NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations
NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations are meticulously crafted to assist students in their exam preparations and assignments. These solutions provide comprehensive guidance for students aiming to excel in Mathematics and secure impressive grades.
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The NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Students can access the NCERT Solutions for Class 7 Maths Chapter 4 - Simple Equations. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Maths much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise 4.1
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 |
(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 |
Solve the following equations by trial and error method.
(i) 5p + 2 = 17
(ii) 3m – 14 = 4
(i) 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) 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.
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 are 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).
(i) 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) 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) 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) From the question, it is given that,
We know that the sum of angles of a triangle is 180o
Let the base angle be b.
Then,
Vertex angle = 2 × base angle = 2b
= b + b + 2b = 180o
= 4b = 180o
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)
(a) 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) 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.
(c) 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) 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) 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) 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.
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.
(i) The above statement can be written in the equation form as,
= x + 4 = 9
(ii) The above statement can be written in the equation form as,
= y – 2 = 8
(iii) The above statement can be written in the equation form as,
= 10a = 70
(iv) The above statement can be written in the equation form as,
= (b/5) = 6
(v) The above statement can be written in the equation form as,
= ¾t = 15
(vi) The above statement can be written in the equation form as,
Seven times m is 7m.
= 7m + 7 = 77
(vii) 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) The above statement can be written in the equation form as,
6 times y is 6y.
= 6y – 6 = 60
(ix) The above statement can be written in the equation form as,
One-third of z is z/3.
= 3 + z/3 = 30
Write the following equations in statement forms.
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) m/5 = 3
(v) (3m)/5 = 6
(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) p/2 + 2 = 8
(i) The sum of numbers p and 4 is 15.
(ii) 7 subtracted from m is 3.
(iii) Twice of number m is 7.
(iv) The number m divided by 5 gives 3.
(v) Three-fifth of m is 6.
(vi) Three times p plus 4 gives you 25.
(vii) Four times p minus 2 gives you 18.
(viii) If you add half of a number p to 2, you get 8.
Exercise 4.2
Give first the step you will use to separate the variable and then solve the equation.
(a) 3l = 42
(b) b/2 = 6
(c) p/7 = 4
(d) 4x = 25
(e) 8y = 36
(f) (z/3) = (5/4)
(g) (a/5) = (7/15)
(h) 20t = – 10
(a) Now, we have to divide both sides of the equation by 3.
Then, we get
= 3l/3 = 42/3
= l = 14
(b) Now, we have to multiply both sides of the equation by 2.
Then, we get
= b/2 × 2= 6 × 2
= b = 12
(c) Now, we have to multiply both sides of the equation by 7.
Then, we get
= p/7 × 7= 4 × 7
= p = 28
(d) Now, we have to divide both sides of the equation by 4.
Then, we get
= 4x/4 = 25/4
= x = 25/4
(e) Now, we have to divide both sides of the equation by 8.
Then, we get
= 8y/8 = 36/8
= x = 9/2
(f) Now, we have to multiply both sides of the equation by 3.
Then, we get
= (z/3) × 3 = (5/4) × 3
= x = 15/4
(g) Now, we have to multiply both sides of the equation by 5.
Then, we get
= (a/5) × 5 = (7/15) × 5
= a = 7/3
(h) Now, we have to divide both sides of the equation by 20.
Then, we get
= 20t/20 = -10/20
= x = – ½
Give the steps you will use to separate the variable and then solve the equation.
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) 20p/3 = 40
(d) 3p/10 = 6
(a) First, we have to add 2 to both sides of the equation.
Then, we get
= 3n – 2 + 2 = 46 + 2
= 3n = 48
Now,
We have to divide both sides of the equation by 3.
Then, we get
= 3n/3 = 48/3
= n = 16
(b) First, we have to subtract 7 from both sides of the equation.
Then, we get
= 5m + 7 – 7 = 17 – 7
= 5m = 10
Now,
We have to divide both sides of the equation by 5.
Then, we get
= 5m/5 = 10/5
= m = 2
(c) First, we have to multiply both sides of the equation by 3.
Then, we get
= (20p/3) × 3 = 40 × 3
= 20p = 120
Now,
We have to divide both sides of the equation by 20.
Then, we get
= 20p/20 = 120/20
= p = 6
(d) First, we have to multiply both sides of the equation by 10.
Then, we get
= (3p/10) × 10 = 6 × 10
= 3p = 60
Now,
We have to divide both sides of the equation by 3.
Then, we get
= 3p/3 = 60/3
= p = 20
Solve the following equations.
