NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5

Mathematics is one of the most important fields of study. It is concerned with logical reasoning and quantitative calculations. Mathematics plays a substantial role in everyday life. The knowledge of this subject can prove to be very useful in the everyday life of students. Students must pay close attention to the concepts that are taught in Class 10 Maths Chapter 3 Exercise 3.5. Mathematics is applied in almost all fields of life. Students particularly looking to make a career in Engineering must work on the development of their mathematical skills right from their schooling years to get a jump start. Students must pay attention to the mathematical concepts introduced in Class 10 since most of the competitive examinations are based on these mathematical concepts. Most of these competitive exams require logical reasoning abilities and quantitative aptitude. The concepts required to develop logical reasoning abilities and quantitative aptitude are mostly covered in the Mathematics Class 10 syllabus. Students planning to appear for competitive examinations like JEE Mains, JEE Advance, CAT, and NEET, should pay special attention to the mathematical concepts that are covered in Class 10. Students may utilise the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5

Students are advised to take note of the fact that the Central Board of Secondary Education is responsible for conducting annual examinations for Class 10 and Class 12. Not just for the conduct of examinations, but CBSE is also responsible for the prescription and authorisation of conditions of examinations. The CBSE board examination is sourced from the NCERT curriculum. It is highly recommended that students use the NCERT books to study for the CBSE board examination. Students can also access the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 from the Extramarks website to study for the CBSE Board Examination of Mathematics. The marks that students secure in the CBSE board examination play a huge part in availing opportunities to students in the field of their choice. Students may use the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 for their preparation. It is advisable to refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 since the CBSE board examination is highly based on the NCERT curriculum.

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NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 – Pair of Linear Equations in Two Variables

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Access NCERT Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

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Class 10 Chapter 3 Exercise 3.5 Solution: A Brief Overview

When the degree of an equation is one, it is called a linear equation. A linear equation is an equation in which the power of all the variables involved is one. Mathematics chapter 3 of Class 10, Pair of Linear Equations in Two Variables covers the technique of how to determine the value of a particular variable through various methods. Pair of Linear Equations in Two Variables is one of the most significant parts of the Class 10 Mathematics NCERT curriculum. This chapter holds a lot of weightage in the CBSE board examination. Pair of a Linear Equation in Two Variables is a chapter that covers a lot of mathematical concepts that are beneficial even in the future for students. Students must pay attention to the details of exercise 3.5 Maths Class 10 chapter 3 Pair of Linear Equations in Two Variables. It is essential to understand the basics of each topic to prepare the chapter effectively. Students must study the topics that are covered in the syllabus following their weightage in the CBSE board examination. This can help students prioritise the important topics and help them avoid wasting time and effort on topics that do not hold value in the CBSE board examination.

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What is a Cross Multiplication Method To Solve Linear Equations in One Variable?

Students must thoroughly understand the cross multiplication method to solve linear equations in one variable, as it is one of the most important topics that have been covered in the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5. Students must diligently go through the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 to thoroughly understand the mathematical concepts covered in the Mathematics chapter 3 of Class 10. Students should practice the NCERT exercise problems and solutions. Students may refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 from the Extramarks website or mobile application to study for the Class 10 CBSE Board.

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NCERT Solutions for Class 10 Maths Chapter 3 Exercises

Students must perform well in the CBSE Board examination since it lays the groundwork for a bright future. While preparing for the CBSE Board Examination of Mathematics, students are advised to practice the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5. Students of Class 10 CBSE must focus on frequently practising Mathematics topics. Students must practice solving the NCERT exercise questions occasionally. Regularly solving the exercise problems can help students prepare well for the CBSE Board Examination. Students may refer to the NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5 to solve the NCERT exercise problems.

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Q.1

$\begin{array}{l} Which of the following pairs of linear equations has unique solution, \\ no solution, or infinitely many solutions. In case there is a unique \\ solution, find it by using cross multiplication method. \\ (i) x - 3y - 3 = 0 (ii) 2x + y = 5 \\ 3x - 9y - 2 = 0 3x + 2y = 8 \\ (iii) 3x - 5y = 20 (iv) x - 3y - 7 = 0 \\ 6x - 10y = 40 3x - 3y - 15 = 0 \end{array}$

Ans.

