simple equations case study class 7

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations and Class 7 Maths Chapter 4 Try These Solutions in Hindi and English Medium updated for session 2024-25. Class 7 Maths Chapter 4, which deals with simple equations, is a fundamental topic that has real-life applications and serves as a foundation for more complex algebraic concepts. According to new syllabus and revised books for 2024-25, the class 7th mathematics chapter 4 simple equations has only three exercises in course.

7th Maths Chapter 4 Solutions in English Medium

• Class 7 Maths Chapter 4 Try These
• Class 7 Maths Exercise 4.1 in English
• Class 7 Maths Exercise 4.2 in English
• Class 7 Maths Exercise 4.3 in English

7th Maths Chapter 4 Solutions in Hindi Medium

• Class 7 Maths Exercise 4.1 in Hindi
• Class 7 Maths Exercise 4.2 in Hindi
• Class 7 Maths Exercise 4.3 in Hindi
• Class 7 Maths Chapter 4 NCERT Book
• Class 7 Maths Solutions Page
• Class 7 all Subjects Solutions

Simple equations are used to solve a wide range of everyday problems. Whether it’s calculating expenses, determining quantities, or finding missing values, equations help you solve various real-life situations. Working with equations develops your mathematical thinking and problem-solving skills. Class 7 Maths chapter 4 teaches students how to translate word problems into mathematical expressions and equations.

7th Maths Exercise 4.1, Exercise 4.2 and Exercise 4.3 in English Medium updated for new academic session based on new NCERT Books. Download Solutions of Prashnavali 4.1, Prashnavali 4.2 and Prashnavali 4.3 in Hindi Medium free to use in PDF format. Class 7 Maths NCERT (https://ncert.nic.in/) Solutions are updated according to NCERT Books 2024-25. View these solutions in Video Format to study online or in PDF file format for offline use. Simple equations lay the groundwork for more advanced algebraic concepts you’ll encounter in higher grades. They provide a bridge between arithmetic and algebra, helping you transition to more abstract mathematical thinking. Download NCERT Solutions Offline Apps for class 7 all subjects in Hindi and English Medium.

Free App for Class 7 all Subjects

As students work with equations, they will start recognizing patterns and relationships between different variables. This skill is valuable not only in math but also in other subjects and analytical tasks. Equations are used to manage budgets, calculate expenses, and plan savings. They help learners make informed financial decisions by understanding how different variables affect your finances.

Important Questions on Class 7 Maths Chapter 4

Check whether the value given in the brackets is a solution to the given equation or not: n + 5 = 19 [ n = 1]..

n + 5 = 19 Putting n = 1 in L.H.S., 1 + 5 = 6 L.H.S. R.H.S., n = 1 is not the solution of given equation.

Write equations for the following statement: The sum of numbers x and 4 is 9.

Write the following equations in statement form: 4 p – 2 = 18..

If you take away 2 from 4 times p you get 18.

Class 7 Maths all three exercises of Chapter 4 Simple Equations solutions with step by step complete explanation are given below. Equations are used extensively in science and engineering fields to model and analyze various phenomena. From physics to chemistry to engineering, equations play a crucial role in understanding the natural world. No login or password is required to access these solutions for the session 2024-25. Class 7 Maths chapter 4 Simple Equations are used to analyze and interpret data. They help in making predictions, drawing conclusions, and understanding trends in various data sets.

In 7 Maths Chapter 4 Simple Equations, we will study about the formation of linear equations in one variable (A variable takes on different numerical values; its value is not fixed. Variables are denoted usually by letters of the alphabets, such as x, y, z, l, m, n, p, etc.). In an equation there is always an equality sign. In computer programming, equations are used to write algorithms, create simulations, and solve computational problems. The equality sign shows that the value of expression to the left hand side or LHS is equal to the value of the expression to the right hand side or RHS. Remember the following things for an equation: 1. If we subtract the same number from both sides of an equality, it still holds. 2. If we multiply or divide both sides of the equality by the same non-zero number, it still holds. 3. Transposing a number (changing the side of the number) is the same as adding or subtracting the number from both sides. Note: If we fail to do the same mathematical operation on both sides of an equality, the equality does not hold. Learning to solve equations involves abstracting real-life situations into mathematical symbols and relationships, which is a valuable cognitive skill. To solve the practical problems based on equations, first convert the situation into equation and then apply the mathematical operation on it.

Solving equations involves trial and error, deduction, and careful analysis of the problem. This cultivates your critical thinking skills. Hindi Medium NCERT Solutions 2024-25 are now prepared for the new session 2024-25 and available for all the users. The skills and concepts you learn in this chapter set the stage for more advanced algebraic topics you’ll encounter in higher grades. Hindi & English Medium solutions are in Online as well as offline mode. Working with equations enhances your logical thinking by requiring you to follow a step-by-step process to arrive at a solution. These all are as per students suggestions. Download Class 7 Offline App for offline use. In essence, understanding simple equations is about developing problem-solving skills, logical reasoning, and an appreciation for how math is applied in various contexts. While the chapter might deal with “simple” equations, the skills you acquire here are foundational and will continue to be relevant as you progress in your mathematical journey.

« Chapter 3: Data Handling

Chapter 5: lines and angles ».

Copyright 2024 by Tiwari Academy | A step towards Free Education

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. It helps the students to understand slowly and to get practice well to become perfect and again a good score in their examination. Below we have listed NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.1, Ex 4.2, Ex 4.3 and Ex 4.4.

These materials are prepared based on Class 7 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 7 Solutions Maths Chapter 4 Simple Equations are in accordance with the latest CBSE guidelines and marking schemes

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

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

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Class 7 Maths Chapter 4 Simple Equations Textbook Solutions

Chapter 4 – Simple Equations of NCERT Solutions for Class 7 Maths contains 4 exercises. Let’s now look at the important topics covered in this chapter are mentioned below.

• What is Equation
• Solving an Equation
• More Equations
• From Solution to Equation
• Applications of Simple Equations To Practical Situations

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Revision Notes on Simple Equations

Algebraic expressions.

It is an expression involving constant, variable and some operations like addition, multiplication etc.

Variable is an unknown number which could have a different numerical value. It is called Variable as it can vary.

It is represented by different letters like x, y, a, b etc.

An equation is a condition on a variable. It says that two expressions are equal.

Important Points Related to the Equation

One of the expressions must have a variable.

LHS of the equation is equal to the RHS of the equation.

An expression does not have equality sign but an equation always has an equality sign.

If we interchange the position of the expression from LHS to RHS or vice versa, the equation remains the same.

Both the above equations are same.

How to form equations using statements?

1. The sum of four times of x and 12 is equal to 35.

4x + 12 = 35

2. Half of a number is 3 more than 8.

Balanced Equation

When the LHS = RHS of an equation, then it is said to be a balanced equation.

1. If we add the same number to both the sides .

We can add the same number on both the sides of a balanced equation, the equation will remain the same.

12 – 8 = 3 + 1

If we add 3 to both the sides,

12 – 8 + 3 = 7

3 + 1 + 3 = 7

2. If we subtract the same number from both sides .

We can subtract the same number from both the sides of a balanced equation, the equation will remain the same.

If we subtract 3 from both the sides,

12 – 8 - 3 = 1

3 + 1 -3 = 1

3. If we multiply the same number to both the sides .

We can multiply the same number on both the sides of a balanced equation, the equation will remain the same.

If we multiply 3 to both the sides,

(12 – 8) × 3 = 36 – 24 = 12

(3 + 1) × 3 = 9 + 3 = 12

4. If we divide the same number from both sides .

We can divide the same number from both the sides of a balanced equation, the equation will remain the same.

If we divide both the sides by 2,

(12 – 8) ÷ 2 = 4 ÷ 2 = 2

(3 + 1) ÷ 2 = 4 ÷ 2 = 2

The Solution of an Equation

Any value of the variable which satisfies the equation is the solution of the equation.

There are two methods to solve an equation

1. By adding or subtracting the same number to both the sides of the equation as we have above seen that the equation will remain the same .

x + 11 = 35

Subtract 11 from both the sides.

x + 11 – 11 = 35 – 11

Here, x = 24 is the solution of the given equation.

Divide both the sides by 25.

1. Transposing Method

In this method, we transpose the numbers from one side of the equation to the other side so that all the terms with variable come on one side and all the constants come on another side.

While transposing the numbers the sign of the terms will get changed. i.e. Negative will become positive and positive will become negative.

Now we will transfer 11 from LHS to RHS and its sign will get reversed.

x = 35 – 11

From a Solution to the Equation

As we solve the equation to get the solution, we can get the equation also if we have the solution.

Any equation has only one solution but if we make an equation from a solution then there could be many equations.

Given x = 7

3x = 21 (multiply both sides by 3)

3x + 8 = 29 (add 8 to both the sides)

This is not the only possible equation. There could be other equations also.

Applications of Simple Equations to Practical Situations

If we have statements related to a practical situation then first we have to convert it in the form of the equation then solve it to find the solution.

Radha’s Mother’s age is 5 years more than three times Shikha’s age. Find Shikha’s age, if her mother is 44 years old.

Let Shikha’s age = y years

Her mother’s age is 3y + 5 which is 44.

 Hence, the equation for Shikha’s age is 3y + 5 =44

3y + 5  = 44

3y = 44 – 5 (by transposing 5)

y = 13 (by dividing both sides by 3)

 Hence, Shikha’s age = 13 years

A number consists of two digits. The digit in the tens place is twice the digit in the units place. If 18 be subtracted from the number, the digits are reversed. Find the number.

Let the digit at units place = y

So, the digit in the tens place = 2y

So, the number is (2y) y.

As it is given that if 18 is subtracted from the number, the digits are reversed. 

So, we have

(2y) y - 18 = y(2y)

10 × (2y) + 1 × y - 18 = 10 × y + 1 × (2y)

20y + y - 18 = 10y + 2y

21y - 18 = 12y

9y = 18 (By dividing both sides by 9)

y = 2

The digit at the units place is y = 2.

And the digit at the tens place is 2y

= 2 × 2

Hence the required number is 42.

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• NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations
• NCERT Solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - PDF Download

Important topics under ncert solutions for class 7 maths chapter 4 simple equations.

Chapter 4 of the class 7 maths syllabus is on ‘Simple Equations’. This chapter on Simple Equations is one of the key components of the class 7 maths syllabus since it introduces and nourishes the concept of algebraic equations. This very crucial chapter in mathematics covered in class 7 is divided into 5 major sections or topics. The following is a list of the important topics covered under NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations. We advise students to sincerely go through each of these topics to get a clear understanding of the concepts that are related to this chapter. This will surely help students reap the full benefits of the solutions provided by Vedantu for this chapter on Simple Equations. 

An Equation

Solving an equation, forming an equation, application of simple equations to practical situations.

NCERT Solutions for Class 7 Maths Chapter 4 provides you the study material that covers the important facts and basic concepts of equations. Students can avail of this material on Vedantu’s platform easily and can revise and practice more sums in this chapter.

Our NCERT Solutions for Class 7 Maths Chapter 4 provides relevant information and problems with accurate solutions. If students have any queries, they can get in touch with the instructors through live chat on Vedantu’s official website. Subjects like Science, Maths, English, Social Science, Hindi will become easy to study if you have access to NCERT Solution for Class 7 Science , Maths solutions and solutions of other subjects.

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

1. Complete the last column of the table:

2. Check whether the value given in the brackets is a solution to the given equation or not:

(a) $ {n + 5 = 19(n = 1)}$ 

Ans: Putting $ {\text{n}} = 1$ in the above equation, we get:

$  1 + 5 = 19 $ 

$ \Rightarrow 6 = 19 $  

$ \therefore 6 \ne 19$ 

$ \because {\text{n}} = 1$  is not a solution to the given equation. (b) $ {7n + 5 = 19(n =  - 2)}$ 

Ans: Putting $ {\text{n}} =  - 2$  in the above equation, we get:

$  7( - 2) + 5 = 19 $ 

$ \Rightarrow  - 14 + 5 = 19 $  

$ \Rightarrow  - 9 = 19 $ 

$ \therefore  - 9 \ne 19 $ 

$ \because {\text{n}} =  - 2$  is not a solution to the given equation. (c) $ {7n + 5 = 19(n = 2)}$ 

Ans: Putting $ {\text{n}} = 2$  in the above equation, we get:

$ 7(2) + 5 = 19 $ 

$  \Rightarrow 14 + 5 = 19 $

$  \Rightarrow 19 = 19 $

$ \because $ The above relation holds to be true, $ {\text{n}} = 2$  is a solution to the given equation. (d) $ {4p - 3 = 13(p = 1)}$ 

Ans: Putting $ {\text{p}} = 1$ in the above equation, we get:

$  4(1) - 3 = 13 $ 

$    \Rightarrow 4 - 3 = 13 $ 

$  \Rightarrow 1 = 13 $

$   \because 1 \ne 13 $

$ \therefore {\text{p}} = 1$  is not a solution to the given equation. (e) $ {4p - 3 = 13(p =  - 4)}$ 

Ans: Putting $ {\text{p}} =  - 4$  in the above equation, we get:

$   4( - 4) - 3 = 13 $ 

$    \Rightarrow  - 16 - 3 = 13 $ 

$    \Rightarrow  - 19 = 13 $

$   \because  - 19 \ne 13 $

$ \therefore {\text{p}} =  - 4$ is not a solution to the given equation. (f) $ {4p - 3 = 13(p = 0)}$

Ans: Putting $ {\text{p}} = 0$  in the above equation, we get:

$  4(0) - 3 = 13 $ 

$  \Rightarrow 0 - 3 = 13 $ 

$  \Rightarrow  - 3 = 13 $ 

$ \because  - 3 \ne 13 $  

$ \therefore {\text{p}} = 0$ is not a solution to the given equation. 3. Solve the following equations by trial and error method:

(i) $ {5p + 2 = 17}$ 

Ans: Putting $ {\text{p}} =  - 3$  in L.H.S. $ 5( - 3) + 2 =  - 15 + 2 =  - 13$ 

$ \because  - 13 \ne 17$ $ \therefore {\text{p}} =  - 3$  is not the solution.

Putting $ {\text{p}} =  - 2$  in L.H.S. $ 5( - 2) + 2 =  - 10 + 2 =  - 8$ 

$ \because  - 8 \ne 17$  $ \therefore {\text{p}} =  - 2$  is not the solution.

Putting $ {\text{p}} =  - 1$  in L.H.S. $ 5( - 1) + 2 =  - 5 + 2 =  - 3$ 

$ \because  - 3 \ne 17$ $ \therefore {\text{p}} =  - 1$  is not the solution.

Putting $ {\text{p}} = 0$  in L.H.S. $ 5(0) + 2 = 0 + 2 = 2$ 

$ \because 2 \ne 17$ $ \therefore {\text{p}} = 0$  is not the solution. Putting $ {\text{p}} = 1$  in L.H.S. $ 5(1) + 2 = 5 + 2 = 7$ 

$ \because 7 \ne 17$  $ \therefore {\text{p}} = 1$  is not the solution.

Putting $ {\text{p}} = 2$  in L.H.S. $ 5(2) + 2 = 10 + 2 = 12$ 

$ \because 12 \ne 17$ $ \therefore {\text{p}} = 2$  is not the solution. Putting $ {\text{p}} = 3$  in L.H.S. $ 5(3) + 2 = 15 + 2 = 17$ 

$ \because 17 = 17$ $ \therefore {\text{p}} = 3$  is the solution.

(ii) $ {3m - 14 = 4}$

Ans: Putting $ {\text{m}} =  - 2$  in L.H.S. $ 3( - 2) - 14 =  - 6 - 14 =  - 20$ 

$ \because  - 20 \ne 4$  $ \therefore {\text{m}} =  - 2$  is not the solution.

Putting $ {\text{m}} =  - 1$  in L.H.S. $ 3( - 1) - 14 =  - 3 - 14 =  - 17$ 

$ \because  - 17 \ne 4$  $ \therefore {\text{m}} =  - 1$  is not the solution.

Putting $ {\text{m}} = 0$  in L.H.S. $ 3(0) - 14 = 0 - 14 =  - 14$ 

$ \because  - 14 \ne 4$ $ \therefore {\text{m}} = 0$  is not the solution.

Putting $ {\text{m}} = 1$  in L.H.S. $ 3(1) - 14 = 3 - 14 =  - 11$ 

$ \because  - 11 \ne 4$  $ \therefore {\text{m}} = 1$  is not the solution.

Putting $ {\text{m}} = 2$  in L.H.S. $ 3(2) - 14 = 6 - 14 =  - 8$ 

$ \because  - 8 \ne 4$ ∴m=2 is not the solution.

Putting $ {\text{m}} = 3$  in L.H.S. $ 3(3) - 14 = 9 - 14 =  - 5$ 

$ \because  - 5 \ne 4$  $ \therefore {\text{m}} = 3$ is not the solution.

Putting $ {\text{m}} = 4$  in L.H.S. $ 3(4) - 14 = 12 - 14 =  - 2$ 

$ \because  - 2 \ne 4$ $ \therefore {\text{m}} = 4$  is not the solution.

Putting $ {\text{m}} = 5$  in L.H.S. $ 3(5) - 14 = 15 - 14 =  - 1$ 

$ \because 1 \ne 4$ $ \therefore {\text{m}} = 5$  is not the solution.

Putting $ {\text{m}} = 6$ in L.H.S. $ 3(6) - 14 = 18 - 14 = 4$ 

$ \because 4 = 4$ $ \therefore {\text{m}} = 6$ is the solution.

4. Write the equations for the following statements:

(i) The sum of numbers x and 4 is 9.

Ans: $ {\text{x}} + 4 = 9$

(ii) 2 subtracted from y is 8.

Ans: $ {\text{y}} - 2 = 8$

(iii) Ten times a is 70.

Ans: $ 10{\text{a}} = 70$

(iv) The number b divided by 5 gives 6.

Ans: $ \dfrac{{\text{b}}}{5} = 6$ (v) Three-fourth of t is 15.

Ans: $ \dfrac{{\text{3}}}{4}{\text{t}} = 15$ (vi) Seven times m plus 7 gets you 77.

Ans: $ 7{\text{m}} + 7 = 77$ (vii) One-fourth of a number x minus 4 gives 4.

Ans: $ \dfrac{{\text{1}}}{4}{\text{x}} - 4 = 4$

(viii) If you take away 6 from 6 times y , you get 60.

Ans: $ 6{\text{y}} - 6 = 60$ (ix) If you add 3 to one-third of z , you get 30.

Ans: $ \dfrac{1}{3}{\text{z}} + 3 = 30$

5. Write the following equations in statement form:

(i) $ {p + 4 = 15}$ 

Ans: Sum of numbers p and 4 gives you 15.

(ii) $ {m - 7 = 3}$ 

Ans: 7 subtracted from m gives number 3.

(iii) $ {2m = 7}$ 

Ans: Two times m equals to 7.

(iv) $ \dfrac{{m}}{{5}}{ = 3}$ 

Ans: One fifth of number m equals 5.

(v) $ \dfrac{{{3m}}}{{5}}{ = 6}$ 

Ans: Three fifth of m gives 6.

(vi) $ {3p  +  4  =  25}$ 

Ans: 4 added to three times p gives 25.

(vii) $ {4p - 2  = 18}$ 

Ans: 2 subtracted from four times p equals to 18.

(viii) $ \dfrac{{p}}{{2}}{ + 2 = 8}$ 

Ans: 2 added to half of p gives 8.

6. Set up an equation in the following cases:

i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Tale m to be the number of Parmit’s marbles.)

Ans: Let m be the number of Parmit’s marbles:

$ \therefore 5{\text{m}} + 7 = 37$

ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

Ans: Let age of Laxmi be y:

$ \therefore 3{\text{y}} + 4 = 49$

iii) 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. (Take the lowest score to be l .)

Ans: Let the lowest score in class be l:

$ \therefore 2{\text{l}} + 7 = 87$

iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180°.)

Ans: Let the base angle of isosceles triangle be b:

⇒ the vertex angle is 2b

$ \therefore {\text{b}} + {\text{b}} + 2{\text{b}} = 180^\circ $ 

$  \Rightarrow 4{\text{b}} = 180^\circ $

Class –VII Mathematics (Ex. 4.2)

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

a) $ {x - 1 = 0}$ 

Ans:  

$   {\text{x}} - 1 + 1 = 0 + 1 $ 

$   \Rightarrow {\text{x}} = 1  $  

(Adding 1 both sides) b) $ {x + 1 = 0}$ 

$  {\text{x}} + 1 - 1 = 0 - 1 $ 

$  \Rightarrow {\text{x}} =  - 1 $  

(Subtracting 1 from both sides)

c) $ {x - 1 = 5}$ 

Ans: $   {\text{x}} + 1 + 1 = 5 + 1 $ 

$   \Rightarrow {\text{x}} = 6  $  

(Adding 1 both sides)

d) $ {x + 6 = 2}$ 

$   {\text{x}} + 6 - 6 = 2 - 6 $ 

$    \Rightarrow {\text{x}} =  - 4 $  

(Subtracting 6 from both sides)

e) $ {y - 4 =  - 7}$ 

$   {\text{y}} - 4 + 4 =  - 7 + 4 $ 

$    \Rightarrow {\text{y}} =  - 3 $  

(Adding 4 both sides)

f) $ {y - 4 = 4}$ 

$   {\text{y}} - 4 + 4 = 4 + 4 $

$    \Rightarrow {\text{y}} = 8 $  

g) $ \operatorname{y}  + 4 = 4$ 

$   {\text{y}} + 4 - 4 = 4 - 4 $ 

$    \Rightarrow {\text{y}} = 0 $  

(Subtracting 4 from both sides)

h) $ \operatorname{y}  + 4 =  - 4$ 

$   {\text{y}} + 4 - 4 =  - 4 - 4 $ 

$    \Rightarrow {\text{y}} =  - 8 $  

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

a) $ 3l = 42$ 

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

$    \Rightarrow {\text{l}} = 14 $  

(Dividing both sides by 3)

b) $ \dfrac{b}{2} = 6$ 

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

$ \Rightarrow {\text{b}} = 12 $  

(Multiplying both sides by 2)

c) $ \dfrac{p}{7} = 4$ 

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

$  \Rightarrow {\text{p}} = 28 $  

(Multiplying both sides by 7)

d) $ 4{\text{x}} = 25$ 

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

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

(Dividing both sides by 4)

e) $ 8{\text{y}} = 36$ 

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

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

(Dividing both sides by 8)

f) $ \dfrac{{\text{z}}}{3} = \dfrac{5}{4}$ 

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

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

(Multiplying both sides by 3)

g) $ \dfrac{a}{5} = \dfrac{7}{{15}}$ 

$  \dfrac{{\text{a}}}{5} \times 5 = \dfrac{7}{{15}} \times 5 $ 

$    \Rightarrow {\text{a}} = \dfrac{7}{3} $  

(Multiplying both sides by 5)

h) $ 20t =  - 10$ 

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

$ \Rightarrow {\text{t}} = \dfrac{{ - 1}}{2} $  

(Dividing both sides by 20)

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

(a) $ {3n - 2 = 46}$ 

Ans: $ 3{\text{n}} - 2 + 2 = 46 + 2$ (Adding 2 both sides)

$  \Rightarrow 3{\text{n}} = 48$

$  \Rightarrow \dfrac{{3{\text{n}}}}{3} = \dfrac{{48}}{3}$ (Dividing both sides by 3)

