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

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 are the part of NCERT Solutions for Class 7 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1.

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 (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 = 2l + 7
But the highest score is given by 87.
Thus, the required equation is: 2l + 7 = 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°

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