$\begin{array}{l} (i) x - 3 y - 3 = 0 \\ 3 x - 9 y - 2 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 9} = \frac{1}{3}, \frac{c_{1}}{c_{2}} = \frac{- 3}{- 2} = \frac{3}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ Thus, the given pair of linear equations have no solution. \\ (ii) 2 x + y = 5 i.e., 2 x + y - 5 = 0 \\ 3 x + 2 y = 8 i.e., 3 x + 2 y - 8 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{2}{3}, \frac{b_{1}}{b_{2}} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{5}{8} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Thus, the given pair of linear equations have unique solution. \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{- 8 - (- 10)} = \frac{y}{- 15 - (- 16)} = \frac{1}{4 - 3} \\ \Rightarrow \frac{x}{2} = \frac{y}{1} = \frac{1}{1} \\ \Rightarrow x = 2 and y = 1 \end{array}$ $\begin{array}{l} (iii) 3 x - 5 y = 20 i.e., 3 x - 5 y - 20 = 0 \\ 6 x - 10 y = 40 i.e., 6 x - 10 y - 40 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{3}{6} = \frac{1}{2}, \frac{b_{1}}{b_{2}} = \frac{- 5}{- 10} = \frac{1}{2}, \frac{c_{1}}{c_{2}} = \frac{- 20}{- 40} = \frac{1}{2} \\ and so \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ Thus, the given pair of linear equations have infinitely \\ many solutions. \\ (iv) x - 3 y - 7 = 0 \\ 3 x - 3 y - 15 = 0 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{1}{3}, \frac{b_{1}}{b_{2}} = \frac{- 3}{- 3} = 1, \frac{c_{1}}{c_{2}} = \frac{- 7}{- 15} = \frac{7}{15} \\ and so \\ \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \\ Thus, the given pair of linear equations have unique solution. \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{45 - 21} = \frac{y}{- 21 - (- 15)} = \frac{1}{- 3 - (- 9)} \\ \Rightarrow \frac{x}{24} = \frac{y}{- 6} = \frac{1}{6} \\ \Rightarrow x = 4 and y = - 1 \end{array}$

Q.2

$\begin{array}{l} (i) For which values of a and b does the following pair of linear equations have an \\ infinite number of solutions? \\ 2x + 3y =7 \\ (a – b)x + (a + b)y = 3a + b - 2 \\ (ii) For which value of k will the following pair of linear equations have no solution? \\ 3x + y = 1 \\ (2k – 1)x + (k – 1) y = 2k + 1 \end{array}$

Ans.

$\begin{array}{l} (i) \\ Given pair of linear equations are: \\ 2 x + 3 y = 7 \\ (a - b) x + (a + b) y = 3 a + b - 2 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{2}{a - b}, \frac{b_{1}}{b_{2}} = \frac{3}{a + b}, \frac{c_{1}}{c_{2}} = \frac{7}{3 a + b - 2} \\ For a pair of linear equations to have infinitely many solutions: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \\ So, we need \\ \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3 a + b - 2} \\ or \frac{2}{a - b} = \frac{7}{3 a + b - 2} and \frac{2}{a - b} = \frac{3}{a + b} \\ or 2 (3 a + b - 2) = 7 (a - b) and 2 (a + b) = 3 (a - b) \\ or 6 a + 2 b - 4 = 7 a - 7 b and 2 a - 3 a = - 3 b - 2 b \\ or - a + 9 b - 4 = 0 and a = 5 b \\ Now, we put a = 5 b in - a + 9 b - 4 = 0 . \\ Thus, \\ - 5 b + 9 b = 4 \\ or 4 b = 4 \\ or b = 1 \\ Also, \\ a = 5 b = 5 \times 1 = 5 \end{array}$ $\begin{array}{l} (ii) \\ Given pair of linear equations are: \\ 3 x + y = 1 \\ (2 k - 1) x + (k - 1) y = 2 k + 1 \\ Here, \\ \frac{a_{1}}{a_{2}} = \frac{3}{2 k - 1}, \frac{b_{1}}{b_{2}} = \frac{1}{k - 1}, \frac{c_{1}}{c_{2}} = \frac{1}{2 k + 1} \\ For a pair of linear equations to have no solution: \\ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}} \\ So, the given pair of linear equations have no solution when \\ \frac{3}{2 k - 1} = \frac{1}{k - 1} \\ or 3 k - 3 = 2 k - 1 \\ or 3 k - 2 k = - 1 + 3 \\ or k = 2 \end{array}$

Q.3

$\begin{array}{l} Solve the following pair of linear equations by the \\ substitution and cross-multiplication methods: \\ 8 x + 5 y = 9 \\ 3 x + 2 y = 4 \end{array}$

Ans.

