# Important Questions Class 7 Maths Chapter 4

## Important Questions Class 7 Mathematics Chapter 4 – Simple Equations

Mathematics is a very important subject that you study in school. It is not a mere subject in school, but we need Mathematics to solve our real-life problems. In your early classes, you have learned about variables.

In this chapter, you will learn about simple equations. Simple equations include only one variable and the value of the variable is calculated by solving the equations. In higher classes, equations are largely used with multiple variables, and students must build the concepts. They must practice questions regularly to clear their doubts.

Extramarks is a well-known company providing study materials related to CBSE and NCERT. Our experts believe in practice, and for this purpose, they have made the Important Questions Class 7 Mathematics Chapter 4. They collected different types of questions from various sources. They solved the questions like the textbook exercises, CBSE sample papers, CBSE question papers of past years, NCERT Exemplars and important reference books. Experienced professionals have further checked the answers to ensure the best quality of the content.

Extramarks provide all the necessary study materials for your exam preparation. You can download these study materials after registering on our official website, and you will find the CBSE syllabus, CBSE question papers of past years, and CBSE sample papers. CBSE extra questions, CBSE revision notes, NCERT books, NCERT Exemplars, NCERT solutions, NCERT important questions, vital formulas and many more.

## Important Questions Class 7 Mathematics Chapter 4 with Solutions

Practice is very important for students to score better in exams. The experts of Extramarks understand the importance of practice and they have made this question series. They have collected the question from various sources such as the textbook exercises, CBSE sample papers, CBSE question papers of past years, NCERT Exemplars and important reference books. Thus, the important questions Class 7 Mathematics Chapter 4 will help students to practice the question regularly. The important questions are given below-

Question 1. Check whether the given value in the brackets is the correct solution to the below-given equation or not:

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

LHS = n + 5

By substituting the value of n = 1

Then we get,

LHS = n + 5

= 1 + 5

= 6

By comparing the LHS and RHS

6 ≠ 19

LHS ≠ RHS

Hence we get that the value of n = 1 is not the solution to the given equation above n + 5 = 19.

(b) 7n + 5 = 19 (n = – 2)

LHS = 7n + 5

Now, by substituting the value of n = -2

Then we get,

LHS = 7n + 5

= (7 × (-2)) + 5

= – 14 + 5

= – 9

By comparing the LHS and RHS

-9 ≠ 19

LHS ≠ RHS

Hence we know that the value of n = -2 is not the solution to the above-given equation 7n + 5 = 19.

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

LHS = 7n + 5

By substituting the value of n as = 2

Then,

LHS = 7n + 5

= (7 × (2)) + 5

= 14 + 5

= 19

By comparing both the LHS and the RHS

19 = 19

LHS = RHS

Hence, the value of n = 2 is the solution to the given equation is 7n + 5 = 19.

(d) 4p – 3 = 13 (p = 1)

LHS = 4p – 3

By substituting the value of p = 1

Then,

LHS = 4p – 3

= (4 × 1) – 3

= 4 – 3

= 1

By comparing both the LHS and the RHS

1 ≠ 13

LHS ≠ RHS

Hence then, the value of p = 1 is not the solution to the above-given equation 4p – 3 = 13.

(e) 4p – 3 = 13 (p = – 4)

LHS = 4p – 3

By substituting the value of p as = – 4

Then,

LHS = 4p – 3

= (4 × (-4)) – 3

= -16 – 3

= -19

By comparing both the LHS and the RHS

-19 ≠ 13

LHS ≠ RHS

Hence then, the value of p = -4 is not the solution to the above-given equation 4p – 3 = 13.

(f) 4p – 3 = 13 (p = 0)

LHS = 4p – 3

By substituting the value of p = 0

Then,

LHS = 4p – 3

= (4 × 0) – 3

= 0 – 3

= -3

By comparing the LHS and RHS

– 3 ≠ 13

LHS ≠ RHS

Hence then, the value of p = 0 is not the solution to the above-given equation 4p – 3 = 13.

Question 2. Solve the following equations by the trial and error method:

(i) 5p + 2 = 17

LHS = 5p + 2

By substituting the value of p as = 0

Then,

LHS = 5p + 2

= (5 × 0) + 2

= 0 + 2

= 2

By comparing both the LHS and RHS

2 ≠ 17

LHS ≠ RHS

Hence then, the value of p = 0 is not the solution to the above-given equation.

Let, p = 1

LHS = 5p + 2

= (5 × 1) + 2

= 5 + 2

= 7

By comparing the LHS and the RHS

7 ≠ 17

LHS ≠ RHS

Hence then, the value of p = 1 is not the solution to the above-given equation.

