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Introduction

If two linear equations have the two same variables, they are called a pair of linear equations in two variables. Following is the most general form of linear equations:
Class ten Math cbse - Pair of Linear Equations in Two Variables_1
Here, a1, a2, b1, b2, c1 and c2 are real numbers such that;
Class ten Math cbse - Pair of Linear Equations in Two Variables_2
A pair of linear equations can be represented and solved by the following methods:
  1. Graphical method
  2. Algebraic method

Graphical Method:

For a given pair of linear equations in two variables, the graph is represented by two lines.
  1. If the lines intersect at a point, that point gives the unique solution for the two equations. If there is a unique solution of the given pair of equations, the equations are called consistent.
  2. If the lines coincide, there are indefinitely many solutions for the pair of linear equations. In this case, each point on the line is a solution. If there are infinitely many solutions of the given pair of linear equations, the equations are called dependent (consistent).
  3. If the lines are parallel, there is no solution for the pair of linear equations. If there is no solution of the given pair of linear equations, the equations are called inconsistent.

Algebraic Method:

There are following methods for finding the solutions of the pair of linear equations:
  1. Substitution method
  2. Elimination method
  3. Cross-multiplication method
If a pair of linear equations is given by Class ten Math cbse - Pair of Linear Equations in Two Variables_3 then following situations can arise.
Situation 1:
Class ten Math cbse - Pair of Linear Equations in Two Variables_4
In this case, the pair of linear equations is consistent. This means there is unique solution for the given pair of linear equations. The graph of the linear equations would be two intersecting lines.
Situation 2:
Class ten Math cbse - Pair of Linear Equations in Two Variables_5
In this case, the pair of linear equations is inconsistent. This means there is no solution for the given pair of linear equations. The graph of linear equations will be two parallel lines.
Situation 3:
Class ten Math cbse - Pair of Linear Equations in Two Variables_6
In this case, the pair of linear equations is dependent and consistent. This means there are infinitely many solutions for the given pair of linear equations. The graph of linear equations will be coincident lines.

NCERT Solution

Exercise 3.1 (NCERT)

Question – 1 - Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” Represent this situation algebraically and graphically.
Solution: Let us assume Aftab’s current age = x and his daughter’s current age = y
Seven years ago:
Aftab’s age = x - 7 and daughter’s age = y - 7
As per question;
Class ten Math cbse - Pair of Linear Equations in Two Variables_7
This equation gives following values for x and y
Class ten Math cbse - Pair of Linear Equations in Two Variables_8
Three years from now:
Aftab’s age = x + 3 and daughter’s age = y + 3
As per question;
Class ten Math cbse - Pair of Linear Equations in Two Variables_9
This equation gives following values for x and y:
Class ten Math cbse - Pair of Linear Equations in Two Variables_10
The following graph is plotted for the given pair of linear equations:
Class ten Math cbse - Pair of Linear Equations in Two Variables_11
Daughter’s age = 12 years
Aftab’s age = 42 years
Question – 2 - The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebraically and geometrically.
Solution: Let us assume that price of one bat = x and price of one ball = y. Following equations can be written as per the question:
Class ten Math cbse - Pair of Linear Equations in Two Variables_12
This equation will give following values for x and y
Class ten Math cbse - Pair of Linear Equations in Two Variables_13
This equation will give following values for x and y
Class ten Math cbse - Pair of Linear Equations in Two Variables_14
The following graph is plotted for the given pair of linear equations.
Class ten Math cbse - Pair of Linear Equations in Two Variables_15
Price of one bat = Rs. 1300
Price of one ball = zero
Question – 3 - The cost of 2 kg apples and 1 kg grapes was found to be Rs. 160. After a month, the cost of 4 kg apples and 2 kg grapes is Rs. 300. Represent this situation algebraically and graphically.
Answer: Let us assume that cost of 1 kg apple = x and cost of 1 kg grapes = y. Following equations can be written as per the question.
Class ten Math cbse - Pair of Linear Equations in Two Variables_16
This equation gives following values for x and y
Class ten Math cbse - Pair of Linear Equations in Two Variables_17
This equation will give following values for x and y:
Class ten Math cbse - Pair of Linear Equations in Two Variables_18
The following graph is plotted for the given pair of linear equations.
Class ten Math cbse - Pair of Linear Equations in Two Variables_19
Since we get parallel lines so there will be no solution for this pair of linear equations.

