NCERT Solutions for Class 8 Maths Exercise 2.2 Chapter 2- Linear Equations in One Variable
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The Class 8 Maths Chapter 2.2 topics are covered in the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. The books that NCERT publishes offer an in-depth understanding of a variety of areas. NCERT does not only offer books as educational resources. The government established the NCERT, or National Council of Educational Research and Teaching, to supply educational and instructional materials to schools. In order to provide their students with educational resources and train new teachers, many schools have partnered with NCERT. Students frequently use the NCERT study materials to get ready for numerous tests, including the Class 10 and Class 12 board exams and various entrance exams, such as JEE, NEET, CUET, NDA, etc., to get into the programmes and institutions of their choice. The answers to the problems in NCERT Class 8 Maths Chapter 2 Exercise 2.2 are provided in the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. For students, these NCERT solutions are helpful. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, is a useful resource since it gives students the solutions they need to address the problems in the NCERT books. Students may learn more about the chapter, the right answers, and the best ways to respond to each question by using the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2.
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NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable (EX 2.2) Exercise 2.2
On Extramarks’ website and app, students can find NCERT solutions for Class 8 for every chapter. NCERT Solutions are essential for students to study because they will help them understand the subjects. Students should keep in mind that working through NCERT textbooks can help them perform better on entrance exams like JEE, CUET, and others as well as in board exams. NCERT books can help students improve their fundamental understanding of the chapter as well as how the questions should be answered. Students can obtain assistance from NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, to understand various topics.
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In order to fully understand the chapter, students must carefully review all the example questions and exercise questions provided in the NCERT books. This is a crucial component of getting ready for an exam, whether it is an entrance exam or an end-of-term test. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 by Extramarks is a study guide that students can use if they are having trouble understanding the methods and solutions that NCERT has supplied through the sample questions. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, will be of great help to students preparing for exams.
Access NCERT Solutions for Maths Chapter 2 – Linear Equations in One Variable
The goal of NCERT exercises is to gauge how well students comprehend the concepts covered in a given chapter. NCERT solutions are crucial for board examinations. If students are having difficulty understanding linear equations in one variable, they should refer to the study materials for Mathematics Chapter 2 Class 8, such as the NCERT Solutions for Class 8 Maths Chapter 2 Exercise 2.2, which are available on the Extramarks website and application. Many students find this chapter difficult and turn to the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, which offers thorough solutions.
There are numerous significant topics in the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 for exam preparation. Students will benefit from this by receiving higher marks on the questions in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, during their exams. The objective of NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, is to assist students in getting past a problem they are facing so they can go on to learn and practice other concepts from different textbooks. With the aid of the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, students could finish the NCERT exercises for the chapter, giving them extra time to study for examinations and practice exams. They may perform better on their final exams if they use the solutions to Class 8 Maths Chapter 2 Exercise 2.2.
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What You’ll Learn In NCERT Class 8 Maths Chapter 2 Exercise 2.2
Through the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, it can be concluded that this chapter begins with a brief introduction before covering a number of concepts, including how to solve equations with linear expressions on one side and numbers on the other, how to apply the aforementioned exercises, and how to solve equations with variables on both sides, equations that can be transformed into a simpler form or a linear form. The second exercise helps students put what they learned in the first exercise into practice. It provides the basis for NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Therefore, it is crucial for students to comprehend the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 in order to achieve a higher grade.
NCERT Maths Class 8 Chapter 2 Exercise 2.2: Get Detailed Solutions
Students can understand the concepts in Exercise 2.2 by using the exact solutions and explanations provided in the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Students should remember that every NCERT exercise, including Exercise 2.2, has solutions like the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Students must fully understand the content presented in Exercise 2.2 in order to choose how to respond to various questions. When addressing problems where the query only offers scant information, they can benefit from this. Students could benefit from being aware of the various processes. To score well on exams, students must consequently absorb and retain this information. They could consult the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, if they require additional assistance. Any queries students may have regarding Exercise 13.2 can be answered by reading the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. The ideas in Exercise 2.2 are crucial for students’ exams. As a result, it is crucial for students to comprehend how certain questions should be answered because doing so can improve their academic achievement. If necessary, students must use Extramarks to assist them. For assistance for their exam, they might consult the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2. Hence, the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, is a helpful tool for students that they must use to improve their scores in board exams or year-end exams.
About Chapter 2 of NCERT Textbook: Linear Equations in One Variable
Chapter 2 of NCERT Class 8 Mathematics is named Linear Equations in One Variable. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 will provide an overview of the chapter. This chapter contains a brief introduction to the chapter followed by various concepts such as solving equations that have linear expressions on one side and numbers on the other, some applications of the above exercise, and solving equations that have a variable on both sides. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, is based on the second exercise that helps students apply the things they have learned in the previous exercise. Hence, the NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2 is important for students to understand in order to get a good score in the exams.
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NCERT Solutions Class 8 Maths Chapter 2 All Exercises
The topics covered in NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, include the applications of the formulas and theories learned in the first exercise. The NCERT Solutions For Class 8 Maths Chapter 2 Exercise 2.2, will help students answer the questions in the exercise. Students must thoroughly study each exercise to be able to solve the next one. It will help them improve their overall performance and understanding of the concepts. Hence, students must make use of the resources provided to them in order to perform to the best of their abilities.
NCERT Solutions for Class 8
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Q.1
Ans
Q.2 The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and the breadth of the pool?
Ans
The perimeter of a rectangular swimming pool is 154 m.
Let the breadth be x m. The length will be (2x + 2) m.
According to the question the equation becomes,
Hence, the length and breadth of the pool are 52 m and 25 m.
