Swiflearn > NCERT Solutions > NCERT Solutions for Class 7 > NCERT Solutions for Class 7 Maths > NCERT Solutions for Class 7 Maths Chapter 4: Simple Equations

NCERT Solutions for Class 7 Maths Chapter 4: Simple Equations

Chapter 4 Simple Equations
Click to rate this post!
[Total: 19 Average: 4.6]

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations are provided to aid the students while preparing for their exams as well as assignments. For students who feel stressed about seeking the most comprehensive and detailed NCERT Solutions for Class 7 Maths, we at Swiflearn have prepared step by step solutions with detailed descriptions. We suggest students who aspire to score good marks in Maths, go through these solutions, and increase their knowledge. The important topics covered in this chapter are mentioned below.

NCERT Solutions for Class 7 Maths Chapter 4 by Swiflearn are by far the best and most reliable NCERT Solutions that you can find on the internet. These NCERT Solutions for Class 7 Maths Chapter 4 are designed as per the CBSE Class 7 Maths Syllabus. These NCERT Solutions will surely make your learning convenient & fun. Students can also Download FREE PDF of NCERT Solutions Class 7 Chapter 4.

Ncert solutions

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations: EX 4.1 PDF

Question 1:
Complete the last column of the table.

Solution:-
(i) x + 3 = 0
LHS = x + 3
Putting the value of x=3
Then,
LHS = 3 + 3 = 6
Comparing LHS and RHS
LHS ≠ RHS

∴No, equation not satisfied.
(ii) x + 3 = 0
LHS = x + 3
Putting the value of x=0
Then,
LHS = 0 + 3 = 3
Comparing LHS and RHS
LHS ≠ RHS
∴No, equation not satisfied.
(iii) x + 3 = 0
LHS = x + 3
Putting the value of x=– 3
Then,
LHS = – 3 + 3 = 0
Comparing LHS and RHS
LHS = RHS
∴Yes, equation is satisfied

(iv) x – 7 = 1
LHS = x – 7
Putting the value of x= = 7
Then,
LHS = 7 – 7 = 0
Comparing LHS and RHS
LHS ≠ RHS
∴No, equation not satisfied
(v) x – 7 = 1
LHS = x – 7
Putting the value of x= 8
Then,
LHS = 8 – 7 = 1
Comparing LHS and RHS
LHS = RHS
∴Yes, equation is satisfied.
(vi) 5x = 25
LHS = 5x
Putting the value of x=0
Then,
LHS = 5 × 0 = 0
Comparing LHS and RHS
LHS ≠ RHS
∴No, the equation is not satisfied.

(vii) 5x = 25
LHS = 5x
Putting the value of x= 5
Then,
LHS = 5 × 5 = 25
Comparing LHS and RHS
LHS = RHS
∴Yes, equation is satisfied.
(viii) 5x = 25
LHS = 5x
Putting the value of x= -5
Then,
LHS = 5 × (-5) = – 25
Comparing LHS and RHS
LHS ≠ RHS
∴No, equation not satisfied.

(ix) ?/3
= 2
LHS = ?/3
Putting the value of m= – 6
Then,
LHS = −6/3
= – 2
Comparing LHS and RHS
LHS ≠ RHS
∴No, equation not satisfied.
(x) ?/3 = 2
LHS = ?/3
Putting the value of m= 0
Then,
LHS = 0/3
= 0
Comparing LHS and RHS
LHS ≠ RHS
∴No, equation not satisfied.

(xi) ?/3
= 2
LHS = ?/3
Putting the value of m= 6
Then,
LHS = 6/3
= 2
Comparing LHS and RHS
LHS = RHS
∴Yes, equation is satisfied.

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)
LHS = n + 5
Putting the value of n = 1
Then,
LHS = n + 5

= 1 + 5
= 6
Comparing LHS and RHS
6 ≠ 19
LHS ≠ RHS
Therefore, the value of n = 1 is not a solution of the equation n + 5 = 19.
(b)
LHS = 7n + 5
Putting the value of n = -2
Then,
LHS = 7n + 5
= ((7 × (-2)) + 5
= – 14 + 5
= – 9
Comparing LHS and RHS
-9 ≠ 19
LHS ≠ RHS
Therefore, the value of n = -2 is not a solution of the equation 7n + 5 = 19

c)
LHS = 7n + 5
Putting the value of n = 2
Then,
LHS = 7n + 5
= ((7 × (2)) + 5
= 14 + 5
= 19
Comparing LHS and RHS
19 = 19
LHS = RHS
Therefore, the value of n = 2 is a solution of the 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
Comparing LHS and RHS
1 ≠ 13
LHS ≠ RHS
Therefore, the value of p = 1 is not a solution of the equation 4p – 3 = 13.
(e)
LHS = 4p – 3

Putting the value of p = – 4
Then,
LHS = 4p – 3
= ((4 × (-4)) – 3
= -16 – 3
= -19
Comparing LHS and RHS
-19 ≠ 13
LHS ≠ RHS
Therefore, the value of p = -4 is not a solution of the equation 4p – 3 = 13.
(f)
LHS = 4p – 3
Putting the value of p = 0
Then,
LHS = 4p – 3
= (4 × 0) – 3
= 0 – 3
= -3
Comparing LHS and RHS
– 3 ≠ 13
LHS ≠ RHS
Therefore, the value of p = 0 is not a solution of the equation 4p – 3 = 13.

