This chapter focuses on solving pairs of linear equations with two variables and their real-life applications.
Pair of Linear Equations in Two Variables - Practice Worksheet
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This worksheet covers essential long-answer questions to help you build confidence in Pair of Linear Equations in Two Variables from Mathematics for Class X (Mathematics).
Basic comprehension exercises
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Questions
Explain the graphical method of solving a pair of linear equations in two variables. What does the point of intersection represent?
The graphical method involves plotting two linear equations on the same graph. Each equation represents a straight line. The point where these two lines intersect is the solution to the pair of equations, representing the values of x and y that satisfy both equations simultaneously. If the lines are parallel, there is no solution, indicating the equations are inconsistent. If the lines coincide, there are infinitely many solutions, meaning the equations are dependent. This method is useful for visualizing the relationship between two variables and understanding the nature of their solutions.
Describe the substitution method for solving a pair of linear equations in two variables with an example.
The substitution method involves solving one equation for one variable and then substituting this expression into the other equation. For example, consider the equations 2x + y = 5 and 3x - 2y = 4. Solve the first equation for y: y = 5 - 2x. Substitute this into the second equation: 3x - 2(5 - 2x) = 4. Simplify to find x = 2, then substitute back to find y = 1. This method is straightforward when one equation can easily be solved for one variable.
How does the elimination method work in solving a pair of linear equations? Provide a step-by-step explanation.
The elimination method involves adding or subtracting the equations to eliminate one variable, making it possible to solve for the other. For example, take the equations x + y = 4 and x - y = 2. Adding them eliminates y, giving 2x = 6, so x = 3. Substituting x back into one equation gives y = 1. This method is efficient when the coefficients of one variable are opposites or can be made so by multiplication.
What are the conditions for a pair of linear equations to have a unique solution, no solution, or infinitely many solutions?
A pair of linear equations has a unique solution if the lines intersect at one point, meaning the ratios of the coefficients of x and y are not equal (a1/a2 ≠ b1/b2). No solution exists if the lines are parallel (a1/a2 = b1/b2 ≠ c1/c2), indicating the equations are inconsistent. Infinitely many solutions occur if the lines coincide (a1/a2 = b1/b2 = c1/c2), meaning the equations are dependent. These conditions help determine the nature of the solutions without graphing.
Solve the pair of equations 3x + 2y = 12 and 6x + 4y = 24 using the elimination method. What do you observe?
To solve 3x + 2y = 12 and 6x + 4y = 24, multiply the first equation by 2 to get 6x + 4y = 24. Subtracting this from the second equation gives 0 = 0, which is always true. This indicates that the two equations are equivalent and represent the same line. Therefore, there are infinitely many solutions, as any point on the line satisfies both equations.
A shopkeeper sells two types of pens. The cost of 5 pens of type A and 3 pens of type B is Rs. 79, while the cost of 2 pens of type A and 7 pens of type B is Rs. 89. Find the cost of each type of pen.
Let the cost of type A pen be x and type B pen be y. The equations are 5x + 3y = 79 and 2x + 7y = 89. Multiply the first equation by 2 and the second by 5 to get 10x + 6y = 158 and 10x + 35y = 445. Subtract the first new equation from the second to eliminate x: 29y = 287, so y = 9.89. Substitute y back into the first original equation to find x = 10.26. Thus, type A pen costs approximately Rs. 10.26 and type B pen costs approximately Rs. 9.89.
Explain how to form a pair of linear equations from a word problem. Use the example of two numbers whose sum is 50 and difference is 10.
To form a pair of linear equations from a word problem, identify the variables and the relationships between them. For example, let the two numbers be x and y. Their sum is 50: x + y = 50. Their difference is 10: x - y = 10. These two equations form a system that can be solved using substitution or elimination. Solving them gives x = 30 and y = 20. This approach can be applied to various real-life situations by carefully defining variables and translating words into mathematical equations.
What is the significance of the graphical representation of a pair of linear equations in understanding their solutions?
The graphical representation of a pair of linear equations provides a visual understanding of their solutions. The point of intersection represents the unique solution if the lines intersect. Parallel lines indicate no solution, while coinciding lines represent infinitely many solutions. This method helps in comprehending the nature of the solutions, whether they are consistent, inconsistent, or dependent, without algebraic manipulation. It also aids in verifying the solutions obtained by other methods.
Solve the pair of equations 0.4x + 0.3y = 1.7 and 0.7x - 0.2y = 0.8 using the substitution method.