Exercise 3.2 

Question – 1 - Form a pair of linear equations in the following problems, and find their solutions graphically.
(a) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution: Let us assume that number of boys = x and number of girls = y. We get following equations as per question:
class ten math cbse - Pair of Linear equations in two variables_1, exercise 3.2 - ncert solution
This equation will give following values for x and y;
class ten math cbse - Pair of Linear equations in two variables_2, exercise 3.2 - ncert solution
This equation will give following values for x and y;
class ten math cbse - Pair of Linear equations in two variables_3, exercise 3.2 - ncert solution
Following graph is plotted for the given pair of linear equations:
class ten math cbse - Pair of Linear equations in two variables_4, exercise 3.2 - ncert solution
(b) 5 pencils and 7 pens 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.
Solution: Let us assume that price of a pencil is x and that of a pen is y. We get following equations as per question:
class ten math cbse - Pair of Linear equations in two variables_5, exercise 3.2 - ncert solution
This equation will give following values for x and y;
class ten math cbse - Pair of Linear equations in two variables_6, exercise 3.2 - ncert solution
This equation will give following values for x and y;
class ten math cbse - Pair of Linear equations in two variables_7, exercise 3.2 - ncert solution
Following graph is plotted for the given pair of linear equations.
class ten math cbse - Pair of Linear equations in two variables_8, exercise 3.2 - ncert solution
Price of one pencil = Rs. 3
Price of one pen = Rs. 5
Question – 2- On comparing the ratios class ten math cbse - Pair of Linear equations in two variables_9, exercise 3.2 - ncert solution find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
class ten math cbse - Pair of Linear equations in two variables_10, exercise 3.2 - ncert solution
Solution: In the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_11, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_12, exercise 3.2 - ncert solution
Hence the lines representing the given pair of linear equations intersect at a point.
class ten math cbse - Pair of Linear equations in two variables_13, exercise 3.2 - ncert solution
Solution: In the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_14, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_15, exercise 3.2 - ncert solution
Hence the lines representing the given pair of linear equations will be coincident.
class ten math cbse - Pair of Linear equations in two variables_16, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_17, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_18, exercise 3.2 - ncert solution
Hence the lines representing the given pair of linear equations will be parallel.
Question – 3 - On comparing the ratios class ten math cbse - Pair of Linear equations in two variables_19, exercise 3.2 - ncert solution find out whether the following pairs of linear equations are consistent or inconsistent.
class ten math cbse - Pair of Linear equations in two variables_20, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_21, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_22, exercise 3.2 - ncert solution
Hence, the given pair of linear equations is consistent.
class ten math cbse - Pair of Linear equations in two variables_23, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_24, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_25, exercise 3.2 - ncert solution
Hence the given pair of linear equations is inconsistent.
class ten math cbse - Pair of Linear equations in two variables_26, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_27, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_28, exercise 3.2 - ncert solution
Hence the given pair of linear equations is consistent.
class ten math cbse - Pair of Linear equations in two variables_29, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_30, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_31, exercise 3.2 - ncert solution
Hence the given pair of linear equations is dependent and consistent.

Question – 4 - Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.
class ten math cbse - Pair of Linear equations in two variables_32, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_33, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_34, exercise 3.2 - ncert solution
Hence the given pair of linear equations is dependent and consistent. The lines for the equations will be coincident. This means there would be infinitely many solutions for the given pair of linear equations.
class ten math cbse - Pair of Linear equations in two variables_35, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_36, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_37, exercise 3.2 - ncert solution
Hence the given pair of linear equations is inconsistent.
class ten math cbse - Pair of Linear equations in two variables_38, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_39, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_40, exercise 3.2 - ncert solution
Hence the given pair of linear equations is consistent.
Following graph can be plotted with the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_41, exercise 3.2 - ncert solution
Value of x = 2 and that of y = 2
class ten math cbse - Pair of Linear equations in two variables_42, exercise 3.2 - ncert solution
Solution: For the given pair of linear equations;
class ten math cbse - Pair of Linear equations in two variables_43, exercise 3.2 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_44, exercise 3.2 - ncert solution
Hence the given pair of linear equations is inconsistent.
Question – 5 - Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution: Let us assume that width of garden = x and length = y. Then as per question;
class ten math cbse - Pair of Linear equations in two variables_45, exercise 3.2 - ncert solution
Substitution the value of y from equation (1) in equation (2) we get;
class ten math cbse - Pair of Linear equations in two variables_46, exercise 3.2 - ncert solution
Hence, length = 20 m and breadth = 16 m
Question – 6 - Given the linear equation class ten math cbse - Pair of Linear equations in two variables_47, exercise 3.2 - ncert solution write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(a) Intersecting lines
Solution: For intersecting line, the linear equations should meet following condition:
class ten math cbse - Pair of Linear equations in two variables_48, exercise 3.2 - ncert solution
For getting another equation to meet this criterion, multiply the coefficient of x with any number and multiply the coefficient of y with any other number. A possible equation can be as follows:
class ten math cbse - Pair of Linear equations in two variables_49, exercise 3.2 - ncert solution
(b) Parallel lines
Solution: For parallel lines, the linear equations should meet following condition:
class ten math cbse - Pair of Linear equations in two variables_50, exercise 3.2 - ncert solution
For getting another equation to meet this criterion, multiply the coefficients of x and y with the same number and multiply the constant term with any other number. A possible equation can be as follows:
class ten math cbse - Pair of Linear equations in two variables_51, exercise 3.2 - ncert solution
(c) Coincident lines
Solution: For getting coincident lines, the equations should meet following condition;
class ten math cbse - Pair of Linear equations in two variables_52, exercise 3.2 - ncert solution
For getting another equation to meet this criterion, multiply the whole equation with any number. A possible equation can be as follows:
class ten math cbse - Pair of Linear equations in two variables_53, exercise 3.2 - ncert solution
Question – 7 - Draw the graphs of the equations class ten math cbse - Pair of Linear equations in two variables_54, exercise 3.2 - ncert solutionDetermine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution: Following graph can be plotted with the given pair of linear equations:
class ten math cbse - Pair of Linear equations in two variables_55, exercise 3.2 - ncert solution
The coordinates of vertices of the triangle are; (-1, 0), (2, 3) and (4, 0)

