lines are parallel. Setting the coefficient ratios equal gives \( k = 6 \).

Exam Tip: Make sure both equations are written in standard form (\( ax + by + c = 0 \)) before comparing constants to avoid sign errors.

Question 4. Check whether given pair of lines is consistent or not 5x – 1 = 2y, y = -1/2 + 5/2 x
Answer: Let's express both equations in standard form \( ax + by + c = 0 \).
First equation:
\( 5x - 2y - 1 = 0 \)
Second equation:
\( y = -\frac{1}{2} + \frac{5}{2}x \)
Multiply by 2:
\( 2y = -1 + 5x \implies 5x - 2y - 1 = 0 \)
Since both equations simplify to the exact same relation, they represent coincident lines. Coincident lines have infinitely many solutions, so the system is consistent (dependent).
In simple words: If you rearrange the second equation, it is identical to the first. Because they represent the exact same line, there are infinite solutions, making the system consistent.

Exam Tip: Coincident lines have a consistent pair because "consistent" simply means having at least one solution.

Question 5. Determine the value of ‘a’ if the system of linear equations 3x+2y -4 =0 and ax – y – 3 = 0 will represent intersecting lines.
Answer: The condition for two lines to be intersecting (having a unique solution) is:
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Given \( 3x + 2y - 4 = 0 \) and \( ax - y - 3 = 0 \):
\( a_1 = 3, b_1 = 2 \)
\( a_2 = a, b_2 = -1 \)
Using the intersecting condition:
\( \frac{3}{a} \neq \frac{2}{-1} \)
\( \implies \frac{3}{a} \neq -2 \)
\( \implies a \neq -\frac{3}{2} \)
Therefore, \( a \) can be any real number except \( -\frac{3}{2} \).
In simple words: The lines will cross as long as the ratios of their x and y coefficients are not equal. This means \( a \) can be anything except \( -1.5 \).

Exam Tip: Express the final answer as "all real values except \( -\frac{3}{2} \)" rather than just writing the inequality to ensure you get full credit.

Question 6. Write any one equation of the line which is parallel to 2x – 3y =5
Answer: For a line to be parallel to \( \sqrt{2}x - \sqrt{3}y = 5 \), the ratio of the coefficients of \( x \) and \( y \) must remain identical, while the constant term is different.
One such equation is:
\( 5\sqrt{2}x - 5\sqrt{3}y = 5\sqrt{5} \)
Other simpler options like \( \sqrt{2}x - \sqrt{3}y = 10 \) are also perfectly correct.
In simple words: Parallel lines must have the same slope, meaning the x and y parts stay in the same ratio, but the constant at the end is different.

Exam Tip: The easiest way to write a parallel line is to keep the left-hand side identical and just change the constant on the right-hand side.

Question 7. Find the point of intersection of line -3x + 7y =3 with x-axis
Answer: The intersection point of any line with the x-axis always has a y-coordinate equal to 0.
Substituting \( y = 0 \) into \( -3x + 7y = 3 \):
\( -3x + 7(0) = 3 \)
\( \implies -3x = 3 \)
\( \implies x = -1 \)
Therefore, the coordinates of the intersection point are \( (-1, 0) \).
In simple words: Since the line crosses the x-axis, the height (y) is 0. Putting y = 0 into the equation gives x = -1.

Exam Tip: Remember that points on the x-axis are always in the form \( (x, 0) \), and points on the y-axis are in the form \( (0, y) \).

Question 8. For what value of k the following pair has infinite number of solutions. (k-3)x + 3y = k k(x+y)=12
Answer: For a system to have an infinite number of solutions, the lines must be coincident, which requires:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Rewriting the second equation \( k(x + y) = 12 \) as \( kx + ky = 12 \).
Our coefficients are:
\( a_1 = k-3, b_1 = 3, c_1 = k \)
\( a_2 = k, b_2 = k, c_2 = 12 \)
Setting up the ratios:
\( \frac{k-3}{k} = \frac{3}{k} = \frac{k}{12} \)
From the equation \( \frac{k-3}{k} = \frac{3}{k} \):
\( k - 3 = 3 \implies k = 6 \)
Checking with the remaining ratio \( \frac{k}{12} \) when \( k = 6 \):
\( \frac{3}{6} = \frac{6}{12} = \frac{1}{2} \), which is consistent.
Therefore, \( k = 6 \).
In simple words: For infinite solutions, the ratios of the coefficients must all be equal. Solving this gives us \( k = 6 \).

Exam Tip: Always substitute your final value of \( k \) back into all three ratios to make sure they are indeed equal.

Question 9. Write the condition sothat a1x + b1y = c1 and a2x + b2y = c2 have unique solution.
Answer: For the system of linear equations \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \) to have a unique (single) solution, the lines must intersect.
The algebraic condition for intersecting lines is:
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
In simple words: For there to be exactly one solution, the ratio of the x-coefficients must not equal the ratio of the y-coefficients.

Exam Tip: Keep this condition in mind, as it forms the basis of all unique solution problems in linear equations.