Q.3
Ans
Q.4 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 get
x = 40
Therefore, x+15 = 40+15 = 55
Hence the numbers are 40 and 55.
Q.5 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?
Ans
Let the ratio between these numbers be x.
Therefore, the numbers will be 5x and 3x respectively.
Difference between these numbers = 18
According, to the question the equation becomes
5x − 3x = 18
2x = 18
Therefore, the first number is 5x=5×9=45 and,
The second number is 3x=3×9=27.
Q.6 Three consecutive integers add up to 51. What are these integers?
Ans
Let three consecutive integers be x, x + 1, and x + 2.
Sum of these numbers = x+ x + 1 + x + 2 = 51
3x + 3 = 51
On transposing 3 to R.H.S, we obtain
3x = 51 − 3
3x = 48
Hence, the integers are 16, 17 and 18.
Q.7 The sum of three consecutive multiples of 8 is 888. Find the multiples.
Ans
Let the three consecutive multiples of 8 be 8x, 8(x + 1), 8(x + 2).
Sum of these numbers = 8x + 8(x + 1) + 8(x + 2) = 888
8(x + x + 1 + x + 2) = 888
8(3x + 3) = 888
Therefore, 8x = 8×36 = 288
8(x+1) = 8(36+1) = 8×37 = 296
8(x+2) = 8(36+2) = 8×38 = 304
Hence, the numbers are 288,296 and 304.
Q.8 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.
Ans
Let three consecutive integers be x, x + 1, x + 2. According to the question,
2x + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
On transposing 11 to R.H.S, we obtain
9x = 74 − 11
9x = 63
Q.9 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?
Ans
Let the ratio between Rahul’s age and Haroon’s age be x.
Therefore, the age of Rahul and Haroon will be 5x years and 7x years.
Four years later, the age of Rahul and Haroon will be (5x + 4) years and (7x + 4) years.
According to the given question, the equation becomes,
(5x + 4 + 7x + 4) = 56
12x + 8 = 56
On transposing 8 to R.H.S, we obtain
12x = 56 − 8
12x = 48
Q.10 The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?
Ans
Let the ratio between the number of boys and numbers of girls be x.
Then, number of boys = 7x
Number of girls = 5x
According to the given question,
Number of boys = Number of girls + 8
7x = 5x + 8
On transposing 5x to L.H.S, we obtain
7x − 5x = 8
2x = 8
Q.11 Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?
Ans
Let Baichung’s father’s age be x years.
Therefore, Baichung’s age will be (x − 29) years and Baichung’s grandfather’s age will be (x + 26) years.
According to the given question, the equation becomes:
x + x − 29 + x + 26 = 135
3x − 3 = 135
On transposing 3 to R.H.S, we obtain
3x = 135 + 3
3x = 138
Q.12 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?
Ans
Let Ravi’s present age be x years.
Fifteen years later, Ravi’s age = 4 × His present age
x + 15 = 4x
On transposing x to R.H.S, we obtain
15 = 4x − x
15 = 3x
Q.13
Ans
Q.14 Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?
Ans
Let the ratio between the numbers of notes of different denominations be x.
Therefore, numbers of ₹ 100 notes, ₹ 50 notes, and ₹ 10 notes will be 2x, 3x, and 5x respectively.
Amount of ₹ 100 notes = ₹ (100×2x) = ₹ 200x
Amount of ₹ 50 notes = ₹ (50×3x) = ₹ 150x
Amount of ₹ 10 notes = ₹ (10×5x) = ₹ 50x
Total cash = ₹ 400000.
Therefore,
200x + 150x + 50x = 400000
⇒ 400x = 400000
On dividing both sides by 400, we obtain
x = 1000
Therefore,
Number of ₹ 100 notes = 2x = 2 × 1000 = 2000 notes
Number of ₹ 50 notes = 3x = 3 × 1000 = 3000 notes
Number of ₹ 10 notes = 5x = 5 × 1000 = 5000 notes
Q.15 I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of Rs 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?
Ans
Let the number of ₹ 5 coins be x.
Since the number of ₹ 2 coins is 3 times the number of ₹ 5 coins, so the number of ₹ 2 coins = 3x
Therefore, the number of ₹1 coins = 160 − (Number of coins of ₹ 5 and of ₹ 2)
= 160 − (3x + x) = 160 − 4x
Now, Amount of ₹ 1 coins = ₹ [1 × (160 − 4x)] = ₹ (160 − 4x)
Amount of ₹ 2 coins = ₹ (2 × 3x) = ₹ 6x
Amount of ₹ 5 coins = ₹ (5 × x) = ₹ 5x
Total amount is ₹300.
Therefore, 160 − 4x + 6x + 5x =300
160 + 7x = 300
On transposing 160 to R.H.S., we obtain
7x = 300 – 160 = 140
On dividing both sides by 7, we get
x = 20No. of ₹ 1 coins = 160 – 4x = 160 – 4×20 = 80
No. of ₹ 2 coins = 3x = 3×20 = 60
No. of ₹ 5 coins = x = 20
Therefore, number of ₹ 1 coins is 80, ₹ 2 coins is 60 and ₹ 5 coins is 20.
Q.16 The organisers of an essay competition decide that a winner of the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.
Ans
Let the number of winners be x.
Therefore, the number of participants who did not win will be 63 − x.
Amount given to the winners = ₹ (100 × x) = ₹ 100x
Amount given to the participants who did not win = ₹ [25(63 − x)]
= ₹ (1575 − 25x)
According to the given question,
100x + 1575 − 25x = 3000
On transposing 1575 to R.H.S, we obtain
75x = 3000 − 1575
75x = 1425
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