Exercise 3.3

Solving by substitution method:
  • One of the equations is picked and in this, one variable is expressed in terms of another variable.
  • The value of one variable (in terms of another variable) is substituted in the second equation to calculate the value of the remaining variable.
Question – 1 - Solve the following pair of linear equations by the substitution method.
class ten math cbse - Pair of Linear equations in two variables_1, exercise 3.3 - ncert solution
Solution: Let us use the first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_2, exercise 3.3 - ncert solution
Substituting the value of x in second equation we get;
class ten math cbse - Pair of Linear equations in two variables_3, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_4, exercise 3.3 - ncert solution
Solution: Let us use the first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_5, exercise 3.3 - ncert solution
Substituting the value of s in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_6, exercise 3.3 - ncert solution
Substituting the value of t in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_7, exercise 3.3 - ncert solution
Solution: Let us use first equation to express one variable in terms of another variable.
class ten math cbse - Pair of Linear equations in two variables_8, exercise 3.3 - ncert solution
The second equation is same as the first equation;
class ten math cbse - Pair of Linear equations in two variables_9, exercise 3.3 - ncert solution
Dividing the equation by 3, we get;
class ten math cbse - Pair of Linear equations in two variables_10, exercise 3.3 - ncert solution
Hence, the given pair of linear equations has infinitely many solutions.
class ten math cbse - Pair of Linear equations in two variables_11, exercise 3.3 - ncert solution
Solution: Let us use the first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_12, exercise 3.3 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_13, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_14, exercise 3.3 - ncert solution
Solution: Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_15, exercise 3.3 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_16, exercise 3.3 - ncert solution
This can be possible only when the value of y is zero. Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_17, exercise 3.3 - ncert solution
Solution: Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_18, exercise 3.3 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_19, exercise 3.3 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_20, exercise 3.3 - ncert solution
Question – 2 - Solve class ten math cbse - Pair of Linear equations in two variables_21, exercise 3.3 - ncert solution and hence find the value of ‘m’ for which class ten math cbse - Pair of Linear equations in two variables_22, exercise 3.3 - ncert solution
Solution: Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_23, exercise 3.3 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_24, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_25, exercise 3.3 - ncert solution
Now; we have to find the value of m in following equation;
class ten math cbse - Pair of Linear equations in two variables_26, exercise 3.3 - ncert solution 
Question – 2 - Form a pair of linear equations for the following problems and find their solution by substitution method.
(a) The difference between two numbers is 26 and one number is three times the other. Find them.
Solution: Let us assume the larger number is x and smaller number is y. Then we have following equations, as per question:
class ten math cbse - Pair of Linear equations in two variables_27, exercise 3.3 - ncert solution
Substituting the value of x from second equation in the first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_28, exercise 3.3 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_29, exercise 3.3 - ncert solution
(b) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution: Let us assume the larger angle is x and smaller angle is y. Then we have following equations;
class ten math cbse - Pair of Linear equations in two variables_30, exercise 3.3 - ncert solution
Substituting the value of x from first equation in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_31, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_32, exercise 3.3 - ncert solution
(c) The coach of a cricket team buys 7 bats and 6 balls for Rs, 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Solution: Let us assume, price of a bat is x and that of a ball is y. Then we have following equations;
class ten math cbse - Pair of Linear equations in two variables_33, exercise 3.3 - ncert solution
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_34, exercise 3.3 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_35, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_36, exercise 3.3 - ncert solution
Hence price of a bat=Rs.500 and price of a ball=Rs.50
(d) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for the journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution: Let us assume the fixed charge is x and charge for 1 km journey is y. Then we have following equations;
class ten math cbse - Pair of Linear equations in two variables_37, exercise 3.3 - ncert solution
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_38, exercise 3.3 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_39, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_40, exercise 3.3 - ncert solution
(e) A fraction becomes class ten math cbse - Pair of Linear equations in two variables_41, exercise 3.3 - ncert solution if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes class ten math cbse - Pair of Linear equations in two variables_42, exercise 3.3 - ncert solution Find the fraction.
Solution: Let us assume, numerator is x and denominator is y. Then we have following equations;
class ten math cbse - Pair of Linear equations in two variables_43, exercise 3.3 - ncert solution
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_44, exercise 3.3 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_45, exercise 3.3 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_46, exercise 3.3 - ncert solution
(f) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Solution: Let us assume Jacob’s current age is x and his son’s current age is y.
Five years hence, Jacob’s age = x + 5 and his son’s age = y + 5
As per question;
class ten math cbse - Pair of Linear equations in two variables_47, exercise 3.3 - ncert solution
Five years ago, Jacob’s age = x – 5 and his son’s age = y – 5
As per question;
class ten math cbse - Pair of Linear equations in two variables_48, exercise 3.3 - ncert solution
Substituting the value of x from first equation in this equation, we get;
class ten math cbse - Pair of Linear equations in two variables_49, exercise 3.3 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_50, exercise 3.3 - ncert solution
Hence, Jacob’s age = 40 years and son’s age = 10 years
Elimination Method:
In elimination method, one of the equations is multiplied by a suitable number so that the coefficient of one of the variables in both the equations becomes equal. After that, one equation is subtracted or added to another equation to eliminate one of the variables.