Level - II

Question 1. 5 pencils and 7pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
Answer: Let the price of a single pencil be \( x \) and that of a single pen be \( y \).
Based on the problem:
\( 5x + 7y = 50 \) - (Equation 1)
\( 7x + 5y = 46 \) - (Equation 2)
Adding Equation 1 and Equation 2:
\( 12x + 12y = 96 \implies x + y = 8 \) - (Equation 3)
Subtracting Equation 2 from Equation 1:
\( -2x + 2y = 4 \implies -x + y = 2 \) - (Equation 4)
Adding Equation 3 and Equation 4:
\( 2y = 10 \implies y = 5 \)
Substituting \( y = 5 \) into Equation 3:
\( x + 5 = 8 \implies x = 3 \)
Therefore, the cost of one pencil is Rs. 3 and the cost of one pen is Rs. 5.
In simple words: We can make two equations using the costs. Adding and subtracting them simplifies the math, showing that a pencil costs Rs. 3 and a pen costs Rs. 5.

Exam Tip: For symmetric equations where the coefficients of x and y are swapped (like 5 and 7 here), adding and subtracting the equations first makes them much easier to solve.

Question 2. Solve the equations: 3x – y = 3 7x + 2y = 20
Answer: Let's solve the system:
\( 3x - y = 3 \) - (Equation 1)
\( 7x + 2y = 20 \) - (Equation 2)
From Equation 1, we can express \( y \) in terms of \( x \):
\( y = 3x - 3 \)
Substituting this into Equation 2:
\( 7x + 2(3x - 3) = 20 \)
\( \implies 7x + 6x - 6 = 20 \)
\( \implies 13x = 26 \)
\( \implies x = 2 \)
Now, substituting \( x = 2 \) back to find \( y \):
\( y = 3(2) - 3 = 3 \)
Thus, the solution is \( x = 2 \) and \( y = 3 \).
In simple words: We rewrite the first equation to express y as 3x - 3, then substitute this into the second equation to get x = 2 and y = 3.

Exam Tip: Substitution is highly efficient when one of the variables has a coefficient of 1 or -1.

Question 3. Find the fraction which becomes to 2/3 when the numerator is increased by 2 and equal to 4/7 when the denominator is increased by 4
Answer: Let the required fraction be \( \frac{x}{y} \).
Based on the given conditions:
1. When the numerator is increased by 2, the fraction is \( \frac{2}{3} \):
\( \frac{x + 2}{y} = \frac{2}{3} \implies 3(x + 2) = 2y \implies 3x - 2y = -6 \) - (Equation 1)
2. When the denominator is increased by 4, the fraction is \( \frac{4}{7} \):
\( \frac{x}{y + 4} = \frac{4}{7} \implies 7x = 4(y + 4) \implies 7x - 4y = 16 \) - (Equation 2)
Multiply Equation 1 by 2:
\( 6x - 4y = -12 \) - (Equation 3)
Subtract Equation 3 from Equation 2:
\( (7x - 4y) - (6x - 4y) = 16 - (-12) \)
\( \implies x = 28 \)
Substitute \( x = 28 \) into Equation 1:
\( 3(28) - 2y = -6 \)
\( \implies 84 - 2y = -6 \)
\( \implies 2y = 90 \implies y = 45 \)
So, the fraction is \( \frac{28}{45} \).
In simple words: We assume the fraction is x/y. Setting up equations from the word description and solving them gives the numerator as 28 and the denominator as 45.

Exam Tip: Always clearly define the numerator as x and denominator as y at the start of your solution to keep the steps clear.

Question 4. Solve the equation: px + qy = p – q qx – py = p + q
Answer: Let's solve the system of literal equations:
\( px + qy = p - q \) - (Equation 1)
\( qx - py = p + q \) - (Equation 2)
To eliminate \( y \), multiply Equation 1 by \( p \) and Equation 2 by \( q \):
\( p^2x + pqy = p^2 - pq \) - (Equation 3)
\( q^2x - pqy = pq + q^2 \) - (Equation 4)
Add Equation 3 and Equation 4:
\( (p^2 + q^2)x = p^2 + q^2 \)
\( \implies x = \frac{p^2 + q^2}{p^2 + q^2} = 1 \)
Substitute \( x = 1 \) into Equation 1:
\( p(1) + qy = p - q \)
\( \implies qy = p - q - p \)
\( \implies qy = -q \implies y = -1 \)
Thus, \( x = 1 \) and \( y = -1 \).
In simple words: We multiply the equations to make the y-coefficients match but with opposite signs. Adding them cancels out y, leaving x = 1, which then easily gives y = -1.

Exam Tip: Do not be intimidated by variables like p and q instead of numbers. Treat them as constants and follow standard elimination steps.

Question 5. Solve the equation using the method of substitution: 3x – 5y = –1 x – y = –1
Answer: We solve using the substitution method:
\( 3x - 5y = -1 \) - (Equation 1)
\( x - y = -1 \) - (Equation 2)
From Equation 2, express \( x \) in terms of \( y \):
\( x = y - 1 \)
Substitute this expression into Equation 1:
\( 3(y - 1) - 5y = -1 \)
\( \implies 3y - 3 - 5y = -1 \)
\( \implies -2y - 3 = -1 \)
\( \implies -2y = 2 \implies y = -1 \)
Substitute \( y = -1 \) back to find \( x \):
\( x = -1 - 1 = -2 \)
So, the solution is \( x = -2 \) and \( y = -1 \).
In simple words: We rearrange the second equation to get x = y - 1, then plug this into the first equation to solve for y. This gives y = -1 and x = -2.

