**GSEB Solutions for Class 7 Mathematics – Equation (English Medium)**

GSEB SolutionsMathsScience

**Exercise**

**Solution 1:**

**Solution 2:**

1. m + 3 > 9

It is not an equation but an inequality.

2. 2x + 5 = (-3)

It is an equation.

3. 3z + 4 < 9

It is not an equation but an inequality.

4. 3n = 36

It is an equation.

5. 7p + 4 = 12

It is an equation.

6. 4b – 3 = 21

It is an equation.

**Solution 3(1):**

x – 4 = (-2)

L.H.S. = x – 4

= 2 – 4 ….(Substituting x = 2)

= (-2)

= R.H.S.

Here, L.H.S. = R.H.S.

Thus, the equation has 2 as a solution.

**Solution 3(2):**

2x – 3 = 5

L.H.S. = 2x – 3

= 2(2) – 3 …(Substituting x = 2)

= 4 – 3

= 1

≠R.H.S.

Here, L.H.S. ≠ R.H.S.

Thus, the equation does not have 2 as a solution.

**Solution 3(3):**

3x – 6 = 0

L.H.S. = 3x – 6

= 3(2) – 6 …(Substituting x = 2)

= 6 – 6

= 0

= R.H.S.

Here, L.H.S. = R.H.S.

Thus, the equation has 2 as a solution.

**Solution 3(4):**

a + 3 = 5

L.H.S. = a + 3

= 2 + 3 …(Substituting a = 2)

= 5

= R.H.S.

Here, L.H.S. = R.H.S.

Thus, the equation has 2 as a solution.

**Solution 3(5):**

6 – m = 10

L.H.S. = 6 – m

= 6 – 2 … (Substituting m = 2)

= 4

≠ R.H.S.

Here, L.H.S. ≠ R.H.S.

Thus, the equation does not have 2 as a solution.

**Solution 3(6):**

-2b + 1 = (-3)

L.H.S. = -2b + 1

= -4 + 1 …(Substituting b = 2)

= -3

= R.H.S

Here, L.H.S. = R.H.S.

Thus, the equation has 2 as a solution.

**Solution 3(7):**

14x = 28

L.H.S. = 14x

= 14 × 2 …(Substituting x = 2)

= 28

= R.H.S.

Here, L.H.S. = R.H.S.

Thus, the equation has 2 as a solution.

**Solution 3(8):**

6y – 1 = 11

L.H.S. = 6y – 1

= 6 × 2 – 1 …(Substituting y = 2)

= 12 – 1

= 11

= R.H.S

Here, L.H.S. = R.H.S.

Thus, the equation has 2 as a solution.

**Solution 3(9):**

5y + 4 = 20

L.H.S. = 5y + 4

= 5 × 2 + 4 …(Substituting y = 2)

= 10 + 4

= 14

≠ R.H.S.

Here, L.H.S. ≠ R.H.S.

Thus, the equation does not have 2 as a solution.

**Solution 4(1):**

3 + m = 8

∴ m = 8 – 3

∴ m = 5

Hence, the solution of the equation is m = 5.

**Solution 4(2):**

**Solution 4(3):**

**Solution 4(4):**

**Solution 4(5):**

**Solution 4(6):**

**Solution 4(7):**

**Solution 4(8):**

**Solution 4(9):**

**Solution 5(1):**

**Solution 5(2):**

**Solution 5(3):**

Let the present age of Reeta be x years.

∴ Beena’s present age = (x + 2) years

Hence, present age of Teena = (x + 2) + 3 = (x + 5) years

Now, sum of the ages of Reeta, Beena and Teena

= [x + (x + 2) + (x + 5)] years

= x + x + 2 + x + 5

= (3x + 7) years

But the sum of the ages of Reeta, Beena and Teena = 79 years … (given)

Hence, 3x + 7 = 79

∴ 3x = 79 – 7

∴ 3x = 72

∴ x = 24

Thus, we have

Present age of Reeta = x years = 24 years

Present age of Beena = (x + 2) = (24 + 2) = 26 years

And, present age of Teena = (x + 5) = (24 + 5) = 29 years

Hence, the present age of Reeta is 24 years, present age of Beena is 26 years and present age Teena is 29 years.