Exercise 3.4 

Question – 1 - Solve the following pair of linear equations by the elimination method and the substitution method.
class ten math cbse - Pair of Linear equations in two variables_1, exercise 3.4 - ncert solution
Solution: Elimination Method:
Multiply the first equation by 3 as follows:
class ten math cbse - Pair of Linear equations in two variables_2, exercise 3.4 - ncert solution
Now, add second equation to this equation;
class ten math cbse - Pair of Linear equations in two variables_3, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_4, exercise 3.4 - ncert solution
Substitution method;
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_5, exercise 3.4 - ncert solution
Substitute the value of x in second equation;
class ten math cbse - Pair of Linear equations in two variables_6, exercise 3.4 - ncert solution
Substitute the value of y in first equation;
class ten math cbse - Pair of Linear equations in two variables_7, exercise 3.4 - ncert solution
Solution: Elimination method:
Multiply second equation by 2;
class ten math cbse - Pair of Linear equations in two variables_8, exercise 3.4 - ncert solution
Add this equation with first equation;
class ten math cbse - Pair of Linear equations in two variables_9, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_10, exercise 3.4 - ncert solution
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_11, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_12, exercise 3.4 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_13, exercise 3.4 - ncert solution
Solution: Elimination Method:
Multiply the first equation by 3;
class ten math cbse - Pair of Linear equations in two variables_14, exercise 3.4 - ncert solution
Subtract the second equation from this equation;
class ten math cbse - Pair of Linear equations in two variables_15, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_16, exercise 3.4 - ncert solution
Substitution Method;
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_17, exercise 3.4 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_18, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_19, exercise 3.4 - ncert solution
Solution: Elimination method:
Equation 1:
class ten math cbse - Pair of Linear equations in two variables_20, exercise 3.4 - ncert solution
Equation 2:
class ten math cbse - Pair of Linear equations in two variables_21, exercise 3.4 - ncert solution
Subtracting second equation from first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_22, exercise 3.4 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_23, exercise 3.4 - ncert solution
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_24, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_25, exercise 3.4 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_26, exercise 3.4 - ncert solution 