Exam Tip: If the question specifies a particular method (like "method of substitution"), you must use that exact method to get full marks.

Question 6. Solve the equations: 1/(2x) - 1/y = -1 1/x + 1/(2y) = 8 Where, x ≠ 0, y ≠ 0
Answer: Let \( \frac{1}{x} = p \) and \( \frac{1}{y} = q \).
Our equations become:
\( \frac{p}{2} - q = -1 \implies p - 2q = -2 \) - (Equation 1)
\( p + \frac{q}{2} = 8 \implies 2p + q = 16 \) - (Equation 2)
Multiply Equation 2 by 2:
\( 4p + 2q = 32 \) - (Equation 3)
Add Equation 1 and Equation 3:
\( p - 2q + 4p + 2q = -2 + 32 \)
\( \implies 5p = 30 \implies p = 6 \)
Substitute \( p = 6 \) into Equation 2:
\( 2(6) + q = 16 \implies 12 + q = 16 \implies q = 4 \)
Now, convert back to \( x \) and \( y \):
\( x = \frac{1}{p} = \frac{1}{6} \)
\( y = \frac{1}{q} = \frac{1}{4} \)
Thus, the solution is \( x = \frac{1}{6} \) and \( y = \frac{1}{4} \).
In simple words: We replace 1/x and 1/y with simpler letters like p and q. After finding p = 6 and q = 4, we flip them back to get x = 1/6 and y = 1/4.

Exam Tip: Substituting variables to reduce equations to linear form is the standard approach for equations where variables are in the denominator.

Question 7. Solve the equations by using the method of cross multiplication: x + y = 7 5x + 12y =7
Answer: Let's write both equations in the standard form \( ax + by + c = 0 \):
\( 1x + 1y - 7 = 0 \)
\( 5x + 12y - 7 = 0 \)
Using the cross-multiplication formula:
\( \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \)
Here, we have:
\( a_1 = 1, b_1 = 1, c_1 = -7 \)
\( a_2 = 5, b_2 = 12, c_2 = -7 \)
Substituting the values:
\( \frac{x}{(1)(-7) - (12)(-7)} = \frac{y}{(-7)(5) - (-7)(1)} = \frac{1}{(1)(12) - (5)(1)} \)
\( \implies \frac{x}{-7 + 84} = \frac{y}{-35 + 7} = \frac{1}{12 - 5} \)
\( \implies \frac{x}{77} = \frac{y}{-28} = \frac{1}{7} \)
This gives:
\( x = \frac{77}{7} = 11 \)
\( y = \frac{-28}{7} = -4 \)
So, the solution is \( x = 11 \) and \( y = -4 \).
In simple words: We rearrange the equations so they equal zero, then use the cross-multiplication formula to find x = 11 and y = -4.

Exam Tip: Double check your signs when using cross-multiplication, especially since \( c_1 \) and \( c_2 \) are negative here.

Level - III

Question 1. Draw the graph of the equations 4x – y = 4 and 4x + y = 12. Determine the vertices of the triangle formed by the lines representing these equations and the x- axis. Shade the triangular region so formed
Answer: First, let's determine a few points for each line to plot them:
For \( 4x - y = 4 \):
- If \( x = 1 \), \( y = 0 \). Point is \( (1, 0) \).
- If \( x = 2 \), \( y = 4 \). Point is \( (2, 4) \).

For \( 4x + y = 12 \):
- If \( x = 3 \), \( y = 0 \). Point is \( (3, 0) \).
- If \( x = 2 \), \( y = 4 \). Point is \( (2, 4) \).

The intersection point of the two lines is \( (2, 4) \).
The lines intersect the x-axis where \( y = 0 \), which gives the points \( (1, 0) \) and \( (3, 0) \).
Therefore, the vertices of the triangle formed by these lines and the x-axis are \( (2, 4) \export \), \( (1, 0) \), and \( (3, 0) \).
In simple words: We plot both lines on a graph. The point where they cross is (2, 4). The points where they touch the flat x-axis are (1, 0) and (3, 0). These three coordinates form the corners of the triangle.

Exam Tip: Always label the axes, write down the equations on the drawn lines, and explicitly list the three vertices of the shaded triangle.

Question 2. Solve Graphically x – y = -1 and 3x + 2y = 12. Calculate the area bounded by these lines and the x- axis,
Answer: Let's find points to graph each line:
For \( x - y = -1 \):
- If \( x = -1 \), \( y = 0 \). Point is \( (-1, 0) \).
- If \( x = 2 \), \( y = 3 \). Point is \( (2, 3) \).

For \( 3x + 2y = 12 \):
- If \( x = 4 \), \( y = 0 \). Point is \( (4, 0) \).
- If \( x = 2 \), \( y = 3 \). Point is \( (2, 3) \).

Graphing these lines shows they intersect at \( (2, 3) \).
The triangle formed by these lines and the x-axis has vertices at \( (2, 3) \), \( (-1, 0) \), and \( (4, 0) \).
- Base along the x-axis: \( 4 - (-1) = 5 \) units.
- Height of the triangle: 3 units (the y-coordinate of the intersection).
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 3 = 7.5 \text{ sq. units} \)
In simple words: Graphing the lines shows they cross at (2,3). The base of the triangle on the x-axis goes from -1 to 4, which is 5 units long. The height is 3 units. The area is \( 0.5 \times 5 \times 3 = 7.5 \) square units.

