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## 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) 𝑚/3

= 2

LHS = 𝑚/3

Putting the value of m= – 6

Then,

LHS = −6/3

= – 2

Comparing LHS and RHS

LHS ≠ RHS

∴No, equation not satisfied.

(x) 𝑚/3 = 2

LHS = 𝑚/3

Putting the value of m= 0

Then,

LHS = 0/3

= 0

Comparing LHS and RHS

LHS ≠ RHS

∴No, equation not satisfied.

(xi) 𝑚/3

= 2

LHS = 𝑚/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

𝑏/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

= 𝑥/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 + 𝑧/3 = 30

**Question 5:**

**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**

**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) 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**

**Solution**:-

(a) 3l = 42

3𝑙/3 = 42/3{Divide both sides of the equation by 3}

l = 14

(b) 𝑏/2 = 6

𝑏/2 × 2 = 6 × 2

{Multiply both sides of the equation by 2}

b = 12

(c) 𝑝/7 = 4

𝑝/7 × 7= 4 × 7

{Multiply both sides of the equation by 7}

p = 28

(d) 4x = 25

4𝑥/4 = 25/4

{Divide both sides of the equation by 4}

x = 25/4

(e) 8y =36

8𝑦/8 = 36/8

{Divide both sides of the equation by 8}

x = 9/4

(f) 𝑧/3 = 5/4

𝑧/3 × 3 = (5/4 ) × 3

{Multiply both sides of the equation by 3}

x = 15/4

(g) ( 𝑎/5 )= 7/15 ( 𝑎/5 ) × 5 = ( 7/15) × 5

{Multiply both sides of the equation by 5}

a = 7/3

(h) 20𝑡 = −10

20𝑡/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) 𝟐𝟎𝒑/𝟑 = 40**

**(d) 𝟑𝒑/𝟏𝟎 = 6**

**Solution**:-

(a) 3n – 2 = 46

3n – 2 + 2 = 46 + 2

{Adding 2 to both sides of the equation}

3n = 48

Then,

3𝑛/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,

5𝑚/5 = 10/5

{Divide both sides of the equation by 5}

m = 2

(c) 20𝑝/3 = 40 ( 20𝑝/3 ) × 3 = 40 × 3

{Multiply both sides of the equation by 3}

20p = 120

Then,

20𝑝/20 = 120 /20

{Divide both sides of the equation by 20}

p = 6

(d) 3𝑝/10 = 6

( 3𝑝/10 ) × 10 = 6 × 10

{Multiply both sides of the equation by 10}

3p = 60

Then,

3𝑝/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) 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**

**Solution**:-

(a)10p = 100

10𝑝/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,

10𝑝/10 = 90/10

{Divide both sides of the equation by 10}

p = 9

(c) 𝑝/4 = 5 𝑝/4 × 4 = 5 × 4

{Multiply both sides of the equation by 4}

p = 20

(d) – 𝑝/3 = 5

– 𝑝/3 × (- 3) = 5 × (- 3)

{Multiply both sides of the equation by -3}

p = – 15

(e) 3𝑝/4 = 6 ( 3𝑝/4 ) × (4) = 6 × 4

{Multiply both sides of the equation by 4}

3p = 24

Then,

3𝑝/3 = 24/3

{Divide both sides of the equation by 3}

p = 8

(f) 3s = -9

3𝑠/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,

3𝑠/3 = − 12/3

{Divide both sides of the equation by 3}

s = – 4

(h) 3s = 0

3𝑠/3 = 0/3

{Divide both sides of the equation by 3}

s = 0

(i) 2q = 6

2𝑞/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,

2𝑞/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,

2𝑞/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,

2𝑞/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