$  \Rightarrow {\text{n}} = 16$

(b) $ {5m + 7 = 17}$ 

Ans: $ 5{\text{m}} + 7 - 7 = 17 - 7$ (Subtracting 7 both sides)

$  \Rightarrow 5{\text{m}} = 10$ 

$  \Rightarrow \dfrac{{5{\text{m}}}}{5} = \dfrac{{10}}{5}$ (Dividing 5 both sides)

$  \Rightarrow {\text{m}} = 2$

(c) $ \dfrac{{{20p}}}{{3}}{ = 40}$ 

Ans: $ \dfrac{{20{\text{p}}}}{3} \times 3 = 40 \times 3$ (Multiplying both sides by 3)

$  \Rightarrow 20{\text{p}} = 120$ 

$  \Rightarrow \dfrac{{20p}}{{20}} = \dfrac{{120}}{{20}}$ (Dividing both sides by 20)

$  \Rightarrow {\text{p}} = 6$

(d) $ \dfrac{{{3p}}}{{{10}}}{ = 6}$ 

Ans: $ \dfrac{{3{\text{p}}}}{{10}} \times 10 = 6 \times 10$ (Multiplying both sides by 10)

$  \Rightarrow 3{\text{p}} = 60$ 

$  \Rightarrow \dfrac{{3{\text{p}}}}{3} = \dfrac{{60}}{3}$ (Dividing both sides by 3)

$  \Rightarrow {\text{p}} = 20$

4. Solve the following equation:

(a) $ {10p = 100}$ 

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

$  \Rightarrow {\text{p}} = 10$ (b) $ {10p + 10 = 100}$ 

Ans: $ 10{\text{p}} + 10 - 10 = 100 - 10$ $  \Rightarrow 10{\text{p}} = 90 $ 

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

$  \Rightarrow {\text{p}} = 9 $

(c) $ \dfrac{{p}}{{4}}{ = 5}$ 

Ans: $ \dfrac{{\text{p}}}{4} \times 4 = 5 \times 4$ 

(d) $ \dfrac{{{ - p}}}{{3}}{ = 5}$ 

Ans: $ \dfrac{{ - {\text{p}}}}{3} \times 3 = 5 \times 3$

$  \Rightarrow  - {\text{p}} = 15 $

$  \Rightarrow {\text{p}} =  - 15 $

(e) $ \dfrac{{{3p}}}{{4}}{ = 6}$ 

Ans: $ \dfrac{{{\text{3p}}}}{4} \times 4 = 6 \times 4$ 

$  \Rightarrow 3{\text{p}} = 24$ 

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

$  \Rightarrow {\text{p}} = 8$

(f) $ {3s =  - 9}$ 

Ans: $ \dfrac{{3{\text{s}}}}{3} = \dfrac{{ - 9}}{3}$ 

$  \Rightarrow {\text{s}} =  - 3$

(g) $ {3s + 12 = 0}$ 

Ans: $ 3{\text{s}} + 12 - 12 = 0 - 12$ 

$  \Rightarrow 3{\text{s}} =  - 12$ 

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

$  \Rightarrow {\text{s}} =  - 4$

(h) $ {3s = 0}$ 

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

$  \Rightarrow {\text{s}} = 0$

(i) $ {2q = 6}$ 

Ans: $ \dfrac{{2{\text{q}}}}{2} = \dfrac{6}{2}$ 

$  \Rightarrow {\text{q}} = 3$

(j) $ {\mathbf{2q - 6 = 0}}$ 

Ans: $ 2{\text{q}} - 6 + 6 = 0 + 6$ 

$  \Rightarrow 2{\text{q}} = 6$ 

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

(k) $ {\mathbf{2q + 6 = 0}}$ 

Ans: $ 2{\text{q}} + 6 - 6 = 0 - 6$ 

$  \Rightarrow 2{\text{q}} =  - 6$ 

$  \Rightarrow \dfrac{{2{\text{q}}}}{2} = \dfrac{{ - 6}}{2}$ 

$  \Rightarrow {\text{q}} =  - 3$

(l) $ {\mathbf{2q + 6 = 12}}$ 

Ans: $ 2{\text{q}} + 6 - 6 = 12 - 6$ 

Class –VII Mathematics (Ex. 4.3)

1. Solve the following equations:

(a) $ {\mathbf{2y + }}\dfrac{{\mathbf{5}}}{{\mathbf{2}}}{\mathbf{ = }}\dfrac{{{\mathbf{37}}}}{{\mathbf{2}}}$ 

Ans: $ 2{\text{y}} + \dfrac{5}{2} = \dfrac{{37}}{2}$ 

$  \Rightarrow 2{\text{y}} = \dfrac{{37}}{2} - \dfrac{5}{2}$ 

$  \Rightarrow 2y = \dfrac{{37 - 5}}{2}$ 

$  \Rightarrow 2y = \dfrac{{32}}{2}$ 

$  \Rightarrow 2y = 16$ 

$  \Rightarrow y = \dfrac{{16}}{2}$ 

$  \Rightarrow y = 8$

(b) $ {\mathbf{5t + 28 = 10}}$ 

Ans: $  \Rightarrow 5{\text{t}} = 10 - 28$ 

$  \Rightarrow 5{\text{t}} =  - 18$ 

$  \Rightarrow {\text{t}} = \dfrac{{ - 18}}{5}$

(c) $ \dfrac{{\mathbf{a}}}{{\mathbf{5}}}{\mathbf{ + 3 = 2}}$ 

Ans: $ \dfrac{{\text{a}}}{5} + 3 = 2$ 

$  \Rightarrow \dfrac{{\text{a}}}{5} = 2 - 3$ 

$  \Rightarrow \dfrac{{\text{a}}}{5} =  - 1$ 

$  \Rightarrow {\text{a}} =  - 1 \times 5$ 

$  \Rightarrow {\text{a}} =  - 5$

(d) $ \dfrac{{\mathbf{q}}}{{\mathbf{4}}}{\mathbf{ + 7 = 5}}$ 

Ans: $  \Rightarrow \dfrac{{\text{q}}}{4} = 5 - 7$ 

$  \Rightarrow \dfrac{{\text{q}}}{4} =  - 2$ 

$  \Rightarrow {\text{q}} =  - 2 \times 4$ 

$  \Rightarrow {\text{q}} =  - 8$

(e) $ \dfrac{{\mathbf{5}}}{{\mathbf{2}}}{\mathbf{x = 10}}$ 

Ans: $ \dfrac{5}{2}{\text{x}} = 10$ 

$  \Rightarrow 5{\text{x}} = 10 \times 2$ 

$  \Rightarrow 5{\text{x}} = 20$ 

$  \Rightarrow {\text{x}} = \dfrac{{20}}{5}$ 

$  \Rightarrow {\text{x}} = 4$

(f) $ \dfrac{{\mathbf{5}}}{{\mathbf{2}}}{\mathbf{x = }}\dfrac{{{\mathbf{25}}}}{{\mathbf{4}}}$ 

Ans: $ \dfrac{5}{2}{\text{x}} = \dfrac{{25}}{4}$ 

$  \Rightarrow 5{\text{x}} = \dfrac{{25}}{4} \times 2$ 

$  \Rightarrow 5{\text{x}} = 25 \times 2$ 

$  \Rightarrow 5{\text{x}} = 50$

(g) $ {\mathbf{7m + }}\dfrac{{{\mathbf{19}}}}{{\mathbf{2}}}{\mathbf{ = 13}}$ 

Ans: $ 7{\text{m}} + \dfrac{{19}}{2} = 13$ 

$  \Rightarrow 7{\text{m}} = 13 - \dfrac{{19}}{2}$ 

$  \Rightarrow 7{\text{m}} = \dfrac{{26 - 19}}{2}$ 

$  \Rightarrow 7{\text{m}} = \dfrac{7}{2}$ 

$  \Rightarrow {\text{m}} = \dfrac{7}{2} \times \dfrac{1}{7}$ 

$  \Rightarrow {\text{m}} = \dfrac{1}{2}$

(h) $ {\mathbf{6z + 10 =  - 2}}$ 

Ans: $ 6z + 10 =  - 2$ 

$  \Rightarrow 6z =  - 2 - 10$ 

$  \Rightarrow 6z =  - 12$ 

$  \Rightarrow z = \dfrac{{ - 12}}{6}$ 

$  \Rightarrow z =  - 2$

(i) $\dfrac{{3l}}{2} = \dfrac{2}{3}$

Ans: $\dfrac{{3l}}{2} = \dfrac{2}{3}$

$  \Rightarrow 3l = \dfrac{2}{3} \times 2 $

$ \Rightarrow 3l = \dfrac{4}{3} $ 

$ \Rightarrow l = \dfrac{4}{3} \times \dfrac{1}{3} $ 

$ \Rightarrow l = \dfrac{4}{9} $

(j) $\dfrac{{{2b}}}{{3}} - {5 = 3}$

Ans: $\dfrac{{2b}}{3} - 5 = 3$

$ \Rightarrow \dfrac{{2b}}{3} = 3 + 5 $ 

$\Rightarrow \dfrac{{2b}}{3} = 8 $ 

$  \Rightarrow 2b = 8 \times 3 $ 

$   \Rightarrow 2b = 24 $ 

$  \Rightarrow b = 12 $

2. Solve the following equations:

(a) ${2(x + 4) = 12}$

Ans: $2(x + 4) = 12$

$ \Rightarrow x + 4 = \dfrac{{12}}{2} $ 

$ \Rightarrow x + 4 = 6 $ 

$  \Rightarrow x = 6 - 4 $ 

$ \Rightarrow x = 2 $

(b) ${3(n} - {5) = 21}$

Ans: $3(n - 5) = 21$

$  \Rightarrow n - 5 = \dfrac{{21}}{3} $ 

$   \Rightarrow n - 5 = 7 $ 

$ \Rightarrow n = 7 + 5 $ 

$ \Rightarrow n = 12 $

(c) ${3(n} - {5) = } - {21}$

Ans: $ 3(n - 5) =  - 21$

$    \Rightarrow n - 5 = \dfrac{{ - 21}}{3} $

$\Rightarrow n - 5 =  - 7 $ 

$  \Rightarrow n =  - 7 + 5 $ 

$   \Rightarrow n =  - 2 $

(d) ${3 - 2(2 - y) = 7}$

Ans: $3 - 2(2 - y) = 7$

$  \Rightarrow  - 2(2 - y) = 7 - 3 $

$ \Rightarrow (2 - y) = \dfrac{4}{{ - 2}} $ 

$  \Rightarrow 2 - y =  - 2 $

$ \Rightarrow  - y =  - 2 - 2 $ 

$ \Rightarrow  - y =  - 4 $

$  \Rightarrow y = 4 $

(e) ${ - 4(2 - x) = 9}$

Ans: $ - 4(2 - x) = 9$

$  \Rightarrow 2 - x = \dfrac{9}{{ - 4}} $ 

$ \Rightarrow  - x = \dfrac{{ - 9}}{4} - 2 $ 

$  \Rightarrow  - x = \dfrac{{ - 9 - 8}}{4} $ 

$  \Rightarrow  - x = \dfrac{{ - 17}}{4} $ 

$  \Rightarrow x = \dfrac{{17}}{4} $

(f) ${4(2 - x) = 9}$

Ans: $4(2 - x) = 9$

$   \Rightarrow 2 - x = \dfrac{9}{4} $ 

$  \Rightarrow  - x = \dfrac{9}{4} - 2 $ 

$ \Rightarrow  - x = 2\dfrac{1}{4} - 2 $ 

$ \Rightarrow  - x = \dfrac{1}{4} $ 

$   \Rightarrow x =  - \dfrac{1}{4} $

(g) ${4 + 5(p - 1) = 34}$

Ans: $4 + 5(p - 1) = 34$

$   \Rightarrow 5(p - 1) = 34 - 4 $ 

$   \Rightarrow 5(p - 1) = 30 $ 

$  \Rightarrow p - 1 = \dfrac{{30}}{5} $ 

$  \Rightarrow p - 1 = 6 $ 

$  \Rightarrow p = 6 + 1 $ 

$ \Rightarrow p = 7 $

(h) ${34 - 5(p - 1) = 4}$$$

Ans: $34 - 5(p - 1) = 4$

$    \Rightarrow  - 5(p - 1) = 4 - 34 $ 

$ \Rightarrow  - 5(p - 1) =  - 30 $ 

$ \Rightarrow p - 1 = \dfrac{{ - 30}}{{ - 5}} $ 

$ \Rightarrow p - 1 = 6 $ 

$ \Rightarrow p = 6 + 1 $ 

3. Solve the following equations:

(a) ${4 = 5(p - 2)}$

Ans: $4 = 5(p - 2)$

$  \Rightarrow \dfrac{4}{5} = p - 2 $ 

$  \Rightarrow \dfrac{4}{5} + 2 = p $ 

$  \Rightarrow \dfrac{{4 + 10}}{5} = p $ 

$ \Rightarrow \dfrac{{14}}{5} = p $

(b) ${ - 4 = 5(p - 2)}$

Ans: $ - 4 = 5(p - 2)$

$   \Rightarrow \dfrac{{ - 4}}{5} = (p - 2) $ 

$ \Rightarrow \dfrac{{ - 4}}{5} + 2 = p $ 

$\Rightarrow \dfrac{{ - 4 + 10}}{5} = p $ 

$  \Rightarrow \dfrac{6}{5} = p $

(c) ${ - 16 =  - 5(2 - p)}$

Ans: $ - 16 =  - 5(2 - p)$  

$ \Rightarrow \dfrac{{ - 16}}{{ - 5}} = (2 - p) $ 

$ \Rightarrow \dfrac{{16}}{5} - 2 = p $ 

$  \Rightarrow \dfrac{{16 - 10}}{5} = p $ 

(d) ${10 = 4 + 3(t + 2)}$

Ans: $10 = 4 + 3(t + 2)$

$   \Rightarrow 10 - 4 = 3(t + 2) $ 

$ \Rightarrow \dfrac{6}{3} = t + 2 $ 

$   \Rightarrow 2 = t + 2 $ 

$  \Rightarrow 2 - 2 = t $ 

$ \Rightarrow t = 0 $

(e) ${28 = 4 + 3(t + 5)}$

Ans: $28 = 4 + 3(t + 5)$

$  \Rightarrow 28 - 4 = 3(t + 5) $ 

$ \Rightarrow \dfrac{{24}}{3} = t + 5 $ 

$ \Rightarrow 8 - 5 = t $ 

$  \Rightarrow 3 = t $

(f) ${0 = 16 + 4(m - 6)}$

Ans: $0 = 16 + 4(m - 6)$

$   \Rightarrow 0 - 16 = 4(m - 6) $ 

$  \Rightarrow \dfrac{{ - 16}}{4} = m - 6 $ 

$\Rightarrow  - 4 + 6 = m $ 

$ \Rightarrow 2 = m $

4. (a) Construct 3 equations starting with ${x = 2}$

(i) $x = 2$

Adding 5 to both sides, we get:

$  x + 5 = 2 + 5 $ 

$ \Rightarrow x + 5 = 7 $

(ii) $x = 2$

Multiplying both sides by 11, we get:

(iii) $x = 2$

$11x + 5 = 27$

(b) Construct 3 equations starting with ${x =  - 2}$

(i) $x =  - 2$

Adding 3 to both sides, we get:

$x + 3 = 1$

Dividing both sides by 5, we get

$\dfrac{{x + 3}}{5} = \dfrac{1}{5}$

(ii) $x =  - 2$

Adding 2 to both sides, we get:

$x + 2 = 0$

Multiplying both side by 100, we get:

$  100(x + 2) = 100 \times 0 $ 

$ \Rightarrow 100x + 200 = 0 $

(iii) $x =  - 2$

Subtracting 4 from both sides, we get:

$  x - 4 =  - 2 - 4 $

$  \Rightarrow x - 4 =  - 6 $

Class-VII Mathematics (Ex 4.4)

1. Setup 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 the number be x, we get the equation:

$  8x + 4 = 60 $ 

$ \Rightarrow 8x = 60 - 4 $ 

$ \Rightarrow 8x = 56 $ 

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

$   \Rightarrow x = 7 $

(b) One fifth of a number minus 4 gives 3.

$  \dfrac{x}{5} - 4 = 3 $ 

$\Rightarrow \dfrac{x}{5} = 3 + 4 $ 

$\Rightarrow \dfrac{x}{5} = 7 $

$\Rightarrow x = 7 \times 5 $ 

$\Rightarrow x = 35 $

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

$  \dfrac{3}{4}x + 3 = 21 $ 

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

$ \Rightarrow \dfrac{3}{4}x = 18 $ 

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

$  \Rightarrow x = 24 $

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

$  2x - 11 = 15 $ 

$  \Rightarrow 2x = 15 + 11 $ 

$\Rightarrow 2x = 26 $ 

$ \Rightarrow x = \dfrac{{26}}{2} $ 

$ \Rightarrow x = 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 be x, we get the equation:

$  50 - 3x = 8 $ 

$  \Rightarrow  - 3x = 8 - 50 $ 

$  \Rightarrow  - 3x =  - 42 $ 

$   \Rightarrow x = \dfrac{{ - 42}}{{ - 3}} $ 

$  \Rightarrow x = 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 be x, we get the following equation:

$  \dfrac{{x + 19}}{5} = 8 $ 

$  \Rightarrow x + 19 = 8 \times 5 $ 

$  \Rightarrow x + 19 = 40 $ 

$   \Rightarrow x = 40 - 19 $ 

$   \Rightarrow x = 21 $

(g) Answer thinks of a number. If he takes away ${7}$ from $\dfrac{{5}}{{2}}$of the number, the result is $\dfrac{{{11}}}{{2}}$.

$  \dfrac{{5x}}{2} - 7 = \dfrac{{11}}{2} $ 

$\Rightarrow \dfrac{{5x}}{2} = \dfrac{{11}}{2} + 7 $ 

$  \Rightarrow \dfrac{{5x}}{2} = \dfrac{{11 + 14}}{2} $ 

$ \Rightarrow \dfrac{{5x}}{2} = \dfrac{{25}}{2} $ 

$  \Rightarrow 5x = 25 $ 

$ \Rightarrow x = \dfrac{{25}}{5} $ 

$ \Rightarrow x = 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

Ans: Let the lowest score in class be x:

$ \therefore 2{\text{x}} + 7 = 87$

$   \Rightarrow 2x = 87 - 7 $

$ \Rightarrow 2x = 80 $ 

$\Rightarrow x = \dfrac{{80}}{2} $

$ \Rightarrow x = 40 $

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

Ans: Let one of the base angles be x.

$  \therefore x + x + 40^\circ  = 180^\circ  $

$ \Rightarrow 2x = 180^\circ  - 40^\circ  $ 

$  \Rightarrow 2x = 140^\circ  $ 

$  \Rightarrow x = \dfrac{{140^\circ }}{2} $ 

$ \Rightarrow x = 70^\circ  $

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

Ans: Let Rahul’s score be x.

$ \Rightarrow $Sachin’s score is 2x.

$  \therefore x + 2x = 200 - 2 $

$  \Rightarrow 3x = 198 $

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

$  \Rightarrow x = 66 $

$  \therefore 2x = 132 $

Rahul’ score is 66

Sachin’s score is 132.

3. Solve the following:

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

Ans: Let the number of marbles Parmit has, be x

$  \therefore 5x + 7 = 37 $

$   \Rightarrow 5x = 37 - 7 $

$\Rightarrow 5x = 30 $

$  \Rightarrow x = \dfrac{{30}}{5} $ 

$  \Rightarrow x = 6 $

Hence, Parmit has a total of 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,

$ \Rightarrow 3x + 4 = 49 $ 

$ \Rightarrow 3x = 49 - 4 $ 

$  \Rightarrow 3x = 45 $ 

$ \Rightarrow x = \dfrac{{45}}{3} $ 

$ \Rightarrow x = 15 $

Hence, Laxmi is 15 years old.

(c) People of Sundergram planted a total of 102 trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted?

Ans: Let the number of fruit trees be x,

$\therefore $number of non-fruit trees is 3x+2

$  \Rightarrow x + 3x + 2 = 102 $

$ \Rightarrow 4x + 2 = 102 $ 

$  \Rightarrow 4x = 102 - 2 $ 

$  \Rightarrow 4x = 100 $ 

$ \Rightarrow x = \dfrac{{100}}{4} $ 

$  \Rightarrow x = 25 $

Hence, number of fruit trees planted is 25.

4. Solve the following riddle:

I am a number, Tell my identity!

Take me seven times over, And add a fifty!

To reach a triple century, You still need forty!

Ans: Let the number be x,

“ Take me seven times over, And add a fifty ”

$ \Rightarrow 7x + 50$

“ To reach a triple century, You still need forty!”

$ \Rightarrow $Result of the equation is :

$300 - 40 = 260$

Hence the final equation becomes:

$7x + 50 = 260$

Solving for the equation:

$7x+50=260$

$\Rightarrow 7x=260-50$

$\Rightarrow 7x=210$

$ \Rightarrow x = \dfrac{{210}}{7} $ 

$ \Rightarrow x = 30 $

NCERT Solutions for Class 7 Maths Chapter 4  Free PDF Download

NCERT Solutions for Class 7 Maths Chapter 4 Exercises

A variable takes on different numerical values whereas a constant has a fixed value.

An equation is a statement of a variable in which two expressions in the variable should have equal value.

An equation remains unchanged if its L.H.S and R.H.S are interchanged.

Transposing a number means moving it to the other side.

The equations remain unchanged when we:

Add the same number to both sides.

Subtract the same number from both sides.

Multiply both sides by the same number.

Divide both sides by the same number.

When we transpose a number from one side of the equation to the other side its sign is changed.

As we just read that variable is not fixed which means the numerical value of the variable changes. These variables are denoted by letters of the alphabet such as l, m, n, p, q, r, s, u, v, x, y, z, etc. When we perform operations like addition, subtraction, multiplication, and division on variables then expressions are formed.

The value of an expression depends upon the chosen value of the variable. If there is only one term in an expression then it is called a monomial expression.

If there are two terms in an expression then it is a binomial expression.

If there are three terms in an expression then it is a trinomial expression.

A polynomial expression is an expression that has four terms.

Note: A polynomial expression can have many terms but none of the terms can have a negative exponent for any variable.

An equation is a mathematical statement on a variable where two expressions on either side of the equality sign should have equal value. At least one of the expressions must contain the variable.

Note: An equation does not change when the expression on the left-hand side and on the right-hand side is interchanged.

In an equation, there is always an equality sign between two expressions.

Example:   Write the following statements in the form of equations.

The difference of five times x and 11 is 28.

One-fourth of a member minus 8 is 18.

i) We have, five times x that is 5x.

Difference of 5x – 11 is 5x -11

∴ 5x – 11 = 28

Thus, the required equation is: 5x – 11 = 28

ii) Let the number be x.

∴ One-fourth of x is ¼ (x).

Now, one-fourth of x minus 8 is ¼(x) – 8.

Thus, the required equation is: ¼  (x) – 8 = 18

Let us see one more example, which will help you with the Exercise 4.1 of NCERT Solutions Chapter 4.

Example: Write a statement for the equations: 2x – 5 = 15

Solution: 2x – 5 = 15

Taking away 5 from twice a number is 15.

An equation is like a weighing balance having equal weights on both of its pans such that the arm of the balance is exactly horizontal. The horizontal arm is not disturbed if the same amount of weights are added to both pans (or the same amount of weights are removed from both the pans).