Question – 2 - Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method.
(a) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes class ten math cbse - Pair of Linear equations in two variables_27, exercise 3.4 - ncert solution if we only add 1 to the denominator. What is the fraction?
Solution: Let us assume that numerator is x and denominator is y. Then as per question;
class ten math cbse - Pair of Linear equations in two variables_28, exercise 3.4 - ncert solution
The second condition can give following equation;
class ten math cbse - Pair of Linear equations in two variables_29, exercise 3.4 - ncert solution
Adding first and second equations, we get;
class ten math cbse - Pair of Linear equations in two variables_30, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_31, exercise 3.4 - ncert solution
(b) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Solution: Let us assume Nuri’s age = x and Sonu’s age = y. Then we can get following equations;
Five years ago:
Nuri’s age = x – 5 and Sonu’s age = y – 5
class ten math cbse - Pair of Linear equations in two variables_32, exercise 3.4 - ncert solution
Five years from now;
Nuri’s age = x + 5 and Sonu’s age = y + 5
class ten math cbse - Pair of Linear equations in two variables_33, exercise 3.4 - ncert solution
Subtracting the second equation from first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_34, exercise 3.4 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_35, exercise 3.4 - ncert solution
(c) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
Solution: Let us assume, one of the digits is x and another digit is y.
class ten math cbse - Pair of Linear equations in two variables_36, exercise 3.4 - ncert solution
The number can be given as;
class ten math cbse - Pair of Linear equations in two variables_37, exercise 3.4 - ncert solution
And the number obtained after reversing the digits is;
class ten math cbse - Pair of Linear equations in two variables_38, exercise 3.4 - ncert solution
As per question;
class ten math cbse - Pair of Linear equations in two variables_39, exercise 3.4 - ncert solution
Multiplying the first equation by 11, we get;
class ten math cbse - Pair of Linear equations in two variables_40, exercise 3.4 - ncert solution
Adding this to the second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_41, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_42, exercise 3.4 - ncert solution
Hence, the required number = 18
(d) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find out how many notes of Rs. 50 and Rs. 100 she received.
Solution: Let us assume the number of Rs. 50 notes is x and that of Rs. 100 notes is y.
class ten math cbse - Pair of Linear equations in two variables_43, exercise 3.4 - ncert solution
Multiplying the first equation by 50 we get;
class ten math cbse - Pair of Linear equations in two variables_44, exercise 3.4 - ncert solution
Subtracting this equation from second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_45, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_46, exercise 3.4 - ncert solution
Hence, number or Rs. 50 notes = 10 and number of Rs. 100 notes = 15
(e) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution: Let us assume fixed charge is x and per day charge after three days is y.
Amount paid by Saritha:
class ten math cbse - Pair of Linear equations in two variables_47, exercise 3.4 - ncert solution
Amount paid by Susy:
class ten math cbse - Pair of Linear equations in two variables_48, exercise 3.4 - ncert solution
Subtracting the second equation from first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_49, exercise 3.4 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_50, exercise 3.4 - ncert solution
Hence, fixed charge for three days = Rs. 15 and per day charge after three days = Rs. 3

Exercise 3.4 - Part 

Question – 1 - Solve the following pair of linear equations by the elimination method and the substitution method.
class ten math cbse - Pair of Linear equations in two variables_1, exercise 3.4 - ncert solution
Solution: Elimination Method:
Multiply the first equation by 3 as follows:
class ten math cbse - Pair of Linear equations in two variables_2, exercise 3.4 - ncert solution
Now, add second equation to this equation;
class ten math cbse - Pair of Linear equations in two variables_3, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_4, exercise 3.4 - ncert solution
Substitution method;
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_5, exercise 3.4 - ncert solution
Substitute the value of x in second equation;
class ten math cbse - Pair of Linear equations in two variables_6, exercise 3.4 - ncert solution
Substitute the value of y in first equation;
class ten math cbse - Pair of Linear equations in two variables_7, exercise 3.4 - ncert solution
Solution: Elimination method:
Multiply second equation by 2;
class ten math cbse - Pair of Linear equations in two variables_8, exercise 3.4 - ncert solution
Add this equation with first equation;
class ten math cbse - Pair of Linear equations in two variables_9, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_10, exercise 3.4 - ncert solution
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_11, exercise 3.4 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_12, exercise 3.4 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_13, exercise 3.4 - ncert solution
Solution: Elimination Method:
Multiply the first equation by 3;
class ten math cbse - Pair of Linear equations in two variables_14, exercise 3.4 - ncert solution
Subtract the second equation from this equation;
class ten math cbse - Pair of Linear equations in two variables_15, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_16, exercise 3.4 - ncert solution
Substitution Method;
Let us use first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_17, exercise 3.4 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_18, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_19, exercise 3.4 - ncert solution
Solution: Elimination method:
Equation 1:
class ten math cbse - Pair of Linear equations in two variables_20, exercise 3.4 - ncert solution
Equation 2:
class ten math cbse - Pair of Linear equations in two variables_21, exercise 3.4 - ncert solution
Subtracting second equation from first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_22, exercise 3.4 - ncert solution
Substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_23, exercise 3.4 - ncert solution
Substitution Method:
Let us use second equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_24, exercise 3.4 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_25, exercise 3.4 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_26, exercise 3.4 - ncert solution 

