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NCERT Solutions for Class 7 Maths Chapter 4: Simple Equations - Exercise 4.1

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Last updated date: 28th Mar 2024
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MVSAT 2024

NCERT Solutions for Class 7 Maths Chapter 4 (EX 4.2)

Free PDF download of NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.2 (EX 4.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 7 Maths Chapter 4 Simple Equations Exercise 4.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails. Every NCERT Solution is provided to make the study simple and interesting on Vedantu. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 7 Science, Maths solutions and solutions of other subjects.


Class:

NCERT Solutions for Class 7

Subject:

Class 7 Maths

Chapter Name:

Chapter 4 - Simple Equations

Exercise:

Exercise - 4.2

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes

Access NCERT Solutions for Class 7 Mathematics Chapter 4 - Simple Equations

Exercise 4.2

1. Give first the step you will use to separate the variable and then solve the equation:

a. $x - 1 = 0$

Ans: The first step is to add 1 to both sides.

$x - 1 + 1 = 0 + 1$

$x = 1$

Thus, the value of $x$ is 1.

b. $x + 1 = 0$

Ans: The first step is to subtract 1 from both sides.

$x + 1 - 1 = 0 - 1$

$x =  - 1$

Thus, the value of $x$ is $ - 1$.

c. $x - 1 = 5$

Ans: The first step is to add 1 to both sides.

$x - 1 + 1 = 5 + 1$

$x = 6$

Thus, the value of $x$ is 6.

d. $x + 6 = 2$

Ans: The first step is to subtract 6 from both sides.

$x + 6 - 6 = 2 - 6$

$x =  - 4$

Thus, the value of $x$ is $ - 4$.

e. $y - 4 =  - 7$

Ans: The first step is to add 4 to both sides.

$y - 4 + 4 =  - 7 + 4$

$y =  - 3$

Thus, the value of $y$ is $ - 3$.

f. $y - 4 = 4$

Ans: The first step is to add 4 to both sides.

$y - 4 + 4 = 4 + 4$

$y = 8$

Thus, the value of $y$ is 8.

g. $y + 4 = 4$

Ans: The first step is to subtract 4 from both sides.

$y + 4 - 4 = 4 - 4$

$y = 0$

Thus, the value of $y$ is 0.

h. $y + 4 =  - 4$

Ans: The first step is to subtract 4 from both sides.

$y + 4 - 4 =  - 4 - 4$

$y =  - 8$

Thus, the value of $y$ is $ - 8$.


2. Give first the step you will use to separate the variable and then solve the equations:

a. $3l = 42$

Ans: The first step is to divide both sides by 2.

$l = \dfrac{{42}}{3}$

$l = 14$

Hence, the value of \[l\] is 14.

b. $\dfrac{b}{2} = 6$

Ans: The first step is to multiply both sides by 2.

$\dfrac{b}{2} \times 2 = 2 \times 6$

$b = 2 \times 6$

$b = 12$

Hence, the value of $b$ is 12.

c. $\dfrac{p}{7} = 4$

Ans: The first step is to multiply both sides by 7.

$\dfrac{p}{7} \times 7 = 4 \times 7$

$p = 28$

Hence, the value of $p$ is 28.

d. $4x = 25$

Ans: The first step is to divide both sides by 4.

$x = \dfrac{{25}}{4}$

The value of $x$ is $\dfrac{{25}}{4}$.

e. $8y = 36$

Ans: The first step is to divide both sides by 8.

$\dfrac{{8y}}{8} = \dfrac{{36}}{8}$$$

$y = \dfrac{{36}}{8}$

$y = \dfrac{9}{2}$

The value of $y$ is $\dfrac{9}{2}$.

f. $\dfrac{z}{3} = \dfrac{5}{4}$

Ans: The first step is to multiply both sides by 3.

$\dfrac{z}{3} \times 3 = \dfrac{5}{4} \times 3$

$z = \dfrac{{15}}{4}$

The value of $z$ is $\dfrac{{15}}{4}$.

g. $\dfrac{a}{3} = \dfrac{7}{{15}}$

Ans: The first step is to multiply both sides by 3.

