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

Exercise- 4.1

1. Complete the last column of the table:

Ans:

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

a. \[n+5=19\left( n=1 \right)\]

Ans: $ n+5=19 $ 

Putting  $ n=1 $  in L.H.S.,

$ 1+5=6 $

$ \because L.H.S.\ne R.H.S.$ 

$ \therefore n=1 $ is not the solution of given equation.

b. $ 7n+5=19\left( n=-2 \right) $ 

Ans: $ 7n+5=19 $ 

Putting  $ n=-2 $  in L.H.S.,

$ 7\left( -2 \right)+5=-14+5=-9 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore n=-2 $ is not the solution of given equation.

c. $ 7n+5=19\left( n=2 \right) $ 

Ans: $ 7n+5=19 $ 

Putting  $ n=2 $  in L.H.S.,

$ 7\left( 2 \right)+5=14+5=19 $

$ \because L.H.S.=R.H.S.$

$ \therefore n=2 $ is the solution of given equation.

d. $ 4p-3=13\left( p=1 \right) $ 

Ans: $ 4p-3=13 $ 

Putting  $ p=1 $  in L.H.S.,

$ 4\left( 1 \right)-3=4-3=1 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore $ $ p=1 $  is not the solution of given equation.

e. $ 4p-3=13\left( p=-4 \right) $ 

Ans: $ 4p-3=13 $ 

Putting  $ p=-4 $  in L.H.S.,

$ 4\left( -4 \right)-3=-16-3=-19 $

$ \because L.H.S.\ne R.H.S. $

$ \therefore $ $ p=-4 $  is not the solution of given equation.

f. $ 4p-3=13\left( p=0 \right) $ 

Ans: $ 4p-3=13 $ 

Putting  $ p=0 $  in L.H.S.,

$ 4\left( 0 \right)-3=0-3=-3 $ 

$ \because L.H.S.\ne R.H.S. $

$ \therefore $ $ p=0 $ is not the solution of given equation.

3. Solve the following equations by trial and error method:

a. $ 5p+2=17 $ 

Ans: Putting  $ p=-3 $ in L.H.S.  $ 5(-3)+2=-15+2=-13 $     $ \because -13\ne 17 $ 

Therefore, \[p=-3\text{ }\] is not the solution. 

Putting  $ p=-2 $ in L.H.S.  $ 5(-2)+2=-10+2=-8 $    $ \because -8\ne 17\quad  $ 

Therefore,  $ p=-2 $  is not the solution. 

Putting  $ p=-1 $ in L.H.S. $ 5(-1)+2=-5+2=-3 $   $ \because -3\ne 17 $ 

Therefore,  $ p=-1 $ is not the solution.

Putting  $ p=0 $ in L.H.S.  $ 5(0)+2=0+2=2 $    $ \because 2\ne 17\quad  $ 

Therefore,  $ p=0 $ is not the solution.

Putting  $ p=1 $ in L.H.S.  $ 5(1)+2=5+2=7 $    $ \because 7\ne 17\quad  $ 

Therefore,  $ p=1 $ is not the solution. 

Putting  $ p=2 $  in L.H.S.  $ 5(2)+2=10+2=12 $   $ \because 12\ne 17\quad  $ 

Therefore,  $ p=2 $ is not the solution.

Putting  $ p=3 $ in L.H.S.  $ 5(3)+2=15+2=17 $   $ \because 17=17\quad  $ 

Therefore,  $ p=3 $ is the solution.

b. $ 3m-14=4 $ 

Ans: Putting  $ m=-2 $ in L.H.S.  $ \quad 3(-2)-14=-6-14=-20\,\,\,\because -20\ne 4\quad  $ 

Therefore,  $ m=-2 $  is not the solution. 

Putting  $ m=-1 $ in L.H.S.  $ \quad 3(-1)-14=-3-14=-17\,\,\,\,\because -17\ne 4\quad  $ 

Therefore,  $ m=-1 $  is not the solution.

Putting  $ m=0 $  in L.H.S.  $ 3(0)-14-0-14=-14 $   $ \because -14\ne 4\quad  $ 

Therefore,  $ m=0 $  is not the solution.

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

$ \because -11\ne 4\quad  $ Therefore,  $ m=1 $ is not the solution. 

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

$ \because -8\ne 4\quad  $ Therefore,  $ m=2 $ is not the solution. 

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

$ \because -5\ne 4\quad  $ Therefore,  $ m=3 $ is not the solution. 

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

$ \because -2\ne 4\quad  $ Therefore,  $ m=4 $ is not the solution. 

Putting  $ m=5 $ in L.H.S.  $ \quad 3(5)-14=15-14=1 $    $ \because 1\ne 4 $ 

Therefore,  $ m=5 $  is not the solution.

4. Write equations for the following statements: 

i. The sum of numbers x and 4 is 9. 

Ans: $ x+4=9 $ 

ii. 2 subtracted from y is 8. 

Ans: $ y-2=8 $ 

iii. Ten times a is 70. 

Ans: $ 10a=70 $ 

iv. The number b divided by 5 gives 6. 

Ans: $ \dfrac{b}{5}=6 $ 

v. Three-fourth of t is 15. 

Ans: $ \dfrac{3}{4}t=15 $ 

vi. Seven times m plus 7 gets you 77. 