We use this principle when we solve an equation. The equality sign between the L.H.S and R.H.S corresponds to the horizontal beam (arm) of a balance.

An equation remains undisturbed or unchanged:

If L.H.S and R.H.S are interchanged.

To both the sides, if the same number is added. 

From both the sides, if the same number is subtracted.

When both L.H.S and R.H.S are multiplied by the same number.

If both L.H.S and R.H.S  are divided by the same number.

To understand the concept  better, let us try an example.  This will help you with Exercise 4.2 of NCERT Solutions Chapter 4.

Example: Solve 5x – 3 = 12

Solution: 5x – 3 = 12

Adding 3 to both sides, we have

5x – 3 + 3 = 12 + 3

5x  = 15

Dividing both side by 5, we have: 5x/5 = 15/5

x = 3, which is the required solution.

Note: For checking the corrections of the answer, we substitute the value of the variable in the given equation.

i.e., L.H.S = (5 x 3) - 3 = 15 - 3 = 12 = R.H.S

or   L.H.S = R.H.S

Example: Solve ½ (x) + 5 = 65

Solution: ½ (x) + 5 = 65

Subtracting 5 from both sides, we have

½ (x) + 5 - 5 = 65 - 5

Multiplying both sides by 2, we have

[½ (x)] x 2 = 60 x 2 or x = 120

x = 120 is the required solution.

So far we have learned to solve an equation. Now, we shall form (or construct) the equation when its solution (root) is given. Let us know the following successive steps:

Start with x = 9

Multiply both side by 3,

Subtract 2 from both sides,

3x -2 = 27 - 2

3x - 2 = 25, which is an equation.

Note: For a given equation, you get one solution; but for a given solution one can make many equations.

Let us understand this with some more examples so that you can solve the Exercise 4.3 of NCERT Solutions Chapter 4.

Example: Solve 5 ( x - 3 ) = 25

Solution: 5 (x - 3) = 25

(or) x - 3   = 25/5    (Dividing both side by 5)

(or) x - 3   =  5

(or) x =  5 + 3        (Transposing -3 to R.H.S)

          =  8

Thus, x = 8 is the required solution.

Example: Solve 3 (x + 1)/2 = 18

Solution: 3 ( x + 1)/2 = 18

(or) (x + 1)/2 = 18/3 (Dividing both sides by 3)

(or) (x + 1)/2 = 6

(or) x/2 = (6 - 1)/2  (Transposing  1 to R.H.S)

(or) x = (12 - 1)/2 = 11/2

∴ x = 11/2 is the required solution.

Let us understand this with some more examples so that you can solve the Exercise 4.4 of NCERT Solutions Chapter 4.

Example: The sum of five times a number and 18 is 63. Find the number.

Solution: Let the required number be ‘ x ‘.

5 times the number = 5x

∴ According to the condition, we have

5x + 18 = 63

5x = 63 – 18 (Transposing 18 from L.H.S to R.H.S)

(or) Dividing both sides by 5, we have

5x/5 = 45/5

Thus, the required number = 9

Significances of NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations

The significance of NCERT Solutions for Class 7 Maths Chapter 4, "Simple Equations, " is paramount in the realm of mathematics education. These solutions play a pivotal role in helping students grasp the fundamental concepts of solving equations, a skill that forms the basis of more complex mathematical concepts in later grades. They provide step-by-step explanations and worked-out examples, ensuring that students understand the underlying principles and techniques required to solve equations confidently. Moreover, these solutions offer clarity and practice, enabling students to build a strong foundation in mathematics. Beyond acade mic excellence, they nurture problem-solving skills, critical thinking, and logical reasoning, which are valuable assets both inside and outside the classroom. In essence, NCERT Solutions for this chapter empower students to navigate the world of mathematics with confidence and competence.

Conclusion 

NCERT Solutions for Class 7 Maths Chapter 4 , "Simple Equations," are indispensable resources in the journey of mathematical learning. These solutions provide a structured and comprehensive approach to understanding the essential concepts of equation-solving. They offer step-by-step guidance and practical examples that aid students in developing a strong foundation in mathematical problem-solving. Beyond the classroom, these solutions promote critical thinking and logical reasoning skills, which are invaluable in various aspects of life. Moreover, they enhance students' confidence and competence in tackling mathematical challenges. In essence, NCERT Solutions for this chapter serve as a catalyst for mathematical proficiency, equipping students with essential skills for their academic journey and beyond.

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

1. If Both Sides of an Equation are Divided by the same (Non-Qero) Quantity, the Equality.

• Does not change.

May or may not change.

None of these.

2. If 2x – 3 = 5 , then

3. How do you Avail of the Study Materials for NCERT Solutions for Class 7 Maths?

NCERT Solutions for Class 7 Maths and its study material per topic are available on Vedantu’s portal. They are available in pdf format and you can download them on your phone, computers, laptop, and any other device. The teachers have crafted the NCERT Solutions Maths in step-to-step method and they are self-explanatory. The students can revise and solve many questions to master the topic.

4. What are the Important Topics Covered in NCERT Solutions Class 7 Maths Chapter 4?

Vedantu provides solutions of Class 7 Maths Chapter 4 in its portal covering all the concepts in a precise as well as a detailed manner. The content is prepared by Maths subject matter experts and is easy to grasp. The syllabus and exercises are completely set according to the CBSE guidelines. A few of the important topics covered under Chapter 7 are given below :

Constants and Variables

L.H.S and R.H.S

5. How Many Questions are there in NCERT Solutions Class 7 Maths Chapter 4  Simple equations?

There are a variety of questions found in the NCERT Solutions for  Class 7 Maths, Chapter 4, Simple Equations. This chapter consists of four exercises. In Exercise 4.1, there are a total of six questions, Exercise 4.2 has four questions, Exercise 4.3 has four questions and lastly Exercise 4.4 has  four questions. These questions are available on the Vedantu portal, which can be accessed easily as it is free of cost. You can download NCERT Solutions of Class 7 Maths Chapter 4 for future use and can learn anytime.

6. In Chapter 4 Simple Equations of Class 7 Maths, is there any theory asked?

No, in Chapter 4 Simple Equations there is no theory asked. The series of questions asked from this Chapter is mainly practical. Even if a theory comes, it won’t be in long format, it could be either one-liner statements or one-word type. So, your main focus should be on solving, not on learning the theory. Numericals are of more weightage in Simple Equations. Practise is the best solution to have a grip of the topics in an accurate way.  Since the numericals are to be solved accurately the concepts have to be cleared.

7. How to remember the difficult formulae of simple equations?

Simple Equations holds maximum marks in the examination. If you want to remember the difficult formulas, then try to understand them. Solve a question and analyse it by yourself. If you mug up all the formulas, then you won’t gain anything. Start with knowing how the formulas have been structured from step one. This way you cannot forget and if you forget you will know the method of derivation. When you learn in this method then you will never forget it.

8. What should I do to solve the numerical problems of Chapter 4 of Class 7 Maths quickly?

To solve numerical problems quickly, there is only one solution, which is ‘practice’. Make numbers of your daily friends, and try to play with them regularly. Devote an ample number of hours to create a friendship with them. Don’t just think priorly that numerical problems are tough, instead be confident that you can easily solve them. Also, you can have a look at Vedantu’s solutions available on the Vedantu app or website, where numerical problems are solved in simple steps.

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NCERT Solutions Class 7 Maths Chapter 4

Home » NCERT Solutions » NCERT Solutions Class 7 Maths Chapter 4

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NCERT Solutions for Class 7 Mathematics Chapter 4 

Ncert solutions for class 7 mathematics chapter 4 simple equations .

The NCERT Solutions for Class 7 Mathematics Chapter 4 by Extramarks is a compilation of detailed step-by-step solutions to all exercises included in this chapter.

Quick Links

Students should practise all the exercise questions to get in-depth knowledge about the topics. The solutions are crafted by the subject matter experts, who have framed solutions in a systematic and organised manner  which is easy to understand. Students who refer to these materials will be able to prepare confidently  for their exams and achieve desired  results. 

NCERT Solutions for Class 7 Mathematics Chapter 4 Simple equations

Access ncert solutions for class 7 mathematics chapter 4 – simple equations .

Chapter 4 of Class 7 Mathematics is on simple equations divided into five major sections. It is one of the most important chapters in Class 7 Mathematics, as it brushes-up  basic concepts of algebraic equations. 

Students are advised to go through the chapter to get a clear understanding of the concepts in simple equations. The solutions to the questions in this chapter provided by Extramarks will help students to clarify the concepts and they will be able to  solve any questions in the term tests and exams confidently. 

 Following are the important topics covered under NCERT Class 7 Mathematics Chapter 4.

• Stepping up of an equation 
• Review of what we know
• What Equation is?
• Solving an equation
• More Equations
•  From Solution to Equation
• Application of Simple Equations to practical situations

NCERT Solutions for Class 7 Mathematics Chapter 4 Exercises 

The total number of questions in each of the chapter’s exercises are given in the table below.

• A variable takes on different numerical values whereas a constant has a fixed value. 
• An equation is a statement of a variable in which two expressions of the variable should have equal value.
• An equation remains unchanged if its LHS and RHS are interchanged.
• Transposing a number means moving it to the other side.
• The equations remain unchanged when we:
• Add the same number to both sides. 
• Subtract the same number from both sides.
• Multiply  and divide both sides by the same number.
• When we transpose a number from one side of the equation to the other its sign changes

A variable does not have a fixed value. The numerical value of the variable changes. These variables are denoted by letters of the alphabet such as l, m, n, p, q, r, s, t, u, v, w, x, y, z, etc. Expressions are formed when we perform operations such as addition, subtraction, multiplication, and division on variables. 

• The value of an expression depends upon the chosen value of the variable. If there is only one term in an expression then it is called a monomial expression.
• If there are two terms in an expression then it is called a binomial expression. 
• If there are three terms in an expression then it is called a trinomial expression. 
• A polynomial expression is an expression that has four terms.

Note : A polynomial expression can have many terms but none of the terms can have a negative exponent for any variable.

An Equation

An equation is a mathematical statement on a variable where two expressions on either side of the equal sign should have equal value. At least one of the expressions must contain the variable. 

Note : An equation does not change when the expression on the left-hand side or the right-hand side is interchanged. 

In an equation, there is always an equality sign between two expressions.

Example : Write the following statements in the form of equations.

• The difference of five times x and 11 is 28.
• One-fourth of a number minus 8 is 18.
• We have five times x that is 5x

The difference of 5x-11 is 5x-11

Thus, the required equation is 5x-11=28

• Let the number be x

One-fourth of x is ¼(x)

Now, one-fourth of x minus 8 is 1/4(x) – 8

Thus, the required equation is ¼(x) – 8=18

Let us see one more example which will help you with Exercise 4.1 of NCERT solutions chapter 4 

Example : Write a statement for the equation 2x-5=15

Solution : 2x-5=15

Taking away 5 from twice a number is 15

Solving an Equation 

We use this principle when we solve an equation. The equality sign between the LHS and RHS corresponds to the horizontal beam of the balance. 

An equation remains undisturbed or unchanged:

• If LHS and RHS are interchanged.
• To both the sides, if the same number is added
• From both sides if the same number is subtracted.
• When both LHS and RHS are multiplied by the same number 
• When both LHS and RHS are divided by the same number

To understand the concept better, let us try to solve an example. This will help you with exercise 4.2 of NCERT Solutions Chapter 4.

Example : Solve 5x-3=12

Adding 3 to both sides, we get 

5x-3+3=12+3

Dividing both sides by 5, we get 5x/5=15/5

x=3, which is the required solution.

Note : For checking  the answer, we substitute the value of the variable in the given equation

i.e., L.H.S = (5*3)-3= 15-3= 12= R.H.S

or L.H.S = R.H.S

Example : ½(x) + 5= 65

Subtracting 5 from both sides we have,

½(x) +5-5 = 65-5

Multiplying 2 on both sides, we have

½(x) *2 = 60*2

x = 120, is the required solution.

Forming an Equation

We have learned how to solve an equation. Now we shall form or construct the equation when the solution(root) is given. Let us know the following successive steps:

• Start with x = 9
• Multiply both sides by 3 
• Subtract 2 from both sides 

3x – 2 = 27-2

3x – 2 = 25, which is an equation.

Note : For a given equation, you get one solution; but for a given solution, one can make many equations. 

Let us understand this with  more examples so that you can solve exercise 4.3 of NCERT Solutions Chapter 4.

Example: Solve 5(x-3) = 25

(Or) x-3 = 25/5 (Dividing both sides by 5)

(Or) x – 3 = 5

(or) x = 5+3  (Transposing -3 to R.H.S)

x = 8, which is the required solution.

Example: Solve 3(x+1)/2 = 18

Solution : 3(x+1)/2 = 18

(or) (x+1)/2 = 18/2 (Dividing both sides by 2)

(or) (x+1)/2 = 6

(or) x/2 = (6-1)/2 (Transposing 1 to R.H.S)

(or) x = (12-1)/2 = 11/2, which is the required solution.

Application of Simple Equations to Practical Situations

Let us understand this with more examples so that you can solve exercise 4.4 of NCERT Solutions Chapter 4. 

Example : The sum of five times a number and 18 is 63. Find the number 

Let the required number be x 

5 times the number is 5x

According to the condition, we have 

5x + 18 = 63

5x = 63 – 18    (Transposing 18 form L.H.S to R.H.S)

(or) dividing both sides by 5, we have 

5x/5  = 45/5

Thus, the required number is = 9

Related Questions 

Question: If 2x-3 = 5, then 

Solution: x = 4

Question: If both sides of the equation are divided by the same (non–zero) quantity, the equality –

• Does not change 
• May or may not change 
• None of these 

Solutions: Does not change

NCERT Solutions Class 7 Maths Chapter-wise List

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Q.1 Complete the last column of the table .

Q.2 Check whether the value given in the brackets is a solution to the given equation or not : ( a ) n + 5 = 19 ( n = 1 ) ( b ) 7 n + 5 = 19 ( n = – 2 ) ( c ) 7 n + 5 = 19 ( n = 2 ) ( d ) 4 p – 3 = 13 ( p = 1 ) ( e ) 4 p – 3 = 13 ( p = – 4 ) ( f ) 4 p – 3 = 13 ( p = 0 )

( a ) n + 5 = 19 ( n = 1 ) Put n = 1 in L .H .S to get n+5 = 1 + 5 = 6 ≠ R .H .S So, the given value in the bracket is not a solution to the given equation . ( b ) 7n + 5 = 19 ( n = − 2 ) Put n = − 2 in L .H .S to get 7n+5 = 7 ( − 2 ) + 5 = − 14 + 5 = − 9 ≠ R .H .S So, the given value in the bracket is not a solution to the given equation . ( c ) 7n + 5 = 19 ( n = 2 ) Put n = 2 in L .H .S to get 7n+5 = 7 ( 2 ) + 5 = 14 + 5 = 19 = R .H .S So, the given value in the bracket is a solution to the given equation . ( d ) 4p − 3 = 13 ( p = 1 ) Put p = 1 in L .H .S to get 4p − 3 = 4 ( 1 ) − 3 = 4 − 3 = 1 ≠ R .H .S So, the given value in the bracket is not a solution to the given equation . ( e ) 4p − 3 = 13 ( p = − 4 ) Put p = − 4 in L .H .S to get 4p − 3 = 4 ( − 4 ) − 3 = − 16 − 3 = − 19 ≠ R .H .S So, the given value in the bracket is not a solution to the given equation . ( f ) 4p − 3 = 13 ( p = 0 ) Put p = 0 in L .H .S to get 4p − 3 = 4 ( 0 ) − 3 = 0 − 3 = − 3 ≠ R .H .S So, the given value in the bracket is not a solution to the given equation . 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Q.3 Solve the following equations by trial and error method : ( i ) 5 p + 2 = 17 ( ii ) 3 m – 14 = 4

( i ) 5p + 2 = 17 Put p = 1 to L .H .S to get 5 ( 1 ) +2 = 5+2 = 7 ≠ R .H .S Put p = 2 to L .H .S to get 5 ( 2 ) +2 = 10+2 = 12 ≠ R .H .S Put p = 3 to L .H .S to get 5 ( 3 ) +2 = 15+2 = 17 = R .H .S Thus, p = 3 is a solution to the given equation ( ii ) 3m − 14 = 4 Put m = 4 in L .H .S to get 3 ( 4 ) − 14 = 12 − 14 = − 2 ≠ R .H .S Put m = 5 in L .H .S to get 3 ( 5 ) − 14 = 15 − 14 = − 1 ≠ R .H .S Put m = 6 in L .H .S to get 3 ( 4 ) − 14 = 18 − 14 = 4 = R .H .S Thus, m = 6 is a solution to the given equation MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caqGGaGaaeynaiaabchacqGHRaWkcaqGGaGaaeOmaiabg2da9iaabg dacaqG3aaabaGaaeiuaiaabwhacaqG0bGaaeiiaiaabchacqGH9aqp caqGXaGaaeiiaiaabshacaqGVbGaaeiiaiaabYeacaqGUaGaaeisai aab6cacaqGtbGaaeiiaiaabshacaqGVbGaaeiiaiaabEgacaqGLbGa aeiDaaqaaiaabwdadaqadaqaaiaaigdaaiaawIcacaGLPaaacaqGRa GaaeOmaiabg2da9iaabwdacaqGRaGaaeOmaiabg2da9iaabEdacqGH GjsUcaqGsbGaaeOlaiaabIeacaqGUaGaae4uaaqaaiaabcfacaqG1b GaaeiDaiaabccacaqGWbGaeyypa0JaaeOmaiaabccacaqG0bGaae4B aiaabccacaqGmbGaaeOlaiaabIeacaqGUaGaae4uaiaabccacaqG0b Gaae4BaiaabccacaqGNbGaaeyzaiaabshaaeaacaqG1aWaaeWaaeaa caaIYaaacaGLOaGaayzkaaGaae4kaiaabkdacqGH9aqpcaqGXaGaae imaiaabUcacaqGYaGaeyypa0JaaeymaiaabkdacqGHGjsUcaqGsbGa aeOlaiaabIeacaqGUaGaae4uaaqaaiaabcfacaqG1bGaaeiDaiaabc cacaqGWbGaeyypa0Jaae4maiaabccacaqG0bGaae4BaiaabccacaqG mbGaaeOlaiaabIeacaqGUaGaae4uaiaabccacaqG0bGaae4Baiaabc cacaqGNbGaaeyzaiaabshaaeaacaqG1aWaaeWaaeaacaaIZaaacaGL OaGaayzkaaGaae4kaiaabkdacqGH9aqpcaqGXaGaaeynaiaabUcaca qGYaGaeyypa0JaaeymaiaabEdacqGH9aqpcaqGsbGaaeOlaiaabIea caqGUaGaae4uaaqaaiaabsfacaqGObGaaeyDaiaabohacaqGSaGaae iiaiaabchacqGH9aqpcaqGZaGaaeiiaiaabMgacaqGZbGaaeiiaiaa bggacaqGGaGaae4Caiaab+gacaqGSbGaaeyDaiaabshacaqGPbGaae 4Baiaab6gacaqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaabIgacaqG LbGaaeiiaiaabEgacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaabw gacaqGXbGaaeyDaiaabggacaqG0bGaaeyAaiaab+gacaqGUbGaaeii aaqaamaabmaabaGaaeyAaiaabMgaaiaawIcacaGLPaaacaqGGaGaae 4maiaab2gacqGHsislcaqGXaGaaeinaiabg2da9iaabsdaaeaacaqG qbGaaeyDaiaabshacaqGGaGaaeyBaiabg2da9iaabsdacaqGGaGaae yAaiaab6gacaqGGaGaaeitaiaab6cacaqGibGaaeOlaiaabofacaqG GaGaaeiDaiaab+gacaqGGaGaae4zaiaabwgacaqG0baabaGaae4mam aabmaabaGaaGinaaGaayjkaiaawMcaaiabgkHiTiaaigdacaaI0aGa eyypa0JaaGymaiaaikdacqGHsislcaaIXaGaaGinaiabg2da9iabgk HiTiaaikdacqGHGjsUcaqGsbGaaeOlaiaabIeacaqGUaGaae4uaaqa aiaabcfacaqG1bGaaeiDaiaabccacaqGTbGaeyypa0Jaaeynaiaabc cacaqGPbGaaeOBaiaabccacaqGmbGaaeOlaiaabIeacaqGUaGaae4u aiaabccacaqG0bGaae4BaiaabccacaqGNbGaaeyzaiaabshaaeaaca qGZaWaaeWaaeaacaaI1aaacaGLOaGaayzkaaGaeyOeI0IaaGymaiaa isdacqGH9aqpcaaIXaGaaGynaiabgkHiTiaaigdacaaI0aGaeyypa0 JaeyOeI0IaaGymaiabgcMi5kaabkfacaqGUaGaaeisaiaab6cacaqG tbaabaGaaeiuaiaabwhacaqG0bGaaeiiaiaab2gacqGH9aqpcaqG2a GaaeiiaiaabMgacaqGUbGaaeiiaiaabYeacaqGUaGaaeisaiaab6ca caqGtbGaaeiiaiaabshacaqGVbGaaeiiaiaabEgacaqGLbGaaeiDaa qaaiaabodadaqadaqaaiaaisdaaiaawIcacaGLPaaacqGHsislcaaI XaGaaGinaiabg2da9iaaigdacaaI4aGaeyOeI0IaaGymaiaaisdacq GH9aqpcaaI0aGaeyypa0JaaeOuaiaab6cacaqGibGaaeOlaiaabofa aeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccacaqGTbGaey ypa0JaaeOnaiaabccacaqGPbGaae4CaiaabccacaqGHbGaaeiiaiaa bohacaqGVbGaaeiBaiaabwhacaqG0bGaaeyAaiaab+gacaqGUbGaae iiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqG NbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGLbGaaeyCaiaabw hacaqGHbGaaeiDaiaabMgacaqGVbGaaeOBaaaaaa@76BD@

Q.4 Write equations for the following statements : ( i ) The sum of numbers x and 4 is 9 . ( ii ) The difference between y and 2 is 8 . ( iii ) Ten times a is 70 . ( iv ) The number b divided by 5 gives 6 . ( v ) Three fourth of t is 15 . ( vi ) Seven times m plus 7 gets you 77 . ( vii ) One fourth of a number minus 4 gives 4 . ( viii ) If you take away 6 from 6 times y , you get 60 . ( ix ) If you add 3 toone third of z , you get 30 .

