**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.2****NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.3**

**NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 (Rationalised Contents)**

**Ex 4.1 Class 7 Maths Question 1.**

Complete the last column of the table:**Solution:**

**Ex 4.1 Class 7 Maths Question 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)

**Solution:
**(a) n + 5 = 19

Putting n = 1 in LHS, we get

1 + 5 = 6 ≠ 19 (RHS)

Since LHS ≠ RHS

Thus, n = 1 is not the solution of the given equation.

(b) 7n + 5 = 19

Putting n = –2 in LHS, we get

7 × (-2) + 5 = -14 + 5 = -9 ≠ 19 (RHS)

Since LHS ≠ RHS

Thus, n = -2 is not the solution of the given equation.

(c) 7n+ 5 = 19

Putting n = 2 in LHS, we get

7 × 2 + 5 = 14 + 5 = 19 = 19 (RHS)

Since LHS = RHS

Thus, n = 2 is the solution of the given equation.

(d) 4p – 3 = 13

Putting p = 1 in LHS, we get

4 × 1 – 3 = 4 – 3 = 1 ≠ 13 (RHS)

Since LHS ≠ RHS

Thus, p = 1 is not the solution of the given equation.

(e) 4p – 3 = 13

Putting p = -4 in LHS, we get

4 × (-4) – 3 = -16 – 3 = -19 ≠ 13 (RHS)

Since LHS ≠ RHS

Thus, p = -4 is not the solution of the given equation.

(f) 4p – 3 = 13

Putting p = 0 in LHS, we get

4 × (0) – 3 = 0 – 3 = -3 ≠ 13 (RHS)

Since LHS ≠ RHS

Thus, p = 0 is not the solution of the given equation.

**Ex 4.1 Class 7 Maths Question 3.**

Solve the following equations by trial and error
method:(i) 5p + 2 = 17

(ii) 3m – 14 = 4

**Solution:
**(i) 5p + 2 = 17

Putting p = 1, LHS = 5 × 1 + 2 = 5 + 2 = 7 ≠ 17 (RHS)

Putting p = 2, LHS = 5 × 2 + 2 = 10 + 2 = 12 ≠ 17 (RHS)

Putting p = 3, LHS = 5 × 3 + 2 = 15 + 2 = 17 = 17 (RHS)

Since the given equation is satisfied for p = 3.

Thus, p = 3 is the solution of the given equation.

(ii) 3m – 14 = 4

Putting m = 1, LHS = 3 × 1 – 14 = 3 – 14 = -11 ≠ 4 (RHS)

Putting m = 2, LHS = 3 × 2 – 14 = 6 – 14 = -8 ≠ 4 (RHS)

Putting m = 3, LHS = 3 × 3 – 14 = 9 – 14 = -5 ≠ 4 (RHS)

Putting m = 4, LHS = 3 × 4 – 14 = 12 – 14 = -2 ≠ 4 (RHS)

Putting m = 5, LHS = 3 × 5 – 14 = 15 – 14 = -1 ≠ 4 (RHS)

Putting m = 6, LHS = 3 × 6 – 14 = 18 – 14 = 4 (=) 4 (RHS)

Since, the given equation is satisfied for m = 6.

Thus, m = 6 is the solution of the given equation.

**Ex 4.1 Class 7 Maths Question 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.

**Solution:**

**Ex 4.1 Class 7 Maths Question 5.**

Write the following equations in statement forms:

**Solution:**

**Ex 4.1 Class 7 Maths Question 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).

**Solution:
**(i) Let

*m*be the number of Parmit’s marbles.

∴ Irfan’s marble = 5m + 7

Total number of Irfan’s marble is given by 37.

Thus, the required equation is: 5m + 7 = 37

(ii) Let Laxmi’s age be *y* years.

∴ Laxmi’s father’s age = (3y + 4) years

But the Laxmi’s father’s age is given by 49 years.

Thus, the required equation is: 3y + 4 = 49

(iii) Let the lowest score be *l*.

∴ The highest score = 2*l* + 7

But the highest score is given by 87.

Thus, the required equation is: 2*l* + 7 = 87

(iv) Let each base angle be ‘*b*’ degrees.

∴ Vertex angle of the triangle = 2*b*

Sum of the angles of a triangle = 180°

∴ Required equation is: *b* + *b* + 2*b* = 180° or
4*b* = 180°

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