NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable – Here are all the NCERT solutions for Class 8 Maths Chapter 2. This solution contains questions, answers, images, explanations of the complete chapter 2 titled Linear Equations in One Variable of Maths taught in class 8. If you are a student of class 8 who is using NCERT Textbook to study Maths, then you must come across chapter 2 Linear Equations in One Variable. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable in one place.
NCERT Solutions Class 8 Maths Chapter 2 Linear Equations in One Variable
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For a better understanding of this chapter, you should also see summary of Chapter 2 Linear Equations in One Variable , Maths, Class 8.
| Class | 8 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 2 |
| Chapter Name |
Linear Equations in One Variable |
NCERT Solutions Class 8 Maths chapter 2 Linear Equations in One Variable
Class 8, Maths chapter 2, Linear Equations in One Variable solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: Solve the following equations.
1.1 x – 2 = 7
1.2 y + 3 = 10
1.3 6 = z + 2
1.4 \(\frac{3}{7}+x=\frac{17}{7}\)
1.5 6 x=12
1.6 \(\frac{t}{5}=10\)
1.7 \(\frac{2 x}{3}=18\)
1.8 \(1.6=\frac{y}{1.5}\)
1.9 \(7 x-9=16\)
1.10 \(14 y-8=13\)
1.11 \(17+ 6p = 9\)
1.12 \(\frac{x}{3}+1=\frac{7}{15}\)
Ans : 1.1 x – 2 = 7 Transposing 2 to R.H.S, we obtain x = 7+2 = 9 1.2 y + 3 = 10 Transposing 3 to R.H.S, we obtain y = 10 - 3 = 7 1.3 6 = z + 2 Transposing 2 to L.H.S, we obtain 6 - 2 = z z = 4 1.4 \(\frac{3}{7}+x=\frac{17}{7}\) Transposing \(\frac{3}{7}\)to R. H.S, we obtain \(x=\frac{17}{7}-\frac{3}{7}=\frac{14}{7}=2\) 1.5 6 x=12 Dividing both sides by 6, we obtain \(\begin{array}{l}{\frac{6 x}{6}=\frac{12}{6}} \\ {x=2}\end{array}\) 1.6 \(\frac{t}{5}=10\) \(\begin{array}{l}{\text { Multiplying both sides by } 5, \text { we obtain }} \\ {\frac{g}{5} \times 5=10 \times 5} \\ {t=50}\end{array}\) 1.7 \(\frac{2 x}{3}=18\) \(\begin{array}{l}{\text { Multiplying both sides by }3/ 2 \text { , we obtain }} \\ {\frac{2 x}{3} \times \frac{3}{2}=18 \times \frac{3}{2}} \\ {x=27}\end{array}\) 1.8 \(1.6=\frac{y}{1.5}\) \(\begin{array}{l}{\text { Multiplying both sides by } 1.5, \text { we obtain }} \\ {1.6 \times 1.5=\frac{y}{1.5} \times 1.5} \\ {2.4=y}\end{array}\) 1.9 \(7 x-9=16\) \(\begin{array}{l}{\text { Transposing } 9 \text { to R.H.S, we obtain }} \\ {7 x=16+9} \\ {7 x=25} \\ {\text { Dividing both sides by } 7, \text { we obtain }} \\ {\frac{7 x}{7}=\frac{25}{7}} \\ {x=\frac{25}{7}}\end{array}\) 1.10 \(14 y-8=13\) Transposing 8 to R.H.S, we obtain 14y = 13 + 8 14y = 21 Dividing both sides by 14, we obtain \(\begin{array}{l}{\frac{14 y}{14}=\frac{21}{14}} \\ {y=\frac{3}{2}}\end{array}\) 1.11 \(17+ 6p = 9\) Transposing 17 to R.H.S, we obtain 6p = 9 - 17 6p = -8 \(\begin{array}{l}{\text { Dividing both sides by } 6, \text { we obtain }} \\ {\frac{6 p}{6}=-\frac{8}{6}} \\ {p=-\frac{4}{3}}\end{array}\) 1.12 \(\frac{x}{3}+1=\frac{7}{15}\) Transposing 1 to R.H.S, we obtain \(\begin{array}{l}{\frac{x}{3}=\frac{7}{15}-1} \\ {\frac{x}{3}=\frac{7-15}{15}} \\ {\frac{x}{3}=-\frac{8}{15}} \\ {\text { Multiplying both sides by } 3, \text { we obtain }} \\ {\frac{x}{3} \times 3=-\frac{8}{15} \times 3} \\ {x=-\frac{8}{5}}\end{array}\)
Q.2: If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get \(\frac{1}{2}\) . What is the number?
