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

NCERT Solutions For Class 7 Maths Ex 4.1 PDF Free Download

NCERT Solutions for Class 7 Maths Exercise 4.1 Chapter 4 Simple Equations in simple PDF are given here. In this exercise of NCERT Solutions for Class 7 Chapter 4, students learn about what actually equation means. In an equation there is always an equality sign. The equality sign shows that the value of the expression to the left of the sign is equal to the value of the expression to the right of the sign. Learn more about these topics by solving the questions of NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations with the help of solutions provided here.

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 4 Simple Equations – Exercise 4.1

 

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ncert solution class 7 maths chapter 4 exercise 4.1
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ncert solution class 7 maths chapter 4 exercise 4.1
ncert solution class 7 maths chapter 4 exercise 4.1

 

Access answers to Maths NCERT Solutions for Class 7 Chapter 4 – Simple Equations Exercise 4.1

1. Complete the last column of the table.

S. No.

Equation

Value

Say, whether the equation is satisfied. (Yes/No)

(i)

x + 3 = 0

x = 3

(ii)

x + 3 = 0

x = 0

(iii)

x + 3 = 0

x = -3

(iv)

x – 7 = 1

x = 7

(v)

x – 7 = 1

x = 8

(vi)

5x = 25

x = 0

(vii)

5x = 25

x = 5

(viii)

5x = 25

x = -5

(ix)

(m/3) = 2

m = – 6

(x)

(m/3) = 2

m = 0

(xi)

(m/3) = 2

m = 6

Solution:-

(i) x + 3 = 0

LHS = x + 3

By substituting the value of x = 3

Then,

LHS = 3 + 3 = 6

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied.

(ii) x + 3 = 0

LHS = x + 3

By substituting the value of x = 0

Then,

LHS = 0 + 3 = 3

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied.

(iii) x + 3 = 0

LHS = x + 3

By substituting the value of x = – 3

Then,

LHS = – 3 + 3 = 0

By comparing LHS and RHS

LHS = RHS

∴Yes, the equation is satisfied

(iv) x – 7 = 1

LHS = x – 7

By substituting the value of x = 7

Then,

LHS = 7 – 7 = 0

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied

(v) x – 7 = 1

LHS = x – 7

By substituting the value of x = 8

Then,

LHS = 8 – 7 = 1

By comparing LHS and RHS

LHS = RHS

∴Yes, the equation is satisfied.

(vi) 5x = 25

LHS = 5x

By substituting the value of x = 0

Then,

LHS = 5 × 0 = 0

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied.

(vii) 5x = 25

LHS = 5x

By substituting the value of x = 5

Then,

LHS = 5 × 5 = 25

By comparing LHS and RHS

LHS = RHS

∴Yes, the equation is satisfied.

(viii) 5x = 25

LHS = 5x

By substituting the value of x = -5

Then,

LHS = 5 × (-5) = – 25

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied.

(ix) m/3 = 2

LHS = m/3

By substituting the value of m = – 6

Then,

LHS = -6/3 = – 2

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied.

(x) m/3 = 2

LHS = m/3

By substituting the value of m = 0

Then,

LHS = 0/3 = 0

By comparing LHS and RHS

LHS ≠ RHS

∴No, the equation is not satisfied.

(xi) m/3 = 2

LHS = m/3

By substituting the value of m = 6

Then,

LHS = 6/3 = 2

By comparing LHS and RHS

LHS = RHS

∴Yes, the equation is satisfied.

S. No.

Equation

Value

Say, whether the equation is satisfied. (Yes/No)

(i)

x + 3 = 0

x = 3

No

(ii)

x + 3 = 0

x = 0

No

(iii)

x + 3 = 0

x = -3

Yes

(iv)

x – 7 = 1

x = 7

No

(v)

x – 7 = 1

x = 8

Yes

(vi)

5x = 25

x = 0

No

(vii)

5x = 25

x = 5

Yes

(viii)

5x = 25

x = -5

No

(ix)

(m/3) = 2

m = – 6

No

(x)

(m/3) = 2

m = 0

No

(xi)

(m/3) = 2

m = 6

Yes

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

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

Solution:-

LHS = n + 5

By substituting the value of n = 1

Then,

LHS = n + 5

= 1 + 5

= 6

By comparing LHS and RHS

6 ≠ 19

LHS ≠ RHS

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

(b) 7n + 5 = 19 (n = – 2)

Solution:-

LHS = 7n + 5

By substituting the value of n = -2

Then,

LHS = 7n + 5

= (7 × (-2)) + 5

= – 14 + 5

= – 9

By comparing LHS and RHS

-9 ≠ 19

LHS ≠ RHS

Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.