( i )   x + 4 = 9 ( ii)   y − 2 = 8 (iii)   10 a = 70 (iv)   b 5 = 6 (v)   3 4 t = 15 ( vi )   7 m + 7 = 77 ( vii )   x 4 − 4 = 4 ( viii )   6 y − 6 = 60 ( ix )   z 3 + 3 = 30 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabMgaaiaawIcacaGLPaaa caaMe8UaamiEaiabgUcaRiaaisdacqGH9aqpcaaI5aaabaGaaiikai aabMgacaqGPbGaaeykaiaaysW7caWG5bGaeyOeI0IaaGOmaiabg2da 9iaaiIdaaeaacaqGOaGaaeyAaiaabMgacaqGPbGaaeykaiaaysW7ca aIXaGaaGimaiaadggacqGH9aqpcaaI3aGaaGimaaqaaiaabIcacaqG PbGaaeODaiaabMcacaaMe8+aaSaaaeaacaWGIbaabaGaaGynaaaacq GH9aqpcaaI2aaabaGaaeikaiaabAhacaqGPaGaaGjbVpaalaaabaGa aG4maaqaaiaaisdaaaGaamiDaiabg2da9iaaigdacaaI1aaabaWaae WaaeaacaqG2bGaaeyAaaGaayjkaiaawMcaaiaaysW7caaI3aGaamyB aiabgUcaRiaaiEdacqGH9aqpcaaI3aGaaG4naaqaamaabmaabaGaae ODaiaabMgacaqGPbaacaGLOaGaayzkaaGaaGjbVpaalaaabaGaamiE aaqaaiaaisdaaaGaeyOeI0IaaGinaiabg2da9iaaisdaaeaadaqada qaaiaabAhacaqGPbGaaeyAaiaabMgaaiaawIcacaGLPaaacaaMe8Ua aGOnaiaadMhacqGHsislcaaI2aGaeyypa0JaaGOnaiaaicdaaeaada qadaqaaiaabMgacaqG4baacaGLOaGaayzkaaGaaGjbVpaalaaabaGa amOEaaqaaiaaiodaaaGaey4kaSIaaG4maiabg2da9iaaiodacaaIWa aaaaa@9828@

Write the following equations in statement forms : ( i )     p + 4 = 15 ( ii )    m – 7 = 3 ( iii )    2 m = 7 ( iv ) m 5 = 3 ( v )   3 5 m = 6 ( vi ) 3 p + 4 = 25 ( vii ) 4 p – 2 = 18 ( viii ) p 2 + 2 = 8

( i ) The sum of p and 4 is 15 . (ii) 7 subtracted from m is 3 . (iii) Twice of a number m is 7 . (iv) One-fifth of m is 3 . (v) Three-fifth of m is 6 . (vi) Three times of a number p, when add to 4 gives 25 . (vii) When 2 is subtracted from four times of a number p,             it gives 18 . (viii) When 2 is added to half of a number p, it gives 8 . MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaGGOaGaamyAaiaacMcacaqGGaGaaeiv aiaabIgacaqGLbGaaeiiaiaabohacaqG1bGaaeyBaiaabccacaqGVb GaaeOzaiaabccacaqGWbGaaeiiaiaabggacaqGUbGaaeizaiaabcca caqG0aGaaeiiaiaabMgacaqGZbGaaeiiaiaabgdacaqG1aGaaeOlaa qaaiaabIcacaqGPbGaaeyAaiaabMcacaqGGaGaae4naiaabccacaqG ZbGaaeyDaiaabkgacaqG0bGaaeOCaiaabggacaqGJbGaaeiDaiaabw gacaqGKbGaaeiiaiaabAgacaqGYbGaae4Baiaab2gacaqGGaGaaeyB aiaabccacaqGPbGaae4CaiaabccacaqGZaGaaeOlaaqaaiaabIcaca qGPbGaaeyAaiaabMgacaqGPaGaaeiiaiaabsfacaqG3bGaaeyAaiaa bogacaqGLbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaae OBaiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaab2gacaqG GaGaaeyAaiaabohacaqGGaGaae4naiaab6caaeaacaqGOaGaaeyAai aabAhacaqGPaGaaeiiaiaab+eacaqGUbGaaeyzaiaab2cacaqGMbGa aeyAaiaabAgacaqG0bGaaeiAaiaabccacaqGVbGaaeOzaiaabccaca qGTbGaaeiiaiaabMgacaqGZbGaaeiiaiaabodacaqGUaaabaGaaeik aiaabAhacaqGPaGaaeiiaiaabsfacaqGObGaaeOCaiaabwgacaqGLb GaaeylaiaabAgacaqGPbGaaeOzaiaabshacaqGObGaaeiiaiaab+ga caqGMbGaaeiiaiaab2gacaqGGaGaaeyAaiaabohacaqGGaGaaeOnai aab6caaeaacaqGOaGaaeODaiaabMgacaqGPaGaaeiiaiaabsfacaqG ObGaaeOCaiaabwgacaqGLbGaaeiiaiaabshacaqGPbGaaeyBaiaabw gacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaaeOB aiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaabchacaqGSa GaaeiiaiaabEhacaqGObGaaeyzaiaab6gacaqGGaGaaeyyaiaabsga caqGKbGaaeiiaiaabshacaqGVbGaaeiiaiaabsdacaqGGaGaae4zai aabMgacaqG2bGaaeyzaiaabohacaqGGaGaaeOmaiaabwdacaqGUaaa baGaaeikaiaabAhacaqGPbGaaeyAaiaabMcacaqGGaGaae4vaiaabI gacaqGLbGaaeOBaiaabccacaqGYaGaaeiiaiaabMgacaqGZbGaaeii aiaabohacaqG1bGaaeOyaiaabshacaqGYbGaaeyyaiaabogacaqG0b GaaeyzaiaabsgacaqGGaGaaeOzaiaabkhacaqGVbGaaeyBaiaabcca caqGMbGaae4BaiaabwhacaqGYbGaaeiiaiaabshacaqGPbGaaeyBai aabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGa aeOBaiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaabchaca qGSaaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaeyA aiaabshacaqGGaGaae4zaiaabMgacaqG2bGaaeyzaiaabohacaqGGa GaaeymaiaabIdacaqGUaaabaGaaeikaiaabAhacaqGPbGaaeyAaiaa bMgacaqGPaGaaeiiaiaabEfacaqGObGaaeyzaiaab6gacaqGGaGaae OmaiaabccacaqGPbGaae4CaiaabccacaqGHbGaaeizaiaabsgacaqG LbGaaeizaiaabccacaqG0bGaae4BaiaabccacaqGObGaaeyyaiaabY gacaqGMbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGGaGaaeOB aiaabwhacaqGTbGaaeOyaiaabwgacaqGYbGaaeiiaiaabchacaqGSa GaaeiiaiaabMgacaqG0bGaaeiiaiaabEgacaqGPbGaaeODaiaabwga caqGZbGaaeiiaiaabIdacaqGUaaaaaa@52BE@

Q.6 Set up an equation in the following cases : i Irfan says that he has 7 marbles more than five times the marbles Parmit has .  Irfan has 37 marbles . ( Take m to be the number of Parmit ’ s marbles . ( ii ) Laxmi ’ s father is 49 years old . He is 4 years older than three times Laxmi ’ s age .( Take Laxmi ’ s age to be y years .) iii The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. ( Take the lowest score to be l ). iv In an isosceles triangle , the vertex angle is twice either base angle .( Let the base angle be b in degrees . Remember that the sum of angles of a triangle is 180​  degrees ).

(i) Let Parmit has m marbles . Then, according to the question, we have 5 × Number of marbles Parmit has +7 = Number of marbles Irfan has     5 × m + 7 = 37 So, we get       5 m + 7 = 37 (ii) Let Laxmi be y years old Then, according to the question, we have 3 × Laxmi’s age + 4 = Laxmi’s father age 3 × y +4 = 49 3 y + 4 = 49 (iii) Let the lowest marks be l . Then, according to the question, we have 2 × lowest marks + 7 = Highest marks 2 × l + 7 = 87 2 l + 7 = 87 (iv) An isoceles triangle has two angles equal . Let the base angle be b . Then, according to the question, we have b + b + 2 b = 180 ° 4 b = 180 ° MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyAaiaabMcacaqGGaGaaeit aiaabwgacaqG0bGaaeiiaiaabcfacaqGHbGaaeOCaiaab2gacaqGPb GaaeiDaiaabccacaqGObGaaeyyaiaabohacaqGGaGaamyBaiaabcca caqGTbGaaeyyaiaabkhacaqGIbGaaeiBaiaabwgacaqGZbGaaeOlaa qaaiaabsfacaqGObGaaeyzaiaab6gacaqGSaGaaeiiaiaabggacaqG JbGaae4yaiaab+gacaqGYbGaaeizaiaabMgacaqGUbGaae4zaiaabc cacaqG0bGaae4BaiaabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeyC aiaabwhacaqGLbGaae4CaiaabshacaqGPbGaae4Baiaab6gacaqGSa GaaeiiaiaabEhacaqGLbGaaeiiaiaabIgacaqGHbGaaeODaiaabwga aeaacaqG1aGaey41aqRaaeOtaiaabwhacaqGTbGaaeOyaiaabwgaca qGYbGaaeiiaiaab+gacaqGMbGaaeiiaiaab2gacaqGHbGaaeOCaiaa bkgacaqGSbGaaeyzaiaabohacaqGGaGaaeiuaiaabggacaqGYbGaae yBaiaabMgacaqG0bGaaeiiaiaabIgacaqGHbGaae4CaiaabccacaqG RaGaae4naiabg2da9iaab6eacaqG1bGaaeyBaiaabkgacaqGLbGaae OCaiaabccacaqGVbGaaeOzaiaabccacaqGTbGaaeyyaiaabkhacaqG IbGaaeiBaiaabwgacaqGZbaabaGaaCzcaiaaxMaacaWLjaGaaCzcai aaxMaacaWLjaGaaCzcaiaaxMaacaWLjaGaaeysaiaabkhacaqGMbGa aeyyaiaab6gacaqGGaGaaeiAaiaabggacaqGZbaabaGaaCzcaiaaxM aacaWLjaGaaCzcaiaaxMaacaWLjaGaaGjbVlaaysW7caqG1aGaey41 aqRaamyBaiabgUcaRiaabEdacqGH9aqpcaqGZaGaae4naaqaaiaabo facaqGVbGaaeilaiaabccacaqG3bGaaeyzaiaabccacaqGNbGaaeyz aiaabshaaeaacaWLjaGaaCzcaiaaxMaacaWLjaGaaCzcaiaaxMaaca aMe8UaaGjbVlaaysW7daqjEaqaaiaabwdacaWGTbGaey4kaSIaae4n aiabg2da9iaabodacaqG3aaaaaqaaiaabIcacaqGPbGaaeyAaiaabM cacaqGGaGaaeitaiaabwgacaqG0bGaaeiiaiaabYeacaqGHbGaaeiE aiaab2gacaqGPbGaaeiiaiaabkgacaqGLbGaaeiiaiaadMhacaqGGa GaaeyEaiaabwgacaqGHbGaaeOCaiaabohacaqGGaGaae4BaiaabYga caqGKbaabaGaaeivaiaabIgacaqGLbGaaeOBaiaabYcacaqGGaGaae yyaiaabogacaqGJbGaae4BaiaabkhacaqGKbGaaeyAaiaab6gacaqG NbGaaeiiaiaabshacaqGVbGaaeiiaiaabshacaqGObGaaeyzaiaabc cacaqGXbGaaeyDaiaabwgacaqGZbGaaeiDaiaabMgacaqGVbGaaeOB aiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaaeiAaiaabggacaqG2b GaaeyzaaqaaiaabodacqGHxdaTcaqGmbGaaeyyaiaabIhacaqGTbGa aeyAaiaabEcacaqGZbGaaeiiaiaabggacaqGNbGaaeyzaiabgUcaRi aabsdacqGH9aqpcaqGmbGaaeyyaiaabIhacaqGTbGaaeyAaiaabEca caqGZbGaaeiiaiaabAgacaqGHbGaaeiDaiaabIgacaqGLbGaaeOCai aabccacaqGHbGaae4zaiaabwgaaeaacaqGZaGaey41aqRaamyEaiaa bUcacaqG0aGaeyypa0JaaeinaiaabMdaaeaadaqjEaqaaiaabodaca WG5bGaey4kaSIaaeinaiabg2da9iaabsdacaqG5aaaaaqaaiaabIca caqGPbGaaeyAaiaabMgacaqGPaGaaeiiaiaabYeacaqGLbGaaeiDai aabccacaqG0bGaaeiAaiaabwgacaqGGaGaaeiBaiaab+gacaqG3bGa aeyzaiaabohacaqG0bGaaeiiaiaab2gacaqGHbGaaeOCaiaabUgaca qGZbGaaeiiaiaabkgacaqGLbGaaeiiaiaadYgacaqGUaaabaGaaeiv aiaabIgacaqGLbGaaeOBaiaabYcacaqGGaGaaeyyaiaabogacaqGJb Gaae4BaiaabkhacaqGKbGaaeyAaiaab6gacaqGNbGaaeiiaiaabsha caqGVbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGXbGaaeyDai aabwgacaqGZbGaaeiDaiaabMgacaqGVbGaaeOBaiaabYcacaqGGaGa ae4DaiaabwgacaqGGaGaaeiAaiaabggacaqG2bGaaeyzaaqaaiaaik dacqGHxdaTcaqGSbGaae4BaiaabEhacaqGLbGaae4CaiaabshacaqG GaGaaeyBaiaabggacaqGYbGaae4AaiaabohacaqGGaGaey4kaSIaae 4naiabg2da9iaabIeacaqGPbGaae4zaiaabIgacaqGLbGaae4Caiaa bshacaqGGaGaaeyBaiaabggacaqGYbGaae4AaiaabohaaeaacaqGYa Gaey41aqRaamiBaiabgUcaRiaabEdacqGH9aqpcaqG4aGaae4naaqa amaaL4babaGaaGOmaiaadYgacqGHRaWkcaaI3aGaeyypa0JaaGioai aaiEdaaaaabaGaaeikaiaabMgacaqG2bGaaeykaiaabccacaqGbbGa aeOBaiaabccacaqGPbGaae4Caiaab+gacaqGJbGaaeyzaiaabYgaca qGLbGaae4CaiaabccacaqG0bGaaeOCaiaabMgacaqGHbGaaeOBaiaa bEgacaqGSbGaaeyzaiaabccacaqGObGaaeyyaiaabohacaqGGaGaae iDaiaabEhacaqGVbGaaeiiaiaabggacaqGUbGaae4zaiaabYgacaqG LbGaae4CaiaabccacaqGLbGaaeyCaiaabwhacaqGHbGaaeiBaiaab6 caaeaacaqGmbGaaeyzaiaabshacaqGGaGaaeiDaiaabIgacaqGLbGa aeiiaiaabkgacaqGHbGaae4CaiaabwgacaqGGaGaaeyyaiaab6gaca qGNbGaaeiBaiaabwgacaqGGaGaaeOyaiaabwgacaqGGaGaamOyaiaa b6caaeaacaqGubGaaeiAaiaabwgacaqGUbGaaeilaiaabccacaqGHb Gaae4yaiaabogacaqGVbGaaeOCaiaabsgacaqGPbGaaeOBaiaabEga caqGGaGaaeiDaiaab+gacaqGGaGaaeiDaiaabIgacaqGLbGaaeiiai aabghacaqG1bGaaeyzaiaabohacaqG0bGaaeyAaiaab+gacaqGUbGa aeilaiaabccacaqG3bGaaeyzaiaabccacaqGObGaaeyyaiaabAhaca qGLbaabaGaamOyaiabgUcaRiaadkgacqGHRaWkcaWGYaGaamOyaiab g2da9iaabgdacaqG4aGaaeimaiabgclaWcqaamaaL4babaGaaGinai aadkgacqGH9aqpcaaIXaGaaGioaiaaicdacqGHWcaSaaaaaaa@1B6D@

Q.7 Give first the step you will use to separate the variable and then solve the equation : ( a ) x – 1 = 0 ( b ) x + 1 = 0 ( c ) x – 1 = 5 ( d ) x + 6 = 2 ( e ) y – 4 = – 7 ( f ) y – 4 = 4 ( g ) y + 4 = 4 ( h ) y + 4 = – 4

( a ) x − 1 = 0 Add 1 to both sides to get x = 1 ( b ) x + 1 = 0 Subtract 1 from both sides to get x = − 1 ( c ) x − 1 = 5 Add 1 to both sides to get x = 6 ( d ) x + 6 = 2 Subtract 6 from both sides to get x = − 4 ( e ) y − 4 = − 7 Add 4 to both sides to get y = − 3 ( f ) y − 4 = 4 Add 4 to both sides to get y = 0 ( g ) y + 4 = 4 Subtract 4 from both sides to get y = 0 ( h ) y + 4 = − 4 Subtract 4 from both sides to get y = − 8 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caWG4bGaeyOeI0Iaaeymaiabg2da9iaaicdaaeaacaqGbbGaaeizai aabsgacaqGGaGaaeymaiaabccacaqG0bGaae4BaiaabccacaqGIbGa ae4BaiaabshacaqGObGaaeiiaiaabohacaqGPbGaaeizaiaabwgaca qGZbGaaeiiaiaabshacaqGVbGaaeiiaiaabEgacaqGLbGaaeiDaaqa amaaL4babaGaamiEaiabg2da9iaaigdaaaaabaWaaeWaaeaacaqGIb aacaGLOaGaayzkaaGaaeiEaiabgUcaRiaabgdacqGH9aqpcaaIWaaa baGaae4uaiaabwhacaqGIbGaaeiDaiaabkhacaqGHbGaae4yaiaabs hacaqGGaGaaeymaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeii aiaabkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKb GaaeyzaiaabohacaqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaabwga caqG0baabaWaauIhaeaacaqG4bGaeyypa0JaeyOeI0Iaaeymaaaaae aadaqadaqaaiaabogaaiaawIcacaGLPaaacaqG4bGaeyOeI0Iaaeym aiabg2da9iaabwdaaeaacaqGbbGaaeizaiaabsgacaqGGaGaaeymai aabccacaqG0bGaae4BaiaabccacaqGIbGaae4BaiaabshacaqGObGa aeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaabshaca qGVbGaaeiiaiaabEgacaqGLbGaaeiDaaqaamaaL4babaGaamiEaiab g2da9iaaiAdaaaaabaWaaeWaaeaacaqGKbaacaGLOaGaayzkaaGaae iEaiabgUcaRiaabccacaqG2aGaaeiiaiabg2da9iaabccacaqGYaaa baGaae4uaiaabwhacaqGIbGaaeiDaiaabkhacaqGHbGaae4yaiaabs hacaqGGaGaaeOnaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeii aiaabkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKb GaaeyzaiaabohacaqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaabwga caqG0baabaWaauIhaeaacaqG4bGaeyypa0JaeyOeI0Iaaeinaaaaae aadaqadaqaaiaabwgaaiaawIcacaGLPaaacaqG5bGaeyOeI0Iaaeii aiaabsdacqGH9aqpcqGHsislcaqGGaGaae4naaqaaiaabgeacaqGKb GaaeizaiaabccacaqG0aGaaeiiaiaabshacaqGVbGaaeiiaiaabkga caqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzai aabohacaqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaabwgacaqG0baa baWaauIhaeaacaqG5bGaeyypa0JaeyOeI0IaaG4maaaaaeaadaqada qaaiaabAgaaiaawIcacaGLPaaacaqG5bGaeyOeI0Iaaeinaiabg2da 9iaabsdaaeaacaqGbbGaaeizaiaabsgacaqGGaGaaeinaiaabccaca qG0bGaae4BaiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiaiaa bohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaabshacaqGVbGaae iiaiaabEgacaqGLbGaaeiDaaqaamaaL4babaGaamyEaiabg2da9iaa icdaaaaabaWaaeWaaeaacaqGNbaacaGLOaGaayzkaaGaaeyEaiabgU caRiaabsdacqGH9aqpcaqG0aaabaGaae4uaiaabwhacaqGIbGaaeiD aiaabkhacaqGHbGaae4yaiaabshacaqGGaGaaeinaiaabccacaqGMb GaaeOCaiaab+gacaqGTbGaaeiiaiaabkgacaqGVbGaaeiDaiaabIga caqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaaeiDai aab+gacaqGGaGaae4zaiaabwgacaqG0baabaWaauIhaeaacaWG5bGa eyypa0JaaGimaaaaaeaadaqadaqaaiaabIgaaiaawIcacaGLPaaaca qG5bGaey4kaSIaaeinaiabg2da9iabgkHiTiaabccacaqG0aaabaGa ae4uaiaabwhacaqGIbGaaeiDaiaabkhacaqGHbGaae4yaiaabshaca qGGaGaaeinaiaabccacaqGMbGaaeOCaiaab+gacaqGTbGaaeiiaiaa bkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGaae yzaiaabohacaqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaabwgacaqG 0baabaWaauIhaeaacaWG5bGaeyypa0JaeyOeI0IaaGioaaaaaaaa@605F@

Q.8 Give first the step you will use to separate the variable and then solve the equation : ( a ) 3 l = 42 ( b ) b 2 = 6 ( c ) p 7 = 4 ( d ) 4 x = 25 ( e ) 8 y = 36 ( f ) z 3 = 5 4 ( g ) a 5 = 7 15 ( h ) 2 t = – 10

( a ) 3 l = 42 Divide both sides by 3 to get l = 42 3 = 14 ( b ) b 2 = 6 Multiply both sides by 3 to get b = 12 ( c ) p 7 = 4 Multiply both sides by 7 to get p = 28 ( d ) 4x = 25 Divide both sides by 4 to get x = 25 4 ( e ) 8y = 36 Divide both sides by 8 to get y = 36 8 = 9 2 ( f ) z 3 = 5 4 Multiply both sides by 3 to get z = 15 4 ( g ) a 5 = 7 15 Multiply both sides by 5 to get a= 7 × 5 15 = 7 × 5 3 × 5 = 7 3 ( h ) 2 t = − 10 Divide both sides by 2 to get t= − 10 2 = − 5 × 2 2 = − 5

Q.9 Give the steps you will use to separate the variable and then solve the equation : ( a ) 3 n -2 = 46 ( b ) 5 m +7 = 17 ( c ) 20 p 3 = 40 ( d )  3 p 10 = 6