Exam Tip: Calculate the base length as \( x_{\text{right}} - x_{\text{left}} \) to avoid coordinate sign confusion.

Question 3. Solve :- for u & v: 4u – v = 14uv 3u + 2v = 16uv where u≠0, v≠ 0
Answer: Since \( u \neq 0 \) and \( v \neq 0 \), we divide both equations by \( uv \) to simplify them:
First equation:
\( \frac{4u - v}{uv} = \frac{14uv}{uv} \implies \frac{4}{v} - \frac{1}{u} = 14 \)
Second equation:
\( \frac{3u + 2v}{uv} = \frac{16uv}{uv} \implies \frac{3}{v} + \frac{2}{u} = 16 \)
Now, substitute \( \frac{1}{v} = x \) and \( \frac{1}{u} = y \):
\( 4x - y = 14 \) - (Equation 1)
\( 3x + 2y = 16 \) - (Equation 2)
Multiply Equation 1 by 2:
\( 8x - 2y = 28 \) - (Equation 3)
Add Equation 2 and Equation 3:
\( 11x = 44 \implies x = 4 \)
Substitute \( x = 4 \) into Equation 1:
\( 4(4) - y = 14 \implies 16 - y = 14 \implies y = 2 \)
Substitute back to find \( u \) and \( v \):
\( v = \frac{1}{x} = \frac{1}{4} \)
\( u = \frac{1}{y} = \frac{1}{2} \)
Thus, the solution is \( u = \frac{1}{2} \) and \( v = \frac{1}{4} \).
In simple words: We divide both equations by \( uv \export \), which turns them into a standard system. After solving for the substituted variables, we take the reciprocals to find \( u = 1/2 \) and \( v = 1/4 \).

Exam Tip: Remember that dividing by \( uv \) is only mathematically valid because the question explicitly states \( u \neq 0 \) and \( v \neq 0 \).

Question 4. Ritu can row downstream 20 km in 2 hr , and upstream 4 km in 2 hr . Find her speed of rowing in still water and the speed of the current. (HOTS)
Answer: Let the speed of Ritu in still water be \( x\text{ km/h} \) and the speed of the water current be \( y\text{ km/h} \).
- Downstream speed = \( (x + y)\text{ km/h} \)
- Upstream speed = \( (x - y)\text{ km/h} \)

Using the relation \( \text{Speed} = \frac{\text{Distance}}{\text{Time}} \export \):
1. Downstream motion:
\( x + y = \frac{20}{2} \implies x + y = 10 \) - (Equation 1)
2. Upstream motion:
\( x - y = \frac{4}{2} \implies x - y = 2 \) - (Equation 2)

Adding Equation 1 and Equation 2:
\( 2x = 12 \implies x = 6 \)
Substituting \( x = 6 \) into Equation 1:
\( 6 + y = 10 \implies y = 4 \)

Therefore, Ritu's speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h.
In simple words: Let her speed be x and water speed be y. When going with the water, speeds add up to 10 km/h. When fighting the water, speed is 2 km/h. Solving these gives her speed as 6 km/h and the water as 4 km/h.

Exam Tip: Set up speed equations directly from the downstream and upstream formulas, making sure to define your variables clearly first.

Question 5. In a Δ ABC, ∠C = 3∠B = 2 (∠A + ∠B ) find the these angle. (HOTS)
Answer: In \( \Delta ABC \), the sum of all interior angles is \( 180^\circ \):
\( \angle A + \angle B + \angle C = 180^\circ \) - (Equation 1)
We are given:
\( \angle C = 3\angle B \) - (Equation 2)
and
\( 3\angle B = 2(\angle A + \angle B) \)
\( \implies 3\angle B = 2\angle A + 2\angle B \)
\( \implies \angle B = 2\angle A \implies \angle A = \frac{\angle B}{2} \) - (Equation 3)
Now, substitute Equation 2 and Equation 3 into Equation 1:
\( \frac{\angle B}{2} + \angle B + 3\angle B = 180^\circ \)
\( \implies 4.5\angle B = 180^\circ \)
\( \implies \angle B = 40^\circ \)
Using our values to find the remaining angles:
\( \angle A = \frac{40^\circ}{2} = 20^\circ \)
\( \angle C = 3(40^\circ) = 120^\circ \)
Thus, the angles of the triangle are \( \angle A = 20^\circ \), \( \angle B = 40^\circ \), and \( \angle C = 120^\circ \).
In simple words: We know the angles add up to 180 degrees. Using the relationships given, we write all angles in terms of angle B. This allows us to solve and find the angles as 20, 40, and 120 degrees.

Exam Tip: Expressing all angles in terms of a single angle (like angle B here) is a highly reliable way to solve triangle angle relationship problems.