Let, p = 2

LHS = 5p + 2

= (5 × 2) + 2

= 10 + 2

= 12

By comparing both the LHS and the RHS

12 ≠ 17

LHS ≠ RHS

Hence then, the value of p = 2 is not the solution to the above-given equation.

Let, p = 3

LHS = 5p + 2

= (5 × 3) + 2

= 15 + 2

= 17

By comparing both the LHS and the RHS

17 = 17

LHS = RHS

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

(ii) 3m – 14 = 4

LHS = 3m – 14

By substituting the value of m as = 3

Then,

LHS = 3m – 14

= (3 × 3) – 14

= 9 – 14

= – 5

By comparing both the LHS and the RHS

-5 ≠ 4

LHS ≠ RHS

Hence then, the value of m = 3 is not the solution to the above-given equation.

Let, m = 4

LHS = 3m – 14

= (3 × 4) – 14

= 12 – 14

= – 2

By comparing both the LHS and the RHS

-2 ≠ 4

LHS ≠ RHS

Hence then, the value of m = 4 is not the solution to the above-given equation.

Let, m = 5

LHS = 3m – 14

= (3 × 5) – 14

= 15 – 14

= 1

By comparing both the LHS and the RHS

1 ≠ 4

LHS ≠ RHS

Hence then, the value of m = 5 is not the solution to the above-given equation.

Let, m = 6

LHS = 3m – 14

= (3 × 6) – 14

= 18 – 14

= 4

By comparing both the LHS and RHS

4 = 4

LHS = RHS

Hence, the value of m = 6 is the solution to the above-given equation.

Question 3. Write the equations for the following statements:

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

The above statement can also be written in the equation form as,

= x + 4 = 9

(ii) 2 subtracted from y is 8.

The above statement can also be written in the equation form as,

= y – 2 = 8

(iii) Ten times a is 70.

The above statement can also be written in the equation form,

= 10a = 70

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

The above statement can also be written in the equation form as,

= (b/5) = 6

(v) Three-fourths of t is 15.

The above statement can also be written in the equation form,

= ¾t = 15

(vi) Seven times m plus seven will get you 77.

The above statement can also be written in the equation form,

Seven times m will be 7m

= 7m + 7 = 77

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

The above statement can also be written in the equation form,

One-fourth of the number x is x/4.

= x/4 – 4 = 4

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

The above statement can also be written in the equation form as,

Six times y is 6y.

= 6y – 6 = 60

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

The above statement can also be written in the equation form as,

One-third of z is z/3.

= 3 + z/3 = 30

Question 4. Set up an equation form in the following cases given below:

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

From the question, it is given that,

Number of Permit’s marbles = m

Then,

Irfan has seven marbles, more than five times the marbles Permit has

= 5 × Number of Permit’s marbles + 7 = Total number of the marbles Irfan having

= (5 × m) + 7 = 37

= 5m + 7 = 37

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

From the question, it is given that,

Let Laxmi’s age be = y years old.

Then,

Lakshmi’s father is four years older than three times her age.

= 3 × Laxmi’s age + 4 = Age of her father

= (3 × y) + 4 = 49

= 3y + 4 = 49

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

From the question, it is given that,

Highest score in the class = 87

Let the lowest score be equal to the l

= 2 × Lowest score + 7 = Highest score in class

= (2 × l) + 7 = 87

= 2l + 7 = 87

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

From the above question, it is given that,

We know that the total sum of angles of a triangle is 180o.

Let the base angle be b

Then,

Vertex angle = 2 × base angle = 2b

= b + b + 2b = 180o

= 4b = 180o

Question 5. Write the following equations in statement forms:

(i) p + 4 = 15

The sum of the numbers p and 4 is 15.

(ii) m – 7 = 3

Seven subtracted from m is 3.

(iii) 2m = 7

Twice of number m is 7.

(iv) m/5 = 3

The number m divided by the number 5 gives 3.

(v) (3m)/5 = 6

Three-fifth of m is 6.

(vi) 3p + 4 = 25

Three times p plus four will give you 25.

(vii) 4p – 2 = 18

Four times p minus 2 gives you the number 18.

(viii) p/2 + 2 = 8

If you add half of the number p to 2, you get 8.

