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# NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations Exercise 4.3

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Maths Class 7 Chapter 4 Exercise 4.3 Solutions - FREE PDF Download

Class 7 Maths Chapter 4 Exercise 4.3 Solutions focuses on solving simple equations. This exercise will help you understand how to find the value of the unknown variable by performing operations like addition, subtraction, multiplication, and division on both sides of the equation. Students will learn to balance the equation to isolate the variable and find its correct value. Students can access the revised Class 7 Maths NCERT Solutions from our page which is prepared so that you can understand it easily.

Table of Content
1. NCERT Solutions for Maths Class 7 Chapter 4 Exercise 4.3 Solutions - FREE PDF Download
2. Glance on NCERT Solutions Maths Chapter 4 Exercise 4.3 Class 7 | Vedantu
3. Access NCERT Solutions for Maths Class 7 Chapter 4 - Simple Equations
3.1Exercise 4.3
4. Class 7 Maths Chapter 4: Exercises Breakdown
5. CBSE Class 7 Maths Chapter 4 Other Study Materials
6. Chapter-Specific NCERT Solutions for Class 7 Maths
FAQs

These solutions are aligned with the updated Class 7 Maths Syllabus guidelines for Class 7, ensuring students are well-prepared for exams. Class 7 Chapter 4 Maths Exercise 4.3 Questions and Answers PDF provides accurate answers to textbook questions and assists in effective exam preparation and better performance.

## Glance on NCERT Solutions Maths Chapter 4 Exercise 4.3 Class 7 | Vedantu

• Class 7 Maths Chapter 4 Exercise 4.3 Solutions topics such as More Equations and Application Of Simple Equations To Practical Solutions.

• The More Equations section likely deals with constructing new equations based on a given value. Students might be given a number (x = a) and asked to create new equations by adding, subtracting, multiplying, or dividing by other numbers. This helps solidify your grasp of how basic operations affect equations.

• The Application of Simple Equations to Practical Situations section is about using simple equations to solve real-world problems. Students will be presented with word problems that can be translated into mathematical equations. Students will find the answer to the real-life situation by solving these equations.

• NCERT Class 7 Maths Exercise 4.3 Solution contains 4 Questions and Solutions.

## Access NCERT Solutions for Maths Class 7 Chapter 4 - Simple Equations

### Exercise 4.3

1. Set up equations and solve them to find the unknown numbers in the following cases.

a. Add 4 to eight times a number’ you get 60.

Ans: Let us assume that the number is $x$.

According to the question,

$8x + 4 = 60$

To solve this equation, subtract 4 from both sides and divide by 8.

$8x + 4 - 4 = 60 - 4$

$8x = 56$

$x = \dfrac{{56}}{8}$

$x = 7$

Hence, the number is 7.

b. One-fifth of a number minus 4 gives 3.

Ans: Let the number be $y$.

According to the question,

$\dfrac{y}{5} - 4 = 3$

To solve this equation, add 4 to both sides and multiply both sides by 5.

$\dfrac{y}{5} = 3 + 4$

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

$y = 35$

Hence, the number is 35.

c. If I take three fourth of a number and add 3 to it, I get 21.

Ans: Let the number be $x$.

According to the question,

$\dfrac{3}{4}x + 3 = 21$

To solve this equation, subtract 3 from both sides, then multiply both sides by 4.

$\dfrac{3}{4}x = 21 - 3$

$\dfrac{3}{4}x = 18$

$3x = 72$

Divide both sides by 3.

$x = \dfrac{{72}}{3}$

$x = 24$

Hence, the number is 24.

d. When I subtracted 11 from twice a number, the result was 15.

Ans: Let the number be $x$.

According to the question,

$2x - 11 = 15$

Add 11 to both sides and divide both sides by 2.

$2x = 11 + 15$

$2x = 26$

$x = \dfrac{{26}}{2}$

$x = 13$

Hence, the number is 13.

e. Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.

Ans: Let the number of notebooks with Munna be $x$.

According to the question,

$50 - 3m = 8$

Subtract 50 from both sides.

$- 3m = 8 - 50$

$- 3m = - 42$

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

$m = 14$

Hence, the number of notebooks with Munna are 14.

f. Ibenhal thinks of a number. If she adds 19 to it divides the sum by 5, she will get 8.

Ans:  Let the number he thinks of be $x$.