Exercise 3.5 (NCERT)

Question – 1 - Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.
class ten math cbse - Pair of Linear equations in two variables_1, exercise 3.5 - ncert solution
Solution:
class ten math cbse - Pair of Linear equations in two variables_2, exercise 3.5 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_3, exercise 3.5 - ncert solution
Hence, there will be no solution for the given pair of linear equations.
class ten math cbse - Pair of Linear equations in two variables_4, exercise 3.5 - ncert solution
Solution:
class ten math cbse - Pair of Linear equations in two variables_5, exercise 3.5 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_6, exercise 3.5 - ncert solution
Hence, there will be unique solution for the given pair of linear equations.
From cross-multiplication method, we know;
class ten math cbse - Pair of Linear equations in two variables_7, exercise 3.5 - ncert solution
Solution:
class ten math cbse - Pair of Linear equations in two variables_8, exercise 3.5 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_9, exercise 3.5 - ncert solution
Hence, there will be infinitely many solutions for the given pair of linear equations.
class ten math cbse - Pair of Linear equations in two variables_10, exercise 3.5 - ncert solution
Solution:
class ten math cbse - Pair of Linear equations in two variables_11, exercise 3.5 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_12, exercise 3.5 - ncert solution
Hence, there will be unique solution for the given pair of linear equations.
From cross-multiplication method, we know;
class ten math cbse - Pair of Linear equations in two variables_13, exercise 3.5 - ncert solution
Question – 2 - For which values of a and b does the following pair of linear equations have an infinite number of solutions?
class ten math cbse - Pair of Linear equations in two variables_14, exercise 3.5 - ncert solution
Solution:
class ten math cbse - Pair of Linear equations in two variables_15, exercise 3.5 - ncert solution
For infinite number of solutions, the equations should fulfill following criterion:
class ten math cbse - Pair of Linear equations in two variables_16, exercise 3.5 - ncert solution
From equations (1) and (2), it is clear;
class ten math cbse - Pair of Linear equations in two variables_17, exercise 3.5 - ncert solution
Substituting the value of b in equation (1), we get;
class ten math cbse - Pair of Linear equations in two variables_18, exercise 3.5 - ncert solution
Question – 3 - For which value of k will the following pair of linear equations have no solution?
class ten math cbse - Pair of Linear equations in two variables_19, exercise 3.5 - ncert solution
Solution:
class ten math cbse - Pair of Linear equations in two variables_20, exercise 3.5 - ncert solution
For no solution, the equations should fulfill following criterion:
class ten math cbse - Pair of Linear equations in two variables_21, exercise 3.5 - ncert solution
Question – 4 - Solve the following pair of linear equations by the substitution and cross-multiplication methods:
class ten math cbse - Pair of Linear equations in two variables_22, exercise 3.5 - ncert solution
Solution: Substitution method:
Let us use the first equation to express one variable in terms of another variable;
class ten math cbse - Pair of Linear equations in two variables_23, exercise 3.5 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_24, exercise 3.5 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_25, exercise 3.5 - ncert solution
Cross-multiplication Method:
class ten math cbse - Pair of Linear equations in two variables_26, exercise 3.5 - ncert solution
From cross-multiplication method, we know;
class ten math cbse - Pair of Linear equations in two variables_27, exercise 3.5 - ncert solution