First, solve the first equation for y: 0.3y = 1.7 - 0.4x, so y = (1.7 - 0.4x)/0.3. Substitute this into the second equation: 0.7x - 0.2[(1.7 - 0.4x)/0.3] = 0.8. Multiply through by 0.3 to eliminate denominators: 0.21x - 0.2(1.7 - 0.4x) = 0.24. Expand and simplify: 0.21x - 0.34 + 0.08x = 0.24, leading to 0.29x = 0.58, so x = 2. Substitute x back into the expression for y: y = (1.7 - 0.8)/0.3 = 3. Thus, the solution is x = 2, y = 3.
A fraction becomes 1/2 when 1 is subtracted from the numerator and 1 is added to the denominator. It becomes 1/3 when 1 is added to the numerator and 1 is subtracted from the denominator. Find the original fraction.
Let the numerator be x and the denominator be y. The first condition gives (x - 1)/(y + 1) = 1/2, leading to 2x - 2 = y + 1 or 2x - y = 3. The second condition gives (x + 1)/(y - 1) = 1/3, leading to 3x + 3 = y - 1 or 3x - y = -4. Subtract the first equation from the second: x = -7. Substitute x into the first equation: -14 - y = 3, so y = -17. Thus, the original fraction is -7/-17, which simplifies to 7/17.
Question 1 of 10
Explain the graphical method of solving a pair of linear equations in two variables. What does the point of intersection represent?
Pair of Linear Equations in Two Variables - Mastery Worksheet
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This worksheet challenges you with deeper, multi-concept long-answer questions from Pair of Linear Equations in Two Variables to prepare for higher-weightage questions in Class X Mathematics.
Intermediate analysis exercises
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Questions
Solve the pair of linear equations graphically: 2x + 3y = 8 and 4x + 6y = 7. What do you observe about their solutions?
The equations 2x + 3y = 8 and 4x + 6y = 7 are parallel lines since their coefficients are proportional (a1/a2 = b1/b2 ≠ c1/c2). Therefore, they have no solution, indicating an inconsistent pair of equations.
A fraction becomes 9/11 when 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Let the fraction be x/y. The equations formed are (x+2)/(y+2) = 9/11 and (x+3)/(y+3) = 5/6. Solving these, we find x = 7 and y = 9. Thus, the fraction is 7/9.
The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number.
Let the digits be x and y. The equations are 10x + y + 10y + x = 66 and |x - y| = 2. Solving gives two possible numbers: 42 and 24.
Compare the graphical and algebraic methods of solving a pair of linear equations. Which method is more efficient and why?
Graphical method provides a visual representation and is useful for understanding the nature of solutions, but it's less precise for non-integral solutions. Algebraic methods (substitution, elimination) are more efficient for exact solutions, especially with non-integral values.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for seven days, while Susy paid ₹21 for five days. Find the fixed and additional charges.
Let fixed charge be ₹x and additional charge per day be ₹y. The equations are x + 4y = 27 and x + 2y = 21. Solving gives x = 15 and y = 3.
Explain the conditions under which a pair of linear equations has no solution, a unique solution, or infinitely many solutions.
No solution: Lines are parallel (a1/a2 = b1/b2 ≠ c1/c2). Unique solution: Lines intersect (a1/a2 ≠ b1/b2). Infinitely many solutions: Lines coincide (a1/a2 = b1/b2 = c1/c2).
The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and ball.
Let cost of a bat be ₹x and a ball be ₹y. The equations are 7x + 6y = 3800 and 3x + 5y = 1750. Solving gives x = 500 and y = 50.
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?
Let Jacob's age be x and son's age be y. The equations are x + 5 = 3(y + 5) and x - 5 = 7(y - 5). Solving gives x = 40 and y = 10.
A rectangular garden's length is 4m more than its width. Half the perimeter is 36m. Find the dimensions of the garden.
Let width be x and length be x + 4. The equation is (2x + 2(x + 4))/2 = 36. Solving gives x = 16. Thus, dimensions are 16m and 20m.
The taxi charges in a city consist of a fixed charge plus a charge per km. For 10km, the charge is ₹105, and for 15km, it's ₹155. Find the fixed charge and the rate per km.
Let fixed charge be ₹x and rate per km be ₹y. The equations are x + 10y = 105 and x + 15y = 155. Solving gives x = 5 and y = 10.
Question 1 of 10
Solve the pair of linear equations graphically: 2x + 3y = 8 and 4x + 6y = 7. What do you observe about their solutions?
Pair of Linear Equations in Two Variables - Challenge Worksheet
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The final worksheet presents challenging long-answer questions that test your depth of understanding and exam-readiness for Pair of Linear Equations in Two Variables in Class X.
Advanced critical thinking
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Questions
Akhila went to a fair and spent ` 20 on rides and games. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs ` 3, and a game of Hoopla costs ` 4. Formulate the situation as a pair of linear equations and solve it graphically.
Let the number of rides be x and the number of Hoopla games be y. The equations are y = (1/2)x and 3x + 4y = 20. Plotting these equations on a graph will give the solution where they intersect.