Question 6. Give the steps which are used to separate the variable and then solve the equation:

(a) 3n – 2 = 46

First, we have to add two to both the side of the equation,

= 3n – 2 + 2 = 46 + 2

= 3n = 48

Now,

We have to divide both of these sides of the equation by 3,

Then, we get,

= 3n/3 = 48/3

= n = 16

(b) 5m + 7 = 17

First, we have to subtract seven from the both sides of the equation,

= 5m + 7 – 7 = 17 – 7

= 5m = 10

Now,

We have to divide both these sides of the equation by 5,

= 5m/5 = 10/5

= m = 2

(c) 20p/3 = 40

First, we have to multiply both of these sides of the equation by 3,

Then, we get,

= (20p/3) × 3 = 40 × 3

= 20p = 120

Now,

We have to divide both of these sides of the equation by 20,

= 20p/20 = 120/20

= p = 6

(d) 3p/10 = 6

First, we have to multiply both of these sides of the equation by 10,

= (3p/10) × 10 = 6 × 10

= 3p = 60

Now,

We have to divide both of these sides of the equation by 3,

Then, we get,

= 3p/3 = 60/3

= p = 20

Question 7. Solve the following equations:

(a) 10p = 100

Now,

We have to divide both of these sides of the equation by 10,

Then, we get,

= 10p/10 = 100/10

= p = 10

(b) 10p + 10 = 100

Firstly we have to subtract 10 from both sides of the following equation,

Then, we get,

= 10p + 10 – 10 = 100 – 10

= 10p = 90

Now,

We have to divide both of these sides of the equation by 10,

Then, we get,

= 10p/10 = 90/10

= p = 9

(c) p/4 = 5

Now,

We have to multiply both of these sides of the equation by 4,

Then, we get,

= p/4 × 4 = 5 × 4

= p = 20

(d) – p/3 = 5

Now,

We have to multiply both of these sides of the equation by – 3,

= – p/3 × (- 3) = 5 × (- 3)

= p = – 15

(e) 3p/4 = 6

First, we have to multiply both of these sides of the equation by 4,

Then, we get,

= (3p/4) × (4) = 6 × 4

= 3p = 24

Now,

We have to divide both of these sides of the equation by 3,

Then, we get,

= 3p/3 = 24/3

= p = 8

(f) 3s = – 9

Now,

We have to divide both of these sides of the equation by 3,

Then, we get,

= 3s/3 = -9/3

= s = -3

(g) 3s + 12 = 0

First, we have to subtract the number 12 from the both sides of the equation,

Then, we get,

= 3s + 12 – 12 = 0 – 12

= 3s = -12

Now,

We have to divide both of these sides of the equation by 3,

Then, we get,

= 3s/3 = -12/3

= s = – 4

(h) 3s = 0

Now,

We have to divide both of these sides of the equation by 3,

Then, we get,

= 3s/3 = 0/3

= s = 0

(i) 2q = 6

Now,

We have to divide both of these sides of the equation by 2,

Then, we get,

= 2q/2 = 6/2

= q = 3

(j) 2q – 6 = 0

Firstly we have to add 6 to both the sides of the following equation,

Then, we get,

= 2q – 6 + 6 = 0 + 6

= 2q = 6

Now,

We have to divide both of these sides of the equation by 2,

Then, we get,

= 2q/2 = 6/2

= q = 3

(k) 2q + 6 = 0

First, we have to subtract the number 6 from the both sides of the following equation,

Then, we get,

= 2q + 6 – 6 = 0 – 6

= 2q = – 6

Now,

We have to divide both of these sides of the equation by the number 2,

Then, we get,

= 2q/2 = – 6/2

= q = – 3

(l) 2q + 6 = 12

First, we have to subtract the number 6 to the both sides of the following equation,

Then, we get,

= 2q + 6 – 6 = 12 – 6

= 2q = 6

Now,

We have to divide both of these sides of the equation by 2,

Then, we get,

= 2q/2 = 6/2

= q = 3

Question 8. Solve the following equations:

(a) 2y + (5/2) = (37/2)

By transposing the fraction (5/2) from LHS to RHS, so it becomes -5/2

Then,

= 2y = (37/2) – (5/2)

= 2y = (37-5)/2

= 2y = 32/2

Now,

Divide both the side by the number 2,

= 2y/2 = (32/2)/2

= y = (32/2) × (1/2)