Question 6. 8 men and 12 boys can finish a piece of work in 10 days while 6 men and 8 boys can finish it in 14 days. Find the time taken by 1 man alone and that by one boy alone to finish the work. (HOTS)
Answer: Let the time taken by 1 man alone to finish the work be \( x \) days, and by 1 boy alone be \( y \) days.
- Work done by 1 man in 1 day = \( \frac{1}{x} \)
- Work done by 1 boy in 1 day = \( \frac{1}{y} \)

Based on the given data:
1. 8 men and 12 boys do the work in 10 days:
\( \frac{8}{x} + \frac{12}{y} = \frac{1}{10} \) - (Equation 1)
2. 6 men and 8 boys do the work in 14 days:
\( \frac{6}{x} + \frac{8}{y} = \frac{1}{14} \) - (Equation 2)

To eliminate the \( x \) term, multiply Equation 1 by 3 and Equation 2 by 4:
\( \frac{24}{x} + \frac{36}{y} = \frac{3}{10} \) - (Equation 3)
\( \frac{24}{x} + \frac{32}{y} = \frac{2}{7} \) - (Equation 4)

Subtracting Equation 4 from Equation 3:
\( \frac{4}{y} = \frac{3}{10} - \frac{2}{7} \)
\( \implies \frac{4}{y} = \frac{21 - 20}{70} = \frac{1}{70} \)
\( \implies y = 280 \)

Substitute \( y = 280 \) into Equation 1:
\( \frac{8}{x} + \frac{12}{280} = \frac{1}{10} \)
\( \implies \frac{8}{x} + \frac{3}{70} = \frac{1}{10} \)
\( \implies \frac{8}{x} = \frac{1}{10} - \frac{3}{70} = \frac{4}{70} = \frac{2}{35} \)
\( \implies 2x = 280 \implies x = 140 \)

Therefore, 1 man alone takes 140 days to complete the work, and 1 boy alone takes 280 days.
In simple words: We write equations for the daily work rate. Solving them reveals that one man working completely on his own would need 140 days, while one boy would take 280 days.

Exam Tip: In rate/work problems, always model the equations using the work done in one single day to establish linear-like relationships.

Question 7. Find the value of K for which the system of linear equations 2x+5y = 3, (k +1 )x + 2(k + 2) y = 2K will have infinite number of solutions. (HOTS)
Answer: For a pair of linear equations to have infinitely many solutions, they must represent coincident lines:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Given \( 2x + 5y = 3 \) and \( (k + 1)x + 2(k + 2)y = 2k \).
Comparing coefficients:
\( \frac{2}{k + 1} = \frac{5}{2(k + 2)} = \frac{3}{2k} \)
From the first and third ratios:
\( \frac{2}{k + 1} = \frac{3}{2k} \)
\( \implies 4k = 3(k + 1) \)
\( \implies 4k = 3k + 3 \implies k = 3 \)
Let's verify by substituting \( k = 3 \) back into the ratios:
- \( \frac{2}{3+1} = \frac{2}{4} = \frac{1}{2} \)
- \( \frac{5}{2(3+2)} = \frac{5}{10} = \frac{1}{2} \)
- \( \frac{3}{2(3)} = \frac{3}{6} = \frac{1}{2} \)
Since all three ratios are equal, \( k = 3 \) is the correct value.
In simple words: For infinite solutions, the ratios of the coefficients of x, y, and the constants must all be equal. Comparing these fractions leads us to \( k = 3 \).

Exam Tip: Using the first and last terms often yields a simpler equation to solve than using the middle term.

SELF EVALUTION

Question 1. Solve for x and y: x+y=a+b ax – by= a^2 – b^2
Answer: Let's solve the literal equations:
\( x + y = a + b \) - (Equation 1)
\( ax - by = a^2 - b^2 \) - (Equation 2)
From Equation 1, we can write:
\( y = a + b - x \)
Substitute this expression into Equation 2:
\( ax - b(a + b - x) = a^2 - b^2 \)
\( \implies ax - ab - b^2 + bx = a^2 - b^2 \)
\( \implies (a + b)x - ab - b^2 = a^2 - b^2 \)
Add \( b^2 \) to both sides and simplify:
\( \implies (a + b)x = a^2 + ab \)
\( \implies (a + b)x = a(a + b) \)
Dividing both sides by \( (a + b) \):
\( \implies x = a \)
Substitute \( x = a \) into Equation 1:
\( a + y = a + b \implies y = b \)
So, the solution is \( x = a \) and \( y = b \).
In simple words: We write y in terms of x from the first equation, then plug it into the second. Simplifying the algebraic terms gives \( x = a \) and \( y = b \).

Exam Tip: Factoring \( a^2 + ab \) as \( a(a+b) \) is a helpful step to simplify the division.

Question 2. For what value of k will the equation x +5y-7=0 and 4x +20y +k=0 represent coincident lines?
Answer: For the two equations to represent coincident lines, they must have equal coefficient ratios:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Given \( x + 5y - 7 = 0 \) and \( 4x + 20y + k = 0 \):
\( a_1 = 1, b_1 = 5, c_1 = -7 \)
\( a_2 = 4, b_2 = 20, c_2 = k \)
Substituting into the condition:
\( \frac{1}{4} = \frac{5}{20} = \frac{-7}{k} \)
\( \implies \frac{1}{4} = \frac{-7}{k} \)
\( \implies k = -28 \)
Thus, the value of \( k \) is \( -28 \).
In simple words: For the lines to lie exactly on top of each other, the equations must be multiples of each other. Since the second equation has coefficients four times larger than the first, the constant must also be four times larger, so \( k = 4 \times (-7) = -28 \).

