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Introduction:
Welcome to Online Tutorial Classes! We are excited to present a valuable resource for CBSE students studying Class 7 Mathematics. In this post, we will delve into the solving the exercise 4.1 of NCERT class 7 Maths chapter 4 ‘Simple Equations’ from the NCERT curriculum and provide you with a comprehensive notes and worksheets to help you master this important topic.
Exercise 4.1 of NCERT Class 7 Maths Chapter 4, ‘Simple Equations’, introduces the concept of equations and variables, and explains how to form and solve equations. The chapter also discusses the properties of equations and their applications in practical situations.
NCERT Maths book:
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NCERT class 7 Maths Chapter 4 ‘Simple Equations’ Exercise 4.1 Overview
NCERT Class 7 Maths Chapter 4, Exercise 4.1 of ‘Simple Equations’, introduces students to the concept of equations and variables. An equation is a condition on a variable, which can vary and take on different numerical values. Variables are usually denoted by letters of the alphabet, such as x, y, z, l, m, n, p, etc. Expressions are formed by performing operations like addition, subtraction, multiplication, and division on the variables. The chapter explains that the value of an expression depends on the chosen value of the variable.
The chapter further elaborates on the properties of equations, including the fact that an equation remains the same when the expressions on the left and right are interchanged. It also discusses methods of solving equations, such as performing the same operations on both sides of the equation or transposing terms from one side to the other. The chapter concludes with applications of simple equations to practical situations.
CBSE class 7 Maths Chapter 4 ‘Simple Equations’ Exercise 4.1 Solutions
Question 1
Complete the last column of the table.
Solution:
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 (n = 1)
Solution:
LHS = n + 5
By substituting the value of n = 1,
LHS = n + 5
= 1 + 5
= 6
By comparing LHS and RHS,
6 ≠ 19
LHS ≠ RHS
Hence, the value of n = 1 is not a solution to the given equation n + 5 = 19.
(b) 7n + 5 = 19 (n = – 2)
Solution:
LHS = 7n + 5
By substituting the value of n = -2,
LHS = 7n + 5
= (7 × (-2)) + 5
= – 14 + 5
= – 9
By comparing LHS and RHS,
-9 ≠ 19
LHS ≠ RHS
Hence, the value of n = -2 is not a solution to the given equation 7n + 5 = 19.
(c) 7n + 5 = 19 (n = 2)
Solution:
LHS = 7n + 5
By substituting the value of n = 2,
LHS = 7n + 5
= (7 × (2)) + 5
= 14 + 5
= 19
By comparing LHS and RHS,
19 = 19
LHS = RHS
Hence, the value of n = 2 is a solution to the given equation 7n + 5 = 19.
(d) 4p – 3 = 13 (p = 1)
Solution:
LHS = 4p – 3
By substituting the value of p = 1,
LHS = 4p – 3
= (4 × 1) – 3
= 4 – 3
= 1
By comparing LHS and RHS,
1 ≠ 13
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation 4p – 3 = 13.
(e) 4p – 3 = 13 (p = – 4)
Solution:
LHS = 4p – 3
By substituting the value of p = – 4,
LHS = 4p – 3
= (4 × (-4)) – 3
= -16 – 3
= -19
By comparing LHS and RHS,
-19 ≠ 13
LHS ≠ RHS
Hence, the value of p = -4 is not a solution to the given equation 4p – 3 = 13.
(f) 4p – 3 = 13 (p = 0)
Solution:
LHS = 4p – 3
By substituting the value of p = 0,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
By comparing LHS and RHS,
– 3 ≠ 13
LHS ≠ RHS
Hence, the value of p = 0 is not a solution to the given equation 4p – 3 = 13.
Question 3
Solve the following equations by trial and error method:
(i) 5p + 2 = 17 (ii) 3m – 14 = 4
Solution:
LHS = 5p + 2
Let, p = 1
LHS = 5p + 2
= (5 × 1) + 2
= 5 + 2
= 7
By comparing LHS and RHS,
7 ≠ 17
LHS ≠ RHS
Hence, the value of p = 1 is not a solution to the given equation.
