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

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

Edited By Ramraj Saini | Updated on Feb 07, 2024 05:00 PM IST

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - In this chapter, students will be introduced by a new concept 'Equation'. This topic not only builds knowledge of algebra for maths students of class 7 but also helps to develop analytical thinking. Solutions of NCERT Class 7 Maths chapter 4 Simple Equations have 4 exercises with 18 questions in them. The NCERT solutions for Maths chapter 4 Simple Equations class 7 also discuss some topic-wise questions.

This Story also Contains
  1. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Formulae
  2. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Points
  3. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation
  4. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation (Intext Questions and Exercise)
  5. NCERT solutions for maths chapter 4 simple equation class 7 topic 4.7
  6. NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.1
  7. NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.2
  8. NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.3
  9. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4
  10. More About Simple Equations Class 6 Maths NCERT Chapter 4
  11. Simple Equations Class 6 Maths Chapter 4-Topic
  12. NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - Download PDF
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations - Download PDF

The NCERT Solutions can be extremely helpful for the Class 7 Maths students to understand the basics of this chapter and to clear all their doubts easily. Students can use NCERT Solutions for Class 7 as worksheets to prepare for their CBSE final exams. Here you will get solutions to all four exercises of this NCERT chapter.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Formulae

Balancing Equation:

LHS = RHS

(LHS ± a) = (RHS ± a)

(LHS × b) = (RHS × b)

(LHS ÷ c) = (RHS ÷ c)

Solution of an Equation: For a variable 'x', if LHS(x) = RHS(x), then 'x' is a solution.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation - Important Points

Variable: Something that can vary, Its value is not fixed such as x, y, z, p, q, r, etc.

Equation: It is a mathematical statement that uses an equal sign(=) to show that the value of the expression to the left of the sign (LHS) is equal to the value of the expression to the right of the sign (RHS).

Interchangeability: If the LHS and RHS are interchanged, the equation remains the same.

Balanced equation: the value of LHS remains equal to the value of the RHS. If 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.

Solution of an equation: The value of the variable for which the equation is satisfied ( for that variable, LHS = RHS ).

Transposing: Transposing of a number has the same effect as adding the same number to both sides or subtracting the same number from both sides of the equation.

Free download NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation PDF for CBSE Exam.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation

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NCERT Solutions for Class 7 Maths Chapter 4 Simple Equation (Intext Questions and Exercise)

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

Q. The value of the expression (10y – 20) depends on the value of y. Verify this by giving five different values to y and finding for each y the value of (10 y – 20). From the different values of (10y – 20) you obtain, do you see a solution to 10y – 20 = 50? If there is no solution, try giving more values to y and find whether the condition 10y – 20 = 50 is met.

Answer:

(i) Let y = 2 . We have :

10y\ -\ 20\ =\ 10(2)\ -\ 20\ =\ 0

(ii) Let y = 3 . We have :

10y\ -\ 20\ =\ 10(3)\ -\ 20\ =\ 10

(iii) Let y = 4 . We have :

10y\ -\ 20\ =\ 10(4)\ -\ 20\ =\ 20

(iv) Let y = 5 . We have :

10y\ -\ 20\ =\ 10(5)\ -\ 20\ =\ 30

(v) Let y = 6 . We have :

10y\ -\ 20\ =\ 10(6)\ -\ 20\ =\ 40

Hence 10y - 20 depends upon y.

Now, consider 10y\ -\ 20\ =\ 50

Transpose - 20 to the RHS :

10y\ =\ 50\ +\ 20\ =\ 70

or y\ =\ 7

NCERT solutions for maths chapter 4 simple equation class 7 topic 4.7

(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?

(ii) What is that number one third of which added to 5 gives 8?

Answer:

(i) Let the number be n.

Then according to the question, we have :

6n\ -\ 5\ =\ 7

or 6n\ =\ 7\ +\ 5\ =\ 12

or n\ =\ 2


(ii) Let the number be x.