class ten math cbse - Pair of Linear equations in two variables_28, exercise 3.5 - ncert solution 
Question – 5 - Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(a) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Solution: Let us assume that fixed charges is x and per day charges is y
Amount paid by student A:
class ten math cbse - Pair of Linear equations in two variables_29, exercise 3.5 - ncert solution 
Amount paid by student B:
class ten math cbse - Pair of Linear equations in two variables_30, exercise 3.5 - ncert solution 
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_31, exercise 3.5 - ncert solution 
Hence, there will be unique solution for the given pair of linear equations.
Subtracting first equation from second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_32, exercise 3.5 - ncert solution 
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_33, exercise 3.5 - ncert solution 
Hence, fixed charges = Rs. 400 and per day charges = Rs. 30
(b) A fraction becomes class ten math cbse - Pair of Linear equations in two variables_34, exercise 3.5 - ncert solution when 1 is subtracted from the numerator and it becomes class ten math cbse - Pair of Linear equations in two variables_35, exercise 3.5 - ncert solution when 8 is added to its denominator. Find the fraction.
Solution: Let us assume numerator is x and denominator is y.
First condition:
class ten math cbse - Pair of Linear equations in two variables_36, exercise 3.5 - ncert solution
Second condition:
class ten math cbse - Pair of Linear equations in two variables_37, exercise 3.5 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_38, exercise 3.5 - ncert solution
Hence, there will be unique solution for the given pair of linear equations.
Subtracting first equation from second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_39, exercise 3.5 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_40, exercise 3.5 - ncert solution
(c) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Solution: Let us assume that the number of correct answers = x and number of incorrect answers = y
First condition:
class ten math cbse - Pair of Linear equations in two variables_41, exercise 3.5 - ncert solution
Second condition:
class ten math cbse - Pair of Linear equations in two variables_42, exercise 3.5 - ncert solution
It is clear that;
class ten math cbse - Pair of Linear equations in two variables_43, exercise 3.5 - ncert solution
Hence, there will be unique solution for the given pair of linear equations.
Multiply the first equation and subtract second equation from the resultant:
class ten math cbse - Pair of Linear equations in two variables_44, exercise 3.5 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_45, exercise 3.5 - ncert solution
Hence, number of correct answers = 15 and number of incorrect answers = 10
(d) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Solution: Let us assume that the speed of car A is x and that of car B is y.
First condition: When both the cars are going in the same direction, their relative speed is; class ten math cbse - Pair of Linear equations in two variables_46, exercise 3.5 - ncert solution
Hence,
class ten math cbse - Pair of Linear equations in two variables_47, exercise 3.5 - ncert solution
Second condition: When the cars are moving towards each other, their relative speed is; class ten math cbse - Pair of Linear equations in two variables_48, exercise 3.5 - ncert solution
Hence,
class ten math cbse - Pair of Linear equations in two variables_49, exercise 3.5 - ncert solution
Adding the two equations, we get;
class ten math cbse - Pair of Linear equations in two variables_50, exercise 3.5 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_51, exercise 3.5 - ncert solution
Hence, speed of car A = 60 km/h and speed of car B = 40 km/h
(e) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Solution: Let us assume length is x and breadth is y.
Area of rectangle = xy
First condition: Length = x - 5 and breadth = y + 3
Area of rectangle:
class ten math cbse - Pair of Linear equations in two variables_52, exercise 3.5 - ncert solution
Second condition: Length = x +3 and breadth = y + 2
Area of rectangle:
class ten math cbse - Pair of Linear equations in two variables_53, exercise 3.5 - ncert solution
Multiplying the first equation by 2, we get;
class ten math cbse - Pair of Linear equations in two variables_54, exercise 3.5 - ncert solution
Multiplying the second equation by 3, we get;
class ten math cbse - Pair of Linear equations in two variables_55, exercise 3.5 - ncert solution
Subtracting these two equations, we get;
class ten math cbse - Pair of Linear equations in two variables_56, exercise 3.5 - ncert solution
Substituting the value of y in any of the above equations, we get;
class ten math cbse - Pair of Linear equations in two variables_57, exercise 3.5 - ncert solution
Hence, length = 17 unit and breadth = 9 unit

Exercise 3.6 (NCERT)

Question – 1 - Solve the following pairs of equations by reducing them to a pair of linear equations:
class ten math cbse - Pair of Linear equations in two variables_1, exercise 3.6 - ncert solution
Solution: First equation:
class ten math cbse - Pair of Linear equations in two variables_2, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_3, exercise 3.6 - ncert solution
Multiplying the first equation by 4 and subtracting the resultant from second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_4, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_5, exercise 3.6 - ncert solution
Solution: First equation:
class ten math cbse - Pair of Linear equations in two variables_6, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_7, exercise 3.6 - ncert solution
From first and second equation, it is clear;
class ten math cbse - Pair of Linear equations in two variables_8, exercise 3.6 - ncert solution
Substituting the value of class ten math cbse - Pair of Linear equations in two variables_9, exercise 3.6 - ncert solution in any of the above equation, we get;
class ten math cbse - Pair of Linear equations in two variables_10, exercise 3.6 - ncert solution
Substituting the value of class ten math cbse - Pair of Linear equations in two variables_11, exercise 3.6 - ncert solution in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_12, exercise 3.6 - ncert solution
Solution: First equation:
class ten math cbse - Pair of Linear equations in two variables_13, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_14, exercise 3.6 - ncert solution
Multiply first equation by 4 and second equation by 3 and add the resultant equations;
class ten math cbse - Pair of Linear equations in two variables_15, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_16, exercise 3.6 - ncert solution