(a) 10p = 100
(b) 10p + 10 = 100
(c) p/4 = 5
(d) – p/3 = 5
(e) 3p/4 = 6
(f) 3s = – 9
(g) 3s + 12 = 0
(h) 3s = 0
(i) 2q = 6
(j) 2q – 6 = 0
(k) 2q + 6 = 0
(l) 2q + 6 = 12
(a) Now,
We have to divide both sides of the equation by 10.
Then, we get
= 10p/10 = 100/10
= p = 10
(b) First, we have to subtract 10 from both sides of the equation.
Then, we get
= 10p + 10 – 10 = 100 – 10
= 10p = 90
Now,
We have to divide both sides of the equation by 10.
Then, we get
= 10p/10 = 90/10
= p = 9
(c) Now,
We have to multiply both sides of the equation by 4.
Then, we get
= p/4 × 4 = 5 × 4
= p = 20
(d) Now,
We have to multiply both sides of the equation by – 3.
Then, we get
= – p/3 × (- 3) = 5 × (- 3)
= p = – 15
(e) First, we have to multiply both sides of the equation by 4.
Then, we get
= (3p/4) × (4) = 6 × 4
= 3p = 24
Now,
We have to divide both sides of the equation by 3.
Then, we get
= 3p/3 = 24/3
= p = 8
(f) Now,
We have to divide both sides of the equation by 3.
Then, we get
= 3s/3 = -9/3
= s = -3
(g) First, we have to subtract 12 from both sides of the equation.
Then, we get
= 3s + 12 – 12 = 0 – 12
= 3s = -12
Now,
We have to divide both sides of the equation by 3.
Then, we get
= 3s/3 = -12/3
= s = – 4
(h) Now,
We have to divide both sides of the equation by 3.
Then, we get
= 3s/3 = 0/3
= s = 0
(i) Now,
We have to divide both sides of the equation by 2.
Then, we get
= 2q/2 = 6/2
= q = 3
(j) First, we have to add 6 to both sides of the equation.
Then, we get
= 2q – 6 + 6 = 0 + 6
= 2q = 6
Now,
We have to divide both sides of the equation by 2.
Then, we get
= 2q/2 = 6/2
= q = 3
(k) First, we have to subtract 6 from both sides of the equation.
Then, we get
= 2q + 6 – 6 = 0 – 6
= 2q = – 6
Now,
We have to divide both sides of the equation by 2.
Then, we get
= 2q/2 = – 6/2
= q = – 3
(l) First, we have to subtract 6 from both sides of the equation.
Then, we get
= 2q + 6 – 6 = 12 – 6
= 2q = 6
Now,
We have to divide both sides of the equation by 2.
Then, we get
= 2q/2 = 6/2
= q = 3
Give first the step you will use to separate the variable and then solve the equation.
(a) x – 1 = 0
(b) x + 1 = 0
(c) x – 1 = 5
(d) x + 6 = 2
(e) y – 4 = – 7
(f) y – 4 = 4
(g) y + 4 = 4
(h) y + 4 = – 4
(a) We have to add 1 to both sides of the given equation.
Then, we get
= x – 1 + 1 = 0 + 1
= x = 1
(b) We have to subtract 1 from both sides of the given equation.
Then, we get
= x + 1 – 1 = 0 – 1
= x = – 1
(c) We have to add 1 to both sides of the given equation.
Then, we get
= x – 1 + 1 = 5 + 1
= x = 6
(d) We have to subtract 6 from both sides of the given equation.
Then, we get
= x + 6 – 6 = 2 – 6
= x = – 4
(e) We have to add 4 to both sides of the given equation.
Then, we get
= y – 4 + 4 = – 7 + 4
= y = – 3
(f) We have to add 4 to both sides of the given equation.
Then, we get
= y – 4 + 4 = 4 + 4
= y = 8
(g) We have to subtract 4 from both sides of the given equation.
Then, we get
= y + 4 – 4 = 4 – 4
= y = 0
(h) We have to subtract 4 from both sides of the given equation.