Then according to the question,

\frac{x}{3}\ +\ 5\ =\ 8

\frac{x}{3}\ =\ 8\ -\ 5\ =\ 3

x\ =\ 9

Q. There are two types of boxes containing mangoes. Each box of the larger type contains 4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type. Each larger box contains 100 mangoes. Find the number of mangoes contained in the smaller box?

Answer:

Let the number of mangoes in the smaller box be n.

Then according to the question, we have :

8n\ +\ 4\ =\ 100

or 8n\ =\ 100\ -\ 4\ =\ 96

or n\ =\ 12

Hence the number of mangoes in the smaller box is 12.

NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.1

1. Complete the last column of the table.

1643866228721

Answer:

The table is shown below:-

1643866253463

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) 7n + 5 = 19 (n = – 2)

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

(e) 4p – 3 = 13 (p = – 4)

(f) 4p – 3 = 13 (p = 0)

Answer:

(a) Put n = 1 in the equation, we have :

n + 5 = 15

or 1 + 5 = 15

or 6 \neq 15

Thus n = 1 is not a solution.

(b) Put n = - 2, we have :

7n + 5 = 19

or 7(-2) + 5 = - 14 + 5 = - 9 \neq 19.

So, n = - 2 is not a solution to the given equation.

(c) Put n = 2, we have :

7n + 5 = 19

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

Thus n = 2 is the solution for the given equation.

(d) Put p = 1 , we have :

4p - 3 = 13

or 4(1) - 3 = 1 \neq 13 .

Thus p = 1 is not a solution.

(e) Put p = - 4 , we get :

4p - 3 = 13

or 4(1) - 3 = 1 \neq 13 .

Thus p = 1 is not a solution.

(f) Put p = 0 , we get :

4p - 3 = 13

or 4(0) - 3 = - 3 \neq 13 .

Thus p = 0 is not a solution.

3. Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4

Answer:

(i) Put p = 1,

We have : 5(1)\ +\ 2\ =\ 7\ \neq\ 17

Put p = 2,

We have : 5(2)\ +\ 2\ =\ 12\ \neq\ 17

Put p = 3,

we have : 5(3)\ +\ 2\ =\ 17\ =\ 17

Thus the solution is p = 3.

(ii) Put m = 4,

we have : 3(4)\ -\ 14\ =\ -2\ \neq\ 4

Put m = 5,

we have : 3(5)\ -\ 14\ =\ 1\ \neq\ 4

Now, put m = 6,

we have : 3(6)\ -\ 14\ =\ 4\ =\ 4

Thus m = 6 is the solution.

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.

Answer:

The equations are given below :

(i) x\ +\ 4\ =\ 9

(ii) y\ -\ 2\ =\ 8

(iii) 10a\ =\ 70

(iv) \frac{b}{5}\ =\ 6

(v) \frac{3}{4}t\ =\ 15

(vi) 7m\ +\ 7\ =\ 77

(vii) \frac{x}{4}\ -\ 4\ =\ 4

(viii) 6y\ -\ 6\ =\ 60

(ix) \frac{z}{3}\ +\ 3\ =\ 30

5. Write the following equations in statement forms:

(i) p + 4 = 15

(ii) m – 7 = 3

(iii) 2m = 7

(iv) m /5 = 3

(v) 3 m/5 = 6

(vi) 3p + 4 = 25

(vii) 4p – 2 = 18

(viii) p /2 + 2 = 8

Answer:

(i) Add 4 to the number p, we get 15.

(ii) Subtract 7 from m to get 3.

(iii) Twice the number m is 7.

(iv) One-fifth of m is 3.

(v) Three-fifth of m is 6.

(vi) 4 is added to thrice the number p to get 25.

(vii) 2 is subtracted from the product of 4 times p to get 18.

(viii) When 2 is added to half of the number p, we get 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. (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).

Answer:

(a) Let the Parmit's marbles be m.

Then according to the question we have : 5m\ +\ 7\ =\ 37

or 5m\ =\ 30

(b) Let the age of Laxmi be y years.

