NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations
Neha Tanna27 Feb, 2024
Share
NCERT Solutions for Class 7 Maths Chapter 4
NCERT Solutions for Class 7 Maths Chapter 4: Students can use Simple Equations to help them with their homework and exam preparation. We have developed NCERT Solutions for Class 7 Maths Chapter 4 step-by-step solutions with thorough descriptions for students who are anxious about tackling NCERT Solutions for Class 7 Maths Chapter 4. We advise students who want to improve their math grades to review these NCERT Solutions for Class 7 Maths Chapter 4 and broaden their understanding.
NCERT Solutions for Class 7 Maths Chapter 4 PDF
Using the link, you can get the NCERT Solutions for Class 7 Maths Chapter 4 PDF. All of the chapter's exercises and problems have comprehensive answers available in this NCERT Solutions for Class 7 Maths Chapter 4 PDF. These answers are intended to make it easier and more accurate for students to comprehend and answer problems using integers. Students can enhance their mathematical abilities and solidify their comprehension of integers by consulting these NCERT Solutions for Class 7 Maths Chapter 4.NCERT Solutions for Class 7 Maths Chapter 4 PDF
NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations
Below we have provided NCERT Solutions for Class 7 Maths Chapter 4 for students to help them understand the chapter better and to score good marks in their examination.1. Complete the last column of the table.
| S. No. | Equation | Value | Say whether the equation is satisfied. (Yes/No) |
| (i) | x + 3 = 0 | x = 3 | |
| (ii) | x + 3 = 0 | x = 0 | |
| (iii) | x + 3 = 0 | x = -3 | |
| (iv) | x – 7 = 1 | x = 7 | |
| (v) | x – 7 = 1 | x = 8 | |
| (vi) | 5x = 25 | x = 0 | |
| (vii) | 5x = 25 | x = 5 | |
| (viii) | 5x = 25 | x = -5 | |
| (ix) | (m/3) = 2 | m = – 6 | |
| (x) | (m/3) = 2 | m = 0 | |
| (xi) | (m/3) = 2 | m = 6 |
Solution:
(i) x + 3 = 0
LHS = x + 3 By substituting the value of x = 3 Then, LHS = 3 + 3 = 6 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied.(ii) x + 3 = 0
LHS = x + 3 By substituting the value of x = 0, Then, LHS = 0 + 3 = 3 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied.(iii) x + 3 = 0
LHS = x + 3 By substituting the value of x = – 3, Then, LHS = – 3 + 3 = 0 By comparing LHS and RHS, LHS = RHS ∴ Yes, the equation is satisfied.(iv) x – 7 = 1
LHS = x – 7 By substituting the value of x = 7, Then, LHS = 7 – 7 = 0 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied. (v) x – 7 = 1 LHS = x – 7 By substituting the value of x = 8, Then, LHS = 8 – 7 = 1 By comparing LHS and RHS, LHS = RHS ∴ Yes, the equation is satisfied.(vi) 5x = 25
LHS = 5x By substituting the value of x = 0, Then, LHS = 5 × 0 = 0 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied.(vii) 5x = 25
LHS = 5x By substituting the value of x = 5, Then, LHS = 5 × 5 = 25 By comparing LHS and RHS, LHS = RHS ∴ Yes, the equation is satisfied.(viii) 5x = 25
LHS = 5x By substituting the value of x = -5, Then, LHS = 5 × (-5) = – 25 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied. (ix) m/3 = 2 LHS = m/3 By substituting the value of m = – 6, Then, LHS = -6/3 = – 2 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied.(x) m/3 = 2
LHS = m/3 By substituting the value of m = 0, Then, LHS = 0/3 = 0 By comparing LHS and RHS, LHS ≠ RHS ∴ No, the equation is not satisfied.(xi) m/3 = 2
LHS = m/3 By substituting the value of m = 6, Then, LHS = 6/3 = 2 By comparing LHS and RHS, LHS = RHS ∴ Yes, the equation is satisfied.| S. No. | 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 | Yes |
| (viii) | 5x = 25 | x = -5 | No |
| (ix) | (m/3) = 2 | m = – 6 | No |
| (x) | (m/3) = 2 | m = 0 | No |
| (xi) | (m/3) = 2 | m = 6 | Yes |
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:
LHS = n + 5 By substituting the value of n = 1, Then, 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, Then, 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.| CBSE Syllabus Class 7 | |
| CBSE Class 7 English Syllabus | CBSE Class 7 Math Syllabus |
| CBSE Class 7 Social Science Syllabus | CBSE Class 7 Science Syllabus |
(c) 7n + 5 = 19 (n = 2)
Solution:
LHS = 7n + 5 By substituting the value of n = 2, Then, 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, Then, 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, Then, 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, Then, 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.CBSE Board Exam Centre List 2024
3. Solve the following equations by trial and error method.
(i) 5p + 2 = 17
Solution:
LHS = 5p + 2 By substituting the value of p = 0, Then, LHS = 5p + 2 = (5 × 0) + 2 = 0 + 2 = 2 By comparing LHS and RHS, 2 ≠ 17 LHS ≠ RHS Hence, the value of p = 0 is not a solution to the given equation. 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 By substituting the value of m = 3, Then, LHS = 3m – 14 = (3 × 3) – 14 = 9 – 14 = – 5 By comparing LHS and RHS, -5 ≠ 4 LHS ≠ RHS Hence, the value of m = 3 is not a solution to the given equation. Let, m = 4 LHS = 3m – 14 = (3 × 4) – 14 = 12 – 14 = – 2 By comparing LHS and RHS, -2 ≠ 4 LHS ≠ RHS Hence, the value of m = 4 is not a solution to the given equation. 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.Related Links -
4. Write equations for the following statements.
(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. = 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. = 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. = 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. = 3 + z/3 = 305. Write the following equations in statement forms.
(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:
The number m divided by 5 gives 3.(v) (3m)/5 = 6
Solution:
Three-fifth of m is 6.(vi) 3p + 4 = 25
Solution:
Three times p plus 4 gives you 25.(vii) 4p – 2 = 18
Solution:
Four times p minus 2 gives you 18.(viii) p/2 + 2 = 8
Solution:
If you add half of a number p to 2, you get 8.6. 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:
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 = (5 × m) + 7 = 37 = 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:
From the question, it is given that Let Laxmi’s age be = y years old Then, Lakshmi’s father is 4 years older than three times her age. = 3 × Laxmi’s age + 4 = Age of Lakshmi’s father = (3 × y) + 4 = 49 = 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:
From the question, it is given that Highest score in the class = 87 Let the lowest score be l. = 2 × Lowest score + 7 = Highest score in the class = (2 × l) + 7 = 87 = 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:
From the question, it is given that We know that the sum of angles of a triangle is 180 o Let the base angle be b. Then, Vertex angle = 2 × base angle = 2b = b + b + 2b = 180 o = 4b = 180 oNCERT Solutions for Class 7 Maths Chapter 4 FAQs
What is the name of Chapter 4 of Class 7 maths?
What is the rule of simple equation?
What is an example of a simple equation for Class 7?
Live & recorded classes available at easeDashboard for progress trackingMillions of practice questions at your fingertips