Explain the conditions under which a pair of linear equations in two variables has no solution, a unique solution, or infinitely many solutions. Provide examples for each case.
A pair of linear equations has no solution if the lines are parallel (a1/a2 = b1/b2 ≠ c1/c2), a unique solution if they intersect (a1/a2 ≠ b1/b2), and infinitely many solutions if they coincide (a1/a2 = b1/b2 = c1/c2). Examples include x + 2y = 4 and 2x + 4y = 12 for no solution, x + y = 5 and 2x - y = 4 for unique solution, and x + y = 2 and 2x + 2y = 4 for infinitely many solutions.
Champa purchased some pants and skirts. The number of skirts is two less than twice the number of pants. Also, the number of skirts is four less than four times the number of pants. Represent this situation algebraically and find the number of pants and skirts she bought.
Let the number of pants be x and skirts be y. The equations are y = 2x - 2 and y = 4x - 4. Solving these gives x = 1 and y = 0, meaning she bought 1 pant and no skirts.
The sum of a two-digit number and the number obtained by reversing its digits is 66. If the digits differ by 2, find the number. How many such numbers exist?
Let the digits be x and y. The equations are 10x + y + 10y + x = 66 and |x - y| = 2. Solving gives two numbers: 42 and 24.
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ` 27 for seven days, and Susy paid ` 21 for five days. Find the fixed charge and the charge for each extra day.
Let the fixed charge be ` x and the additional charge per day be ` y. The equations are x + 4y = 27 and x + 2y = 21. Solving gives x = 15 and y = 3.
The ratio of incomes of two persons is 9:7, and the ratio of their expenditures is 4:3. If each saves ` 2000 per month, find their monthly incomes.
Let the incomes be 9x and 7x, and expenditures be 4y and 3y. The equations are 9x - 4y = 2000 and 7x - 3y = 2000. Solving gives x = 2000, so incomes are ` 18000 and ` 14000.
A fraction becomes 9/11 if 2 is added to both numerator and denominator. If 3 is added to both, it becomes 5/6. Find the fraction.
Let the fraction be x/y. The equations are (x + 2)/(y + 2) = 9/11 and (x + 3)/(y + 3) = 5/6. Solving gives x = 7 and y = 9, so the fraction is 7/9.
Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. Find their present ages.
Let Nuri's age be x and Sonu's age be y. The equations are x - 5 = 3(y - 5) and x + 10 = 2(y + 10). Solving gives x = 50 and y = 20.
The taxi charges in a city consist of a fixed charge plus a charge per km. For 10 km, the charge is ` 105, and for 15 km, it's ` 155. Find the fixed charge and the charge per km. What will be the charge for 25 km?
Let the fixed charge be ` x and the charge per km be ` y. The equations are x + 10y = 105 and x + 15y = 155. Solving gives x = 5 and y = 10. The charge for 25 km is ` 255.
A cricket team's coach buys 7 bats and 6 balls for ` 3800. Later, she buys 3 bats and 5 balls for ` 1750. Find the cost of each bat and ball.
Let the cost of a bat be ` x and a ball be ` y. The equations are 7x + 6y = 3800 and 3x + 5y = 1750. Solving gives x = 500 and y = 50.
Question 1 of 10
Akhila went to a fair and spent ` 20 on rides and games. The number of times she played Hoopla is half the number of rides she had on the Giant Wheel. Each ride costs ` 3, and a game of Hoopla costs ` 4. Formulate the situation as a pair of linear equations and solve it graphically.
This chapter explores real numbers, focusing on key properties such as the Fundamental Theorem of Arithmetic and the concept of irrational numbers, which are crucial for understanding the number system.
This chapter discusses polynomials, their degrees, and classifications such as linear, quadratic, and cubic. Understanding polynomials is essential for solving various mathematical problems.
This chapter explores quadratic equations, highlighting their forms and significance in real-world applications.
This chapter introduces arithmetic progressions, which are sequences of numbers generated by adding a fixed value to the previous term. Understanding these patterns is crucial for solving real-life mathematical problems.
This chapter focuses on the properties of triangles, specifically their similarity and how it can be applied in various real-world contexts.
This chapter covers the concepts of coordinate geometry, including finding distances between points and dividing line segments. Understanding these concepts is essential for solving geometry problems using algebra.
This chapter focuses on the foundational concepts of trigonometry, particularly the relationships between the angles and sides of right triangles.
This chapter explores how trigonometry is applied in real-life situations, particularly in measuring heights and distances.
This chapter explores the properties of circles, particularly focusing on tangents and their relationship with radii and secants.
This chapter focuses on sectors and segments of circles, essential concepts in geometry. Understanding these helps in solving real-life problems related to areas and measurements.
