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

These NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1 Questions and Answers are prepared by our highly skilled subject experts.

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

Question 1.
Complete the last column of the table.
Answer:

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 No
(viii) \(\frac{m}{3}\) = 2 m = -6 No
(ix) \(\frac{m}{3}\) = 2 m = 0 No
(x) \(\frac{m}{3}\) = 2  m = 6 yes

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

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)
Answer:
(a) n + 5 = 19
1 + 5 =19 (Putting n = 1)
6 ≠ 19
∴ n = 1 is not a solution.

(b) 7n + 5 =19
7(- 2) + 5 =19 (Put n = – 2)
-14 + 5 = 19
-9 ≠ 19
n = -2 is not a solution.

(c) 7n + 5 =19
7(2) + 5=19 (Put n = 2)
14 + 5 = 19
19 = 19
∴ n = 2 is a solution.

(d) 4p – 3 = 13
4(1) – 3 =13 (Put p = 1)
4 – 3 = 13
1 ≠ 13
∴ p = 1 is not a solution.

(e) 4p – 3 = 13
4(- 4) – 3 = 13 (Put p = – 4)
-16 -3 = 13
-19 ≠ 13
∴ p = – 4 is not a solution.

(f) 4p – 3 = 13
4(0) – 3 = 13 (Put p = 0)
0 – 3 = 13
-3 ≠ 13
p = 0 is not a solution.

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

Question 3.
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
Answer:
(i) 5p + 2 = 17
Put p = 0
L.H.S = 5 (0) + 2
= 0 + 2
= 2 ≠ RHS

Put p = 1
L.H.S = 5(1)+ 2
= 5 + 2
= 7 ≠ RHS

Put p = -1
L.H.S = 5 (-1) + 2
= -5 + 2
= -3 ≠ RHS

Put p = 2
L.H.S = 5 (2) + 2
= 10 + 2
= 12 ≠ RHS

Put p = -2
L.H.S = 5 (-2) + 2
= -10 + 2
= -8 ≠ 17 RHS

Put p = 3
L.H.S = 5 (3) + 2
= 15 + 2
= 17 = RHS
p = 3 is the solution of 5P + 2 = 17

(ii) 3m – 14 =4
Put m = 0
L.H.S = 3m – 14
= 3 (0) – 14
= – 14
– 14 ≠ 4
L.H.S ≠ R.H.S

Put m = 1
L.H.S = 3m -14
= 3(1) – 14
= 3 – 14
= – 11 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 2
L.H.S = 3m – 14
= 3 (2) – 14
= 6 – 14
= – 8 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 3
L.H.S = 3(3) – 14
= 9 – 14
= – 5 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 4
L.H.S = 3(4) – 14
= 12 – 14
= – 2 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 5
L.H.S = 3(5) – 14
= 15 – 14
= 1 ≠ R.H.S
L.H.S ≠ R.H.S

Put m = 6
L.H.S = 3(6) – 14
= 18 – 14
= 4 = R.H.S
L.H.S = R.H.S
m = 6 is the solution to 3m – 14 = 4

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

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.
Answer:
(i) x + 4 = 9
(ii) y – 2 = 8
(iii) 10 a = 70
(iv) \(\frac { b }{ 5 }\) = 6
(v) \(\frac { 3t }{ 4 }\) = 15
(vi) 7m + 7 = 77
(vii) \(\frac { x }{ 4 }\) x – 4 = 4
(viii) 6 y – 6 = 60
(ix) \(\frac { z }{ 3 }\) + 3 = 30

Question 5.
Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) \(\frac { m }{ 5 }\) = 3
(v) \(\frac { 3m }{ 5 }\) = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) \(\frac { p }{ 2 }\) + 2 = 8
Answer:
(i) The sum of p and 4 is 15.
(ii) 7 subtracted from m is 3.
(hi) Twice a number m is 7.
(iv) One-fifth of a number m is 3.
(v) Three-fifth of a number m is 6.
(vi) Three times a number p when added to 4 gives 25.
(vii) 2 subtracted from four times a number p is 18.
(viii) 2 added to half of a number p is 8.

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

Question 6.
Set up an equation in the following cases:
(i) Irfan says that he has 7 marbles more than live 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 verte x 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).
Answer:
(i) Let Parmit has m marbles.
Then, five time the marbles Parmit has = 5 m.
Irfan has 7 marbles more than five times the marbles parmit has
So, Irfan has (5m + 7) marbles
But it is given that Irfan has 37 marbles.
5m + 7 = 37

(ii) Let Laxmi’s age = y years
3 times Laxmi’s age = 3y years.
Age of Laxmi’s father = 3 times
Laxmi’s age + 4 years
= 3y + 4 years
But Laxmi’s father is 49 years old.
3y + 4 = 49

(iii) Let the lowest score (marks) = 1
Twice the lowest marks = 2 l
Since highest marks = (twice the lowest marks) + 7 = 2l + 7
But the highest marks = 87
2l + 7 = 87

(iv) Let the base angle be b degrees
The base angle of an isosceles triangle are equal.
The other base angle = b degrees
Since the vertex angle = Twice either base angle = 2b degrees.
Also, the sum of three angles of triangle = 180°
b + b + 2b = 180°
(OR)
4b = 180°

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