Then we have : 3y\ +\ 4\ =\ 49

or 3y\ =\ 45

(c) Let the lowest marks be l, then :

2l \ +\ 7\ =\ 87

or 2l \ =\ 80

(d) Let the base angle of the triangle be b degree.

Then according to the question we have :

b\ +\ b\ +\ 2b\ =\ 180^{\circ}

or 3b\ =\ 180^{\circ}

NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.2

1. 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

Answer:

(a) Add 1 to both the sides, we have :

x\ =\ 1

(b) Transpoing 1 to the RHS, we have :

x\ =\ -\ 1

(c) Transposing - 1 to the RHS, we have :

x\ =\ 6

(d) Transposing 6 to the RHS, we get :

x\ =\ -\ 4

(e) Transposing - 4 to the RHS, we have :

y\ =\ -\ 3

(f) Transposing - 4 to the RHS, we get :

y\ =\ 8

(g) Transposing 4 to the RHS, we get :

y\ =\ 0

(h) Transposing 4 to the RHS, we get :

y\ =\ -\ 8

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

(a) 3l = 42

(b) b / 2 = 6

(c) p /7 = 4

(d) 4x = 25

(e) 8y = 36

(f) z/ 3 = 5 /4

(g) a /5 =7/ 15

(h) 20t = – 10

Answer:

(a) Divide both sides by 3, we get :

l\ =\ 14

(b) Multiply both sides by 2, we get :

b\ =\ 12

(c) Multiply both sides by 7, we get :

p\ =\ 28

(d) Divide both sides by 4, we get :

x\ =\ \frac{25}{4}

(e) Divide both sides by 8, we get :

y\ =\ \frac{9}{2}

(f) Multiply both sides by 3, we get :

z\ =\ \frac{15}{4}

(g) Multiply both sides by 5, we get :

a\ =\ \frac{7}{3}

(h) Divide both sides by 20, we get :

t =\ -\frac{1}{2}

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

(a) 3n – 2 = 46

(b) 5m + 7 = 17

(c) 20p / 3 = 40

(d) 3p/10 = 6

Answer:

(a) We have 3n – 2 = 46.

Transposing - 2 to the RHS, we have :

3n\ =\ 46\ +\ 2\ =\ 48

or n\ =\ 16

(b) We have 5m + 7 = 17

Transposing 7 to the RHS, we have :

5m\ =\ 17\ -\ 7\ =\ 10

or m\ =\ 2

(c) We have 20p / 3 = 40

Multiply both sides by \frac{3}{20} :

\frac{20p}{3}\times \frac{3}{20}\ =\ 40\times \frac{3}{20}

or p\ =\ 6

(d) We have 3p/10 = 6

Multiply both sides by \frac{10}{3} :

\frac{3p}{10}\times \frac{10}{3}\ =\ 6\times \frac{10}{3}

or p\ =\ 20

4. Solve the following equations:

(a) 10p = 100

(b) 10p + 10 = 100

(c) p /4 = 5

(d) – p/3 = 5

(e) 3 p/4 = 6

(f) 3s = –9

(g) 3s + 12 = 0

(h) 3s = 0

(i) 2q = 6

(j) 2q – 6 = 0

(k) 2q + 6 = 0

(l) 2q + 6 = 12

Answer:

(a) Divide both sides by 10, we get :

p\ =\ 10

(b) Transposing 10 to the RHS, we get :

10p\ =\ 100\ -\ 10\ =\ 90

Now, dividing both sides by 10 gives : p\ =\ 9

(c) Multiplying both sides by 4, we have :

p\ =\ 20

(d) Multiplying both sides by - 3 , we have :

p\ =\ 15

(e) Multiplying both sides by \frac{4}{3} , we have :

p\ =\ 8

(f) Dividing both sides by 3, we have :

s\ =\ -3

(g) Transposing 12 to the RHS and then dividing both sides by 3, we have :

s\ =\ -4

(h) Dividing both sides by 3, we get :

s\ =\ 0

(i) Dividing both sides by 2, we get :