class ten math cbse - Pair of Linear equations in two variables_17, exercise 3.6 - ncert solution
Solution: First equation;
class ten math cbse - Pair of Linear equations in two variables_18, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_19, exercise 3.6 - ncert solution
Multiply the second equation by 2 and subtract first equation from resultant:
class ten math cbse - Pair of Linear equations in two variables_20, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_21, exercise 3.6 - ncert solution
Similarly, substituting the value of y in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_22, exercise 3.6 - ncert solution
From above two equations, it is clear;
class ten math cbse - Pair of Linear equations in two variables_23, exercise 3.6 - ncert solution
Solution: First equation:
class ten math cbse - Pair of Linear equations in two variables_24, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_25, exercise 3.6 - ncert solution
Multiplying the first equation by 3 and subtracting the resultant from second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_26, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_27, exercise 3.6 - ncert solution
Solution: Multiplying first equation by 5 and second equation by 6 and subtracting them, we get;
class ten math cbse - Pair of Linear equations in two variables_28, exercise 3.6 - ncert solution
Substituting the value of y in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_29, exercise 3.6 - ncert solution
This is possible only when x=1 or zero
Taking the non-zero value and substituting in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_30, exercise 3.6 - ncert solution


class ten math cbse - Pair of Linear equations in two variables_31, exercise 3.6 - ncert solution
Solution: First equation:
class ten math cbse - Pair of Linear equations in two variables_32, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_33, exercise 3.6 - ncert solution
From first and second equations, it is clear;
class ten math cbse - Pair of Linear equations in two variables_34, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_35, exercise 3.6 - ncert solution


class ten math cbse - Pair of Linear equations in two variables_36, exercise 3.6 - ncert solution
Solution: First equation:
class ten math cbse - Pair of Linear equations in two variables_37, exercise 3.6 - ncert solution
Second equation:
class ten math cbse - Pair of Linear equations in two variables_38, exercise 3.6 - ncert solution
From first and second equation, it is clear;
class ten math cbse - Pair of Linear equations in two variables_39, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_40, exercise 3.6 - ncert solution
Substituting the value of x in second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_41, exercise 3.6 - ncert solution 
Question – 2 - Formulate the following problems as a pair of equations, and hence find their solutions.
(a) 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.
Solution: Let us assume the speed of boat in still water is x and speed of current is y
Hence, downstream speed = x + y and upstream speed = x – y
First condition: Using time = distance/speed;
class ten math cbse - Pair of Linear equations in two variables_42, exercise 3.6 - ncert solution
Second condition: Using time = distance/speed;
class ten math cbse - Pair of Linear equations in two variables_43, exercise 3.6 - ncert solution
Adding first and second equation, we get;
class ten math cbse - Pair of Linear equations in two variables_44, exercise 3.6 - ncert solution
Speed of boat = 6 km/h and speed of current = 4 km/h
(b) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work and also that taken by 1 man alone.
Solution: First condition: One day’s work done by 2 women and 5 men;
class ten math cbse - Pair of Linear equations in two variables_45, exercise 3.6 - ncert solution
Second condition: One day’s work done by 3 women and 6 men;
class ten math cbse - Pair of Linear equations in two variables_46, exercise 3.6 - ncert solution
From first and second equation, it is clear;
class ten math cbse - Pair of Linear equations in two variables_47, exercise 3.6 - ncert solution
This means that 1 woman’s work is equal to 2 men’s work. Substituting the value of w in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_48, exercise 3.6 - ncert solution
This means that 1 woman will take 18 days to finish the work.
Similarly,
class ten math cbse - Pair of Linear equations in two variables_49, exercise 3.6 - ncert solution
This means that 1 man will take 36 days to finish the work.
(c) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution: Let us assume that speed of train is x and speed of bus is y
First condition: She travels 60 km by train and 240 km by bus;
class ten math cbse - Pair of Linear equations in two variables_50, exercise 3.6 - ncert solution
Second condition: She travels 100 km by train and 200 km by bus;
class ten math cbse - Pair of Linear equations in two variables_51, exercise 3.6 - ncert solution
Subtracting first equation from second, we get;
class ten math cbse - Pair of Linear equations in two variables_52, exercise 3.6 - ncert solution
Substituting the value of x in first equation, we get;
class ten math cbse - Pair of Linear equations in two variables_53, exercise 3.6 - ncert solution
Speed of train = 60 km/h and speed of bus = 80 km/h

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