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

Solution:-

LHS = 7n + 5

By substituting the value of n = 2

Then,

LHS = 7n + 5

= (7 × (2)) + 5

= 14 + 5

= 19

By comparing LHS and RHS

19 = 19

LHS = RHS

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

(d) 4p – 3 = 13 (p = 1)

Solution:-

LHS = 4p – 3

By substituting the value of p = 1

Then,

LHS = 4p – 3

= (4 × 1) – 3

= 4 – 3

= 1

By comparing LHS and RHS

1 ≠ 13

LHS ≠ RHS

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

(e) 4p – 3 = 13 (p = – 4)

Solution:-

LHS = 4p – 3

By substituting the value of p = – 4

Then,

LHS = 4p – 3

= (4 × (-4)) – 3

= -16 – 3

= -19

By comparing LHS and RHS

-19 ≠ 13

LHS ≠ RHS

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

(f) 4p – 3 = 13 (p = 0)

Solution:-

LHS = 4p – 3

By substituting the value of p = 0

Then,

LHS = 4p – 3

= (4 × 0) – 3

= 0 – 3

= -3

By comparing LHS and RHS

– 3 ≠ 13

LHS ≠ RHS

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

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

(i) 5p + 2 = 17

Solution:-

LHS = 5p + 2

By substituting the value of p = 0

Then,

LHS = 5p + 2

= (5 × 0) + 2

= 0 + 2

= 2

By comparing LHS and RHS

2 ≠ 17

LHS ≠ RHS

Hence, the value of p = 0 is not a solution to the given equation.

Let, p = 1

LHS = 5p + 2

= (5 × 1) + 2

= 5 + 2

= 7

By comparing LHS and RHS

7 ≠ 17

LHS ≠ RHS

Hence, the value of p = 1 is not a solution to the given equation.

Let, p = 2

LHS = 5p + 2

= (5 × 2) + 2

= 10 + 2

= 12

By comparing LHS and RHS

12 ≠ 17

LHS ≠ RHS

Hence, the value of p = 2 is not a solution to the given equation.

Let, p = 3

LHS = 5p + 2

= (5 × 3) + 2

= 15 + 2

= 17

By comparing LHS and RHS

17 = 17

LHS = RHS

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

(ii) 3m – 14 = 4

Solution:-

LHS = 3m – 14

By substituting the value of m = 3

Then,

LHS = 3m – 14

= (3 × 3) – 14

= 9 – 14

= – 5

By comparing LHS and RHS

-5 ≠ 4

LHS ≠ RHS

Hence, the value of m = 3 is not a solution to the given equation.

Let, m = 4

LHS = 3m – 14

= (3 × 4) – 14

= 12 – 14

= – 2

By comparing LHS and RHS

-2 ≠ 4

LHS ≠ RHS

Hence, the value of m = 4 is not a solution to the given equation.

Let, m = 5

LHS = 3m – 14

= (3 × 5) – 14

= 15 – 14

= 1

By comparing LHS and RHS

1 ≠ 4

LHS ≠ RHS

Hence, the value of m = 5 is not a solution to the given equation.

Let, m = 6

LHS = 3m – 14

= (3 × 6) – 14

= 18 – 14

= 4

By comparing LHS and RHS

4 = 4

LHS = RHS

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

4. Write equations for the following statements:

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

Solution:-

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

= x + 4 = 9

(ii) 2 subtracted from y is 8.

Solution:-

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

= y – 2 = 8

(iii) Ten times a is 70.

Solution:-

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

= 10a = 70

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

Solution:-

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

= (b/5) = 6

(v) Three-fourth of t is 15.

Solution:-

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

= ¾t = 15

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

Solution:-

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

Seven times m is 7m

= 7m + 7 = 77

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

Solution:-

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

One-fourth of a number x is x/4

= x/4 – 4 = 4

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

Solution:-

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

6 times of y is 6y

= 6y – 6 = 60

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

Solution:-

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

One-third of z is z/3

= 3 + z/3 = 30

5. Write the following equations in statement forms:

(i) p + 4 = 15

Solution:-

The sum of numbers p and 4 is 15.

(ii) m – 7 = 3

Solution:-

7 subtracted from m is 3.

(iii) 2m = 7

Solution:-

Twice of number m is 7.

(iv) m/5 = 3

Solution:-

The number m divided by 5 gives 3.

(v) (3m)/5 = 6

Solution:-

Three-fifth of m is 6.

(vi) 3p + 4 = 25

Solution:-

Three times p plus 4 gives you 25.

(vii) 4p – 2 = 18

Solution:-

Four times p minus 2 gives you 18.

(viii) p/2 + 2 = 8

Solution-

If you add half of a number p to 2, you 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.)

Solution:-

From the question it is given that,

Number of Parmit’s marbles = m

Then,

Irfan has 7 marbles more than five times the marbles Parmit has

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

= (5 × m) + 7 = 37

= 5m + 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.)

Solution:-

From the question it is given that,

Let Laxmi’s age to be = y years old

Then,

Lakshmi’s father is 4 years older than three times of her age

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

= (3 × y) + 4 = 49

= 3y + 4 = 49

(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.)

Solution:-

From the question it is given that,

Highest score in the class = 87

Let lowest score be l

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

= (2 × l) + 7 = 87

= 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 180 degrees).

Solution:-

From the question it is given that,

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

Let base angle be b

Then,

Vertex angle = 2 × base angle = 2b

= b + b + 2b = 180o

= 4b = 180o


Access other exercises of NCERT Solutions For Class 7 Chapter 4 – Simple Equations

Exercise 4.2 Solutions

Exercise 4.3 Solutions

Exercise 4.4 Solutions

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