Then, we get
= y + 4 – 4 = – 4 – 4
= y = – 8
Exercise 4.3
Solve the following equations:
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = – 21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8
(a) Let us divide both the sides by 2,
= (2(x + 4))/2 = 12/2
= x + 4 = 6
By transposing 4 from LHS to RHS it becomes -4
= x = 6 – 4
= x = 2
(b) Let us divide both the sides by 3,
= (3(n – 5))/3 = 21/3
= n – 5 = 7
By transposing -5 from LHS to RHS it becomes 5
= n = 7 + 5
= n = 12
(c) Let us divide both the sides by 3,
= (3(n – 5))/3 = – 21/3
= n – 5 = -7
By transposing -5 from LHS to RHS it becomes 5
= n = – 7 + 5
= n = – 2
(d) Let us divide both the sides by -4,
= (-4(2 + x))/ (-4) = 8/ (-4)
= 2 + x = -2
By transposing 2 from LHS to RHS it becomes – 2
= x = -2 – 2
= x = – 4
(e) Let us divide both the sides by 4,
= (4(2 – x))/ 4 = 8/ 4
= 2 – x = 2
By transposing 2 from LHS to RHS it becomes – 2
= – x = 2 – 2
= – x = 0
= x = 0
Solve the following equations:
(a) 4 = 5(p – 2)
(b) – 4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) =34
(e) 0 = 16 + 4(m – 6)
(a) Let us divide both the sides by 5,
= 4/5 = (5(p – 2))/5
= 4/5 = p -2
By transposing – 2 from RHS to LHS it becomes 2
= (4/5) + 2 = p
= (4 + 10)/ 5 = p
= p = 14/5
(b) Let us divide both the sides by 5,
= – 4/5 = (5(p – 2))/5
= – 4/5 = p -2
By transposing – 2 from RHS to LHS it becomes 2
= – (4/5) + 2 = p
= (- 4 + 10)/ 5 = p
= p = 6/5
(c) By transposing 4 from RHS to LHS it becomes – 4
= 16 – 4 = 3(t + 2)
= 12 = 3(t + 2)
Let us divide both the sides by 3,
= 12/3 = (3(t + 2))/ 3
= 4 = t + 2
By transposing 2 from RHS to LHS it becomes – 2
= 4 – 2 = t
= t = 2
(d) By transposing 4 from LHS to RHS it becomes – 4
= 5(p – 1) = 34 – 4
= 5(p – 1) = 30
Let us divide both the sides by 5,
= (5(p – 1))/ 5 = 30/5
= p – 1 = 6
By transposing – 1 from RHS to LHS it becomes 1
= p = 6 + 1
= p = 7
(e) By transposing 16 from RHS to LHS it becomes – 16
= 0 – 16 = 4(m – 6)
= – 16 = 4(m – 6)
Let us divide both the sides by 4,
= – 16/4 = (4(m – 6))/ 4
= – 4 = m – 6
By transposing – 6 from RHS to LHS it becomes 6
= – 4 + 6 = m
= m = 2
(a) Construct 3 equations starting with x = 2
(b) Construct 3 equations starting with x = – 2
(a) First equation is,
Multiply both sides by 6
= 6x = 12 … [equation 1]
Second equation is,
Subtracting 4 from both sides,
= 6x – 4 = 12 -4
= 6x – 4 = 8 … [equation 2]
Third equation is,
Divide both sides by 6
= (6x/6) – (4/6) = (8/6)
= x – (4/6) = (8/6) … [equation 3]
(b) First equation is,
Multiply both sides by 5
= 5x = -10 … [equation 1]
Second equation is,
Subtracting 3 from both sides,
= 5x – 3 = – 10 – 3
= 5x – 3 = – 13 … [equation 2]
Third equation is,
Dividing both sides by 2
= (5x/2) – (3/2) = (-13/2) … [equation 3]
Solve the following equations:
(a) 2y + (5/2) = (37/2)
(b) 5t + 28 = 10
(c) (a/5) + 3 = 2
(d) (q/4) + 7 = 5
(e) (5/2) x = -5
(f) (5/2) x = 25/4
(g) 7m + (19/2) = 13
(h) 6z + 10 = – 2
(i) (3/2) l = 2/3
(j) (2b/3) – 5 = 3
(a) By transposing (5/2) from LHS to RHS it becomes -5/2
Then,
= 2y = (37/2) – (5/2)
= 2y = (37-5)/2
= 2y = 32/2
Now,
Divide both sides by 2,
= 2y/2 = (32/2)/2
= y = (32/2) × (1/2)
= y = 32/4
= y = 8
(b) By transposing 28 from LHS to RHS it becomes -28
Then,
= 5t = 10 – 28
= 5t = – 18
Now,
Divide both sides by 5,
= 5t/5= -18/5
= t = -18/5
(c) By transposing 3 from LHS to RHS it becomes -3
Then,
= a/5 = 2 – 3
= a/5 = – 1
Now,
Multiply both sides by 5,
= (a/5) × 5= -1 × 5
= a = -5
(d) By transposing 7 from LHS to RHS it becomes -7
Then,
= q/4 = 5 – 7
= q/4 = – 2
Now,
Multiply both sides by 4,
= (q/4) × 4= -2 × 4
= a = -8
(e) First we have to multiply both the sides by 2,
= (5x/2) × 2 = – 5 × 2
= 5x = – 10
Now,
We have to divide both the sides by 5,
Then we get,
= 5x/5 = -10/5
= x = -2