$\dfrac{{x + 19}}{5} = 8$

Multiply both sides by 5.

$x + 19 = 8 \times 5$

$x + 19 = 40$

Subtract 19 from both sides.

$x = 40 - 19$

$x = 21$

Hence, the number is 21.

g. Answer thinks of a number. If he takes 7 away from $\dfrac{5}{2}$ of the number, the result is $23$.

Ans: Let the number he thinks of be $x$.

According to the question,

$\dfrac{5}{2}x - 7 = 23$

Add 7 to both sides.

$\dfrac{5}{2}x = 23 + 7$

$\dfrac{5}{2}x = 30$

Multiply both sides by $2$.

$\dfrac{5}{2}x\times 2 = 30\times 2$

$5x=60$

Hence, the number is 5.

2. Solve the Following.

a. The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87. What is the lowest score?

Ans: Let the lowest score be $y$.

According to the question,

$2y + 7 = 87$

Subtract 7 from both sides.

$2y = 87 - 7$

$2y = 80$

Divide both sides by 2.

$y = \dfrac{{80}}{2}$

$y = 40$

b. In an isosceles triangle, the base angles are equal. The vertex angle is $40^\circ$. What are the base angles of the triangle? (Remember the sum of three angles of a triangle is $180^\circ$)

Ans: Let the base angle of the triangle be $b$, and $c$. Since the triangle is isosceles, then, $b = c$.

The sum of interior angles is $180^\circ$.

$40^\circ + b + b = 180^\circ$

$40^\circ + 2b = 180^\circ$

Subtract $40^\circ$ from both sides.

$2b = 180^\circ - 40^\circ$

$2b = 140^\circ$

Divide both sides by 2.

$b = \dfrac{{140^\circ }}{2}$

$b = 70^\circ$

Hence, the base angle measure is $70^\circ$.

c. Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?

Ans: Let the score of Rahul be $x$ runs and Sachin’s score be $2x$.

According to the question,

$x + 2x = 198$

$3x = 198$

Divide both sides by 3.

$x = \dfrac{{198}}{3}$

$x = 66$

Thus, the runs scored by Rahul is 66.

Since, Sachin’s score is twice of Rahul then, Sachin’s score will be as follows.

$2 \times 66 = 132\,\,\,\,{\text{runs}}$.

Thus, the runs scored by Sachin are 132 runs.

3. Solve the Following:

a. Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37marbles. How many marbles does Parmit have?

Ans: Let the number of marbles Parmit has be $m$.

According to the question,

$5m + 7 = 37$

Subtract 7 from both sides.

$5m = 37 - 7$

$5m = 30$

Divide both sides by 5.

$m = \dfrac{{30}}{5}$

$m = 6$

Hence, Parmit has 6 marbles.

b. Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

Ans: Let the age of Laxmi be $x$ years.

According to the question,

$3y + 4 = 49$

Now, to solve this, subtract 4 from both sides and divide the expression obtained by 3.

$3y = 49 - 4$

$3y = 45$

$y = \dfrac{{45}}{3}$

$y = 15$

Hence, the age of Laxmi is 15 years.

c. People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees was two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?

Ans: Let the number of fruit trees be $t$.

Then, according to the question, $3t + 2 = 77$.

To solve the expression, add the like terms first. Then, subtract 2 from both sides.

$3t+2-2=77-2$

$3t=75$

$t=\dfrac{75}{3}$

$t=25$

Thus, the value of $t$ is 25.

4. Solve the Following Riddle:

I am a number, tell my identity!

Take me seven times over, and add fifty!

To reach a triple century, You still need forty.

Ans: Let the number that is being talked about in the riddle be $n$.

The line, take me seven times over suggests that the number is multiplied by 7. So, the number will become $7n$. Now, fifty is added. So, the number will become, $7n + 50$.

Now, according to the question,

$7n + 50 + 40 = 300$

$7n + 90 = 300$

We will solve this equation by subtracting 90 from both sides and dividing by 7.

$7n = 300 - 90$

$7n = 210$

$n = \dfrac{{210}}{7}$

$n = 30$

Therefore, the required number is 30.