Let, p = 2
LHS = 5p + 2
= (5 × 2) + 2
= 10 + 2
= 12
By comparing LHS and RHS,
12 ≠ 17
LHS ≠ RHS
Hence, the value of p = 2 is not a solution to the given equation.
Let, p = 3
LHS = 5p + 2
= (5 × 3) + 2
= 15 + 2
= 17
By comparing LHS and RHS,
17 = 17
LHS = RHS
Hence, the value of p = 3 is a solution to the given equation.
(ii) 3m – 14 = 4
Solution:
LHS = 3m – 14
Let, m = 5
LHS = 3m – 14
= (3 × 5) – 14
= 15 – 14
= 1
By comparing LHS and RHS,
1 ≠ 4
LHS ≠ RHS
Hence, the value of m = 5 is not a solution to the given equation.
Let, m = 6
LHS = 3m – 14
= (3 × 6) – 14
= 18 – 14
= 4
By comparing LHS and RHS,
4 = 4
LHS = RHS
Hence, the value of m = 6 is a solution to the given equation.
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:
(i) The sum of numbers x and 4 is 9.
Solution:
The above statement can be written in the equation form as,
x + 4 = 9
(ii) 2 subtracted from y is 8.
Solution:
The above statement can be written in the equation form as,
y – 2 = 8
(iii) Ten times a is 70.
Solution:
The above statement can be written in the equation form as,
10a = 70
(iv) The number b divided by 5 gives 6.
Solution:
The above statement can be written in the equation form as,
(b/5) = 6
(v) Three-fourths of t is 15.
Solution:
The above statement can be written in the equation form as,
¾t = 15
(vi) Seven times m plus 7 gets you 77.
Solution:
The above statement can be written in the equation form as,
Seven times m is 7m.
∴ the equation can be written as:
7m + 7 = 77
(vii) One-fourth of a number x minus 4 gives 4.
Solution:
The above statement can be written in the equation form as,
One-fourth of a number x is x/4.
∴ the equation can be written as:
x/4 – 4 = 4
(viii) If you take away 6 from 6 times y, you get 60.
Solution:
The above statement can be written in the equation form as,
6 times y is 6y.
∴ the equation can be written as:
6y – 6 = 60
(ix) If you add 3 to one-third of z, you get 30.
Solution:
The above statement can be written in the equation form as,
One-third of z is z/3.
∴ the equation can be written as:
3 + z/3 = 30
Question 5
Write the following equations in statement forms:
Solution:
(i) p + 4 = 15
Solution:
The sum of numbers p and 4 is 15.
(ii) m – 7 = 3
Solution:
7 subtracted from m is 3.
(iii) 2m = 7
Solution:
Twice of number m is 7.
(iv) m/5 = 3
Solution:
One fifth of m is 3.
(v) (3m)/5 = 6
Solution:
Three-fifth of m is 6.
(vi) 3p + 4 = 25
Solution:
Three times p plus 4 is 25.
(vii) 4p – 2 = 18
Solution:
Four times p minus 2 is 18.
(viii) p/2 + 2 = 8
Solution:
If you add half of a number p to 2, you get 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.)
(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:
From the question, it is given that
Number of Parmit’s marbles = m
Then,
Irfan has 7 marbles, more than five times the marbles Parmit has.
∴ 5 × Number of Parmit’s marbles + 7 = Total number of marbles Irfan having
Or (5 × m) + 7 = 37
Or 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 be = y years
∴ Lakshmi’s father is 4 years older than three times her age.
Or 3 × Laxmi’s age + 4 = Age of Lakshmi’s father
Or (3 × y) + 4 = 49
Or 3y + 4 = 49
(iii) The teacher tells the class that the highest marks obtained by a student in her class are twice the lowest marks plus 7. The highest score is 87 (Take the lowest score to be l).
Solution:
Highest score in the class = 87
Let the lowest score be l.
∴ 2 × Lowest score + 7 = Highest score in the class
Or (2 × l) + 7 = 87
Or 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:
The sum of angles of a triangle is 180o
Let the base angle be b.
∴ Vertex angle = 2 × base angle = 2b
Or b + b + 2b = 180o
Or 4b = 180o
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CBSE class 7 Mathematics Notes
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