q\ =\ 3

(j) Transposing - 6 to the RHS and then dividing both sides by 2, we get :

q\ =\ 3

(k) Transposing 6 to the RHS and then dividing both sides by 2, we get :

q\ =\ -\ 3

(l) Transposing 6 to the RHS and then dividing both sides by 2, we get :

q\ =\ 3

NCERT Solutions for Maths Chapter 4 Simple Equation class 7 Exercise 4.3

1. Solve the following equations:

(a) 2y + \frac{5}{2} = \frac{37}{2}

(b) 5t + 28 = 10

(c) a /5 + 3 = 2

(d) q/ 4 + 7 = 5

e) 5 x/2 = –5

(f) \frac{5}{2}x = \frac{25}{4}

(g) 7m + 19/2 = 13

(h) 6z + 10 = –2

(i) \frac{3l}{2} = 2/3

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

Answer:

(a) 2y + \frac{5}{2} = \frac{37}{2}

Transposing \frac{5}{2} to the RHS :

2y\ =\ \frac{37}{2}\ -\ \frac{5}{2}\ =\ 16

y\ =\ 8

(b) 5t + 28 = 10

Transposing 28 to the RHS and then dividing both sides by 5, we get :

5t\ =\ 10\ -\ 28\ =\ -\ 18

t\ =\ -\frac{18}{5}

(c) a /5 + 3 = 2

Transposing 3 to the RHS and multiplying both sides by 5, we get :

\frac{a}{5}\ +\ 3\ =\ 2

\frac{a}{5}\ =\ -\ 1

a\ =\ -\ 5

(d) q/ 4 + 7 = 5

Transposing 7 to the RHS and multiplying both sides by 4:

\frac{q}{4}\ +\ 7\ =\ 5

\frac{q}{4}\ =\ -\ 2

q\ =\ -\ 8

(e) 5 x/2 = – 5

Multiplying both sides by \frac{2}{5} :

x\ =\ -5\times \frac{2}{5}\ =\ -\ 2

(f) \frac{5}{2}x = \frac{25}{4}

Multiplying both sides by \frac{2}{5} :

x\ =\ \frac{25}{4}\times \frac{2}{5}

x\ =\ \frac{5}{2}

(g) 7m + 19/2 = 13

Transposing \frac{19}{2} to the RHS and then dividing both sides by 7 :

7m\ =\ 13\ -\ \frac{19}{2}\ =\ \frac{7}{2}

m\ =\ \frac{1}{2}

(h) 6z + 10 = –2

Transposing 10 to the RHS and then dividing both sides by 6, we get :

6z\ =\ -\ 2\ -\ 10\ =\ -\ 12

z\ =\ -\ 2

(i) \frac{3l}{2} = 2/3

Multiplying both sides by \frac{2}{3} ,

l\ =\ \frac{2}{3}\times \frac{2}{3}\ =\ \frac{4}{9}

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

Transposing 5 to the RHS and then multiplying both sides by \frac{3}{2}

\frac{2b}{3}\ =\ 8

b\ =\ 8\times \frac{3}{2}\ =\ 12

2. Solve the following equations:

(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

Answer:

(a) We have:

2(x + 4) = 12

Dividing both sides by 2, we have :

x\ +\ 4\ =\ 6

Transposing 4 to the RHS, we get :

x\ =\ 6\ -\ 4\ =\ 2

Thus x\ =\ 2

(b) We have:

3(n – 5) = 21

Dividing both sides by 3, we have :

n\ -\ 5\ =\ 7

Transposing - 5 to the RHS, we get :

n\ =\ 7\ +\ 5\ =\ 12

Thus n\ =\ 12

(c) We have :

3(n – 5) = – 21

Dividing both sides by 3, we have :

n\ -\ 5\ =\ -\ 7

Transposing - 5 to the RHS, we get :

n\ =\ -\ 7\ +\ 5\ =\ -\ 2

Thus n\ =\ -\ 2

(d) We have :