(f) First we have to multiply both the sides by 2,
= (5x/2) × 2 = (25/4) × 2
= 5x = (25/2)
Now,
We have to divide both the sides by 5,
Then we get,
= 5x/5 = (25/2)/5
= x = (25/2) × (1/5)
= x = (5/2)
(g) By transposing (19/2) from LHS to RHS it becomes -19/2
Then,
= 7m = 13 – (19/2)
= 7m = (26 – 19)/2
= 7m = 7/2
Now,
Divide both sides by 7,
= 7m/7 = (7/2)/7
= m = (7/2) × (1/7)
= m = ½
(h) By transposing 10 from LHS to RHS it becomes – 10
Then,
= 6z = -2 – 10
= 6z = – 12
Now,
Divide both sides by 6,
= 6z/6 = -12/6
= m = – 2
(i) First we have to multiply both the sides by 2,
= (3l/2) × 2 = (2/3) × 2
= 3l = (4/3)
Now,
We have to divide both the sides by 3,
Then we get,
= 3l/3 = (4/3)/3
= l = (4/3) × (1/3)
= x = (4/9)
(j) By transposing -5 from LHS to RHS it becomes 5
Then,
= 2b/3 = 3 + 5
= 2b/3 = 8
Now,
Multiply both sides by 3,
= (2b/3) × 3= 8 × 3
= 2b = 24
And,
Divide both sides by 2,
= 2b/2 = 24/2
= b = 12
Exercise 4.4
Solve the following:
(a) 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. What is the lowest score?
(b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°.
What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
(c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
(a) Let us assume the lowest score is x
From the question it is given that,
The highest score is = 87
Highest marks obtained by a student in her class is twice the lowest marks plus 7= 2x + 7
5/2 of the number = (5/2) x
The above given statement can be written in the equation form as,
Then,
= 2x + 7 = Highest score
= 2x + 7 = 87
By transposing 7 from LHS to RHS it becomes -7
= 2x = 87 – 7
= 2x = 80
Now,
Divide both the sides by 2
= 2x/2 = 80/2
= x = 40
Hence, the lowest score is 40
(b) From the question it is given that,
We know that, the sum of angles of a triangle is 180o
Let base angle be b
Then,
= b + b + 40o = 180o
= 2b + 40 = 180o
By transposing 40 from LHS to RHS it becomes -40
= 2b = 180 – 40
= 2b = 140
Now,
Divide both the sides by 2
= 2b/2 = 140/2
= b = 70o
Hence, 70o is the base angle of an isosceles triangle.
(c) Let us assume Rahul’s score is x
Then,
Sachin scored twice as many runs as Rahul is 2x
Together, their runs fell two short of a double century,
= Rahul’s score + Sachin’s score = 200 – 2
= x + 2x = 198
= 3x = 198
Divide both the sides by 3,
= 3x/3 = 198/3
= x = 66
So, Rahul’s score is 66
And Sachin’s score is 2x = 2 × 66 = 132
Solve the following:
(i) Irfan says that he has 7 marbles more than five times the marbles Parmit has.
Irfan has 37 marbles. How many marbles does Parmit have?
(ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age.
What is Laxmi’s age?
(iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
(i) Let us assume number of Parmit’s marbles = m
From the question it is given that,
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 has
= (5 × m) + 7 = 37
= 5m + 7 = 37
By transposing 7 from LHS to RHS it becomes -7
= 5m = 37 – 7
= 5m = 30
Divide both the sides by 5
= 5m/5 = 30/5
= m = 6
So, Permit has 6 marbles
(ii) Let Laxmi’s age be = y years old
From the question it is given that,
Lakshmi’s father is 4 years older than three times of her age
= 3 × Laxmi’s age + 4 = Age of Lakshmi’s father
= (3 × y) + 4 = 49
= 3y + 4 = 49
By transposing 4 from LHS to RHS it becomes -4
= 3y = 49 – 4
= 3y = 45
Divide both the sides by 3
= 3y/3 = 45/3
= y = 15
So, Lakshmi’s age is 15 years.
(iii) Let the number of fruit tress be f.
From the question it is given that,
3 × number of fruit trees + 2 = number of non-fruit trees
= 3f + 2 = 77
By transposing 2 from LHS to RHS it becomes -2
=3f = 77 – 2
= 3f = 75
Divide both the sides by 3
= 3f/3 = 75/3
= f = 25
So, number of fruit tree was 25.
Solve the following riddle:
I am a number,
Tell my identity!
Take me seven times over
And add a fifty!