(a) 3 n − 2 = 46 Add 2 to both sides to get 3 n = 48 Now, divide both sides by 3 to get n = 16 (b) 5 m + 7 = 17 Subtract 7 from both sides to get 5 m = 10 Now, divide both sides by 5 to get m = 2 (c) 20 p 3 = 40 Multiply both sides by 3 to get 20 p = 120 Now divide both sides by 20, to get p = 120 20 = 20 × 3 20 Thus ,   p = 3 (d)   3 p 10 = 6 Multiply both sides by 10 to get 3 p = 60 Now divide both sides by 3, to get p = 60 3 = 20 × 3 3 Thus ,   p = 20 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyyaiaabMcacaqGGaGaaG4m aiaad6gacqGHsislcaaIYaGaeyypa0JaaGinaiaaiAdaaeaacaqGbb GaaeizaiaabsgacaqGGaGaaeOmaiaabccacaqG0bGaae4Baiaabcca caqGIbGaae4BaiaabshacaqGObGaaeiiaiaabohacaqGPbGaaeizai aabwgacaqGZbGaaeiiaiaabshacaqGVbGaaeiiaiaabEgacaqGLbGa aeiDaaqaaiaaiodacaWGUbGaeyypa0JaaGinaiaaiIdaaeaacaqGob Gaae4BaiaabEhacaqGSaGaaeiiaiaabsgacaqGPbGaaeODaiaabMga caqGKbGaaeyzaiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiai aabohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaabkgacaqG5bGa aeiiaiaabodacaqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaabwgaca qG0baabaWaauIhaeaacaWGUbGaeyypa0JaaGymaiaaiAdaaaaabaGa aeikaiaabkgacaqGPaGaaeiiaiaaiwdacaWGTbGaey4kaSIaaG4nai abg2da9iaaigdacaaI3aaabaGaae4uaiaabwhacaqGIbGaaeiDaiaa bkhacaqGHbGaae4yaiaabshacaqGGaGaae4naiaabccacaqGMbGaae OCaiaab+gacaqGTbGaaeiiaiaabkgacaqGVbGaaeiDaiaabIgacaqG GaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaaeiDaiaab+ gacaqGGaGaae4zaiaabwgacaqG0baabaGaaGynaiaad2gacqGH9aqp caaIXaGaaGimaaqaaiaab6eacaqGVbGaae4DaiaabYcacaqGGaGaae izaiaabMgacaqG2bGaaeyAaiaabsgacaqGLbGaaeiiaiaabkgacaqG VbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabo hacaqGGaGaaeOyaiaabMhacaqGGaGaaeynaiaabccacaqG0bGaae4B aiaabccacaqGNbGaaeyzaiaabshaaeaadaqjEaqaaiaad2gacqGH9a qpcaaIYaaaaaqaaiaabIcacaqGJbGaaeykaiaabccadaWcaaqaaiaa ikdacaaIWaGaamiCaaqaaiaaiodaaaGaeyypa0JaaGinaiaaicdaae aacaqGnbGaaeyDaiaabYgacaqG0bGaaeyAaiaabchacaqGSbGaaeyE aiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiaiaabohacaqGPb GaaeizaiaabwgacaqGZbGaaeiiaiaabkgacaqG5bGaaeiiaiaaboda caqGGaGaaeiDaiaab+gacaqGGaGaae4zaiaabwgacaqG0baabaGaaG OmaiaaicdacaWGWbGaeyypa0JaaGymaiaaikdacaaIWaaabaGaaeOt aiaab+gacaqG3bGaaeiiaiaabsgacaqGPbGaaeODaiaabMgacaqGKb GaaeyzaiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiaiaaboha caqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaabkgacaqG5bGaaeiiai aabkdacaqGWaGaaeilaiaabccacaqG0bGaae4BaiaabccacaqGNbGa aeyzaiaabshaaeaacaWGWbGaeyypa0ZaaSaaaeaacaaIXaGaaGOmai aaicdaaeaacaaIYaGaaGimaaaacqGH9aqpdaWcaaqaamaaKiaabaGa aGOmaiaaicdaaaGaey41aqRaaG4maaqaamaaKiaabaGaaGOmaiaaic daaaaaaaqaaiaabsfacaqGObGaaeyDaiaabohacaGGSaGaaGjbVpaa L4babaGaamiCaiabg2da9iaaiodaaaaabaGaaeikaiaabsgacaqGPa GaaGjbVpaalaaabaGaaG4maiaadchaaeaacaaIXaGaaGimaaaacqGH 9aqpcaaI2aaabaGaaeytaiaabwhacaqGSbGaaeiDaiaabMgacaqGWb GaaeiBaiaabMhacaqGGaGaaeOyaiaab+gacaqG0bGaaeiAaiaabcca caqGZbGaaeyAaiaabsgacaqGLbGaae4CaiaabccacaqGIbGaaeyEai aabccacaqGXaGaaeimaiaabccacaqG0bGaae4BaiaabccacaqGNbGa aeyzaiaabshaaeaacaaIZaGaamiCaiabg2da9iaaiAdacaaIWaaaba GaaeOtaiaab+gacaqG3bGaaeiiaiaabsgacaqGPbGaaeODaiaabMga caqGKbGaaeyzaiaabccacaqGIbGaae4BaiaabshacaqGObGaaeiiai aabohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaabkgacaqG5bGa aeiiaiaabodacaqGSaGaaeiiaiaabshacaqGVbGaaeiiaiaabEgaca qGLbGaaeiDaaqaaiaadchacqGH9aqpdaWcaaqaaiaaiAdacaaIWaaa baGaaG4maaaacqGH9aqpdaWcaaqaaiaaikdacaaIWaGaey41aqRabG 4mayaawaaabaGabG4mayaawaaaaaqaaiaabsfacaqGObGaaeyDaiaa bohacaGGSaGaaGjbVpaaL4babaGaamiCaiabg2da9iaaikdacaaIWa aaaaaaaa@817B@

Q.10 Solve the following equations : ( a ) 10 p = 100 ( b ) 10 p + 10 = 100 ( c ) p 4 = 5 ( d ) p 3 = 5 ( e ) 3 p 4 = 6 ( f )   3 s = – 9 ( g )   3 s + 12 = 0 ( h )   3 s = 0 ( i )   2 q = 6 ( j )   2 q – 6 = 0 ( k )   2 q + 6 = 0 ( l )   2 q + 6 = 12

( a ) 10 p = 100 p = 100 10 = 10 (b) 10 p + 10 = 100 10 p = 90 p = 90 10 = 9 (c) p 4 = 5 p = 5 × 4 = 20 (d) p 3 = 5 p = 5 × 3 = 15 ( e ) 3 p 4 = 6 3 p = 6 × 4 = 24 p = 24 3 = 8 ( f )   3 s = − 9 s = − 9 3 = − 3 ( g )   3 s + 12 = 0 3 s = − 12 s = − 12 3 = − 4 ( h )   3 s = 0 s = 0 3 = 0 ( i )   2 q = 6 q = 6 2 = 3 ( j )   2 q − 6 = 0 2 q = 6 q = 6 2 = 3 ( k )   2 q + 6 = 0 2 q = − 6 q = − 6 2 = − 3 ( l )   2 q + 6 = 12 2 q = 12 − 6 2 q = 6 q = 6 3 = 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaGGOaGaamyyaiaacMcacaqGGaGaaGym aiaaicdacaWGWbGaeyypa0JaaGymaiaaicdacaaIWaaabaGaamiCai abg2da9maalaaabaGaaGymaiaaicdacaaIWaaabaGaaGymaiaaicda aaGaeyypa0ZaauIhaeaacaaIXaGaaGimaaaaaeaacaqGOaGaaeOyai aabMcacaqGGaGaaGymaiaaicdacaWGWbGaey4kaSIaaGymaiaaicda cqGH9aqpcaaIXaGaaGimaiaaicdaaeaacaaIXaGaaGimaiaadchacq GH9aqpcaaI5aGaaGimaaqaaiaadchacqGH9aqpdaWcaaqaaiaaiMda caaIWaaabaGaaGymaiaaicdaaaGaeyypa0ZaauIhaeaacaaI5aaaaa qaaiaabIcacaqGJbGaaeykaiaabccadaWcaaqaaiaadchaaeaacaaI 0aaaaiabg2da9iaaiwdaaeaacaWGWbGaeyypa0JaaGynaiabgEna0k aaisdacqGH9aqpdaqjEaqaaiaaikdacaaIWaaaaaqaaiaabIcacaqG KbGaaeykaiaabccadaWcaaqaaiaadchaaeaacaaIZaaaaiabg2da9i aaiwdaaeaacaWGWbGaeyypa0JaaGynaiabgEna0kaaiodacqGH9aqp daqjEaqaaiaaigdacaaI1aaaaaqaaiaacIcacaWGLbGaaiykaiaabc cadaWcaaqaaiaaiodacaWGWbaabaGaaGinaaaacqGH9aqpcaaI2aaa baGaaG4maiaadchacqGH9aqpcaaI2aGaey41aqRaaGinaiabg2da9i aaikdacaaI0aaabaGaamiCaiabg2da9maalaaabaGaaGOmaiaaisda aeaacaaIZaaaaiabg2da9maaL4babaGaaGioaaaaaeaacaGGOaGaam OzaiaacMcacaaMe8UaaG4maiaadohacqGH9aqpcqGHsislcaaI5aaa baGaam4Caiabg2da9maalaaabaGaeyOeI0IaaGyoaaqaaiaaiodaaa Gaeyypa0ZaauIhaeaacqGHsislcaaIZaaaaaqaaiaacIcacaWGNbGa aiykaiaaysW7caaIZaGaam4CaiabgUcaRiaaigdacaaIYaGaeyypa0 JaaGimaaqaaiaaiodacaWGZbGaeyypa0JaeyOeI0IaaGymaiaaikda aeaacaWGZbGaeyypa0ZaaSaaaeaacqGHsislcaaIXaGaaGOmaaqaai aaiodaaaGaeyypa0ZaauIhaeaacqGHsislcaaI0aaaaaqaaiaacIca caWGObGaaiykaiaaysW7caaIZaGaam4Caiabg2da9iaaicdaaeaaca WGZbGaeyypa0ZaaSaaaeaacaaIWaaabaGaaG4maaaacqGH9aqpdaqj EaqaaiaaicdaaaaabaGaaiikaiaadMgacaGGPaGaaGjbVlaaikdaca WGXbGaeyypa0JaaGOnaaqaaiaadghacqGH9aqpdaWcaaqaaiaaiAda aeaacaaIYaaaaiabg2da9maaL4babaGaaG4maaaaaeaacaGGOaGaam OAaiaacMcacaaMe8UaaGOmaiaadghacqGHsislcaaI2aGaeyypa0Ja aGimaaqaaiaaikdacaWGXbGaeyypa0JaaGOnaaqaaiaadghacqGH9a qpdaWcaaqaaiaaiAdaaeaacaaIYaaaaiabg2da9maaL4babaGaaG4m aaaaaeaacaGGOaGaam4AaiaacMcacaaMe8UaaGOmaiaadghacqGHRa WkcaaI2aGaeyypa0JaaGimaaqaaiaaikdacaWGXbGaeyypa0JaeyOe I0IaaGOnaaqaaiaadghacqGH9aqpdaWcaaqaaiabgkHiTiaaiAdaae aacaaIYaaaaiabg2da9maaL4babaGaeyOeI0IaaG4maaaaaeaacaGG OaGaamiBaiaacMcacaaMe8UaaGOmaiaadghacqGHRaWkcaaI2aGaey ypa0JaaGymaiaaikdaaeaacaaIYaGaamyCaiabg2da9iaaigdacaaI YaGaeyOeI0IaaGOnaaqaaiaaikdacaWGXbGaeyypa0JaaGOnaaqaai aadghacqGH9aqpdaWcaaqaaiaaiAdaaeaacaaIZaaaaiabg2da9maa L4babaGaaGOmaaaaaaaa@1872@

Q.11 Solve the following equations : ( a ) 2 y + 5 2 = 37 2 ( b ) 5 t + 28 = 10 ( c ) a 5 + 3 = 2 ( d ) q 4 + 7 = 5 ( e ) 5 2 x = 10 ( f ) 5 2 x = 25 4 ( g ) 7 m + 19 2 = 13 ( h ) 6 z + 10 = − 2 ( i ) 3 l 2 = 2 3 ( j ) 2 b 3 − 5 = 3

( a ) 2 y + 5 2 = 37 2 2 y = 37 2 − 5 2 = 37 − 5 2 = 32 2 = 16 2 y = 16 y = 16 2 = 8 T h u s ,   y = 8 ( b ) 5t + 28 = 1 0 5 t = 10 − 28 5 t = − 18 t = − 18 5 ( c ) a 5 + 3 = 2 a 5 = 2 − 3 a 5 = − 1 a = − 5 ( d ) q 4 + 7 = 5 q 4 = 5 − 7 q 4 = − 2 q = − 8 ( e ) 5 2 x = − 10 5 x = − 10 × 2 x = − 10 × 2 5 = − 5 × 2 × 2 5 = − 4 T h u s ,   x = − 4 ( f ) 5 2 x = 25 4 5 x = 25 × 2 4 x = 5 × 5 × 2 2 × 2 × 5 T h u s ,   x = 5 2 ( g ) 7 m + 19 2 = 13 7 m = 13 − 19 2 = 26 − 19 2 = 7 2 7 m = 7 2 m = 7 2 × 7 T h u s ,   m = 1 2 ( h ) 6z + 1 0 = − 2 6 z = − 2 − 10 6 z = − 12 z = − 12 6 = − 2 × 6 6 0 T h u s ,   z = − 2 ( i ) 3 l 2 = 2 3 3 l = 4 3 l = 4 9 ( j ) 2 b 3 − 5 = 3 2 b 3 = 8 2 b = 24 b = 24 2 = 2 × 12 2 T h u s ,   b = 12 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGYaGaamyEaiabgUcaRmaalaaabaGaaGynaaqaaiaaikdaaaGaey ypa0ZaaSaaaeaacaaIZaGaaG4naaqaaiaaikdaaaaabaGaaGOmaiaa dMhacqGH9aqpdaWcaaqaaiaaiodacaaI3aaabaGaaGOmaaaacqGHsi sldaWcaaqaaiaaiwdaaeaacaaIYaaaaiabg2da9maalaaabaGaaG4m aiaaiEdacqGHsislcaaI1aaabaGaaGOmaaaacqGH9aqpdaWcaaqaai aaiodacaaIYaaabaGaaGOmaaaacqGH9aqpcaaIXaGaaGOnaaqaaiaa ikdacaWG5bGaeyypa0JaaGymaiaaiAdaaeaacaWG5bGaeyypa0ZaaS aaaeaacaaIXaGaaGOnaaqaaiaaikdaaaGaeyypa0JaaGioaaqaaiaa dsfacaWGObGaamyDaiaadohacaGGSaGaaGjbVpaaL4babaGaamyEai abg2da9iaaiIdaaaaabaWaaeWaaeaacaqGIbaacaGLOaGaayzkaaGa aeiiaiaabwdacaqG0bGaey4kaSIaaeOmaiaabIdacqGH9aqpcaqGXa GaaGimaaqaaiaaiwdacaWG0bGaeyypa0JaaGymaiaaicdacqGHsisl caaIYaGaaGioaaqaaiaaiwdacaWG0bGaeyypa0JaeyOeI0IaaGymai aaiIdaaeaadaqjEaqaaiaadshacqGH9aqpdaWcaaqaaiabgkHiTiaa igdacaaI4aaabaGaaGynaaaaaaaabaWaaeWaaeaacaqGJbaacaGLOa GaayzkaaWaaSaaaeaacaqGHbaabaGaaGynaaaacqGHRaWkcaaIZaGa eyypa0JaaGOmaaqaamaalaaabaGaamyyaaqaaiaaiwdaaaGaeyypa0 JaaGOmaiabgkHiTiaaiodaaeaadaWcaaqaaiaadggaaeaacaaI1aaa aiabg2da9iabgkHiTiaaigdaaeaadaqjEaqaaiaadggacqGH9aqpcq GHsislcaaI1aaaaaqaamaabmaabaGaaeizaaGaayjkaiaawMcaamaa laaabaGaamyCaaqaaiaaisdaaaGaey4kaSIaaG4naiabg2da9iaaiw daaeaadaWcaaqaaiaadghaaeaacaaI0aaaaiabg2da9iaaiwdacqGH sislcaaI3aaabaWaaSaaaeaacaWGXbaabaGaaGinaaaacqGH9aqpcq GHsislcaaIYaaabaWaauIhaeaacaWGXbGaeyypa0JaeyOeI0IaaGio aaaaaeaadaqadaqaaiaabwgaaiaawIcacaGLPaaadaWcaaqaaiaaiw daaeaacaaIYaaaaiaadIhacqGH9aqpcqGHsislcaaIXaGaaGimaaqa aiaaiwdacaWG4bGaeyypa0JaeyOeI0IaaGymaiaaicdacqGHxdaTca aIYaaabaGaamiEaiabg2da9maalaaabaGaeyOeI0IaaGymaiaaicda cqGHxdaTcaaIYaaabaGaaGynaaaacqGH9aqpdaWcaaqaaiabgkHiTi qaiwdagaGfaiabgEna0kaaikdacqGHxdaTcaaIYaaabaGabGynayaa waaaaiabg2da9iabgkHiTiaaisdaaeaacaWGubGaamiAaiaadwhaca WGZbGaaiilaiaaysW7daqjEaqaaiaadIhacqGH9aqpcqGHsislcaaI 0aaaaaqaamaabmaabaGaaeOzaaGaayjkaiaawMcaamaalaaabaGaaG ynaaqaaiaaikdaaaGaamiEaiabg2da9maalaaabaGaaGOmaiaaiwda aeaacaaI0aaaaaqaaiaaiwdacaWG4bGaeyypa0ZaaSaaaeaacaaIYa GaaGynaiabgEna0kaaikdaaeaacaaI0aaaaaqaaiaadIhacqGH9aqp daWcaaqaaiqaiwdagaGfaiabgEna0kaaiwdacqGHxdaTceaIYaGbay baaeaaceaIYaGbaybacqGHxdaTcaaIYaGaey41aqRabGynayaawaaa aaqaaiaadsfacaWGObGaamyDaiaadohacaGGSaGaaGjbVpaaL4baba GaamiEaiabg2da9maalaaabaGaaGynaaqaaiaaikdaaaaaaaqaamaa bmaabaGaae4zaaGaayjkaiaawMcaaiaaiEdacaWGTbGaey4kaSYaaS aaaeaacaaIXaGaaGyoaaqaaiaaikdaaaGaeyypa0JaaGymaiaaioda aeaacaaI3aGaamyBaiabg2da9iaaigdacaaIZaGaeyOeI0YaaSaaae aacaaIXaGaaGyoaaqaaiaaikdaaaGaeyypa0ZaaSaaaeaacaaIYaGa aGOnaiabgkHiTiaaigdacaaI5aaabaGaaGOmaaaacqGH9aqpdaWcaa qaaiaaiEdaaeaacaaIYaaaaaqaaiaaiEdacaWGTbGaeyypa0ZaaSaa aeaacaaI3aaabaGaaGOmaaaaaeaacaWGTbGaeyypa0ZaaSaaaeaace aI3aGbaybaaeaacaaIYaGaey41aqRabG4nayaawaaaaaqaaiaadsfa caWGObGaamyDaiaadohacaGGSaGaaGjbVpaaL4babaGaamyBaiabg2 da9maalaaabaGaaGymaaqaaiaaikdaaaaaaaqaamaabmaabaGaaeiA aaGaayjkaiaawMcaaiaabccacaqG2aGaaeOEaiabgUcaRiaabgdaca aIWaGaeyypa0JaeyOeI0IaaeOmaaqaaiaaiAdacaWG6bGaeyypa0Ja eyOeI0IaaGOmaiabgkHiTiaaigdacaaIWaaabaGaaGOnaiaadQhacq GH9aqpcqGHsislcaaIXaGaaGOmaaqaaiaadQhacqGH9aqpdaWcaaqa aiabgkHiTiaaigdacaaIYaaabaGaaGOnaaaacqGH9aqpdaWcaaqaai abgkHiTiaaikdacqGHxdaTceaI2aGbaybaaeaaceaI2aGbaybaaaGa aGimaaqaaiaadsfacaWGObGaamyDaiaadohacaGGSaGaaGjbVpaaL4 babaGaamOEaiabg2da9iabgkHiTiaaikdaaaaabaWaaeWaaeaacaqG PbaacaGLOaGaayzkaaWaaSaaaeaacaaIZaGaamiBaaqaaiaaikdaaa Gaeyypa0ZaaSaaaeaacaaIYaaabaGaaG4maaaaaeaacaaIZaGaamiB aiabg2da9maalaaabaGaaGinaaqaaiaaiodaaaaabaWaauIhaeaaca WGSbGaeyypa0ZaaSaaaeaacaaI0aaabaGaaGyoaaaaaaaabaWaaeWa aeaacaqGQbaacaGLOaGaayzkaaWaaSaaaeaacaaIYaGaamOyaaqaai aaiodaaaGaeyOeI0IaaGynaiabg2da9iaaiodaaeaadaWcaaqaaiaa ikdacaWGIbaabaGaaG4maaaacqGH9aqpcaaI4aaabaGaaGOmaiaadk gacqGH9aqpcaaIYaGaaGinaaqaaiaadkgacqGH9aqpdaWcaaqaaiaa ikdacaaI0aaabaGaaGOmaaaacqGH9aqpdaWcaaqaaiqaikdagaGfai abgEna0kaaigdacaaIYaaabaGabGOmayaawaaaaaqaaiaadsfacaWG ObGaamyDaiaadohacaGGSaGaaGjbVpaaL4babaGaamOyaiabg2da9i aaigdacaaIYaaaaaaaaa@981B@

Q.12 Solve the following equations : ( a ) 2 ( x + 4 ) = 12 ( b ) 3 ( n − 5 ) = 21 ( c ) 3 ( n − 5 ) = 21 ( d ) 3 − 2 ( 2 − y ) = 7 ( e ) − 4 ( 2 − x ) = 9 ( f ) 4 ( 2 − x ) = 9 ( g ) 4 + 5 ( p − 1 ) = 34 ( h ) 34 − 5 ( p − 1 ) = 4

(a ) 2 ( x + 4 ) = 12 Divide both sides by 2 to get x + 4 = 6 Thus,   x = 2 ( b ) 3 ( n − 5 ) = − 21 Divide both sides by 3 to get n − 5 = − 7 Thus, n = − 2 ( c ) 3 ( n − 5 ) = − 21 Divide both sides by 2 to get n − 5 = − 7 T h u s ,   n = − 12 ( d ) 3 − 2 ( 2 − y ) = 7 − 2 ( 2 − y ) = 4 Divide both sides by − 2 to get 2 − y = − 2 Multiply both sides by − 1 to get y − 2 = 2 Thus, y = 4 ( e ) − 4 ( 2 − x ) = 9 Divide both sides by − 4 to get 2 − x = − 9 4 Multiply both sides by − 1 to get x − 2 = 9 4 x = 9 4 + 2 = 9 + 8 4 T h u s ,   x = 17 4 ( f ) 4 ( 2 − x ) = 9 Divide both sides by 4 to get 2 − x = 9 4 Multiply both sides by -1 to get x − 2 = − 9 4 x = − 9 4 + 2 = − 9 + 8 4 Thus ,   x = − 1 4 ( g ) 4 + 5 ( p − 1 ) = 34 5 ( p − 1 ) = 30 Divide both sides by 5 to get p − 1 = 6 p = 6 + 1 = 7 Thus ,   p = 7 ( h ) 34 − 5 ( p − 1 ) = 4 − 5 ( p − 1 ) = − 30 Divide both sides by − 5 to get p − 1 = 6 p = 6 + 1 = 7 Thus,   p = 7 MathType@MTEF@5@5@+= 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Q.13 Solve the following equations . ( a ) 4 = 5 ( p − 2 ) ( b ) − 4 = 5 ( p − 2 ) ( c ) − 16 = − 5 ( 2 − p ) ( d ) 10 = 4 + 3 ( t + 2 ) ( e ) 28 = 4 + 3 ( t + 5 ) ( f ) 0 = 16 + 4 ( m − 6 )