Question 3:
Solve the following equations by trial and error method:
(i) 5p + 2 = 17
(ii) 3m – 14 = 4

Solution:-
(i)
LHS = 5p + 2
Putting the value of p = 0
Then,
LHS = 5p + 2
= (5 × 0) + 2
= 0 + 2
= 2
Comparing LHS and RHS
2 ≠ 17
LHS ≠ RHS
Therefore, 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
Comparing LHS and RHS
7 ≠ 17
LHS ≠ RHS
Therefore, 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
Comparing LHS and RHS
12 ≠ 17
LHS ≠ RHS
Therefore, 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
Comparing LHS and RHS
17 = 17
LHS = RHS
Therefore, the value of p = 3 is a solution to the given equation.

(ii)
LHS = 3m – 14
Putting the value of m= 3
Then,
LHS = 3m – 14
= (3 × 3) – 14
= 9 – 14
= – 5
Comparing LHS and RHS
-5 ≠ 4
LHS ≠ RHS
Therefore, 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
Comparing LHS and RHS
-2 ≠ 4
LHS ≠ RHS
Therefore, 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
Comparing LHS and RHS
1 ≠ 4
LHS ≠ RHS
Therefore, 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
Comparing LHS and RHS
4 = 4
LHS = RHS
Therefore, 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.
Therefore, equation will be
x + 4 = 9

(ii)
2 subtracted from y is 8.
Therefore, equation will be
y – 2 = 8

(iii)
Ten times a is 70.
Therefore, equation will be
= 10a = 70

(iv)
The number b divided by 5 gives 6.
Therefore, equation will be
?/5 = 6

(v)
Three-fourth of t is 15.
Therefore, equation will be
=3/4
t = 15

(vi)
Seven times m plus 7 gets you 77.
Therefore, equation will be
Seven times m is 7m
= 7m + 7 = 77

(vii)
One-fourth of a number x minus 4 gives 4.
Therefore, equation will be
One-fourth of a number x is x/4
= ?/4 – 4 = 4

(viii)
If you take away 6 from 6 times y, you get 60.
Solution:-
Therefore, equation will be
6 times of y is 6y
6y – 6 = 60

(ix)
If you add 3 to one-third of z, you get 30.
Therefore, equation will be
One-third of z is z/3
3 + ?/3 = 30

Question 5:
Write the following equations in statement forms:
(i) p + 4 = 15
(ii) m – 7 = 3
(iii) 2m = 7
(iv) ?/? = 3
(v) ??/? = 6

(vi) 3p + 4 = 25
(vii) 4p – 2 = 18
(viii) p/2 + 2 = 8

Solution:-
(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 some 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:-
(i)
According to the question,
Number of marbles permit has= m
Then,
Irfan has seven marbles more than five times the marbles Parmit has
= 5 x (Number of Parmit’s marbles) + 7
Total number of marbles Irfan has (5 × m) + 7 = 37
5m + 7 = 37

(ii)
According to the question,
Let us consider Laxmi’s age to be = y years old
And,
Lakshmi’s father is 4 years older than three times of Lakshmi’s age

3 × Laxmi’s age + 4 = Age of Lakshmi’s Dad
(3 × y) + 4 = 49
3y + 4 = 49

(iii)
According to the question,
Highest score= 87
Let lowest score be l
2 × Lowest score + 7 = Highest score
(2 × l) + 7 = 87
2l + 7 = 87

(iv)
According to the question,
As we know, the sum of angles of a triangle equals to 180o
Taking base angle to be b.
Now,
Vertex angle = 2 x (base angle) = 2b
b + b + 2b = 180o
4b = 180o

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations: EX 4.2 PDF

Question 1:
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

Solution:-
(a)
Step 1: Add 1 to both the side of the equation,
We get,
x – 1 + 1 = 0 + 1
x = 1

(b)
Step 1: Subtract 1 from both side of the equation,
We get,
x + 1 – 1 = 0 – 1
x = – 1

(c)
Step 1: Add 1 to both the side of the equation,
We get,
x – 1 + 1 = 5 + 1
x = 6

(d)
Step 1: Subtract 6 from both side of the equation,
We get,
x + 6 – 6 = 2 – 6
x = – 4

(e)
Step 1: Add 4 to both the side of the equation,
We get,
y – 4 + 4 = – 7 + 4
y = – 3

(f)
Step 1: Add 4 to both the side of the equation,
We get,
y – 4 + 4 = 4 + 4
y = 8

(g)
Step 1: Subtract 4 from both side of the equation,
We get,
y + 4 – 4 = 4 – 4
y = 0

(h)
Step 1: Subtract 4 from both side of the equation,
We get,
y + 4 – 4 = – 4 – 4
y = – 8