$\begin{array}{l} 8 x + 5 y = 9 . .. (1) \\ 3 x + 2 y = 4 . .. (2) \\ By substitution mwthod: \\ From equation (2), we get \\ y = \frac{4 - 3 x}{2} \\ Substituting this value of y in equation (1), we get \\ 8 x + 5 y = 9 \\ or 8 x + 5 \times \frac{4 - 3 x}{2} = 9 \\ or 16 x + 20 - 15 x = 18 \\ or x = 18 - 20 = - 2 \\ Putting this value of x in equation (2), we get \\ 3 \times (- 2) + 2 y = 4 \\ or - 6 + 2 y = 4 \\ or 2y = 4 + 6 = 10 \\ or y = 5 \end{array}$ $\begin{array}{l} By cross-multiplication method : \\ 8 x + 5 y = 9 \\ 3 x + 2 y = 4 \\ Or \\ 8 x + 5 y - 9 = 0 . .. (1) \\ 3 x + 2 y - 4 = 0 . .. (2) \\ {Here, a}_{1} = 8, b_{1} {= 5, c}_{1} = - 9 \\ a_{2} = 3, b_{2} {= 2, c}_{2} = - 4 \\ By cross-multiplication method, we have \\ \frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - a_{1} c_{2}} = \frac{1}{a_{1} b_{2} - b_{1} a_{2}} \\ \Rightarrow \frac{x}{- 20 + 18} = \frac{y}{- 27 + 32} = \frac{1}{16 - 15} \\ \Rightarrow \frac{x}{- 2} = \frac{y}{5} = \frac{1}{1} \\ \Rightarrow x = - 2 and y = 5 \end{array}$

Q.4

$\begin{array}{l} Form the pair of linear equations in the following \\ problems and find their solutions (if they exist) \\ by any algebraic method: \\ (i) A part of monthly hostel charges is fixed and the \\ remaining depends on the number of days one \\ has taken food in the mess. When a student A \\ takes food for 20 days she has to pay ₹ 1000 \\ as hostel charges whereas a student B, who \\ takes food for 26 days, pays ₹ 1180 as hostel \\ charges. Find the fixed charges and the cost of \\ food per day. \\ (ii) A fraction becomes \frac{1}{3} when 1 is subtracted from \\ the numerator and it becomes \frac{1}{4} when 8 is added \\ to its denominator. Find the fraction. \\ (iii) Yash scored 40 marks in a test, getting 3 marks \\ for each right answer and losing 1 mark for each \\ wrong answer. Had 4 marks been awarded for \\ each correct answer and 2 marks been deducted \\ for each incorrect answer, then Yash would have \\ scored 50 marks. How many questions were there \\ in the test? \end{array}$

$\begin{array}{l} (iv) Places A and B are 100 km apart on a highway. One \\ car starts from A and another from B at the same \\ time. If the cars travel in the same direction at \\ different speeds, they meet in 5 hours. If they \\ travel towards each other, they meet in 1 hour. \\ What are the speeds of the two cars? \\ (v) The area of a rectangle gets reduced by 9 square \\ units, if its length is reduced by 5 units and \\ breadth is increased by 3 units. If we increase \\ the length by 3 units and the breadth by 2 units, \\ the area increases by 67 square units . Find the \\ dimensions of the rectangle. \end{array}$

Ans.