Ans: $ 7m+7=77 $ 

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

Ans: $ \dfrac{x}{4}-4=4 $ 

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

Ans: $ 6y-6=60 $ 

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

Ans: $ \dfrac{z}{3}+3=30 $ 

5. Write the following equations in statement form:

i. $ p+4=15 $ 

Ans: The sum of numbers p and 4 is 15.

ii. $ m-7=3 $ 

Ans: 7 subtracted from m is 3

iii. $ 2m=7 $ 

Ans: Two times m is 7.

iv. $ \dfrac{m}{5}=3 $ 

Ans: The number m is divided by 5 gives 3.

v. $ \dfrac{3m}{5}=6 $ 

Ans: Three-fifth of the number m is 6.

vi. $ 3p+4=25 $ 

Ans: Three times p plus 4 gets 25.

vii. $ 4p-2=18 $ 

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

viii. $ \dfrac{p}{2}+2=8 $ 

Ans: If you added 2 to half is p, you get 8.

Set up an equation in the following cases:

1. Irfan says that he has \[\mathbf{7}\] marbles more than five times the marbles Permit has. Irfan has \[\mathbf{37}\] marbles. (Tale m to be the number of Permit’s marbles.) 

Ans: Let  $ m $ be the number of Permit’s marbles.

$ \therefore \quad 5m+7=37 $ 

2. Laxmi’s father is \[\mathbf{49}\] years old. He is \[\mathbf{4}\] years older than three times Laxmi’s age. (Take Laxmi’s age to be y years.) 

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

$ \therefore \quad 3y+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\[\mathbf{7}\]. The highest score is\[\mathbf{87}\]. (Take the lowest score to be l. ) 

Ans: Let the lowest score be \[l.\]   

$ \therefore \quad 2l+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 \[\mathbf{180}{}^\circ \]

Ans: Let the base angle of the isosceles triangle be \[b,\]so vertex angle \[=2\text{ }b.\] $ \therefore 2b+b+b={{180}^{{}^\circ }}\Rightarrow 4b={{180}^{{}^\circ }}\quad  $  [Angle sum property of a $ \Delta  $ ]

Chapter 4 – Simple Equations

The Maths Class 7 Chapter 4 Exercise 4.1 has the following topics to be covered.

• A mind-reading game

• Setting up of an equation

• Review of what we know

• What equation is?

• Exercise 4.1

A Mind-Reading Game

In this section of Chapter 4, there will be a short introduction of what the students have learned in the algebraic section in the previous year. This is where the concept of variables will be recalled by the teachers. This segment will deal with the recollection of variables concepts with a guessing game with the students. The solutions for Class 7 Maths Chapter 4 Exercise 4.1 will help the students to understand what this section was all about. Our experts have formulated Class 7 Maths Chapter 4 Exercise 4.1 Solutions by following the specific guidelines showcased by NCERT so that the students can correlate and understand the concepts without any hassle.

Setting Up of an Equation

On proceeding to the next segment of this chapter, the students will find a complete knowledge of what equation stands for. This part of the chapter will describe the students how an equation is formatted. The use of one variable will be shown by the teacher to set up an equation. In fact, the teacher will also depict the problems prescribed by the NCERT book for a better understanding of the concept. Hence, the Class 7 NCERT Maths Ex 4.1 Solutions will be of utmost importance in this aspect. On fabricating the solutions in a prominent way, our experts have prepared a unique platform for the students. They can easily understand how to approach the problems in this section and develop their concepts properly. Download the NCERT Solutions for Class 7 Maths Chapter 4 Exercise 4.1 in PDF format so that you can refer to it anytime when you face any issue to form equations.

This section will discuss more how to perform arithmetic operations with equations. It will be a step ahead to the next level of forming an equation and performing addition, subtraction, multiplication, and division between numbers and equations. By going through this section, a student will be able to develop an excellent concept of how variables behave in an equation and what happens to them when a mathematical operation is performed. 

Review of what we know

The 3rd segment of NCERT Maths Class 7 Chapter 4 Exercise 4.1 shows what the students have learned so far from the teacher. This section will describe what the students have gained as advanced concepts on equation formation with a single variable. The concept of variables will be explained accordingly. A student will also find how the variables are represented in an equation. In fact, every sign and symbol will be explained in a proper way. This will help the students to inculcate the concept properly so that they can proceed to the next level easily.

The review of mathematical operations will be dealt with in the previous section with examples. In this section, the operations will be recalled with more examples and some deeper insights. The Maths NCERT Solutions Class 7 Chapter 4 Exercise 4.1 for this section and the upcoming exercise will deliver the ideal ways to deal with the problems. A student will be able to understand the right way to find the clues and use them to find the values of variables mentioned in the equation.

What equation is?

The 4th segment of Class 7 Chapter 4 Exercise 4.1 describes all about what equation stands for and its use. There are two segments in an equation. The left hand and the right-hand side are connected by an ‘equal’ sign. It means that for a particular value, both sides will be equal to each other. This is where the advancement in the concept building is noticed. In this part, both sides will have variables. Now that you have learned how to operate mathematical signs with variables, you will solve the equations using your algebraic skills. 

The examples in this section will also depict how the students should understand a problem and approach it. The examples will also tell how to form an equation from the provided information and solve it to find out the correct answer. On availing our NCERT Class 7 Maths Chapter 4 Exercise 4.1 solutions, you will find the ideal ways to approach a problem regarding equation formation and performing mathematical operations to solve them. 

All the examples in this section will deliver knowledge regarding the use of equations with one variable and how to form them. The stepwise presentation of the solution for all problems in Class 7th Maths Chapter 4 Exercise 4.1 will develop a good hold on the new concepts.

About Vedantu

The experts of Vedantu formulated the NCERT solutions for Class 7 Maths Chapter 4 Exercise 4.1 Online PDF in such a way that every student can understand the new concept of simple equations and can solve the problems in exam papers easily. You can download the PDF file and refer the techniques to solve the problems of Exercise 4.1 in Chapter 4 as per their convenience. A solid platform will be constructed regarding the initiation of simple equations. Ace the exams by solving the problems related to this chapter by using our solutions.