$\dfrac{a}{3} \times 3 = \dfrac{7}{{15}} \times 3$

$a = \dfrac{7}{5}$

The value of $a$ is $\dfrac{7}{5}$.

h. $20t =  - 10$

Ans: The first step is to divide both sides by 20.

$\dfrac{{20t}}{{20}} = \dfrac{{ - 10}}{{20}}$

$t =  - \dfrac{1}{2}$

The value of $t$ is $ - \dfrac{1}{2}$.


3. Give first the step you will use to separate the variable and then solve the equations:

a. $3n - 2 = 46$

Ans: The first step is to add 2 to both sides of the equation.

$3n - 2 + 2 = 46 + 2$

$3n = 48$

Now, we will divide both sides by 3 and simplify.

$n = \dfrac{{48}}{3}$

$n = 16$

Thus, the value of $n$ is 16.

b. $5m + 7 = 17$

Ans: The first step is to subtract 7 from both sides.

$5m + 7 - 7 = 17 - 7$

$5m = 10$

Divide both sides by 5.

$m = \dfrac{{10}}{5}$

$m = 2$

Thus, the value of $m$ is 2.

c. $\dfrac{{20p}}{3} = 40$

Ans: The first step is to multiply both sides by 3.

$20p = 40 \times 3$

$20p = 120$

Divide both sides by 20.

$p = \dfrac{{120}}{{20}}$

$p = 6$

Hence, the value of $p$ is 6.

d. $\dfrac{{3p}}{{10}} = 6$

Ans: The first step is to multiply both sides by 10.

$3p = 6 \times 10$

$3p = 60$

Divide both sides by 3.

$p = \dfrac{{60}}{3}$

$p = 30$

Hence, the value of $p$ is 30.


4. Solve the following equation.

a. $10p = 100$

Ans: We will divide both sides by 10, to get the value of $p$.

$p = \dfrac{{100}}{{10}}$

$p = 10$

Thus, the value of $p$ is 10.

b. $10p + 10 = 100$

Ans: To solve this, subtract 10 from both sides and then, divide both sides by 10.

$10p = 100 - 10$

$10p = 90$

$p = \dfrac{{90}}{{10}}$

$p = 9$

Hence, the value of $p$ is 9.

c. $\dfrac{p}{4} = 5$

Ans: To solve this, we will multiply both sides by 4.

$\dfrac{p}{4} \times 4 = 5 \times 4$

$p = 20$

Hence, the value of $p$ is 20.

d. $ - \dfrac{p}{3} = 5$

Ans: First, we will multiply both sides by 3.

$ - \dfrac{p}{3} \times 3 = 5 \times 3$

$ - p = 15$

Now, multiply both sides by $ - 1$.

$p =  - 15$

Hence, the value of $p$ is $ - 15$.

e. $\dfrac{{3p}}{4} = 6$

Ans: First, we will multiply both sides by 4.

$\dfrac{{3p}}{4} \times 4 = 6 \times 4$

$3p = 24$

Now, divide both sides by 3.

$p = \dfrac{{24}}{3}$

$p = 8$

Hence, the value of $p$ is 8.

f. $3s =  - 9$

Ans: Multiply both sides by 3 to get the value of $s$.

$s =  - \dfrac{9}{3}$

$s =  - 3$

Hence, the value of $p$ is $ - 3$.

g. $3s + 12 = 0$

Ans: To solve the given equation, we will subtract 12 from both sides.

$3s + 12 - 12 =  - 12$

$3s =  - 12$

$s = \dfrac{{ - 12}}{3}$

$s =  - 4$

Hence, the value of $s$ is $ - 4$.

h. $3s = 0$

Ans: to solve the given equation, divide both sides by 3.

$3s = 0$

$s = \dfrac{0}{3}$

$s = 0$

Hence, the value of $s$ is 0.

i. $2q = 6$

Ans: To solve the given equation, divide both sides by 2.

$q = \dfrac{6}{2}$

$q = 3$

Hence, the value of $q$ is 3.

j. $2q - 6 = 0$

Ans: To solve the given equation, add 6 to both sides.