( a ) 4 = 5 ( p − 2 ) Divide both sides by 5 to get 4 5 = p − 2 p = 4 5 + 2 = 4 + 10 5 Thus ,   p = 14 5 ( b ) − 4 = 5 ( p − 2 ) Divide both sides by 5 to get − 4 5 = p − 2 p = − 4 5 + 2 = − 4 + 10 5 Thus ,   p = 6 5 ( c ) − 16 = − 5 ( 2 − p ) Divide both sides by − 5 to get 16 5 = 2 − p Multiply both sides by − 1 to get − 16 5 = p − 2 p = − 16 5 + 2 = − 16 + 10 5 Thus ,   p = − 6 5 ( d ) 1 0 = 4 + 3 ( t + 2 ) 6 = 3 ( t + 2 ) Divide both sides by 3 to get 2 = t + 2 Thus, t = 0 ( e ) 28 = 4 + 3 ( t + 5 ) 24 = 3 ( t + 5 ) Divide both sides by 3 to get 8 = t + 5 Thus, t = 3 ( f ) 0 = 16 + 4 ( m − 6 ) − 16 = 4 ( m − 6 ) Divide both sides by 4 to get − 4 = m − 6 Thus, m = 2 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaadaqadaqaaiaabggaaiaawIcacaGLPaaa caqGGaGaaeinaiabg2da9iaabwdacaGGOaGaamiCaiabgkHiTiaabk dacaGGPaaabaGaaeiraiaabMgacaqG2bGaaeyAaiaabsgacaqGLbGa aeiiaiaabkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgaca qGKbGaaeyzaiaabohacaqGGaGaaeOyaiaabMhacaqGGaGaaeynaiaa bccacaqG0bGaae4BaiaabccacaqGNbGaaeyzaiaabshaaeaadaWcaa qaaiaaisdaaeaacaaI1aaaaiabg2da9iaadchacqGHsislcaaIYaaa baGaamiCaiabg2da9maalaaabaGaaGinaaqaaiaaiwdaaaGaey4kaS IaaGOmaiabg2da9maalaaabaGaaGinaiabgUcaRiaaigdacaaIWaaa baGaaGynaaaaaeaacaqGubGaaeiAaiaabwhacaqGZbGaaiilaiaays W7daqjEaqaaiaadchacqGH9aqpdaWcaaqaaiaaigdacaaI0aaabaGa aGynaaaaaaaabaWaaeWaaeaacaqGIbaacaGLOaGaayzkaaGaeyOeI0 Iaaeinaiabg2da9iaabwdacaGGOaGaamiCaiabgkHiTiaaikdacaGG PaaabaGaaeiraiaabMgacaqG2bGaaeyAaiaabsgacaqGLbGaaeiiai aabkgacaqGVbGaaeiDaiaabIgacaqGGaGaae4CaiaabMgacaqGKbGa aeyzaiaabohacaqGGaGaaeOyaiaabMhacaqGGaGaaeynaiaabccaca qG0bGaae4BaiaabccacaqGNbGaaeyzaiaabshaaeaadaWcaaqaaiab gkHiTiaaisdaaeaacaaI1aaaaiabg2da9iaadchacqGHsislcaaIYa aabaGaamiCaiabg2da9maalaaabaGaeyOeI0IaaGinaaqaaiaaiwda aaGaey4kaSIaaGOmaiabg2da9maalaaabaGaeyOeI0IaaGinaiabgU caRiaaigdacaaIWaaabaGaaGynaaaaaeaacaqGubGaaeiAaiaabwha caqGZbGaaiilaiaaysW7daqjEaqaaiaadchacqGH9aqpdaWcaaqaai aaiAdaaeaacaaI1aaaaaaaaeaadaqadaqaaiaabogaaiaawIcacaGL PaaacqGHsislcaaIXaGaaGOnaiabg2da9iabgkHiTiaaiwdacaqGGa GaaiikaiaaikdacqGHsislcaWGWbGaaiykaaqaaiaabseacaqGPbGa aeODaiaabMgacaqGKbGaaeyzaiaabccacaqGIbGaae4Baiaabshaca qGObGaaeiiaiaabohacaqGPbGaaeizaiaabwgacaqGZbGaaeiiaiaa bkgacaqG5bGaaeiiaiabgkHiTiaabwdacaqGGaGaaeiDaiaab+gaca qGGaGaae4zaiaabwgacaqG0baabaWaaSaaaeaacaaIXaGaaGOnaaqa aiaaiwdaaaGaeyypa0JaaGOmaiabgkHiTiaadchaaeaacaqGnbGaae yDaiaabYgacaqG0bGaaeyAaiaabchacaqGSbGaaeyEaiaabccacaqG IbGaae4BaiaabshacaqGObGaaeiiaiaabohacaqGPbGaaeizaiaabw gacaqGZbGaaeiiaiaabkgacaqG5bGaaeiiaiabgkHiTiaabgdacaqG GaGaaeiDaiaab+gacaqGGaGaae4zaiaabwgacaqG0baabaWaaSaaae aacqGHsislcaqGXaGaaeOnaaqaaiaaiwdaaaGaeyypa0JaamiCaiab gkHiTiaaikdaaeaacaWGWbGaeyypa0ZaaSaaaeaacqGHsislcaaIXa GaaGOnaaqaaiaaiwdaaaGaey4kaSIaaGOmaiabg2da9maalaaabaGa eyOeI0IaaGymaiaaiAdacqGHRaWkcaaIXaGaaGimaaqaaiaaiwdaaa aabaGaaeivaiaabIgacaqG1bGaae4CaiaacYcacaaMe8+aauIhaeaa caWGWbGaeyypa0ZaaSaaaeaacqGHsislcaaI2aaabaGaaGynaaaaaa aabaWaaeWaaeaacaqGKbaacaGLOaGaayzkaaGaaeiiaiaabgdacaaI WaGaeyypa0JaaeinaiabgUcaRiaabodacaGGOaGaamiDaiabgUcaRi aaikdacaGGPaaabaGaaGOnaiabg2da9iaaiodadaqadaqaaiaadsha cqGHRaWkcaaIYaaacaGLOaGaayzkaaaabaGaaeiraiaabMgacaqG2b GaaeyAaiaabsgacaqGLbGaaeiiaiaabkgacaqGVbGaaeiDaiaabIga caqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaaeOyai aabMhacaqGGaGaae4maiaabccacaqG0bGaae4BaiaabccacaqGNbGa aeyzaiaabshaaeaacaaIYaGaeyypa0JaamiDaiabgUcaRiaaikdaae aacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccadaqjEaqaaiaa dshacqGH9aqpcaaIWaaaaaqaamaabmaabaGaaeyzaaGaayjkaiaawM caaiaabccacaqGYaGaaeioaiabg2da9iaabsdacqGHRaWkcaqGZaGa aiikaiaadshacqGHRaWkcaqG1aGaaiykaaqaaiaaikdacaaI0aGaey ypa0JaaG4mamaabmaabaGaamiDaiabgUcaRiaaiwdaaiaawIcacaGL PaaaaeaacaqGebGaaeyAaiaabAhacaqGPbGaaeizaiaabwgacaqGGa GaaeOyaiaab+gacaqG0bGaaeiAaiaabccacaqGZbGaaeyAaiaabsga caqGLbGaae4CaiaabccacaqGIbGaaeyEaiaabccacaqGZaGaaeiiai aabshacaqGVbGaaeiiaiaabEgacaqGLbGaaeiDaaqaaiaaiIdacqGH 9aqpcaWG0bGaey4kaSIaaGynaaqaaiaabsfacaqGObGaaeyDaiaabo hacaqGSaGaaeiiamaaL4babaGaamiDaiabg2da9iaaiodaaaaabaWa aeWaaeaacaqGMbaacaGLOaGaayzkaaGaaGimaiabg2da9iaabgdaca qG2aGaey4kaSIaaeinaiaacIcacaWGTbGaeyOeI0IaaeOnaiaacMca aeaacqGHsislcaaIXaGaaGOnaiabg2da9iaaisdadaqadaqaaiaad2 gacqGHsislcaaI2aaacaGLOaGaayzkaaaabaGaaeiraiaabMgacaqG 2bGaaeyAaiaabsgacaqGLbGaaeiiaiaabkgacaqGVbGaaeiDaiaabI gacaqGGaGaae4CaiaabMgacaqGKbGaaeyzaiaabohacaqGGaGaaeOy aiaabMhacaqGGaGaaeinaiaabccacaqG0bGaae4BaiaabccacaqGNb GaaeyzaiaabshaaeaacqGHsislcaaI0aGaeyypa0JaamyBaiabgkHi TiaaiAdaaeaacaqGubGaaeiAaiaabwhacaqGZbGaaeilaiaabccada qjEaqaaiaad2gacqGH9aqpcaaIYaaaaaaaaa@C843@

Q.14 ( a ) Construct 3 equations starting with x = 2 ( b ) Construct 3 equations starting with x = − 2

( a ) Construct 3 equations starting with x = 2 ( b ) Construct 3 equations starting with x = − 2

Q.15 Set up equations and solve them to find the unknown numbers in the following case : a Add 4 to eight times a number ; you get 60. b One fifth of a number minus 4 gives 3. c If I take three fourths of a number and count up 3 more , I get 21. ( d ) When I subtracted 11 from twice a number , the result was 15. ( e ) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8. f Ibenhal thinks of a number . If she adds 19 to it and divides the sum by 5, she will get 8. g Anwar thinks of a number . If he takes away 7 from 52 of the number , the result is 11 2 .

(a) Let the number be x . 8 times of this number = 8x So, we get 8 x + 4 = 6 0 8 x = 56 x= 56 8 T h u s , x = 7 (b) Let the number be x . One-fifth of this number= x 5 So, we get x 5 − 4 = 3 x 5 = 7 x = 35 (c) Let the number be x . Three-fourth of this number = 3 x 4 So,   we   get 3 x 4 + 3 = 21 3 x 4 = 18 3 x = 72 x = 72 3 T h u s ,   x = 24 ( d )   Let the number be x . So, we have 2x-11=15 2x=26 x= 26 13 = 13 T h u s , x = 13 ( e )   Let the number be x Thrice the number of books = 3 x So, we get 50 − 3 x = 8 − 3 x = 8 − 50 = − 42 Divide both sides by -3 to get x= − 42 − 3 = 14 T h u s , x = 14 ( f )   Let the number be x . We, have x + 19 5 = 8 x + 19 = 8 × 5 = 40 x = 40 − 19 = 21 Thus,   x = 21 ( g )   Let the number be x . S o ,   w e   h a v e 5 x 2 − 7 = 23 5 x 2 = 23 + 7 = 30 5 x = 30 × 2 = 60 x = 60 5 = 12 Thus ,   x = 12 MathType@MTEF@5@5@+= feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqeduuDJXwAKbYu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2C G4uz3bIuV1wyUbqeeuuDJXwAKbsr4rNCHbGeaGqiVz0xg9vqqrpepC 0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yq aqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabe qaamaaeaqbaaGceaqabeaacaqGOaGaaeyyaiaabMcacaqGGaGaaeit aiaabwgacaqG0bGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGUb GaaeyDaiaab2gacaqGIbGaaeyzaiaabkhacaqGGaGaaeOyaiaabwga caqGGaGaaeiEaiaab6caaeaacaqG4aGaaeiiaiaabshacaqGPbGaae yBaiaabwgacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabshacaqG ObGaaeyAaiaabohacaqGGaGaaeOBaiaabwhacaqGTbGaaeOyaiaabw gacaqGYbGaaeiiaiaab2dacaqGGaGaaeioaiaabIhaaeaacaqGtbGa ae4BaiaabYcacaqGGaGaae4DaiaabwgacaqGGaGaae4zaiaabwgaca qG0baabaGaaGioaiaadIhacqGHRaWkcaaI0aGaeyypa0JaaGOnaiaa bcdaaeaacaaI4aGaamiEaiabg2da9iaabwdacaqG2aaabaGaaeiEai 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Q.16 Solve the following : a The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score ? b In an isosceles triangle , the base angles are equal . The vertex angle is 40°. What are the base angles of the triangle ? ( Remember , the sum of three angles of a triangle is 180°)

Solve the following : a The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score ? b In an isosceles triangle , the base angles are equal . The vertex angle is 40°. What are the base angles of the triangle ? ( Remember , the sum of three angles of a triangle is 180°)

Q.17 Solve the following : ( i ) Irfan says that he has 7 marbles more than five times the marbles Parmithas . Irfan has 37 marbles . How many marbles does Parmit have ? ( ii ) Laxmi ‘ s father is 49 year sold . He is 4 years older than three times Laxmi ‘ s age . What is Laxmi ‘ sage ? ( iii ) Maya , Madhura and Mohsina are friends studying in the same class . In a class testin geography , Maya got 16 out of 25 . Madhura got 20 . Their average score was 19 . How much did Mohsina score ?

( iv ) People of Sundargram planted a total of 102 trees in the village garden . Some of the trees were fruit trees . The number of non − fruit trees were two more than three times the number of fruit trees . What was the number of fruit trees planted ?

(i) Let Parmit has m marbles . Then, according to the question, we have 5 × Number of marbles Parmit has +7 = Number of marbles Irfan has     5 × m + 7 = 37 So, we get       5 m + 7 = 37 5 m = 37 − 7 = 30 5 m = 30 m = 30 5 = 6 Therefore, ​ Parmit   has   6   marbles . (ii) Let Laxmi be y years old Then, according to the question, we have 3 × Laxmi’s age + 4 = Laxmi’s father age 3 × y + 4 = 49 3 y + 4 = 49 3 y = 49 − 4 = 45 3 y = 45 y = 45 3 = 15 Therefore,   laxmi’s   age   is   15   years . (iii) Let the number of fruit trees be x . So, we have 3 × Number of fruit trees + 2 = Number of non-fruit trees 3 x + 2 = 77 3 x = 77 – 2 = 75 3 x = 75 x = 75 3 = 25 Therefore,   the   number   of   fruit   trees   was   25 . 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Q.18 Solve the following riddle . I am a number , Tell my identity ! Take me seven times over And add a fifty ! To reach a triple century You still need forty

Let the number be x . Then we have ( 7 x + 50 ) + 40 = 300 7 x + 50 + 40 = 300 7 x + 90 + 300 7 x = 300 − 90 7 x = 210 x = 210 7 = 30 . Therefore , the number is 30 .

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Faqs (frequently asked questions), 1. what are the important topics covered in ncert solutions class 7 mathematics chapter 4.

A few of the important topics covered under Chapter 7 are given below:

• Constants and variables 
• L.H.S and R.H.S

2. How do you avail the study materials for NCERT Solutions for Class 7 Mathematics?

NCERT Solutions for Class 7 Mathematics are available on Extramarks. Subject matter experts have crafted the solutions in a step-by-step method that is easy to understand. Students can revise and solve the questions to master this chapter.

3. Are there any theoretical questions in chapter Simple Equations of Class 7 Mathematics?

There aren’t any theoretical questions in this chapter. The questions in this chapter are mostly practical. Even if a theoretical question is asked, the response will be one line or one word long. As a result, your primary focus should be on problem solving rather than learning theory.

4. How many questions are there in NCERT Solutions for class 11 chapter 4?

There are a variety of questions found in NCERT solutions for Class 11 Chapter 4. The chapter is divided into four exercises. Exercise 4.1 has a total of 6 questions, exercise 4.2 has 4 questions, exercise 4.3 has four questions and exercise 4.4 has four questions.

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Simple Equations

• An equation which has only one variable is called Linear Equation

Solving equations which have linear equations on one side and numbers on the other side

EXAMPLE 1: Solve 3x + 15 = 1

• We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport 15 to the right side by changing its sign i.e., -15. => 3x + 15 = 1 => 3x = 1-15

• We need to keep the variable on the left side and numerical terms on the right side. Because of this, we have to transport -12 to the other side by changing its sign i.e., +12.

Solving equations having variables having the variable on both sides

EXAMPLE 1: Solve .

• Simplify the given equation. => =>
• Bring the variable terms on the left side of the equation and the other numerical terms on the right side of the equation. => =>
• Now, to find the value of ‘x’ we need to divide both sides of the equation by 6 to maintain equality.

Application of Linear Equations

Linear equations are used to find the value of an unknown quantity. Have a look at the following examples:

EXAMPLE 1: The sum of the digits of a 2 digit number 13. The numbers obtained by interchanging the digits is 14 more than the given number. Find the number.

SOLUTION: Let the digit at units place be x and the number at tens place be y.

=> y + x = 13 [sum is 13 given] => y = 13 – x [Transposing x to the other side by changing its sign] Thus, the formed number is= [Since x is at ones place and y=13-x is at tens place] After interchanging the digits, the number is= [Now x is at tens place & y=13-x is at one's place] The interchanged number is greater than the original number by 14. [Given]

New number Old number Difference

• Simplify => => => =>
• Transpose the variable term ‘x’ on the left side of the equation and other numerical terms on the right side of the equation by changing their sign. => 18x – 117 = 14 => 18x = 131

EXAMPLE 2: The distance between town A and town B is 123 km. Two buses begin their journey from these towns and move directly toward each other. From town A, the bus is moving at a speed of 45 km per hour and from town B, the bus is moving at 67 km per hour. Assuming the buses start at the same time, find how far is their meeting point from town A. SOLUTION: Let the buses meet after t hours.

We know that distance= speed X time Distance covered by bus 1 = 45 X t Distance covered by bus 2 = 67 X t Therefore, 45t + 67t = 123

The distance travelled by bus 1 from city A to the meeting point= speed of bus 1 X time taken by it to reach the meeting point. =45 X 1.098= 49.41 km Thus, the distance of reaching point from town A is 49.41km. [ANS]

Reducing Equation to Simpler Form

Equations reducible to linear form

• Now, simplify => 6x + 12 + 12x + 15 = 14x + 16 => 18x + 27 = 14x +16
• Transpose the variable term ‘x’ to the left side and the numerical terms on the right side of the equation by changing their sign. => 18x – 14x = 16 – 27 => 4x = 11

Practice these questions

Q3) Three numbers are in the ratio 1:2:3. If the sum of the largest and the smallest equals the second and 45. Find the numbers.

Q4) Find the number whose 1/6 th part decreased 7 equals its 8/9 th part diminished by 1.

Q5) The difference between two numbers is 23. And the quotient obtained by dividing the larger number by the smaller one is 4. Find the numbers.

Q6) A man cycles to the office from his house at a speed of 5km per hour and reaches 6 minutes late. If he cycles at a speed of 7km/hr, he reaches 8 minutes early. What is the distance between the office and his house? Q7) Suraj is now half as old as his father. 20 years ago, Suraj’s father was six times Suraj’s age. What are their ages now? Q8) The perimeter of an isosceles triangle is 91cm. If the length of each equal side is 2cm more than the length of its base. Find the lengths of the sides of the triangle. Q9) The age of a boy in months is equal to the age of his grandfather in years. If the difference between their ages is 66 years, find their ages.

• The basic principle used in solving any linear equation is that any operation performed on one side of the equation must also be performed on the other side of the equation.
• Any term in an equation can be transposed from one side to other side by changing its sign.
• In cross multiplication, we multiply the numerator of LHS by the denominator of RHS and the denominator of LHS by the numerator of RHS and the resultant expression are equal to each other.
• Practical problems are based on the relations between some known and unknown quantities. We convert such problems into equations and then solve them.

Quiz for Simple Equations

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NCERT Exemplar Solutions for Class 7 Maths Chapter 4 Simple Equations

NCERT Exemplar Solutions for Class 7 Maths Chapter 4 Simple Equations are given here in a comprehensive way. Simple Equations solutions are extremely helpful for students to clear all their doubts easily and understand the basics of the chapter more effectively. It is essential to understand various kinds of questions and figure out the best answers for them. In NCERT Exemplar Solutions for Class 7 Maths Chapter 4, students can find and learn more about the equations.

Simple equations can be defined as a mathematical statement consisting of an equal symbol between two algebraic expressions that have the same value. Topics covered in NCERT Exemplar Solutions for Class 7 Maths Chapter 4 are transposing, solutions to equations and applications of simple equations to practical situations.

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Exercise Page: 104

In the Questions 1 to 18, there are four options out of which, one is correct. Choose the correct one.

1. The solution of the equation ax + b = 0 is

(a) a/b (b) –b (c) –b/a (d) b/a

Consider the given equation ax + b = 0

Transpose ‘b’ to right hand side then it becomes –b.

2. If a and b are positive integers, then the solution of the equation ax = b will always be a

(a) positive number (b) negative number (c) 1 (d) 0

(a) positive number

The solution of the given equation ax = b

3. Which of the following is not allowed in a given equation?

(a) Adding the same number to both sides of the equation.

(b) Subtracting the same number from both sides of the equation.

(c) Multiplying both sides of the equation by the same non-zero number.

(d) Dividing both sides of the equation by the same number.

4. The solution of which of the following equations is neither a fraction nor an integer? (a) 2x + 6 = 0 (b) 3x – 5 = 0

(c) 5x – 8 = x + 4 (d) 4x + 7 = x + 2

(d) 4x + 7 = x + 2

Consider the equation 4x + 7 = x + 2

Transpose 7 to right hand side then it becomes –7 and x to Left hand side then it becomes – x.

4x – x = 2 – 7

5. The equation which cannot be solved in integers is

(a) 5y – 3 = – 18 (b) 3x – 9 = 0

(c) 3z + 8 = 3 + z (d) 9y + 8 = 4y – 7

(c) 3z + 8 = 3 + z

Consider the equation 3z + 8 = 3 + z

Transpose 8 to right hand side then it becomes –8 and z to Left hand side then it becomes – z.

3z – z = 3 – 8

6. If 7x + 4 = 25, then x is equal to

(a) 29/7 (b) 100/7 (c) 2 (d) 3

Consider the given equation 7x + 4 = 25

Transpose 4 to right hand side then it becomes –4

7x = 25 – 4

7. The solution of the equation 3x + 7 = – 20 is

(a) 17/7 (b) -9 (c) 9 (d) 13/3

Consider the given equation 3x + 7 = – 20

Transpose 7 to right hand side then it becomes –7

3x = -20 – 7

8. The value of y for which the expressions (y – 15) and (2y + 1) become equal is

(a) 0 (b) 16 (c) 8 (d) – 16

Consider the given equation (y – 15) = (2y + 1)

Transpose -15 to right hand side then it becomes 15 and 2y to Left hand side then it becomes – 2y.

y – 2y = 15 + 1

9. If k + 7 = 16, then the value of 8k – 72 is

(a) 0 (b) 1 (c) 112 (d) 56

Consider the given equation k + 7 = 16

To find out the value of K.

Transpose 7 to right hand side then it becomes -7

Then, the value of 8k – 72 will be,

= (8 × 9) – 72

10. If 43m = 0.086, then the value of m is

(a) 0.002 (b) 0.02 (c) 0.2 (d) 2

Consider the given equation 43m = 0.086

To find out the value of m.

m = 0.086/43

11. x exceeds 3 by 7, can be represented as

(a) x + 3 = 2 (b) x + 7 = 3 (c) x – 3 = 7 (d) x – 7 = 3

(c) x – 3 = 7

12. The equation having 5 as a solution is:

(a) 4x + 1 = 2 (b) 3 – x = 8 (c) x – 5 = 3 (d) 3 + x = 8

(d) 3 + x = 8

Consider the equation 3 + x = 8.

Transpose 3 to right hand side then it becomes -3

13. The equation having – 3 as a solution is:

(a) x + 3 =1 (b) 8 + 2x = 3 (c) 10 + 3x = 1 (d) 2x + 1 = 3

(c) 10 + 3x = 1

Consider the equation 10 + 3x = 1.

Transpose 10 to right hand side then it becomes -10

3x = 1 – 10

14. Which of the following equations can be formed starting with x = 0?

(a) 2x + 1 = – 1 (b) x/ 2 + 5 = 7

(c) 3x – 1 = – 1 (d) 3x – 1 = 1

(c) 3x – 1 = – 1

Consider the equation 3x – 1 = – 1.

Transpose -1 to right hand side then it becomes 1

3x = -1 + 1

15. Which of the following equations cannot be formed using the equation x = 7?

(a) 2x + 1 = 15 (b) 7x – 1 = 50

(c) x – 3 = 4 (d) (x/7) – 1 = 0

(b) 7x – 1 = 50

Consider the equation 7x – 1 = 50.

Transpose -1 to right hand side then it becomes 1.

7x = 50 + 1

16. If (x/2) = 3, then the value of 3x + 2 is

(a) 20 (b) 11 (c) 13/2 (d) 8

Consider the given equation (x/2) = 3

To find out the value of x.