= y = 32/4

= y = 8

(b) 5t + 28 = 10

By transposing 28 from the LHS to RHS, it becomes -28

Then,

= 5t = 10 – 28

= 5t = – 18

Now,

Dividing both these sides by 5,

= 5t/5= -18/5

= t = -18/5

(c) (a/5) + 3 = 2

By transposing three from the LHS to RHS, it becomes -3

Then,

= a/5 = 2 – 3

= a/5 = – 1

Now,

Multiply both these sides by 5,

= (a/5) × 5= -1 × 5

= a = -5

(d) (q/4) + 7 = 5

By transposing the number 7 from the LHS to RHS, it becomes -7

Then,

= q/4 = 5 – 7

= q/4 = – 2

Now,

Multiply both these sides by 4,

= (q/4) × 4= -2 × 4

= a = -8

(e) (5/2) x = -5

First, we have to multiply both these sides by 2,

= (5x/2) × 2 = – 5 × 2

= 5x = – 10

Now,

We have to divide both these sides by 5,

Then we get,

= 5x/5 = -10/5

= x = -2

(f) (5/2) x = 25/4

First, we have to multiply both these sides by 2,

= (5x/2) × 2 = (25/4) × 2

= 5x = (25/2)

Now,

We have to divide both these sides by 5,

Then we get,

= 5x/5 = (25/2)/5

= x = (25/2) × (1/5)

= x = (5/2)

Question 9. Solve the following equations:

(a) 2(x + 4) = 12

Let us divide both these sides by 2,

= (2(x + 4))/2 = 12/2

= x + 4 = 6

By transposing four from the LHS to RHS, it becomes -4

= x = 6 – 4

= x = 2

(b) 3(n – 5) = 21

Let us divide both these sides by 3,

= (3(n – 5))/3 = 21/3

= n – 5 = 7

By transposing -5 from the LHS to the RHS, it becomes 5

= n = 7 + 5

= n = 12

(c) 3(n – 5) = – 21

Let us divide both these sides by 3,

= (3(n – 5))/3 = – 21/3

= n – 5 = -7

By transposing -5 from the LHS to the RHS, it becomes 5

= n = – 7 + 5

= n = – 2

(d) – 4(2 + x) = 8

Let us divide both these sides by -4,

= (-4(2 + x))/ (-4) = 8/ (-4)

= 2 + x = -2

By transposing two from the LHS to the RHS, it becomes – 2

= x = -2 – 2

= x = – 4

(e) 4(2 – x) = 8

Let us divide both these sides by 4,

= (4(2 – x))/ 4 = 8/ 4

= 2 – x = 2

By transposing the number 2 from the LHS to the RHS, it becomes – 2

= – x = 2 – 2

= – x = 0

= x = 0

Question 10. Solve the following equations:

(a) 4 = 5(p – 2)

Let us divide both these sides by 5,

= 4/5 = (5(p – 2))/5

= 4/5 = p -2

By transposing – 2 from the RHS to the LHS, it becomes 2

= (4/5) + 2 = p

= (4 + 10)/ 5 = p

= p = 14/5

(b) – 4 = 5(p – 2)

Let us divide both these sides by 5,

= – 4/5 = (5(p – 2))/5

= – 4/5 = p -2

Now by transposing – 2 from the RHS to the LHS, it becomes 2

= – (4/5) + 2 = p

= (- 4 + 10)/ 5 = p

= p = 6/5

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

By transposing four from the RHS to the LHS, it becomes – 4

= 16 – 4 = 3(t + 2)

= 12 = 3(t + 2)

Let us divide both these side by 3,

= 12/3 = (3(t + 2))/ 3

= 4 = t + 2

By transposing 2 from the RHS to the LHS it becomes – 2

= 4 – 2 = t

= t = 2

(d) 4 + 5(p – 1) =34

By transposing 4 from the LHS to the RHS it becomes – 4

= 5(p – 1) = 34 – 4

= 5(p – 1) = 30

Let us divide both these side by 5,

= (5(p – 1))/ 5 = 30/5

= p – 1 = 6

By transposing – 1 from the RHS to the LHS it becomes 1

= p = 6 + 1

= p = 7

(e) 0 = 16 + 4(m – 6)

By transposing 16 from the RHS to the LHS it becomes – 16

= 0 – 16 = 4(m – 6)

= – 16 = 4(m – 6)

Let us divide both these side by the number 4,

= – 16/4 = (4(m – 6))/ 4

= – 4 = m – 6

Now by transposing – 6 from the RHS to the LHS it becomes 6

= – 4 + 6 = m

= m = 2

Question 11. (a) Construct 3 equations starting with x = 2

First equation is,

Multiply both these side by 6

= 6x = 12 … [equation 1]

Second equation is,

Subtracting the number 4 from both side,

= 6x – 4 = 12 -4

= 6x – 4 = 8 … [equation 2]

Third equation is,

Divide both these sides by 6

= (6x/6) – (4/6) = (8/6)

= x – (4/6) = (8/6) … [equation 3]

First equation is,

Multiply both these side by 5

= 5x = -10 … [equation 1]