– 4(2 + x) = 8

Dividing both sides by - 4, we have :

x\ +\ 2\ =\ -\ 2

Transposing 2 to the RHS, we get :

x\ =\ -\ 2\ -\ 2\ =\ -\ 4

Thus x\ =\ -\ 4

(e) We have :

4(2 - x) = 8

Dividing both sides by 4, we have :

2\ -\ x\ =\ 2

Transposing x to the RHS and 2 to the LHS , we get :

x\ =\ 2\ -\ 2\ =\ 0

Thus x\ =\ 0

Q3 Solve the following equations:

(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)

Answer:

(a) 4 = 5(p – 2)

Dividing both sides by 5, we get :

p\ -\ 2\ =\ \frac{4}{5}

or p\ =\ \frac{4}{5}\ +\ 2\ =\ \frac{14}{5}

(b) – 4 = 5(p – 2)

Dividing both sides by 5, we get :

p\ -\ 2\ =\ \frac{-4}{5}

or p\ =\ \frac{-4}{5}\ +\ 2\ =\ \frac{6}{5}

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

Transposing 4 to the LHS and then dividing both sides by 3, we get :

16\ -\ 4\ =\ 3(t\ +\ 2)

or t\ +\ 2\ =\ 4

or t\ =\ 2

(d) 4 + 5(p – 1) =34

Transposing 4 to the RHS and then dividing both sides by 5, we get :

5(p\ -\ 1)\ =\ 30

or p\ -\ 1\ =\ 6

or p\ =\ 7

(e) 0 = 16 + 4(m – 6)

Transposing 16 to the LHS, we get :

-\ 16\ =\ 4(m\ -\ 6)

or m\ -\ 6\ =\ -4

or m\ =\ 2

4. (a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2

Answer:

(a) The 3 required equations can be :

7x\ =\ 14

7x\ +\ 2 =\ 16

3x\ =\ 6

(b) The required equations are :

7x\ =\ -\ 14

7x\ +\ 2 =\ -\ 12

-\ 3x\ =\ 6

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

1. Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(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.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.

Answer:

Let the number in each case be n.

(a) According to the question: 8n\ +\ 4\ =\ 60

or 8n\ =\ 60\ -\ 4\ =\ 56

or n\ =\ 7

(b) We have :

\frac{n}{5}\ -\ 4\ =\ 3

or \frac{n}{5}\ =\ 7

or n\ =\ 35

(c) The equation is :

\frac{3n}{4}\ +\ 3\ =\ 21

or \frac{3n}{4}\ =\ 18

or n\ =\ 24

(d) We have :

2n\ -\ 11\ =\ 15

or 2n\ =\ 26

or n\ =\ 13

(e) The equation is :

50\ -\ 3n\ =\ 8

or 3n\ =\ 42

or n\ =\ 14

(f) We have :

\frac{n+19}{5}\ =\ 8

or n\ +\ 19\ =\ 40

or n\ =\ 21

(g) We have :

\frac{5n}{2}\ -\ 7\ =\ 23

or \frac{5n}{2}\ =\ 30

or n\ =\ 12

2. 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\degree . What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180\degree ).
(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?

Answer:

(a) Let the lowest score be l.

Then according to the question, we have :

2l\ +\ 7\ =\ 87

or 2l\ =\ 80

or l\ =\ 40

Thus the lowest marks are 40.

(b) Let the base angle of a triangle is \Theta .

Then according to the question, we get :

\Theta \ +\ \Theta\ +\ 40^{\circ}\ =\ 180^{\circ}

or 2\Theta\ +\ 40^{\circ}\ =\ 180^{\circ}

or 2\Theta\ =\ 140^{\circ}

or \Theta \ =\ 70^{\circ}

(c) Let the runs scored by Rahul is x. Then runs by Sachin is 2x.

Further, it is given that their runs fell two short of a double century.

Thus we have : x\ +\ 2x\ =\ 198

or 3x\ =\ 198

or x\ =\ 66 .

Hence runs by Rahul is 66 and the runs scored by Sachin is 132.

3. Solve the following:
(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 years 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 were 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?