Then, the value of 3x + 2 will be,

= (3 × 6) + 2

17. Which of the following numbers satisfy the equation –6 + x = –12 ?

(a) 2 (b) 6 (c) – 6 (d) – 2

Consider the given equation –6 + x = –12.

x = -12 + 6

Left hand side is equal to Right hand side.

18. Shifting one term from one side of an equation to another side with a change of sign is known as

(a) commutativity (b) transposition

(c) distributivity (d) associativity

(b) transposition

In Questions 19 to 48, fill in the blanks to make the statements true.

19. The sum of two numbers is 60 and their difference is 30.

(a) If smaller number is x, the other number is .(use sum)

From the question it is given that smaller number is x,

Then, the other number is 60 – x

(b) The difference of numbers in term of x is .

From the question, smaller number be x, and other number be (60 – x)

Difference between the numbers = (60 – x) – x

= 60 – x – x

(c) The equation formed is .

As per the condition given in the question, difference of the two numbers is 30.

So, 60 – 2x = 30

Transpose, -2x to RHS then it becomes 2x and 30 to LHS it becomes -30,

60 – 30 = 2x

2x – 30 = 0

(d) The solution of the equation is .

(e) The numbers are and .

60 – x = 60 – 15

The numbers are 15 and 45.

20. Sum of two numbers is 81. One is twice the other.

(a) If smaller number is x, the other number is .

From the question it is given that smaller number is x and one is twice the other.

Then, the other number is 2x

(b) The equation formed is .

From the question, smaller number be x, and other number be 2x

Sum of two numbers is 81,

2x + x = 81

(c) The solution of the equation is .

(d) The numbers are and .

2x = 2 × 27

The numbers are 27 and 54.

21. In a test Abha gets twice the marks as that of Palak. Two times Abha’s marks and three times Palak’s marks make 280.

(a) If Palak gets x marks, Abha gets marks.

From the question it is given that Palak gets x marks and Abha gets twice the marks as that of Palak.

Then, Abha gets 2x marks.

As per the condition given in the question, two times Abha’s marks and three times Palak’s marks make 280.

So, (2 × 2x) + (3 × x) = 280

Therefore, 4x + 3x = 280

Consider the equation,

4x + 3x = 280

(d) Marks obtained by Abha are .

Marks obtained by Abha are 80.

2x = 2 × 40

22. The length of a rectangle is two times its breadth. Its perimeter is 60 cm.

(a) If the breadth of rectangle is x cm, the length of the rectangle is .

From the question it is given that the breadth of rectangle is x cm and length of a rectangle is two times its breadth.

Then, length of the rectangle is 2x.

(b) Perimeter in terms of x is .

We know that, perimeter of rectangle = 2(Length + breadth)

= 2(2x + x)

As per the condition given in the question, perimeter is 60 cm

The equation formed is, 4x + 2x = 60

4x + 2x = 60

23. In a bag there are 5 and 2 rupee coins. If they are equal in number and their worth is ₹ 70, then

(a) The worth of x coins of ₹ 5 each .

The worth of x coins of ₹ 5 each 5x.

(b) The worth of x coins of ₹ 2 each .

The worth of x coins of ₹ 2 each 2x.

As per the condition given in the question, If 5 and 2 rupee coins are equal in number and their worth is ₹ 70.

So, the equation is 5x + 2x = ₹ 70

(d) There are 5 rupee coins and 2 rupee coins.

5x + 2x = 70

There are 10 ₹ 5 coins and 10 ₹ 2 coins.

24. In a Mathematics quiz, 30 prizes consisting of 1 st and 2 nd prizes only are to be given. 1 st and 2 nd prizes are worth ₹ 2000 and ₹ 1000 respectively. If the total prize money is ₹ 52,000 then show that:

(a) If 1 st prizes are x in number the number of 2 nd prizes are .

From the question it is given that, 30 prizes consisting of 1 st .

So, If 1 st prizes are x in number the number of 2 nd prizes are 30 – x.

(b) The total value of prizes in terms of x are .

Given, 1 st and 2 nd prizes are worth ₹ 2000 and ₹ 1000 respectively

= (₹ 2000 × x) + (₹ 1000 × (30 – x))

As per the condition given in the question, 30 prizes consisting of 1 st and 2 nd prizes only are to be given, the total prize money is ₹ 52,000.

Then, the equation is = ₹1000x + ₹30,000 = ₹ 52,000

Consider the equation, ₹1000x + ₹30,000 = ₹ 52,000

Transpose, 30000 to RHS then it becomes -30,000

1000x = 52,000 – 30,000

1000x = 22,000

x = 22,000/1000

(e) The number of 1 st prizes are and the number of 2 nd prizes are .

The number of 1 st prize = x = 22

The number of 2 nd prize = 30 – x = 30 – 22 = 8

The number of 1 st prizes are 22 and the number of 2 nd prizes are 8.

25. If z + 3 = 5, then z = .

Consider the equation, z + 3 = 5

Transpose 3 to RHS it becomes -3

26. is the solution of the equation 3x – 2 =7.

Consider the given equation, 3x – 2 = 7

Transpose -2 to RHS then it becomes 2.

x = 3 is the solution of the equation 3x – 2 =7.

27. is the solution of 3x + 10 = 7.

Consider the given equation, 3x + 10 = 7

Transpose 10 to RHS then it becomes -10.

3x = 7 – 10

x = -1 is the solution of the equation 3x + 10 = 7.

28. If 2x + 3 = 5, then value of 3x + 2 is .

Consider the given equation, 2x + 3 = 5

Transpose 3 to RHS then it becomes -3.

2x = 5 – 3

then value of 3x + 2 is = (3 × 1) + 2

29. In integers, 4x – 1 = 8 has solution.

Consider the given integers, 4x – 1 = 8

Transpose -1 to RHS then it becomes 1.

In integers, 4x – 1 = 8 has no solution.

30. In natural numbers, 4x + 5 = – 7 has solution.

Consider the given integers, 4x + 5 = – 7.

Transpose 5 to RHS then it becomes -5.

4x = – 7 – 5

4x = – 12

x = – 3

In natural numbers, 4x + 5 = – 7 has no solution.

31. In natural numbers, x – 5 = – 5 has solution.

Consider the given integers, x – 5 = – 5.

Transpose – 5 to RHS then it becomes 5.

x = – 5 + 5

In natural numbers, x – 5 = – 5 has no solution.

32. In whole numbers, x + 8 = 12 – 4 has solution.

Consider the given integers, x + 8 = 12 – 4.

Transpose 8 to RHS then it becomes – 8.

x = 8 – 8

In natural numbers, x + 8 = 12 – 4 has one solution.

33. If 5 is added to three times a number, it becomes the same as 7 is subtracted from four times the same number. This fact can be represented as .

Let us assume the number be x,

Given, If 5 is added to three times a number = 3x + 5

7 is subtracted from four times the same number = 4x – 7

If 5 is added to three times a number, it becomes the same as 7 is subtracted from four times the same number. This fact can be represented as 3x + 5= 4x – 7.

34. x + 7 = 10 has the solution .

Consider the given equation, x + 7 = 10.

Transpose 7 to RHS then it becomes -7.

x = 10 – 7

35. x – 0 = ; when 3x = 12.

Consider the given equation, 3x = 12

= 4 – 0

36. x – 1= ; when 2x = 2.

Consider the given equation, 2x = 2

= 1 – 1

37. x – = 15; when x/2 = 6.

Consider the given eqaution, x/2 = 6

= 12 – (- 3)

x – (-3) = 15; when x/2 = 6

38. The solution of the equation x + 15 = 19 is .

Consider the given equation, x + 15 = 19

Transpose 15 to RHS then it becomes -15.

x = 19 – 15

39. Finding the value of a variable in a linear equation that the equation is called a of the equation.

Finding the value of a variable in a linear equation that satisfies the equation is called a root of the equation.

40. Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the of the term.

Any term of an equation may be transposed from one side of the equation to the other side of the equation by changing the sign of the term.

41. If (9/5)x = (18/5), then x =

Consider the given equation, (9/5)x = (18/5)

By cross multiplication x = (18/5) × (5/9)

Then, the value of x = 2

42. If 3 – x = -4, then x =

Consider the given equation, 3 – x = -4

Transpose –x to RHS then it becomes x and – 4 to LHS then it becomes 4,

Then, the value of x = 7

43. If x – ½ = – ½, then x = .

Consider the given equation, x – ½ = – ½

Transpose – ½ to RHS then it becomes ½

x = – ½ + ½

Then, the value of x = 0

44. If (1/6) – x = 1/6, then x = .

Consider the given equation, (1/6) – x = 1/6

Transpose – x to RHS then it becomes x and 1/6 to LHS then it becomes – 1/6

x = 1/6 – 1/6

45. If 10 less than a number is 65, then the number is .

Given 10 less than a number is 65

Then, the equation will be x – 10 = 65

Transpose, – 10 to RHS then it becomes 10

x = 65 + 10

46. If a number is increased by 20, it becomes 45. Then the number is .

Let us assume the number be x.

If a number is increased by 20 = x + 20

If a number is increased by 20, it becomes 45, x + 20 = 45

Transpose, 20 to RHS then it becomes -20.

x = 45 – 20

47. If 84 exceeds another number by 12, then the other number is .

If a number is increased by 12 = x + 12

If a number is increased by 12, it becomes 84, x + 12 = 84

Transpose, 12 to RHS then it becomes -12.

x = 84 – 12

48. If x – (7/8) = 7/8, then x = .

Consider the equation, x – (7/8) = 7/8

Transpose, -7/8 to RHS then it becomes 7/8.

Then, x = (7/8) + (7/8)

In Questions 49 to 55, state whether the statements are True or False.

49. 5 is the solution of the equation 3x + 2 = 17.

Consider the equation, 3x + 2 = 17

Transpose 2 to RHS then it becomes – 2.

3x = 17 – 2

50. 9/5 is the solution of the equation 4x – 1 = 8.

Consider the equation, 4x – 1 = 8

51. 4x – 5 = 7 does not have an integer as its solution.

Consider the equation, 4x – 5 = 7

Transpose -5 to RHS then it becomes 5.

52. One third of a number added to itself gives 10, can be represented as (x/3) + 10 = x

One third of a number added to itself = (x/3) + x

One third of a number added to itself gives 10, can be represented as (x/3) + x = 10

53. 3/2 is the solution of the equation 8x – 5 = 7.

Consider the equation, 8x – 5 = 7

Transpose -5 to RHS; then it becomes 5.

54. If 4x – 7 = 11, then x = 4.

Consider the equation, 4x – 7 = 11

Transpose -7 to RHS; then it becomes 7.

4x = 11 + 7

55. If 9 is the solution of variable x in the equation ((5x – 7)/2) = y, then the value of y is 28.

Consider the equation, ((5x – 7)/2) = y

Then, from the question, it is given that the value of x = 9

So, (((5 × 9) – 7)/2) = y

((45 – 7)/2) = y

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Question 1:

Complete the last column of the table.

(i) x + 3 = 0

L.H.S. = x + 3

By putting x = 3,

L.H.S. = 3 + 3 = 6 ≠ R.H.S.

∴ No, the equation is not satisfied.

(ii) x + 3 = 0

By putting x = 0,

L.H.S. = 0 + 3 = 3 ≠ R.H.S.

(iii) x + 3 = 0

By putting x = −3,

L.H.S. = − 3 + 3 = 0 = R.H.S.

∴ Yes, the equation is satisfied.

(iv) x − 7 = 1

L.H.S. = x − 7

By putting x = 7,

L.H.S. = 7 − 7 = 0 ≠ R.H.S.

(v) x − 7 = 1

By putting x = 8,

L.H.S. = 8 − 7 = 1 = R.H.S.

(vi) 5 x = 25

L.H.S. = 5 x

L.H.S. = 5 × 0 = 0 ≠ R.H.S.

(vii) 5 x = 25

By putting x = 5,

L.H.S. = 5 × 5 = 25 = R.H.S.

(viii) 5 x = 25

By putting x = −5,

L.H.S. = 5 × (−5) = −25 ≠ R.H.S.

By putting m = −6,

∴No, the equation is not satisfied.

By putting m = 0,

By putting m = 6,

Page No 81:

Question 2:.

Check whether the value given in the brackets is a solution to the given equation or not:

(a) n + 5 = 19 ( n = 1) (b) 7 n + 5 = 19 ( n = − 2)

(c) 7 n + 5 = 19 ( n = 2) (d) 4 p − 3 = 13 ( p = 1)

(e) 4 p − 3 = 13 ( p = − 4) (f) 4 p − 3 = 13 ( p = 0)

(a) n + 5 = 19 ( n = 1)

Putting n = 1 in L.H.S.,

n + 5 = 1 + 5 = 6 ≠ 19

As L.H.S. ≠ R.H.S.,

Therefore, n = 1 is not a solution of the given equation, n + 5 = 19.

(b) 7 n + 5 = 19 ( n = −2)

Putting n = −2 in L.H.S.,

7 n + 5 = 7 × (−2) + 5 = −14 + 5 = −9 ≠ 19

Therefore, n = −2 is not a solution of the given equation, 7 n + 5 = 19.

(c) 7 n + 5 = 19 ( n = 2)

Putting n = 2 in L.H.S.,

7 n + 5 = 7 × (2) + 5 = 14 + 5 = 19 = R.H.S.

As L.H.S. = R.H.S.,

Therefore, n = 2 is a solution of the given equation, 7 n + 5 = 19.

(d) 4 p − 3 = 13 ( p = 1)

Putting p = 1 in L.H.S.,

4 p − 3 = (4 × 1) − 3 = 1 ≠ 13

As L.H.S ≠ R.H.S.,

Therefore, p = 1 is not a solution of the given equation, 4 p − 3 = 13.

(e) 4 p − 3 = 13 ( p = −4)

Putting p = −4 in L.H.S.,

4 p − 3 = 4 × (−4) − 3 = − 16 − 3 = −19 ≠ 13

Therefore, p = −4 is not a solution of the given equation, 4 p − 3 = 13.

(f) 4 p − 3 = 13 ( p = 0)

Putting p = 0 in L.H.S.,

4 p − 3 = (4 × 0) − 3 = −3 ≠ 13

Therefore, p = 0 is not a solution of the given equation, 4 p − 3 = 13.

Question 3:

Solve the following equations by trial and error method:

(i) 5 p + 2 = 17 (ii) 3 m − 14 = 4

(i) 5 p + 2 = 17

(5 × 1) + 2 = 7 ≠ R.H.S.

Putting p = 2 in L.H.S.,

(5 × 2) + 2 = 10 + 2 = 12 ≠ R.H.S.

Putting p = 3 in L.H.S.,

(5 × 3) + 2 = 17 = R.H.S.

Hence, p = 3 is a solution of the given equation.

(ii) 3 m − 14 = 4

Putting m = 4,

(3 × 4) − 14 = −2 ≠ R.H.S.

Putting m = 5,

(3 × 5) − 14 = 1 ≠ R.H.S.

Putting m = 6,

(3 × 6) − 14 = 18 − 14 = 4 = R.H.S.

Hence, m = 6 is a solution of the given equation.

Question 4:

Write equations for the following statements:

(i) The sum of numbers x and 4 is 9.

(ii) 2 subtracted from y is 8.

(iii) Ten times a is 70.

(iv) The number b divided by 5 gives 6.

(v) Three-fourth of t is 15.

(vi) Seven times m plus 7 gets you 77.

(vii) One-fourth of a number x minus 4 gives 4.

(viii) If you take away 6 from 6 times y , you get 60.

(ix) If you add 3 to one-third of z , you get 30.

(i) x + 4 = 9

(ii) y − 2 = 8

(iii) 10 a = 70

(vi) Seven times of m is 7 m .

7 m + 7 = 77

(viii) Six times of y is 6 y .

6 y − 6 = 60

Video Solution for Simple Equations (Page: 81 , Q.No.: 4)

NCERT Solution for Class 7 math - Simple Equations 81 , Question 4

Question 5:

Write the following equations in statement forms:

(i) p + 4 = 15 (ii) m − 7 = 3

(i) The sum of p and 4 is 15.

(ii) 7 subtracted from m is 3.

(iii) Twice of a number m is 7.

(iv) One-fifth of m is 3.

(v) Three-fifth of m is 6.

(vi) Three times of a number p , when added to 4, gives 25.

(vii) When 2 is subtracted from four times of a number p , it gives 18.

(viii) When 2 is added to half of a number p , it gives 8.

Page No 82:

Question 6:.

Set up an equation in the following cases:

(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. (Take m to be the number of Parmit’s marbles.)

(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.)

(iii) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. (Take the lowest score to be l .)

(iv) In an isosceles triangle, the vertex angle is twice either base angle. (Let the base angle be b in degrees. Remember that the sum of angles of a triangle is 180 degrees.)

(i) Let Parmit has m marbles.

5 × Number of marbles Parmit has + 7 = Number of marbles Irfan has

5 × m + 7 = 37

5 m + 7 = 37

(ii) Let Laxmi be y years old.

3 × Laxmi’s age + 4 = Laxmi’s father’s age

3 × y + 4 = 49

3 y + 4 = 49

(iii) Let the lowest marks be l .

2 × Lowest marks + 7 = Highest marks

2 × l + 7 = 87

2 l + 7 = 87

(iv) An isosceles triangle has two of its angles of equal measure.

Let base angle be b .

Vertex angle = 2 × Base angle = 2 b

Sum of all interior angles of a Δ = 180°

b + b + 2 b = 180°

4 b = 180°

Video Solution for Simple Equations (Page: 82 , Q.No.: 6)

NCERT Solution for Class 7 math - Simple Equations 82 , Question 6

Page No 86:

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

(a) x + 1 = 0 (b) x + 1 = 0 (c) x − 1 = 5

(d) x + 6 = 2 (e) y − 4 = − 7 (f) y − 4 = 4

(g) y + 4 = 4 (h) y + 4 = − 4

(a) x − 1 = 0

Adding 1 to both sides of the given equation, we obtain

x − 1 + 1 = 0 + 1

(b) x + 1 = 0

Subtracting 1 from both sides of the given equation, we obtain

x + 1 − 1 = 0 − 1

x = −1

(c) x − 1 = 5

x − 1 + 1 = 5 + 1

(d) x + 6 = 2

Subtracting 6 from both sides of the given equation, we obtain

x + 6 − 6 = 2 − 6

x = −4

(e) y − 4 = −7

Adding 4 to both sides of the given equation, we obtain

y − 4 + 4 = − 7 + 4

y = −3

(f) y − 4 = 4

y − 4 + 4 = 4 + 4

(g) y + 4 = 4

Subtracting 4 from both sides of the given equation, we obtain

y + 4 − 4 = 4 − 4

(h) y + 4 = −4

y + 4 − 4 = − 4 − 4

y = −8

(a) 3 l = 42

Dividing both sides of the given equation by 3, we obtain

Multiplying both sides of the given equation by 2, we obtain

Multiplying both sides of the given equation by 7, we obtain

(d) 4 x = 25

Dividing both sides of the given equation by 4, we obtain

(e) 8 y = 36

Dividing both sides of the given equation by 8, we obtain

Multiplying both sides of the given equation by 3, we obtain

(h) 20 t = −10

Dividing both sides of the given equation by 20, we obtain

Give the steps you will use to separate the variable and then solve the equation:

(a) 3 n − 2 = 46 Adding 2 to both sides of the given equation, we obtain

3 n  − 2 + 2 = 46 + 2

3 n  = 48

n  = 16

(b) 5 m + 7 = 17

Subtracting 7 from both sides of the given equation, we obtain

5 m + 7 − 7 = 17 − 7

Dividing both sides of the given equation by 5, we obtain

Multiplying both sides of the given equation by 10, we obtain

Solve the following equations:

(g) 3 s + 12 = 0 (h) 3 s = 0 (i) 2 q = 6

(j) 2 q − 6 = 0 (k) 2 q + 6 = 0 (l) 2 q + 6 = 12

(a) 10 p = 100

(b) 10 p + 10 = 100

10 p + 10 − 10 = 100 − 10

(f) 3 s = −9

(g) 3 s + 12 = 0

3 s + 12 − 12= 0 − 12

3 s = −12

(h) 3 s = 0

(i) 2 q = 6

(j) 2 q − 6 = 0

2 q − 6 + 6 = 0 + 6

(k) 2 q + 6 = 0

2 q + 6 − 6 = 0 − 6

2 q = −6

(l) 2 q + 6 = 12

2 q + 6 − 6 = 12 − 6

Page No 89:

Solve the following equations.

Dividing both sides by 2,

(b) 5 t + 28 = 10

5 t = 10 − 28 = −18 (Transposing 28 to R.H.S.)

Dividing both sides by 5,

Multiplying both sides by 5,

a = −1 × 5 = −5

Multiplying both sides by 4,

q = −8

Multiplying both sides by 2,

5 x = −10 × 2 = −20

Dividing both sides by 7,

(h) 6 z + 10 = −2

6 z = − 2 − 10 = −12 (Transposing 10 to R.H.S.)

Dividing both sides by 6,

Dividing both sides by 3,

Multiplying both sides by 3,

2 b = 8 × 3 = 24

(a) 2 ( x + 4) = 12 (b) 3 ( n − 5) = 21

(c) 3 ( n − 5) = − 21 (d) −4 (2 + x ) = 8

(e) 4(2 − x ) = 8

(a) 2 ( x + 4) = 12

x = 6 − 4 = 2 (Transposing 4 to R.H.S.)

(b) 3 ( n − 5) = 21

n = 7 + 5 = 12 (Transposing −5 to R.H.S.)

(c) 3 ( n − 5) = −21

n = − 7 + 5 = −2 (Transposing −5 to R.H.S.)

(d) −4 (2 + x ) = 8

Dividing both sides by −4,

x = − 2 − 2 = −4 (Transposing 2 to R.H.S.)

(e) 4 (2 − x ) = 8

Dividing both sides by 4,

2 − x = 2

− x = 2 − 2 (Transposing 2 to R.H.S.)

− x = 0

(a) 4 = 5 ( p − 2) (b) − 4 = 5 ( p − 2)

(c) 16 = 4 + 3 ( t + 2) (d) 4 + 5 ( p − 1) = 34

(e) 0 = 16 + 4 ( m − 6)

(a) 4 = 5 ( p − 2)

(b) − 4 = 5 ( p − 2)

(c) 16 = 4 + 3 ( t + 2)

16 − 4 = 3 ( t + 2) (Transposing 4 to L.H.S.)

12 = 3 ( t + 2)

4 − 2 = t (Transposing 2 to L.H.S.)

(d) 4 + 5 ( p − 1) = 34

5 ( p − 1) = 34 − 4 = 30 (Transposing 4 to R.H.S.)

p = 6 + 1 = 7 (Transposing −1 to R.H.S.)

0 = 16 + 4 m − 24

0 = −8 + 4 m

4 m = 8 (Transposing −8 to L.H.S)

(a) Construct 3 equations starting with x = 2

(b) Construct 3 equations starting with x = − 2

5 x = 10 (i)

Subtracting 3 from both sides,

5 x − 3 = 10 − 3

5 x − 3 = 7 (ii)

(b) x = −2

Subtracting 2 from both sides,

x − 2 = − 2 − 2

x − 2 = −4 (i)

Again, x = −2

Multiplying by 6,

6 × x = −2 × 6

6 x = −12

Subtracting 12 from both sides,

6 x − 12 = − 12 − 12

6 x − 12 = −24 (ii)

Adding 24 to both sides,

6 x − 12 + 24 = − 24 + 24

6 x + 12 = 0 (iii)

Page No 91:

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.