Exam Tip: Be careful with the sign of the constant terms when setting up the coincident ratio.

Question 3. Solve graphically: 3x +y +1=0 2x -3y +8=0
Answer: Let's find coordinate points to plot both equations on a graph:
For \( 3x + y + 1 = 0 \):
- If \( x = -1 \), \( y = -3(-1) - 1 = 2 \). Point is \( (-1, 2) \).
- If \( x = 0 \), \( y = -1 \). Point is \( (0, -1) \).

For \( 2x - 3y + 8 = 0 \):
- If \( x = -1 \), \( y = \frac{2(-1) + 8}{3} = 2 \). Point is \( (-1, 2) \).
- If \( x = 2 \), \( y = \frac{2(2) + 8}{3} = 4 \). Point is \( (2, 4) \).

Plotting both lines reveals that they intersect at the point \( (-1, 2) \).
Thus, the graphical solution is \( x = -1 \) and \( y = 2 \).
In simple words: Graphing both equations shows that they cross at the coordinates (-1, 2), which is our solution.

Exam Tip: Plotting at least three points for each line is a good way to verify that your points align perfectly on a straight line.

Question 4. The sum of digits of a two digit number is 9. If 27is subtracted from the number, the digits are reversed. Find the number.
Answer: Let the tens digit of the number be \( x \) and the units digit be \( y \).
- Original Number = \( 10x + y \)
- Reversed Number = \( 10y + x \)

Based on the given conditions:
1. The sum of the digits is 9:
\( x + y = 9 \) - (Equation 1)
2. Subtracting 27 reverses the digits:
\( (10x + y) - 27 = 10y + x \)
\( \implies 9x - 9y = 27 \)
Divide by 9:
\( \implies x - y = 3 \) - (Equation 2)

Adding Equation 1 and Equation 2:
\( 2x = 12 \implies x = 6 \)
Substitute \( x = 6 \) into Equation 1:
\( 6 + y = 9 \implies y = 3 \)

Therefore, the required number is \( 10(6) + 3 = 63 \).
In simple words: We represent the digits of the number as x and y. Solving the two simple equations tells us that the tens digit is 6 and the units digit is 3, making the number 63.

Exam Tip: In digit-based word problems, always write the general algebraic form of the number as \( 10x + y \).

Question 5. Draw the graph of x + 2y – 7 =0 and 2x – y -4 = 0. Shade the area bounded by these lines and Y-axis.
Answer: Let's find coordinates to plot each line on a graph:
For \( x + 2y - 7 = 0 \):
- If \( x = 3 \), \( y = 2 \). Point is \( (3, 2) \).
- If \( x = 0 \), \( y = 3.5 \). Point is \( (0, 3.5) \) (Y-intercept).

For \( 2x - y - 4 = 0 \):
- If \( x = 3 \), \( y = 2 \). Point is \( (3, 2) \).
- If \( x = 0 \), \( y = -4 \). Point is \( (0, -4) \) (Y-intercept).

Graphing these lines shows they intersect at \( (3, 2) \).
The triangular region bounded by these lines and the y-axis has vertices at \( (3, 2) \), \( (0, 3.5) \), and \( (0, -4) \).
- Base along the y-axis: \( 3.5 - (-4) = 7.5 \) units.
- Height (perpendicular distance from the y-axis): 3 units.
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 7.5 \times 3 = 11.25 \text{ sq. units} \)
In simple words: Graphing the lines shows they cross at (3,2). The triangle's base along the vertical y-axis goes from -4 to 3.5, which is 7.5 units long. The height of the triangle is 3 units, giving an area of 11.25.

Exam Tip: Pay close attention to whether the problem asks for the area bounded by the x-axis or the y-axis.

Question 6. Students of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows less. If one student is less in a row there would be 3 rows more. Find the number of the students in the class.
Answer: Let the number of students in each row be \( x \), and the number of rows be \( y \).
Total number of students in the class = \( xy \)

Based on the problem conditions:
1. If one student is extra in each row, there are 2 fewer rows:
\( (x + 1)(y - 2) = xy \)
\( \implies xy - 2x + y - 2 = xy \)
\( \implies -2x + y = 2 \) - (Equation 1)

2. If one student is less in each row, there are 3 more rows:
\( (x - 1)(y + 3) = xy \)
\( \implies xy + 3x - y - 3 = xy \)
\( \implies 3x - y = 3 \) - (Equation 2)

Adding Equation 1 and Equation 2:
\( x = 5 \)
Substitute \( x = 5 \) into Equation 1:
\( -2(5) + y = 2 \implies -10 + y = 2 \implies y = 12 \)

Total number of students = \( xy = 5 \times 12 = 60 \).
In simple words: Let x be the students per row, and y be the number of rows. We can construct two simple equations based on the changes. Solving them shows there are 5 students per row and 12 rows, giving a total of 60 students.

Exam Tip: Always remember to multiply your final values for rows and students-per-row together to answer the actual question of "total students".

