Well-designed AP Board Solutions Class 7 Maths Chapter 4 Simple Equations Exercise 4.1 offers step-by-step explanations to help students understand problem-solving strategies.
Simple Equations Class 7 Exercise 4.1 Solutions – 7th Class Maths 4.1 Exercise Solutions
Question 1.
Complete the last column of the table.
Solution:
i) x + 3 = 0
∵ x = 3 ⇒ 3 + 3 = 6 ≠ 0 No
ii) x + 3 = 0
∵ x = 0 ⇒ 0 + 3 = 3 ≠ 0 No
iii) x + 3 = 0
∵ x = – 3 ⇒ – 3 + 3 = 0 Yes
iv) x – 7 = 1
∵ x = 7 ⇒ 7 – 7 = 0 ≠ 1 No
v) x – 7 = 1
∵ x = 8 ⇒ 8 – 7 = 1 Yes
vi) 5x = 25
∵ x = 0 ⇒ 5 × 0 = 5 No
vii) 5x = 25
∵ x = 5 ⇒ 5(5) = 25 Yes
viii) 5x = 25
∵ x = – 5 ⇒ 5(-5) = – 25 ≠ 25 No
ix) \(\frac { m }{ 3 }\) = 2
∵ m = – 6 ⇒ \(\frac { -6 }{ 3 }\) = – 2 ≠ 2 No
x) \(\frac { m }{ 3 }\) = 2
∵ m = 0 ⇒ \(\frac { 0 }{ 3 }\) = 0 ≠ 2 No
xi) \(\frac { m }{ 3 }\) = 2
∵ m = 6 ⇒ \(\frac { 6 }{ 3 }\) = 2 Yes
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)
Solution:
n + 5 = 19
∵ n = 1 ⇒ 1 + 5 = 6 ≠ 19
n = 1 is not the solution.
b) 7n + 5=19(n = – 2)
Solution:
7n + 5 = 19 ∵ n = – 2
7n + 5 = 7(- 2) + 5
= – 14 + 5 = – 9 ≠ 19
∴ n = – 2 is not the solution.
c) 7n + 5 = 19 (n = 2)
Solution:
7n + 5 = 19
∵ n = 2
7(2) + 5 = 14 + 5 = 19
∴ n = 2 is the solution.
d) 4p – 3 = 13 (p = 1)
Solution:
4p – 3 = 13 ∵ p = 1
4p – 3 = 4(1) – 3 = 4 – 3 = 1 ≠ 13
∴ p = 1 is not the solution.
e) 4p – 3 = 13(p = – 4)
Solution:
4p – 3 = 13
4p – 3 = 4(-4) – 3 = – 16 – 3 = – 19 ≠ 13
∴ p = – 4 is not the solution.
f) 4p – 3 = 13(p = 0)
Solution:
4p – 3 = 13 ∵ p = 0
4p – 3 = 4(0) – 3 = – 3 ≠ 13
∴ p = 0 is not the solution.
Question 3.
Solve the following equations by trial and error method:
i) 5p + 2 = 17
Solution:
Trial and error method
Let p = 1
LHS 5p + 2 = 5(1) + 2 = 5 + 2 = 7
RHS = 17
∴ LHS ≠ RHS
p = 1 is not the solution
Let p = 2
LHS 5p + 2 = 5(2) + 2 = 10 + 2 = 12
RHS = 17
LHS ≠ RHS
p = 2 is not the solution
Let p = 3
LHS 5p + 2 = 5(3) + 2 = 15 + 2 = 17
RHS = 17
∴ LHS = RHS
∴ p = 3 is the solution.
ii) 3m – 14 = 4
Solution:
Let m = 1
LHS = 3m – 14 = 3(1) – 14 = 3 – 14 = – 11
RHS = 4
LHS ≠ RHS
∴ m = 1 is not the solution
Let m = 3
LHS = 3m – 14 = 3(3) – 14 = 9 – 14 = – 5
RHS = 4
∴ LHS ≠ RHS
∴ m = 3 is not the solution
Let m = 6
LHS 3m – 14 = 3(6) – 14 = 18 – 14 = 4
RHS = 4
∴ LHS = RHS
∴ m = 6 is the solution
Question 4.
Write equations for the following statements :
i) The sum of numbers x and 4 is 9.
Solution:
x + 4 = 9
ii) 2 subtracted from y is 8.
Solution:
y – 2 = 8
iii) Ten times a is 70.
Solution:
10 a = 70
iv) The number b divided by 5 gives 6 .
Solution:
\(\frac { b }{ 5 }\) = 6
v) Three-fourth of t is 15.
Solution:
\(\frac { 3 }{ 4 }\)(t) = 15
vi) Seven times m plus 7 gets you 77 .
Solution:
7m + 7 = 77
vii) One-fourth of a number x minus 4 gives 4.
Solution:
\(\frac { 1 }{ 4 }\)x – 4 = 4
viii) If you take away 6 from 6 times y, you get 60.
Solution:
6y – 6 = 60
ix) If you add 3 to one-third of z, you get 30 .
Solution:
\(\frac { 1 }{ 3 }\)z + 3 = 30
Question 5.
Write the following equations in statement forms:
i) p + 4 = 15
Solution:
4 is added to p gives 15
ii) m – 7 = 3
Solution:
7 is subtracted from m, the difference is 3 .
iii) 2m = 7
Solution:
Two times of m is 7
iv) \(\frac { m }{ 5 }\) = 3
Solution:
m is divided by 5 gives quotient 3
v) \(\frac { 3m }{ 5 }\) = 6
Solution:
3 times of m divided by 5 gives 6.
vi) 3p + 4 = 25
Solution:
3 times of p is added to 4 to give result 25
vii) 4p – 2 = 18
Solution:
2 is subtracted from 4 times of p gives 18
viii) \(\frac { p }{ 2 }\) + 2 = 8
Solution:
p is divided by 2 and 2 is added the result is 8.
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.)
Solution:
Let m be the number of Parmit marbles
5 times of m = 5m
7 more than 5m = 5m + 7
According to the problem
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 Laxmi’s age to be ‘y’ years 3 times Laxmi’s age = 3y years
4 year older than 3y years = 3y + 4
According to the problem 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 = l
Twice of l = 2l
Twice of l plus 7 = 2l + 7
According to the problem 2l + 7 = 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 angle = b°
Twice the base angle = 2b°
We know that the sum of the angles in a triangle = 180°
2b + b + b = 180°
4b = 180°
b = \(\frac{180}{4}\)
b = 45°