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

(c) If I take three-fourths of a number and add 3 to it, I get 21.

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

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

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

(a) Let the number be x .

8 times of this number = 8 x

8 x + 4 = 60

8 x = 60 − 4 (Transposing 4 to R.H.S.)

Dividing both sides by 8,

(b) Let the number be x .

(c) Let the number be x .

(d) Let the number be x .

Twice of this number = 2 x

2 x − 11 = 15

2 x = 15 + 11 (Transposing −11 to R.H.S.)

(e) Let the number of books be x .

Thrice the number of books = 3 x

50 − 3 x = 8

− 3 x = 8 −50 (Transposing 50 to R.H.S.)

−3 x = −42

Dividing both sides by ­−3,

(f) Let the number be x .

x + 19 = 40

x = 40 − 19 (Transposing 19 to R.H.S.)

(g) Let the number be x .

Solve the following:

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

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

(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?

(a) Let the lowest score be l .

2 l = 87 − 7 (Transposing 7 to R.H.S.)

Therefore, the lowest score is 40.

(b) Let the base angles be equal to b .

The sum of all interior angles of a triangle is 180°.

b + b + 40° = 180°

2 b + 40° = 180°

2 b = 180º − 40º = 140º (Transposing 40º to R.H.S.)

Therefore, the base angles of the triangle are of 70º measure.

(c) Let Rahul’s score be x .

Therefore, Sachin’s score = 2 x

Rahul’s score + Sachin’s score = 200 − 2

2 x + x = 198

Rahul’s score = 66

Sachin’s score = 2 × 66 = 132

Video Solution for Simple Equations (Page: 91 , Q.No.: 2)

NCERT Solution for Class 7 math - Simple Equations 91 , Question 2

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

(ii) Laxmi’s father is 49 year old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?

(iii) 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?

(i) Let Parmit’s marbles equal x .

5 times the number of marbles Parmit has = 5 x

5 x + 7 = 37

5 x = 37 − 7 = 30 (Transposing 7 to R.H.S.)

Therefore, Parmit has 6 marbles.

(ii) Let Laxmi’s age be x years.

3 × Laxmi’s age + 4 = Her father’s age

3 x + 4 = 49

3 x = 49 − 4 (Transposing 4 to R.H.S.)

Therefore, Laxmi’s age is 15 years.

(iii) Let the number of fruit trees be x .

3 × Number of fruit trees + 2 = Number of non-fruit trees

3 x + 2 = 77

3 x = 77 − 2 (Transposing 2 to R.H.S.)

Dividing both sides of the equation by 3,

Therefore, the number of fruit trees was 25.

Video Solution for Simple Equations (Page: 91 , Q.No.: 3)

NCERT Solution for Class 7 math - Simple Equations 91 , Question 3

Page No 92:

Solve the following riddle:

I am a number,

Tell my identity!

Take me seven times over

And add a fifty!

To reach a triple century

You still need forty!

Let the number be x .

(7 x + 50) + 40 = 300

7 x + 90 = 300

7 x = 300 − 90 (Transposing 90 to R.H.S.)

Therefore, the number is 30.

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7th Class Mathematics Simple Equations Question Bank

Done simple equations total questions - 45.

question_answer 1) What is the value of 'x' in  \[\frac{3x-1}{5}-\frac{1+x}{2}=3-\frac{x-1}{2}\]?

A) \[5\]                                      done clear

B)  \[-7\] done clear

C) \[7\]                        done clear

D)        \[-5\] done clear

question_answer 2) If \[0.2(2x-1)-0.5(3x-1)=0.4,\] what is the value of x'?

A)  \[\frac{1}{11}\] done clear

B)                         \[-\frac{1}{11}\]   done clear

C)  \[\frac{3}{11}\]                done clear

D)         \[-\frac{3}{11}\] done clear

question_answer 3) In each of these figures the solution of an equation is given in brackets. Which of them is correct?

question_answer 4) Sunil wrote an equation as \[\frac{m}{5}=4\]. Ravi wrote a statement for Sunil's equation. Which of these is the statement of Ravi if he has written correctly?

A)  One-fifth of m' is 4.   done clear

B)  One-fifth of a number is 5.           done clear

C)  One-fourth of 'm' is 4.               done clear

D)  One-fourth of a number is 4. done clear

question_answer 5) In which of the following cases does an equality NOT hold?

A)  Adding the same number on both the sides. done clear

B)  Not performing the same operation on both the sides.                   done clear

C)  Subtracting the same number from both the sides. done clear

D)  Multiplying both the sides by the same non-zero number. done clear

question_answer 6) What are the two steps involved in solving the equation\[15x+4=26\]?

A)  Subtracting 4 from both the sides and then dividing both sides by 15.       done clear

B)  Adding 4 on both sides & then multiplying both sides by 15. done clear

C)  Subtracting 4 on the L.H.S. and multiplying by 15 on the R.H.S. done clear

D)  Adding 4 on the L.H.S. and dividing by 15 on the R.H.S. done clear

question_answer 7) Which of the following equations can be constructed with \[x=2\]?

A)  \[3x+4=8\]         done clear

B)         \[3x-4=2\]   done clear

C)  \[3x+4=2\]         done clear

D)         \[3x-4=8\] done clear

question_answer 8) 7 subtracted from \[\frac{5}{2}\]of a number results in 23. What is the number?

A)  \[-10\]                 done clear

B)         \[10\] done clear

C)  \[12\]                     done clear

D)         \[-12\] done clear

question_answer 9) In a coconut grove, \[(x+2)\] trees yield 60 coconuts per year, x trees yield 120 coconuts per year and \[(x-2)\] trees yield 180 coconuts per year. If the average yield per year per tree is 100, find x.

A)  \[4\]                       done clear

B)         \[3\] done clear

C)  \[2\]                     done clear

D)         \[1\] done clear

question_answer 10) 4 is added to a number and the sum is multiplied by 5. If 20 is subtracted from the product and the difference is divided by 8, the result is equal to 10. Find the number.

A)  \[16\]                     done clear

B)         \[12\] done clear

C)  \[8\]                     done clear

D)         \[20\] done clear

question_answer 11) A number is 3 less than two times the other. If their sum is increased by 7, the result is 37. Find the numbers.

A)  \[9,\,\,11\]                         done clear

B)         \[11,\,\,13\] done clear

C)  \[11,\,\,19\]                        done clear

D)         \[9,\,\,13\] done clear

question_answer 12) \[\frac{1}{2}\] is subtracted from a number and the difference is multiplied by 4. If 25 is added to the product and the sum is divided by 3, the result is equal to 10. Find the number.

A)  \[\frac{3}{5}\]                                   done clear

B)         \[\frac{7}{4}\]     done clear

C)   \[\frac{6}{7}\]                  done clear

D)         \[\frac{2}{3}\] done clear

question_answer 13) The present age of A is twice that of B. 30  years from now, age of A will be \[1{}^{1}/{}_{2}\]times that of B. Find the present ages (in years) of A and B respectively.

A)  \[60,\,\,30\]                        done clear

B)         \[30,\,\,60\] done clear

C)  \[40,\,\,50\]                      done clear

D)         \[50,\,\,40\] done clear

question_answer 14) A person travelled \[{{\frac{5}{8}}^{th}}\] of the distance by  train, \[{{\frac{1}{4}}^{th}}\] by bus and the remaining 15 km by boat. Find the total distance travelled by him.                    

A)  \[90\,\,km\]                      done clear

B)         \[120\,\,km\]   done clear

C)  \[150\,\,km\]                    done clear

D)         \[180\,\,km\] done clear

question_answer 15) The total cost of three prizes is Rs.2550. If the value of second prize is \[{{\frac{3}{4}}^{th}}\]of the first and the value of 3rd prize is \[\frac{1}{2}\]of the second prize, find the value of first prize.

A)  \[Rs.\,900\]       done clear

B)         \[Rs.\,1500\] done clear

C)  \[Rs.\,1200\]                       done clear

D)         \[Rs.\,450\] done clear

question_answer 16) Which of the following is an equation?

A)  \[4x+5=65\]     done clear

B)  \[4x+5<65\] done clear

C)  \[4x+5>65\] done clear

D)  \[4x+5\ne 65\] done clear

question_answer 17) Which of the following is an algebraic expression for the statement "The sum of \[3x\] and 11 is 32."?

A)  \[11x+3=32\] done clear

B)  \[3x+11=32\]   done clear

C)  \[3x+32=11\] done clear

D)  \[11x+32=11\] done clear

question_answer 18) Choose the statement that best describes the equation \[\frac{1}{4}m=10\].

A)  One - fourth of 10 is m. done clear

B)  One - fourth of m is 3 more than 3. done clear

C)  One - fourth of m is 10.   done clear

D)  Four times m is 10. done clear

question_answer 19) Vinay's father is 44 years old. If he is 5 years older than thrice Vinay's age, which of these equations gives, the age of Vinay's father?

A)  \[3x+5=44\]     done clear

B)  \[44+5x=3x\] done clear

C)  \[44-3y=5+3y\] done clear

D)  \[3x-5=44\] done clear

question_answer 20) Which of the following statements is false?

A)  The solution of \[4x=60\] is 12.   done clear

B)  \[y=7\] satisfies the equation \[y+0=7\]. done clear

C)  \[p=\frac{5}{2}\] is the solution of \[12p-5=25\]. done clear

D)  \[m=\frac{3}{2}\] is the solution of \[4(m+3)=18\]. done clear

question_answer 21) Which of the following does not affect the given equation?

A)  Adding 0 on the L.H.S. and 1 on the R. H. S. done clear

B)  Adding 1 on the L.H.S. and \[(-1)\] on the R.H.S. done clear

C)  Adding the same number on both sides of the equation.   done clear

D)  Adding 0 on the R.H.S. and 1 on the L.H.S. done clear

question_answer 22) P is a linear equation. How many solutions does P have?

A)  1               done clear

B)  0 done clear

C)  3             done clear

D)  Infinitely many done clear

question_answer 23) Ramesh got 5 marks more than Sonu in a test. If the total marks secured by them is 15, how many marks did Ramesh get?

A)  \[25\]                   done clear

B)         \[5\] done clear

C)  \[15\]                   done clear

D)         \[10\]   done clear

question_answer 24) In a test match, Sach in scored twice as many runs as Sehwag. Together, their runs fell two short of a double century. How many runs did Sachin score?

A)  \[66\]                   done clear

B)         \[132\]   done clear

C)  \[198\]                 done clear

D)         \[200\] done clear

question_answer 25) In a math test, the highest marks obtained by a student in the class is twice the lowest marks plus 7. If the highest score is 87, what is the lowest score?

A)  \[42\]                   done clear

B)         \[39\] done clear

C)  \[40\]                     done clear

D)         \[44\] done clear

A)  (i) only          done clear

B)  (ii) only done clear

C)  (iii) only       done clear

D)  Both [b] and [c]. done clear

question_answer 27) On transposing terms from one side of the equation to the other, which of these changes takes place?

A)  Addition becomes subtraction.   done clear

B)  Multiplication becomes addition. done clear

C)  Addition becomes multiplication. done clear

D)  Multiplication becomes subtraction. done clear

question_answer 28) \[M=\frac{2x+5}{7}\]and \[N=\frac{3x-2}{4}\]. What value of x makes\[M=N\]?

A)  \[\frac{-17}{3}\]               done clear

B)         \[\frac{-34}{13}\] done clear

C)  \[\frac{34}{13}\]                                done clear

D)         \[\frac{17}{3}\] done clear

question_answer 29) Given \[A=P(1+rt),\] what is the value of 'r' when \[A=27,\text{ }P=18\]and \[t=5\]?

A)  \[\frac{1}{2}\]                   done clear

B)         \[\frac{1}{5}\] done clear

C)  \[\frac{27}{5}\]                done clear

D)         \[\frac{1}{10}\]   done clear

question_answer 30) Given \[\frac{1}{u}+\frac{1}{v}=\frac{1}{f},\] find the value of 'v' when \[f=20\] and \[u=30\].

A)  \[-20\]                 done clear

B)         \[-60\] done clear

C)  \[60\]                                     done clear

D)  \[-30\] done clear

question_answer 31) The sum of three consecutive integers is 75. Which is the largest among them?

A)  26                            done clear

B)         25 done clear

C)  24                                          done clear

D)  23 done clear

question_answer 32) The lengths of the sides of a triangle are \[(2a+1)\,\,cm,\,\,(3a+2)\,\,cm\] and \[(4a-1)\,\,cm\]. For what value of 'a' is the perimeter of the triangle \[92\,cm\]?

A)  \[5\]                     done clear

B)         \[9\] done clear

C)  \[8\]                                     done clear

D)  \[10\]   done clear

question_answer 33) A father is 26 years older than his son. In 3 years' time, the son's age will be one-third his father's age. What is the present age of the son?

A)  10 years                done clear

B)         13 years done clear

C)  39 years              done clear

D)         29 years done clear

question_answer 34) Pankaj has 96 marbles and Arun has 63 marbles. How many marbles should run give Pankaj so that Pankaj will have twice as many marbles as Arun?

A)  \[9\]                     done clear

C)  \[7\]                                     done clear

question_answer 35) If \[\frac{3p+2}{5}-\frac{4p-3}{7}+\frac{p-1}{35}=4,\] find the  value of p.

A)  \[65\]                   done clear

B)         \[63\] done clear

C)  \[36\]                   done clear

D)         \[56\]   done clear

question_answer 36) The sum of five times a number and 13 is 48. What is the number?

A)  \[3\]                     done clear

B)         \[5\]    done clear

C)  \[7\]                       done clear

D)         \[9\] done clear

question_answer 37) Guru is 20 years older than his son. If the sum of their ages is 50 years, how old is his son?

A)  5 years                 done clear

B)         10 years done clear

C)  15 years                done clear

D)         20 years done clear

question_answer 38) If one-fourth of a number decreased by 12 is 30, what is the number?          

A)  \[168\]                   done clear

B)         \[186\] done clear

C)  \[148\]                 done clear

D)         \[184\] done clear

question_answer 39) lf \[C=\frac{5}{9}(F-32),\] what is the value F?

A)  \[\frac{5C}{9}-32\]          done clear

B)         \[\frac{9C}{5}-32\] done clear

C)  \[\frac{9C}{5}+32\]           done clear

D)         \[\frac{5C}{9}+32\] done clear

A)  1 gram                 done clear

B)         4 grams done clear

C)  5 grams                 done clear

D)         20 grams done clear

A)  \[26\]                   done clear

B)         \[22\]   done clear

C)  \[19\]                                   done clear

D)  \[16\] done clear

question_answer 42) The sum of two-thirds of a number and one-fifth of the same number is 13. Find the number.

A)  \[15\]                     done clear

C)  \[13\]                   done clear

D)         \[5\] done clear

question_answer 43) Evaluate  \[\frac{x-4}{3}-\frac{2x+1}{6}=\frac{5x+1}{2}\]

A)  \[\frac{3}{5}\]                   done clear

B)         \[\frac{4}{5}\] done clear

C)  \[\frac{5}{6}\]                   done clear

D)         \[\frac{-4}{5}\]   done clear

question_answer 44) The denominator of a fraction is 3 more than its numerator. If 2 is added to both the numerator and the denominator, the new fraction is equivalent to \[\frac{2}{3}\] What is the original fraction?

A)  \[\frac{3}{7}\]                   done clear

B)         \[\frac{4}{7}\]   done clear

C)  \[\frac{2}{3}\]                   done clear

D)         \[\frac{3}{5}\] done clear

question_answer 45) 144 beads were shared equally among some children. If there were 3 children fewer, each child would have 16 beads each. How many children were there?

A)  \[8\]                     done clear

B)         \[9\]    done clear

C)  \[12\]                                                     done clear

D)  \[11\] done clear

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Course: Class 7   >   Unit 4

• Unit test Simple Equations

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Simple Equations Class 7 MCQ Test (Online Available)

Free mcq test, table of content, simple equations test - 20.

Duration: 10 Mins

Maximum Marks: 10

Read the following instructions carefully.

1. The test contains 10 total questions.

2. Each question has 4 options out of which only one is correct .

3. You have to finish the test in 10 minutes.

4. You will be awarded 1 mark for each correct answer.

5. You can view your Score & Rank after submitting the test.

6. Check detailed Solution with explanation after submitting the test.

7. Rank is calculated on the basis of Marks Scored & Time

Simple Equations Test - 19

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Objective questions have become the norm now, those students who are studying in class 7 must be well versed with all kinds of MCQ Questions; therefore, the link to access Simple Equations MCQ Class 7 is mentioned here on this page.

Simple Equations Class 7 MCQ questions are curated by subject experts referring to the prescribed NCERT Class 7 Maths Book. Those students who have studied the lesson Simple Equations must practise the CBSE Class 7 MCQ Questions as it helps in deepening the understanding of the topics.

Class 7 Simple Equations MCQ with Answers Online

For the ease of students, the subject experts have simplified the practice process of the class 7 Simple Equations MCQ as they have solved each and every question given. The answers are detailed and easy to grasp. Students who want to practise the questions of Class 7 Simple Equations MCQ with Answers online can use the Selfstudys website.

The MCQ Questions of Simple Equations with answers are also given so that students can understand the methods of solving the concepts based questions. As well as, the solutions are helpful to understand where the students are making mistakes and need to improve.

How to Practise MCQ from NCERT Chapter Simple Equations?

There are several ways to practise the MCQ from NCERT Chapter Simple Equations; one is by solving questions from the chapter’s end and another is by using online medium. In this section, we have mentioned the steps to Practise MCQ from NCERT Chapter Simple Equations online.

• First of all open Selfstudys.com on your Smartphone, PC/Laptop

• Navigate to the CBSE by Tapping/clicking on the navigation bar or button 
• It will open a new lists where you can find MCQ Test - Click on that  

• Then, a new page will load containing the lists of classes; just Tap or click on Class 7. *In Smartphone, you may require to scroll the given classes name towards left.

• Now, after selecting the Class 7, the same page will reload, make sure you select the Maths to access the MCQ Questions of Class 7 Simple Equations.

Note: The online MCQ Questions of Simple Equations can’t be downloaded, those who want to access the PDF of Simple Equations MCQ can refer to the CBSE Class 7 MCQ PDF  section within the CBSE menu.

What is Simple Equations MCQ and How to Use it?

Since class 7 students are in their early stage of academics they may have questions regarding What is Simple Equations MCQ and How to Use it. So, the answer is MCQ Questions are objective questions which contain questions followed by 4 options where only one is considered the correct answer and remaining as a distraction. Why is it so, because MCQ questions are ideal to assess a student’s conceptual knowledge.

Those who want to use the Simple Equations MCQ can use this website to access the online MCQ questions to practise.

Top 5 Benefits of Simple Equations MCQ Class 7

Simple Equations MCQ Class 7 questions benefit students in several ways; however, here we have mentioned a total of 5 benefits that a student will get if they are using the MCQ from NCERT Chapter Simple Equations.

• Helps in Practising Questions: Sometimes, it's hard to get the questions to practise; therefore, the Selfstudys team has curated various sets of MCQ Questions of Class 7 Simple Equations. Having access to the objective questions of Class 7 Simple Equations helps students practising various questions for free of cost.
• Boosts the Critical Thinking Capability: The MCQs or objective questions must be answered in lesser time; therefore, those who will regularly solve Simple Equations MCQ Class 7 will benefit by having a great boost in the critical thinking capability as the questions are in the objective format which can be answered if one has a good command over the concepts of Simple Equations.
• Assists in Covering the Class 7 Maths Syllabus: Simple Equations is a chapter of Class 7 Maths and those who are going to solve the MCQs of Class 7 Simple Equations will be able to practise all the questions as per their Maths Syllabus.
• Helps in Exam Preparation: If a student solves the objective questions from class 7 Simple Equations, they will be able to be prepared for the annual examination too. It is because the questions that are asked in the online MCQ of Simple Equations are asked in the final exam question papers too.
• A Deeper Understanding of Simple Equations: All the important points that are discussed in Simple Equations must be memorised by students as it helps in deepening the understanding of the Delhi Sultans. One of the great benefits of solving Simple Equations MCQ Class 7 is that one can be thorough with the topics and can develop a deeper understanding of Simple Equations chapter of class 7.

Apply These Techniques To Better Answer the MCQ Questions of Simple Equations

Although, there is no wrong or right method to answer the MCQ Questions, those who are interested in knowing the techniques to better answer the MCQ Questions of Simple Equations can follow the below given methods.

Read the Question Carefully:  Questions in Simple Equations MCQ Class 7 can be asked from tricky to hard to understand. In this case, you must read the questions of Simple Equations MCQ carefully. By paying attention to the questions, it will help you connect the dots and assist you recall the studied concepts to answer the MCQs easily.

Eliminate Obviously Wrong Answers:  Many questions of Simple Equations MCQ Class 7 will be so familiar that you can be certain for the wrong answer but uncertain for the right answer, in that situation obviously eliminate wrong answers first. By eliminating irrelevant or incorrect answers, it will help you find the one correct answer from all the given four options.

Look for Clues in the Question:  As we have discussed the first technique is to read the questions carefully, it is vital for looking for the clues in the questions of Simple Equations MCQ Class 7. Every single question contains some kind of clues that help you answer them easily, but due to running out of time many don’t pay attention to it. In order to solve Simple Equations MCQ Class 7 by using this technique you may have to do a thorough practice of Class 7 MCQ Questions.

Use the Process of Elimination:  There is not much difference in the elimination method and eliminating the wrong answer (discussed in point number 2) first, but one difference that makes the elimination process different is you can eliminate the right or wrong answer first. 

This means that when you are confused between two options, you can separate them and then you try to focus on only those 2 options to find out the correct answer of Simple Equations MCQ Class 7. This elimination process works best in most of the scenarios.

Don't Spend too Much Time on One Question:  It is never a good idea to be rigid on one question and spend most of your time answering them. When you are practising Simple Equations MCQ Class 7 questions, you have limited time and you have to make sure that you use your time smartly to attempt all the questions as asked in the Simple Equations Class 7 MCQ.

Double-check your Answer:  Before submitting the Online test of Class 7 Simple Equations MCQ, you should double check your answer if the test time hasn't completed. When you do a double check of your answers, you may find some silly mistakes that you have made due to which you could have lost some marks. Therefore, be conscious and double check your answers before submitting Simple Equations MCQ Class 7.

*As per the Selfstudys Online MCQ Test instructions the time plays a crucial role in calculating your test rank so, be conscious when you use any single minute during your test.

Manage your Time:  Having great time management skills doesn’t only help you quickly answer the questions, but gives you the ability to save time to review the questions or in doing a last minute cross-checking. Thus, when you start solving Simple Equations MCQ Class 7 questions, try to manage time and some of your test time to review the answers you have ticked throughout the test. 

Apart from this, time management skills give you peace of mind and keep you calm.

Stay Calm to Recall Previously Studied Topics:  When you struggle to come up with the correct answer of Simple Equations MCQ Class 7 Questions try to stay calm as it will help you recall previously studied topics. Research says, being calm and relaxed helps in saving energy. Thus, staying calm while solving the MCQ from NCERT Chapter Simple Equations helps you be more focused and answer the questions efficiently. The saved energy can be channelized to increase the focus and concentration to better recall the topics and subtopics of Simple Equations.

There is a high possibility of having more techniques of Simple Equations MCQ Class 7 as mentioned, but these given methods work well in most of the cases.

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