# Mathematics Class 7 – Chapter 4 – Simple Equation – Exercise – 4.1 – NCERT Exercise Solution

1. Complete the last column of the table.

Solution:

(i) x + 3 = 0

Given equation:

X + 3 = 0

Value of X = 3

LHS = x + 3

By substituting the value of x

LHS = 3 + 3 = 6

By comparing LHS and RHS

LHS ≠ RHS

No, the equation is not satisfied.

(ii) x + 3 = 0

Given equation:

X +3 = 0

Value of x = 0

LHS = x + 3

By substituting the value of x

LHS = 0 + 3 = 3

By comparing LHS and RHS

LHS ≠ RHS

No, the equation is not satisfied.

(iii) x + 3 = 0

Given equation:

X + 3 = 0

Value of x = – 3

LHS = x + 3

By substituting the value of x

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

Given equation:

X – 7 = 1

Value of x = 8

LHS = x – 7

By substituting the value of x

LHS = 8 – 7 = 1

By comparing LHS and RHS

LHS = RHS

Yes, the equation is satisfied.

(vi) 5x = 25

Given equation:

5x = 25

Value of x = 0

LHS = 5x

By substituting the value of x

LHS = 5 × 0 = 0

By comparing LHS and RHS

LHS ≠ RHS

No, the equation is not satisfied.

(vii) 5x = 25

Given equation:

5x = 25

Value of x = 5

LHS = 5x

By substituting the value of x

LHS = 5 × 5 = 25

By comparing LHS and RHS

LHS = RHS

Yes, the equation is satisfied.

(viii) 5x = 25

Given equation:

5x = 25

Value of x = -5

LHS = 5x

By substituting the value of x

LHS = 5 × (-5) = – 25

By comparing LHS and RHS

LHS ≠ RHS

No, the equation is not satisfied.

(ix) m/3 = 2

Given equation:

m/3 = 2

value of x = -6

LHS = m/3

By substituting the value of m

LHS = -6/3 = – 2

By comparing LHS and RHS

LHS ≠ RHS

No, the equation is not satisfied.

(x) m/3 = 2

Given equation:

m/3 = 2

value of x = 0

LHS = m/3

By substituting the value of m

LHS = 0/3 = 0

By comparing LHS and RHS

LHS ≠ RHS

No, the equation is not satisfied.

(xi) m/3 = 2

Given equation:

m/3 = 2

value of x = 6

LHS = m/3

By substituting the value of m

LHS = 6/3 = 2

By comparing LHS and RHS

LHS = RHS

Yes, the equation is satisfied.

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:

Given equation:

n+ 5 = 19

Where n = 1

LHS = n + 5

By substituting the value of n = 1

= 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:

Given equation:

7n + 5 = 19

Where n = -2

LHS = 7n + 5

By substituting the value of n = -2

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:

Given equation:

7n + 5 = 19

Where n = 2

LHS = 7n + 5

By substituting the value of n = 2

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:

Given equation:

4p – 3 = 13

Where p = 1

LHS = 4p – 3

By substituting the value of p = 1

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:

Given equation:

4p – 3 = 13

Where p = -4

LHS = 4p – 3

By substituting the value of p = – 4

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:

Given equation:

4p – 3 = 13

Where p = 0

LHS = 4p – 3

By substituting the value of p = 0

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:

Given,

LHS = 5p + 2

Now,

Let p = 0

By substituting the value of p = 0

We get

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.

Now,

Let, p = 1

By substituting the value of 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.

Now,

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.

Now,

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:

Given,

LHS = 3m – 14

By substituting the value of m = 2

LHS = 3m – 14

= (3 × 2) – 14

= 6 – 14

= – 8

By comparing LHS and RHS

-8 ≠ 4

LHS ≠ RHS

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

Now,

Let, m = 3

LHS = 3m – 14

= (3 × 3) – 14

= 9 – 14

= – 5

By comparing LHS and RHS

5 ≠ 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 as,

= x + 4 = 9

(ii) 2 subtracted from y is 8.

Solution:

The above statement can be written as,

= y – 2 = 8

(iii) Ten times a is 70.

Solution:

The above statement can be written as,

= 10a = 70

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

Solution:

The above statement can be written as,

= (b/5) = 6

(v) Three-fourth of t is 15.

Solution:

The above statement can be written as,

= ¾t = 15

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

Solution:

The above statement can be written as,

= 7m + 7 = 77

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

Solution:

The above statement can be written as,

One-fourth of a number x is x/4

Then,

= x/4 – 4 = 4

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

Solution:

The above statement can be written as,

6 times of y is 6y

Now,

= 6y – 6 = 60

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

Solution:

The above statement can be written 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 gives 15.

(ii) m – 7 = 3

Solution:

7 subtracted from m gives 3.

(iii) 2m = 7

Solution:

Twice of number m gives 7.

(iv) m/5 = 3

Solution:

The number m divided by 5 gives 3.

(v) (3m)/5 = 6

Solution:

Three times of m divided by 5 gives 6.

(vi) 3p + 4 = 25

Solution:

Three times p plus 4 gives us 25.

(vii) 4p – 2 = 18

Solution:

Four times p minus 2 gives us 18.

(viii) p/2 + 2 = 8

Solution:

When we add half of a number p to 2, 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.)

Solution:

Let the Parmit’s marble be m

Then, Irfan’s marble = 5m + 7

The total number of marble = 37

Hence, the required equation is:

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:

Let the age of Laxmi is y

Age of Laxmi’s Father = 3y + 4

but the age of Laxmi’s father is given by 49

Hence the required equation:

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:

Let the lowest score be l

then, the highest score = 2l + 1

but the given highest score = 87

Hence, the required equation:

2l + 1 = 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:

Let the base of each angle be ‘b’ degree

then, the vertex angle of the triangle = 2b

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

Hence the required equation is:

b + b + 2b = 180o or 4b

👍👍👍

error: 