## Conclusion

In conclusion, Exercise 4.3 in Chapter 4 of Class 7 Maths provides a solid foundation for solving simple equations. By completing tNCERT Class 7 Maths Exercise 4.3 Solution, you have learned how to perform various operations to isolate and solve for the unknown variable. This practice is essential for building your problem-solving skills and understanding the balance required in equations. Keep practicing these methods to strengthen your mathematical abilities. The concepts you have learned here are fundamental and will be useful in more advanced topics.

## Class 7 Maths Chapter 4: Exercises Breakdown

 Exercise Number of Questions Exercise 4.1 6 Questions & Solutions Exercise 4.2 4 Questions & Solutions

## CBSE Class 7 Maths Chapter 4 Other Study Materials

 S.No Important Links for Chapter 4 Simple Equations 1 Class 7 Simple Equations Important Questions 2 Class 7 Simple Equations Revision Notes 3 Class 7 Simple Equations NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 7 Maths

Given below are the chapter-wise NCERT Solutions for Class 7 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations Exercise 4.3

1. Where can I find NCERT Solutions for Class 7 Maths Exercise 4.3 Solution?​

The students can find solutions for all the exercises separately for Chapter 4 of Class 7 Maths free of cost on the Vedantu website and on the Vedantu app. These questions are solved by professionals to help the students understand each step properly and learn from there. They also include easy explanations of the theory for simplified comprehension of the topics. You should give the theory a read before getting down to the formulas and their applications to grasp the formulas efficiently.

2. How can I download an offline PDF for NCERT Solution for Chapter 4 of Class 7 Maths?

You can download an offline PDF for NCERT Solutions for all the exercises of Chapter 4 of Class 7 Maths from Vedantu. The solutions provide accurate answers and include all the steps for each question. They are immensely helpful for the students in all-encompassing preparation for their exams. Providing the students with carefully framed answers in a manner expected by the board. They follow the CBSE Board guidelines for the course structure which makes them all the more trustworthy.

3. Are the NCERT Solutions for Exercise 4.3 of Chapter 4 Simple Equations of Class 7 Maths helpful?

The NCERT Solutions for Exercise 4.3 of Chapter 4 Simple Equations of Class 7 Maths is the best aid for the students as they provide all the answers solved by subject experts. The students can avail of the PDF from the Vedantu website for offline use. They can either be used for reference or introspection. The solutions make learning accessible and fun with all the study materials available at hand.

4. How many questions are there in Exercise 4.3 of Chapter 4 of Class 7 Maths?

There are a total of four questions in Exercise 4.3. Each of these questions has several subparts except for question 4 which is a fun riddle for the students to solve. You can get absolute knowledge about the Chapter and learn how to write an equation in the NCERT Solutions. These NCERT Solutions also provide the students with elaborate answers to all the questions in each exercise of the chapter.

5. How can I solve the questions in Exercise 4.3 of Chapter 4 of Class 7 Maths?

The fourth exercise in Chapter 4 of Class 7th Mathematics seems easy but can be troublesome for it has a variety of questions. You should try to solve these questions by yourself first after reading the theory. You can avail free PDF for NCERT Solutions that will assist you in comprehensive learning of each topic covered in the exercise. They provide elaborate solutions to aid in the understanding of the problems given in the textbook in case of doubts.

6. What operations are commonly used to solve equations in NCERT Class 7 Maths Exercise 4.3 Solution?

The commonly used operations to solve equations in this exercise are addition, subtraction, multiplication, and division.

7. Can solving simple equations help in real-life situations?

Yes, as we studied in NCERT Class 7 Maths Exercise 4.3 Solution, solving simple equations can help in real-life situations, such as calculating expenses, determining quantities, and solving problems that require logical reasoning.

8. What types of equations will I solve in NCERT Class 7 Maths Exercise 4.3 Solution

You will solve simple linear equations involving one variable. These equations may require you to add, subtract, multiply, or divide to isolate the variable.

9. What are the basic concepts of simple equations in Class 7 Maths Exercise 4.3?

A simple equation is a set of variables, constants, and mathematical operations like addition, subtraction, multiplication, or division which are balanced by an equal sign. The left side of the equation is called the left-hand side (LHS) and the right side of the equation is called the right-hand side (RHS).

10. What is a real-life example of a simple equation of Class 7 Maths Exercise 4.3?

This implies that you might purchase 5 kg of apples and 40 rupees worth of ice cream. This is an example of a general use for basic equations. We can think of any other scenario in which there is an unknown, and we will undoubtedly find a way to apply basic equations in that specific scenario.