Second equation is,

Subtracting the number 3 from both side,

= 5x – 3 = – 10 – 3

= 5x – 3 = – 13 … [equation 2]

Third equation is,

Dividing both these side by 2

= (5x/2) – (3/2) = (-13/2) … [equation 3]

### Benefits of Solving Important Questions Class 7 Mathematics Chapter 4

Practice is very important for students. It helps them in several ways. It not only helps them to clear their doubts but also boosts their confidence. Students need more than textbook exercises and must get help from other sources. The experts recognise the need for practice and have made this question series to help students. They have also solved the questions, and there will be multiple benefits of solving the Important Questions Class 7 Mathematics Chapter 4. These are as follows-

• The textbook exercises have limited questions and students need more questions to improve their preparation. But it will be tiresome for them to collect questions from different sources. The experts have made the task easier for them. They collected the questions from sources such as textbook exercises, CBSE sample papers, CBSE question papers of past years and important reference books. So, students will find different types of questions in the Class 7 Mathematics Chapter 4 Important Questions, which will help them score better in exams.
• The experts have not only collected the questions but have solved the questions too. They have given a step-by-step explanation of every question so that students can easily understand the problems. Thus, students can follow the solutions provided by the experts if they cannot solve the questions. Furthermore, they can check their answers with the experts’ answers. Experienced professionals have further checked the answers to ensure the best quality of the content. So, the Mathematics Class 7 Chapter 4 Important Questions will help them to score better in the exams.
• Practice is very important for students. Many students fear mathematics simply because they need help understanding the subject matter. Practice can help them to understand the subject matter because they will solve more sums. Slowly they can build an interest in the subject and become better at Mathematics. Thus, the Chapter 4 Class 7 Mathematics Important Questions will help them to make the concepts. Therefore, it will boost their confidence and help them score better in exams.

Extramarks is a well-known company in India providing study materials to students. We provide all the important study materials related to CBSE and NCERT, and students can download these articles after registering on our official website. We provide CBSE syllabus, CBSE extra questions, CBSE question papers of past years, CBSE sample papers, CBSE revision notes, NCERT books, NCERT exemplars, NCERT important questions, NCERT exemplars, NCERT solutions, NCERT important questions, vital formulas and many more. Like the Important Questions Class 7 Mathematics Chapter 4, you will also find important questions for other chapters. Links to the study materials are given below-

Q.1 10 more than twice a number x is 83, it is represented as ___________.

Marks:1
1. 10 – 2x = 83

2. 10 + 2x = 83

3. 10 + x = 83

4. 20 + x = 83

Ans

2. 10 + 2x = 83

Explanation

Twice a number x = 2x
10 more than twice a number x = 10 + 2x
It is given that
10 + 2x = 83, which is the required equation.

Q.2 The sum of two consecutive even numbers is 114.

Which one of the following equations shows the given situation?

Marks:1
1. x + 2 = 114

2. 2x + 1 = 114

3. 2x + 2 = 114

4. 2x + 3 = 114

Ans

3. 2x + 2 = 114

Explanation

Let the two consecutive even numbers be x and (x + 2).
Given,
x + (x + 2) = 114
or, 2x + 2 = 114, which is the required equation.

Q.3 If an integer is 5 more than the other integer and their sum is 71. What are the values of these integers?

Marks:1
1. 30 and 41

2. 33 and 38

3. 35 and 36

4. 31 and 40

Ans

2. 33 and 38

Explanation

Let the first integer be x.
Then, other integer = x + 5
Now, x + (x + 5) = 71
or, 2x + 5 = 33

or, x = 33
The other integer is
x + 5 = 38
Hence, required integers are 33 and 38.

Q.4 Give one example of an equation whose solution is 5.

Marks:1
Ans

Let x = 5
On multiplying both sides by 2, we get
2x = 10
On adding 3 to both sides, we get
2x + 3 = 10 + 3
Or 2x + 3 = 13 which is one of the equations whose solution is 5.

Q.5 Set up an equation for the following cases:
1. The age of Kanika is four times that of her daughter. The difference between their ages is 27 years.
2. The interest received by Karim is 30 more than that of Ramesh. The total interest received by them is 70.

Marks:3
Ans

1. Let the age of Kanika’s daughter be x years
So, age of Kanika = 4x years
According to the statement, the difference between their ages is 27 years,
i.e., 4x – x = 27, which is the required equation.
2. Let interest received by Ramesh = x
So, interest received by Karim = (x + 30)
According to the statement, the total interest received is 70
i.e., x + (x+30) = 70, which is the required equation.