Ans : \(\begin{array}{l}{\text { Let the number be } x \text { . According to the question, }} \\ {\left(x-\frac{1}{2}\right) \times \frac{1}{2}=\frac{1}{8}} \\ {\text { On multiplying both sides by } 2, \text { we obtain }} \\ {\left(x-\frac{1}{2}\right) \times \frac{1}{2} \times 2=\frac{1}{8} \times 2} \\ {x-\frac{1}{2}=\frac{1}{4}} \\ {\text { On transposing } 1/2 \text { to R.H.S, we obtain }}\end{array}\) \(\begin{aligned} x &=\frac{1}{4}+\frac{1}{2} \\ &=\frac{1+2}{4}=\frac{3}{4} \end{aligned}\) Therefore, the number is \(\frac{3}{4}\)
Q.3: The perimeter of a rectangular swimming pool is 154m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Ans : Let the breadth be x m. The length will be (2x + 2) m. Perimeter of swimming pool = 2(l + b) = 154 m 2(2x + 2 + x) = 154 2(3x + 2) = 154 \(\begin{array}{l}{\text { Dividing both sides by } 2, \text { we obtain }} \\ {\frac{2(3 x+2)}{2}=\frac{154}{2}} \\ {3 x+2=77}\end{array}\) On transposing 2 to R.H .S, we obtain 3x = 77 - 2 3x = 75 \(\begin{array}{l}{\frac{3 x}{3}=\frac{75}{3}} \\ {x=25} \\ {2 x+2=2 \times 25+2=52}\end{array}\) Hence, the breadth and length of the pool are 25 m and Q2 Which of the following numbers would have digit 6 at unit place. \(\begin{array}{ll}{\text { (i) } 19^{2}} & {\text { (ii) } 24^{2}} \quad {\text { (iii) } 26^{2}}\\ {\text { (iv) } 36^{2}} & {\text { (v) } 34^{2}}\end{array}\) 52 m respectively
Q.4: The base of an isosceles triangle is \(\frac{4}{3}\) cm . The perimeter of the triangle is \(4 \frac{2}{15}\) cm . What is the length of either of the remaining equal sides?
Ans : Let the length of equal sides be x cm. Perimeter = x cm + x cm + Base = \(4 \frac{2}{15}\) cm \(2 x+\frac{4}{3}=\frac{62}{15}\) On transposing \(\frac{4}{3}\) to R. H.S, we obtain \(\begin{array}{l}{2 x=\frac{62}{15}-\frac{4}{3}} \\ {2 x=\frac{62-4 \times 5}{15}=\frac{62-20}{15}} \\ {2 x=\frac{42}{15}} \\ {\text { On dividing both sides by } 2, \text { we obtain }} \\ {\frac{2 x}{2}=\frac{42}{15} \times \frac{1}{2}} \\ {x=\frac{7}{5}=1 \frac{2}{5}}\end{array}\) Therefore, the length of equal sides is \(1 \frac{2}{5}\) cm
Q.5: Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Ans : Let one number be x. Therefore, the other number will be x + 15. According to the question, x + x + 15 = 95 2x + 15 = 95 On transposing 15 to R.H.S, we obtain 2x = 95 -15 2x = 80 On dividing both sides by 2, we obtain \(\begin{array}{l}{\frac{2 x}{2}=\frac{80}{2}} \\ {x=40} \\ {x+15=40+15=55} \\ {\text { Hence, the numbers are } 40 \text { and } 55 .}\end{array}\)
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