Question 7. A man travels 370 km partly by train and remaining by car. If he covers 250 km by train and the rest by the car it takes him 4 hours, but if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and that of the car.
Answer: Let the speed of the train be \( x\text{ km/h} \) and the speed of the car be \( y\text{ km/h} \).
Total distance is 370 km.
- **First Scenario:** 250 km by train and the remaining \( 370 - 250 = 120\text{ km} \) by car. Time taken is 4 hours.
\( \frac{250}{x} + \frac{120}{y} = 4 \) - (Equation 1)

- **Second Scenario:** 130 km by train and the remaining \( 370 - 130 = 240\text{ km} \) by car. Time taken is 4 hours and 18 minutes.
Convert 18 minutes to hours: \( \frac{18}{60} = \frac{3}{10}\text{ hours} \).
Total time = \( 4 + \frac{3}{10} = \frac{43}{10}\text{ hours} \).
\( \frac{130}{x} + \frac{240}{y} = \frac{43}{10} \) - (Equation 2)

Substitute \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \):
\( 250u + 120v = 4 \) - (Equation 3)
\( 130u + 240v = \frac{43}{10} \) - (Equation 4)

Multiply Equation 3 by 2:
\( 500u + 240v = 8 \) - (Equation 5)

Subtract Equation 4 from Equation 5:
\( (500u + 240v) - (130u + 240v) = 8 - \frac{43}{10} \)
\( \implies 370u = \frac{37}{10} \)
\( \implies u = \frac{1}{100} \)

Substitute \( u = \frac{1}{100} \) into Equation 3:
\( 250\left(\frac{1}{100}\right) + 120v = 4 \)
\( \implies 2.5 + 120v = 4 \)
\( \implies 120v = 1.5 \implies v = \frac{1.5}{120} = \frac{1}{80} \)

Converting back to original variables:
\( x = \frac{1}{u} = 100\text{ km/h} \)
\( y = \frac{1}{v} = 80\text{ km/h} \)

Therefore, the speed of the train is 100 km/h and the speed of the car is 80 km/h.
In simple words: We create two equations using travel times (distance divided by speed). By substituting simple variables, we find the speed of the train is 100 km/h and the car is 80 km/h.

Exam Tip: Always convert mixed time units (like hours and minutes) into a single unit (hours) before writing the speed equations.

Question 8. Given linear equation 2x +3y-8=0, write another linear equation such that the geometrical representation of the pair so formed is (i) intersecting lines, (ii) Parallel Lines.
Answer: Given equation is \( 2x + 3y - 8 = 0 \).

(i) **Intersecting Lines:**
For the lines to intersect, we require:
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
One such equation is \( 4x - 3y - 8 = 0 \export \), where \( \frac{2}{4} \neq \frac{3}{-3} \).

(ii) **Parallel Lines:**
For the lines to be parallel, the coefficient ratio must satisfy:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
One such equation is \( 4x + 6y + 16 = 0 \), since \( \frac{2}{4} = \frac{3}{6} \neq \frac{-8}{16} \).
In simple words: To make intersecting lines, change the x and y numbers so their ratios don't match. To make parallel lines, double both the x and y numbers, but use a completely different constant.

Exam Tip: There are infinite possible correct equations; just pick any simple coefficients that satisfy the required ratio condition.

Question 9. Solve for x and y. (a-b)x +(a+b)y = a2 - 2ab – b2 (a+b)(x+y) = a2+b2
Answer: Let's write the given pair of equations:
\( (a - b)x + (a + b)y = a^2 - 2ab - b^2 \) - (Equation 1)
Expand the second equation:
\( (a + b)x + (a + b)y = a^2 + b^2 \) - (Equation 2)
Subtract Equation 2 from Equation 1:
\( [(a - b) - (a + b)]x = (a^2 - 2ab - b^2) - (a^2 + b^2) \)
\( \implies -2bx = -2ab - 2b^2 \)
\( \implies -2bx = -2b(a + b) \)
Dividing by \( -2b \) on both sides:
\( \implies x = a + b \)
Substitute \( x = a + b \) into Equation 2:
\( (a + b)(a + b) + (a + b)y = a^2 + b^2 \)
\( \implies (a + b)^2 + (a + b)y = a^2 + b^2 \)
\( \implies (a + b)y = a^2 + b^2 - (a^2 + 2ab + b^2) \)
\( \implies (a + b)y = -2ab \)
\( \implies y = \frac{-2ab}{a + b} \)
Thus, the solution is \( x = a + b \) and \( y = \frac{-2ab}{a+b} \).
In simple words: We rewrite the second equation to expand the brackets. Since both equations now have the same y-coefficient, subtracting them removes y, allowing us to find x and then y.

Exam Tip: Don't expand the \( (a+b)y \) terms; keeping them intact makes subtraction extremely quick and clean.

Question 10. The sum of two numbers is 8 and the sum of their reciprocal is 8/15. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \).
Based on the problem conditions:
\( x + y = 8 \) - (Equation 1)
and
\( \frac{1}{x} + \frac{1}{y} = \frac{8}{15} \implies \frac{x + y}{xy} = \frac{8}{15} \) - (Equation 2)
Substitute Equation 1 into Equation 2:
\( \frac{8}{xy} = \frac{8}{15} \implies xy = 15 \) - (Equation 3)
From Equation 1, \( y = 8 - x \). Substitute this into Equation 3:
\( x(8 - x) = 15 \)
\( \implies 8x - x^2 = 15 \)
\( \implies x^2 - 8x + 15 = 0 \)
Factorize the quadratic equation:
\( (x - 3)(x - 5) = 0 \)
\( \implies x = 3 \) or \( x = 5 \)
If \( x = 3 \), then \( y = 5 \), and if \( x = 5 \), then \( y = 3 \).
Therefore, the two numbers are 3 and 5.
In simple words: We write two equations based on the sum and the sum of their reciprocals. Substituting the first into the second shows that the numbers multiply to 15, meaning they are 3 and 5.