Answer:

(a) Let the number of Parmit's marble be n.

Then according to question, we have :

5n\ +\ 7\ =\ 37

or 5n\ =\ 37\ -\ 7

or 5n\ =\ 30

or n\ =\ 6

(b) Let the age of Laxmi be x.

Then according to the question, we have :

3x\ +\ 4\ =\ 49

or 3x\ =\ 49\ -\ 4\ =\ 45

or x\ =\ 15

(c) Let the number of fruit trees planted be z.

Then according to the question, we have :

3z\ +\ 2\ =\ 77

or 3z\ =\ 77\ -\ 2\ =\ 75

or z\ =\ 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!

Answer:

Let the number be x.

According to the question the equation is :

7x\ +\ 50\ +\ 40\ =\ 300

or 7x\ +\ 90\ =\ 300

or 7x\ =\ 300\ -\ 90\ =\ 210

or x\ =\ 30

Hence the number is 30.

More About Simple Equations Class 6 Maths NCERT Chapter 4

NCERT Class 7 Maths book chapter 4 Simple Equations begin with a fun game 'mind-reading' and introduce the concept of the equation. So what is an equation? It is a condition on the variable such that two expressions in the variable should have equal value. The value of the variable for which an equation is satisfied is called the solution of the equation. In the equation, there is always an equality(=) sign. The equality sign shows that the value of the expression to the left-hand side (LHS) of the equality sign is equal to the value of the expression to the right-hand side (RHS) of the equality sign. For example, the equation 3x+ 7 =2x - 35 has the expression (3x + 7) on the left of the equality sign and (2x - 35) on the right of the equality sign. But it is not an equation if there is a sign other than the equality sign. For example, 3x + 5 > 75 is not an equation.

Simple Equations Class 6 Maths Chapter 4-Topic

  • A Mind-Reading Game
  • Setting Up Of An Equation
  • Review Of What We Know
  • What Equation Is?
  • More Equations
  • From Solution To Equation
  • Application Of Simple Equations To Practical Situations

NCERT Solutions for Class 7 Maths Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Integers

Chapter 2

Fractions and Decimals

Chapter 3

Data Handling

Chapter 4

Simple Equations

Chapter 5

Lines and Angles

Chapter 6

The Triangle and its Properties

Chapter 7

Congruence of Triangles

Chapter 8Comparing quantities

Chapter 9

Rational Numbers

Chapter 10

Practical Geometry

Chapter 11

Perimeter and Area

Chapter 12

Algebraic Expressions

Chapter 13

Exponents and Powers

Chapter 14

Symmetry

Chapter 15Visualising Solid Shapes

NCERT Solutions for Class 7 Subject Wise

NCERT Solutions for Class 7 Maths

NCERT Solutions for Class 7 Science

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations

As far as the subject Mathematics is concerned, the practising problem is important to score well in exams. It is important to know how to apply the concepts studied in an application-level problem. This is achieved through practice and the CBSE NCERT solutions for Class 7 Maths chapter 4 Simple Equations help for the same. The solutions of NCERT Class 7 Maths chapter 4 Simple Equations are helpful in solving homework problems.

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Questions (FAQs)

1. What is a simple equation, and how do I solve one?

According to chapter 4 maths class 7, A simple equation is a mathematical statement that shows that two expressions are equal. It has an equal sign (=) between the two sides.

To solve a simple equation, aim to isolate the variable (usually represented by a letter) on one side of the equation.

Perform the same operations on both sides of the equation to keep it balanced. This maintains the equality of the expressions.

The goal is to find the value of the variable that satisfies the equation. This value will make both sides of the equation equal.

Also Practice chapter 4 maths class 7 solutions to command the concepts.

 

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Popular Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J
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A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

 Option 2)

 6.45×10−3 kg
Option 3)

 9.89×10−3 kg

 Option 4)

12.89×10−3 kg

 
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An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

 Option 2)

200 \, \, J - 500 \, \, J
Option 3)

2\times 10^{5}J-3\times 10^{5}J

 Option 4)

20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

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less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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