### NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

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

**Complete the given 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; in – -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 (n = 1)

Put n = 1 in LHS

1 + 5 = 6 ≠ 19 (RHS)

Since LHS ≠ RHS

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

(b) 7n + 5 = 19; (n = -2)

Put n = – 2 in LHS

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; (n = 2)

Put n = 2 in LHS

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; (p = 1)

Put p = 1 in LHS

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; (p = -A)

Put p = -4 in LHS

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; (p = 0)

Put p = 0 in LHS

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

For p = 1, LHS

= 5 × 1 + 2 = 5 + 2 = 7 ≠ 17 (RHS)

For p = 2, LHS = 5 × 2 + 2

= 10 + 2 = 12 ≠ 17 (RHS)

For 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 required solution.

(ii) 3m – 14 = 4

For m = 1, LHS = 3 × 1 – 14

= 3 – 14 = -11 ≠ 4 (RHS)

For m = 2, LHS = 3 × 2 – 14 = 6 – 14

= -8 ≠ 4 (RHS)

For m = 3, LHS = 3 × 3 – 14 = 9 – 14

= -5 ≠ 4 (RHS)

Form m = 4, LHS = 3 × 4 – 14

= 12 – 14 = -2 ≠ 4 (RHS)

For m = 5, LHS = 3 × 5 – 14

= 15 – 14 = -1 ≠ 4 (RHS)

For 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 required solution.

**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 1)**

**(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 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 bey years.

∴ Laxmi’s father’s age = 3y + 4

But the Laxmi’s father age is given by 49

Thus the required equation is 3y + 4 = 49

(iii) Let the lowest score be l.

∴ The highest score = 2l + 1

But the highest score is given by 87.

Thus, the required equation is 2l + 1 = 87

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

∴ Vertex angle of the triangle = 2b

Sum of the angles of a triangle = 180°

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