Question 2:
Give first the step you will use to separate the variable and then solve the
equation:
(a) 3l = 42
(b) ?/ ? = 6
(c) ?/? = 4
(d) 4x = 25
(e) 8y = 36
(f) ?/?= ( ?/? )
(g) (?/? ) = (?/??)
(h) 20t = – 10

Solution:-
(a) 3l = 42
3?/3 = 42/3{Divide both sides of the equation by 3}
l = 14

(b) ?/2 = 6
?/2 × 2 = 6 × 2

{Multiply both sides of the equation by 2}
b = 12

(c) ?/7 = 4
?/7 × 7= 4 × 7
{Multiply both sides of the equation by 7}
p = 28

(d) 4x = 25
4?/4 = 25/4

{Divide both sides of the equation by 4}
x = 25/4

(e) 8y =36
8?/8 = 36/8

{Divide both sides of the equation by 8}
x = 9/4

(f) ?/3 = 5/4
?/3 × 3 = (5/4 ) × 3
{Multiply both sides of the equation by 3}
x = 15/4

(g) ( ?/5 )= 7/15 ( ?/5 ) × 5 = ( 7/15) × 5
{Multiply both sides of the equation by 5}
a = 7/3

(h) 20? = −10
20?/20 = − 10/20

{Divide both sides of the equation by 20}
x = – 1/2

Question 3:
Give the steps you will use to separate the variable and then solve the
equation:
(a) 3n – 2 = 46
(b) 5m + 7 = 17
(c) ???/? = 40
(d) ??/?? = 6

Solution:-
(a) 3n – 2 = 46
3n – 2 + 2 = 46 + 2
{Adding 2 to both sides of the equation}
3n = 48
Then,
3?/3 = 48/ 3

{Divide both sides of the equation by 3}
n = 16

(b) 5m + 7 =17
Solution:-
5m + 7 – 7 = 17 – 7
{Subtracting 7 from both sides of the equation}
5m = 10
Then,
5?/5 = 10/5
{Divide both sides of the equation by 5}
m = 2

(c) 20?/3 = 40 ( 20?/3 ) × 3 = 40 × 3
{Multiply both sides of the equation by 3}
20p = 120
Then,
20?/20 = 120 /20

{Divide both sides of the equation by 20}
p = 6

(d) 3?/10 = 6
( 3?/10 ) × 10 = 6 × 10
{Multiply both sides of the equation by 10}
3p = 60
Then,
3?/3 = 60/3

{Divide both sides of the equation by 3}
p = 20

Question 4:
Solve the following equations:
(a) 10p = 100
(b) 10p + 10 = 100
(c) ?/? = 5
(d) – ?/? = 5
(e) ??/? = 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

Solution:-
(a)10p = 100
10?/10 = 100/10

{Divide both sides of the equation by 10}
p = 10

(b) 10p + 10 = 100
10p + 10 – 10 = 100 – 10
{Subtracting 10 from both sides of the equation}
10p = 90
Then,
10?/10 = 90/10

{Divide both sides of the equation by 10}
p = 9

(c) ?/4 = 5 ?/4 × 4 = 5 × 4
{Multiply both sides of the equation by 4}
p = 20

(d) – ?/3 = 5
– ?/3 × (- 3) = 5 × (- 3)
{Multiply both sides of the equation by -3}
p = – 15

(e) 3?/4 = 6 ( 3?/4 ) × (4) = 6 × 4
{Multiply both sides of the equation by 4}
3p = 24
Then,
3?/3 = 24/3

{Divide both sides of the equation by 3}
p = 8

(f) 3s = -9
3?/3 = − 9/3

{Divide both sides of the equation by 3}
s = -3

(g) 3s + 12 = 0
3s + 12 – 12 = 0 – 12
{Subtracting 12 from both sides of the equation}
3s = -12
Then,
3?/3 = − 12/3

{Divide both sides of the equation by 3}
s = – 4

(h) 3s = 0
3?/3 = 0/3

{Divide both sides of the equation by 3}
s = 0

(i) 2q = 6
2?/2 = 6/2

{Divide both sides of the equation by 2}
q = 3

(j) 2q – 6 = 0
2q – 6 + 6 = 0 + 6
{Adding 6 to both sides of the equation}
2q = 6
Then,
2?/2 = 6/2
{Divide both sides of the equation by 2}
q = 3

(k) 2q + 6 = 0
2q + 6 – 6 = 0 – 6
{Subtracting 6 from both sides of the equation}
2q = – 6
Then,
2?/2
= – 6/2

{Divide both sides of the equation by 2}
q = – 3

(l) 2q + 6 = 12
2q + 6 – 6 = 12 – 6
{Subtracting 6 from both sides of the equation}
2q = 6
Then,
2?/2 = 6/2

{Divide both sides of the equation by 2}
q = 3

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations: EX 4.3 PDF

NCERT Solutions for Class 7 Maths Chapter 4 Simple Equations: EX 4.4 PDF

0 0 votes
Article Rating
Subscribe
Notify of
guest
0 Comments
Inline Feedbacks
View all comments
0
Would love your thoughts, please comment.x
()
x