$2q - 6 + 6 = 0 + 6$

$2q = 6$

Divide both sides by 2.

$q = \dfrac{6}{2}$

$q = 3$

Hence, the value of $q$ is 3.

k. $2q + 6 = 0$

Ans: To solve the given equation, subtract 6 from both sides.

$2q + 6 - 6 = 0 - 6$

$2q =  - 6$

Divide both sides by 2.

$q =  - \dfrac{6}{2}$

$q =  - 3$

Hence, the value of $q$ is $ - 3$.

l. $2q + 6 = 12$

Ans: To solve the given equation, subtract 6 from both sides.

$2q + 6 - 6 = 12 - 6$

$2q = 6$

Divide both sides by 2.

$q = \dfrac{6}{2}$

$q = 3$

Hence, the value of $q$ is 3.


NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.2

Opting for the NCERT solutions for Ex 4.2 Class 7 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 4.2 Class 7 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 7 students who are thorough with all the concepts from the Maths textbook and quite well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 7 Maths Chapter 4 Exercise 4.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 7 Maths Chapter 4 Exercise 4.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

Do not delay any more. Download the NCERT solutions for Class 7 Maths Chapter 4 Exercise 4.2 from Vedantu website now for better exam preparation. If you have the Vedantu app in your phone, you can download the same through the app as well. The best part of these solutions is these can be accessed both online and offline as well.

FAQs on NCERT Solutions for Class 7 Maths Chapter 4: Simple Equations - Exercise 4.1

1. How many Questions are there in Class 7 Maths, Chapter 4 Simple Equations?

In Class 7 Maths, Chapter 4 Simple Equations, there are a total of 4 questions. Generally speaking, Chapter 7's questions are all based on solving equations. You can consult India's top online resource, Vedantu if you're looking for NCERT solutions for Class 7 Math. At Vedantu, all of the chapter exercises are collected in one location and solved by a qualified teacher in accordance with the recommendations of the NCERT books. The solutions are complete, step-by-step, and 100% accurate.

2. Is Exercise 4.2 from Class 7 Maths Chapter 4 Simple Equations (EX 4.2) long or short?

Exercise 4.2 from Class 7 Maths Chapter 4 Simple Equations (EX 4.2) is brief. There are four questions and one example. If they devote 1-2 hours per day to exercise 4.2 of the 7th Standard Maths, children can finish it in no more than 2 days. The ability, efficiency, and productivity of the kids as well as the number of other factors determine how long it takes.

3. Is it simple to solve and comprehend Exercise 4.2 from Class 7 Maths Chapter 4 Simple Equations?

Exercise 4.2 for Class 7 Math Chapter 4 Simple equations are not simple or difficult to solve or comprehend. Because some of the exercises' issues are easy and some are difficult, it is somewhere in the middle. Every exercise has a different level of difficulty, though, depending on the student. Therefore, whether or not exercise 4.2 in math class 7 is easy or difficult depends on the students. Some students find it challenging, others find it easy, and still, others find it somewhere in between the two.

4. Where can I get NCERT Solutions for NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations (EX 4.2) Exercise 4.2?

Vedantu, India's top online learning portal, offers NCERT solutions for Class 7 Maths Chapter 4. Exercise 4.2 for Simple Equations has been meticulously produced by highly qualified and experienced teachers in strict accordance with the most recent CBSE rules. These solutions include exact and thorough solutions to every sum in the class 7 NCERT Maths textbook. On Vedantu's official website (Vedantu.com), you may quickly and gratis download PDF versions of these study guides. You can also get the Vedantu mobile app.

5. How What do you mean by Simple equation in class 7?

A simple equation is a mathematical formula that, on both sides of the "equal to" sign, expresses the relationship between two expressions. Equations in this category include a variable, which is typically expressed as x or y. Simple equations often need to be rearranged in order to be solved. In these circumstances, you must introduce all the numbers on one side and all the phrases other than "X" on the other. The procedure of isolating X is the name of this technique. The exact opposite of every mathematical action done on one side of the equation is required on the other side. The relationship between the two sides of the equation must be maintained.