To reach a triple century
You still need forty!
Let us assume the number is x.
Take me seven times over and add a fifty = 7x + 50
To reach a triple century you still need forty = (7x + 50) + 40 = 300
= 7x + 50 + 40 = 300
= 7x + 90 = 300
By transposing 90 from LHS to RHS it becomes -90
= 7x = 300 – 90
= 7x = 210
Divide both sides by 7
= 7x/7 = 210/7
= x = 30
Hence, the number is 30.
Set up equations and solve them to find the unknown numbers in the following cases:
(a) Add 4 to eight times a number; you get 60.
(b) One-fifth of a number minus 4 gives 3.
(c) If I take three-fourths of a number and add 3 to it, I get 21.
(d) When I subtracted 11 from twice a number, the result was 15.
(e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
(f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
(g) Anwar thinks of a number. If he takes away 7 from 5/2 of the number, the result is 23.
(a) Let us assume the required number be x
Eight times a number = 8x
The given above statement can be written in the equation form as,
= 8x + 4 = 60
By transposing 4 from LHS to RHS it becomes – 4
= 8x = 60 – 4
= 8x = 56
Divide both side by 8,
Then we get,
= (8x/8) = 56/8
= x = 7
(b) Let us assume the required number be x
One-fifth of a number = (1/5) x = x/5
The given above statement can be written in the equation form as,
= (x/5) – 4 = 3
By transposing – 4 from LHS to RHS it becomes 4
= x/5 = 3 + 4
= x/5 = 7
Multiply both side by 5,
Then we get,
= (x/5) × 5 = 7 × 5
= x = 35
(c) Let us assume the required number is x
Three-fourths of a number = (3/4) x
The given above statement can be written in the equation form as,
= (3/4) x + 3 = 21
By transposing 3 from LHS to RHS it becomes – 3
= (3/4) x = 21 – 3
= (3/4) x = 18
Multiply both sides by 4,
Then we get,
= (3x/4) × 4 = 18 × 4
= 3x = 72
Then,
Divide both sides by 3,
= (3x/3) = 72/3
= x = 24
(d) Let us assume the required number is x
Twice a number = 2x
The given above statement can be written in the equation form as,
= 2x –11 = 15
By transposing -11 from LHS to RHS it becomes 11
= 2x = 15 + 11
= 2x = 26
Then,
Divide both sides by 2,
= (2x/2) = 26/2
= x = 13
(e) Let us assume the required number is x
Thrice the number = 3x
The given above statement can be written in the equation form as,
= 50 – 3x = 8
By transposing 50 from LHS to RHS it becomes – 50
= – 3x = 8 – 50
= -3x = – 42
Then,
Divide both sides by -3,
= (-3x/-3) = – 42/-3
= x = 14
(f) Let us assume the required number is x
The given above statement can be written in the equation form as,
= (x + 19)/5 = 8
Multiply both sides by 5,
= ((x + 19)/5) × 5 = 8 × 5
= x + 19 = 40
Then,
By transposing 19 from LHS to RHS it becomes – 19
= x = 40 – 19
= x = 21
(g) Let us assume the required number is x
5/2 of the number = (5/2) x
The given above statement can be written in the equation form as,
= (5/2) x – 7 = 23
By transposing -7 from LHS to RHS it becomes 7
= (5/2) x = 23 + 7
= (5/2) x = 30
Multiply both sides by 2,
= ((5/2) x) × 2 = 30 × 2
= 5x = 60
Then,
Divide both the sides by 5
= 5x/5 = 60/5
= x = 12
Frequently Asked Questions
The NCERT solution for Class 7 Chapter 4: Simple Equations is important as it provides a structured approach to learning, ensuring that students develop a strong understanding of foundational concepts early in their academic journey. By mastering these basics, students can build confidence and readiness for tackling more difficult concepts in their further education.
Yes, the NCERT solution for Class 7 Chapter 4: Simple Equations is quite useful for students in preparing for their exams. The solutions are simple, clear, and concise allowing students to understand them better. Simple Equationsally, they can solve the practice questions and exercises that allow them to get exam-ready in no time.
You can get all the NCERT solutions for Class 7 Maths Chapter 4 from the official website of the Orchids International School. These solutions are tailored by subject matter experts and are very easy to understand.
Yes, students must practice all the questions provided in the NCERT solution for Class 7 Maths Chapter 4: Simple Equations as it will help them gain a comprehensive understanding of the concept, identify their weak areas, and strengthen their preparation.
Students can utilize the NCERT solution for Class 7 Maths Chapter 4 effectively by practicing the solutions regularly. Solve the exercises and practice questions given in the solution.