Exam Tip: Converting the reciprocal sum \( \frac{1}{x} + \frac{1}{y} \) directly to \( \frac{x+y}{xy} \) is the fastest algebraic trick to solve this.

Value Based Questions

Question Q1. The owner of a taxi cab company decides to run all the cars he has on CNG fuel instead of petrol/diesel. The car hire charges in city comprises of fixed charges together with the charge for the distance covered. For a journey of 12km, the charge paid Rs.89 and for a journey of 20 km, the charge paid is Rs. 145.
i. What will a person have to pay for travelling a distance of 30 km?
ii. Which concept has been used to find it?
iii. Which values of the owner have been depicted here?
Answer: Let the fixed charge be \( x \) and the rate per kilometer be \( y \).
According to the problem conditions:
\( x + 12y = 89 \) - (Equation 1)
\( x + 20y = 145 \) - (Equation 2)

Subtracting Equation 1 from Equation 2:
\( (x + 20y) - (x + 12y) = 145 - 89 \)
\( \implies 8y = 56 \implies y = 7 \)
Substitute \( y = 7 \) into Equation 1:
\( x + 12(7) = 89 \)
\( \implies x + 84 = 89 \implies x = 5 \)
So, the fixed charge is Rs. 5 and the variable charge is Rs. 7 per km.

Now, address each sub-part:
i. For a journey of 30 km:
\( \text{Total Charge} = x + 30y = 5 + 30(7) = 5 + 210 = \text{Rs. } 215 \).
ii. We used the mathematical concept of solving a pair of linear equations in two variables.
iii. The owner's values of environmental consciousness, caring for public health by reducing air pollution, and using eco-friendly energy sources are depicted here.
In simple words: We calculate a fixed starting fee of Rs. 5 and a rate of Rs. 7 per kilometer using two equations. For a 30 km ride, the cost is Rs. 215. This choice of CNG highlights the owner's eco-friendly mindset.

Exam Tip: Always specify the units (Rs.) clearly in the final answers for word problems.

Question Q2. Riya decides to use public transport to cover a distance of 300 km. She travels this distance partly by train and remaining by bus. She takes 4 hours if she travels 60km by bus and the remaining by train. If she travels 100 km by bus and the remaining by train, she takes 10 minutes more.
i. Find speed of train and bus separately.
ii. Which concept has been used to solve the above problem?
iii. Which values of Riya have been depicted here?
Answer: Let the speed of the bus be \( x\text{ km/h} \) and the speed of the train be \( y\text{ km/h} \).
- **First Scenario:** 60 km by bus and remaining \( 300 - 60 = 240\text{ km} \) by train. Time is 4 hours.
\( \frac{60}{x} + \frac{240}{y} = 4 \) - (Equation 1)

- **Second Scenario:** 100 km by bus and remaining \( 300 - 100 = 200\text{ km} \) by train. Time is 4 hours and 10 minutes.
Convert 10 minutes to hours: \( \frac{10}{60} = \frac{1}{6}\text{ hour} \).
Total time = \( 4 + \frac{1}{6} = \frac{25}{6}\text{ hours} \).
\( \frac{100}{x} + \frac{200}{y} = \frac{25}{6} \) - (Equation 2)

Using substitutions \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \):
\( 60u + 240v = 4 \implies 15u + 60v = 1 \) - (Equation 3)
\( 100u + 200v = \frac{25}{6} \implies 24u + 48v = 1 \) - (Equation 4)

From Equation 3, we have \( u = \frac{1 - 60v}{15} \). Substituting this into Equation 4:
\( 24\left(\frac{1 - 60v}{15}\right) + 48v = 1 \)
\( \implies \frac{8(1 - 60v)}{5} + 48v = 1 \)
\( \implies 8 - 480v + 240v = 5 \)
\( \implies -240v = -3 \implies v = \frac{1}{80} \)

Using \( v = \frac{1}{80} \) to solve for \( u \):
\( u = \frac{1 - 60(1/80)}{15} = \frac{0.25}{15} = \frac{1}{60} \)

Converting back to original speeds:
\( x = \frac{1}{u} = 60\text{ km/h} \) (Bus speed)
\( y = \frac{1}{v} = 80\text{ km/h} \) (Train speed)

Now, addressing the sub-parts:
i. The speed of the train is 80 km/h and the speed of the bus is 60 km/h.
ii. The mathematical concept of a pair of linear equations in two variables (reducible to linear form) has been used.
iii. Riya's values of utilizing public transport to control environmental pollution, conserving fuel, and acting as a responsible citizen are depicted here.
In simple words: We solve the travel equation by substituting variables, which shows the bus moves at 60 km/h and the train moves at 80 km/h. Public transport helps limit car exhaust, showing Riya's care for nature.

Exam Tip: Make sure to convert minutes correctly into fraction form (\( \frac{10}{60} = \frac{1}{6} \)) to keep the arithmetic simple.