$\begin{array}{l} (i) \\ Let the fixed charge be ₹ x and the cost of food per day be ₹ y . \\ According to question, \\ x + 20 y = 1000 . .. (1) \\ and \\ x + 26 y = 1180 . .. (2) \\ Subtracting equation (1) from equation (2), we get \\ \underline{\begin{array}{l} x + 26 y = 1180 \\ x + 20 y = 1000 \\ - - - \end{array}} \\ 6 y = 180 \\ or y = 30 \\ Putting this value of y in equation (1), we get \\ x + 20 \times 30 = 1000 \\ or x = 1000 - 600 = 400 \\ Hence, fixed charge is ₹ 400 and the cost of food per \\ day is ₹ 30 . \end{array}$ $\begin{array}{l} (ii) \\ Let the fraction be \frac{x}{y} . \\ According to question, \\ \frac{x - 1}{y} = \frac{1}{3} \\ \Rightarrow 3 x - 3 = y \\ \Rightarrow 3 x - y = 3 . .. (1) \\ Also, \\ \frac{x}{y + 8} = \frac{1}{4} \\ \Rightarrow x = \frac{y + 8}{4} \\ \Rightarrow 4 x = y + 8 \\ \Rightarrow 4 x - y = 8 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 5 \\ Substituting this value in equation (1), we get \\ 3 \times 5 - y = 3 \\ or - y = 3 - 15 \\ or y = 12 \\ Therefore, \\ \frac{x}{y} = \frac{5}{12} \\ Hence, the fraction is \frac{5}{12} . \end{array}$ $\begin{array}{l} (iii) \\ Let the number of right and the number of wrong \\ answers be x and y respectively. \\ According to questions, \\ 3 x - y = 40 . .. (1) \\ and \\ 4 x - 2 y = 50 \\ \Rightarrow 2 (2 x - y) = 2 \times 25 \\ \Rightarrow 2 x - y = 25 . .. (2) \\ Subtracting equation (2) from equation (1), we get \\ x = 15 \\ Putting this value of x in equation (2), we get \\ 2 \times 15 - y = 25 \\ \Rightarrow - y = 25 - 30 = - 5 \\ \Rightarrow y = 5 \\ Thus, \\ Number of right answers = x = 15 \\ Number of wrong answers = y = 5 \\ and \\ Number of questions = x + y = 15 + 5 = 20 \end{array}$ $\begin{array}{l} (iv) \\ {Let the speeds of I}^{st} {and 2}^{nd} cars be x km/h and y km/h \\ respectively. \\ According to question, \\ x + y = 100 . .. (1) \\ and \\ 5 x - 100 = 5 y \\ \Rightarrow 5 (x - 20) = 5 y \\ \Rightarrow x - 20 = y \\ \Rightarrow x - y = 20 . .. (2) \\ Adding equations (1) and (2), we get \\ 2 x = 120 \Rightarrow x = 60 \\ Putting this value of x in equation (1), we get \\ 60 + y = 100 \\ \Rightarrow y = 100 - 60 = 40 \\ Thus, \\ {Speed of I}^{st} car = x km/h = 60 km/h \\ {Speed of 2}^{nd} car = y km/h = 40 km/h \end{array}$ $\begin{array}{l} \end{array}$ $\begin{array}{l} (v) \\ Let length and breadth of the rectangle be x units and y units \\ respectively. \\ According to question, \\ (x - 5) (y + 3) = xy - 9 \\ \Rightarrow xy - 5 y + 3 x - 15 = xy - 9 \\ \Rightarrow - 5 y + 3 x = 15 - 9 = 6 \\ \Rightarrow 3 x - 5 y - 6 = 0 . .. (1) \\ Also, \\ (x + 3) (y + 2) = xy + 67 \\ \Rightarrow xy + 3 y + 2 x + 6 = xy + 67 \\ \Rightarrow 3 y + 2 x = 67 - 6 = 61 \\ \Rightarrow 2 x + 3 y - 61 = 0 . .. (2) \\ By cross-multiplication method, we have \\ \frac{x}{305 - (- 18)} = \frac{y}{- 12 - (- 183)} = \frac{1}{9 - (- 10)} \\ \Rightarrow x = \frac{323}{19} = 17 and y = \frac{171}{19} = 9 \\ Hence, length and breadth of the rectangle are 17 units and \\ 9 units respectively. \end{array}$

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NCERT Solutions For Class 10 Maths Chapter 3 Exercise 3.5