**Solution 5(4):**

Let the weight of Sachin be x kg.

∴ Weight of Rahul = (x + 5) kg

And, weight of Samir = (2x – 12) kg

Now, total weight of Sachin, Rahul and Samir

= [x + (x + 5) + (2x – 12)]

= x + x + 5 + 2x – 12

= (4x – 7) kg

But total weight of all three boys = 93 kg … (given)

Hence, 4x – 7 = 93

∴ 4x = 93 + 7

∴ 4x = 100

∴ x = 100 ÷ 4

∴ x = 25

Thus, we have

Weight of Sachin = x kg = 25 kg

Weight of Rahul = (x + 5) = (25 + 5) = 30 kg

And, weight of Samir = (2x – 12) = 2(25) – 12 = (50 – 12) = 38 kg

Hence, Sachin’s weight is 25 kg, Rahul’s weight is 30 kg, while Samir weighs 38 kg.

**Solution 5(5):**

Let the number of men be x

∴ Number of women = (x + 89)

And, number of children = (x + 400)

Now, total population of the village

= x + (x + 89) + (x + 400)

= x + x + 39 + x + 400

= 3x + 489

But total population of village = 4989 … (given)

Hence, 3x + 489 = 4989

∴ 3x = 4989 – 489

∴ 3x = 4500

∴ x = 4500 ÷ 3

∴ x = 1500

Thus, we have

Number of men = x = 1500

Number of women = (x + 89) = (1500 + 89) = 1589

And, number of children = (x + 400) = 1500 + 400 = 1900

Hence, there are 1500 men, 1589 women and 1900 children in the village.

**Solution 5(6):**

Let Dhruv have x number of chocolates.

∴ Number of chocolates with Priyanshi = (x – 5)

Now, total number of chocolates = x + (x – 5) = 2x – 5

But the total number of chocolates = 15 … (given)

Hence, 2x – 5 = 15

∴ 2x = 15 + 5

∴ 2x = 20

∴ x = 20 ÷ 2

∴ x = 10

∴ x – 5 = 10 – 5 = 5

Hence, Dhruv has = x = 10 chocolates

Priyanshi has = (x – 5) = (10 – 5) = 5 chocolates

Thus, Dhruv has 10 chocolates and Priyanshi has 5.

**Solution 5(7):**

**Practice – 1**

**Solution 1(1):**

b. x + 4 = 10

Addition of x and 4 gives x + 4.

Thus, x + 4 = 10.

**Solution 1(2):**

a. z – 3 = 8

Subtracting 3 from z gives z – 3.

Further subtraction of 3 from z gives 8, hence z – 3 = 8.

**Solution 1(3):**

c. 4m = 20

Four times of m is 4m.

Four times of m is given to be 20, hence 4m = 20.

**Solution 1(4):**

d. 5x – 3 = 22

Five times of x is 5x.

Subtracting 3 from five times of x gives 5x – 3.

Subtracting 3 from five times of x is given to be 22, hence 5x – 3 = 22.

**Solution 1(5):**

**Solution 2(1):**

x + 4 = 17 (x = 2)

L.H.S. = x + 4

= 2 + 4 ….(Substituting x = 2)

= 6

R.H.S. = 17

∴ L.H.S. ≠ R.H.S.

Hence, the equality of the equation is not maintained.

**Solution 2(2):**

5m + 5 = 20 (m = -3)

L.H.S. = 5m + 5

= 5(-3) + 5 ….(Substituting m = -3)

= -15 + 5

= -10

R.H.S. = 20

∴ L.H.S. ≠ R.H.S.

Hence, the equality of the equation is not maintained.

**Solution 2(3):**

5m + 5 = 20 (m = 3)

L.H.S. = 5m + 5

= 5(3) + 5 ….(Substituting m = 3)

= 15 + 5

= 20

R.H.S. = 20

∴ L.H.S = R.H.S.

Hence, the equality of the equation is maintained.

**Solution 2(4):**

4y – 3 = 17 (y = 2)

L.H.S. = 4y – 3

= 4(2) – 3 ….(Substituting y = 2)

= 8 – 3

= 5

R.H.S. = 17

∴ L.H.S. ≠ R.H.S.

Hence, the equality of the equation is not maintained.

**Solution 2(5):**

4y – 3 = 17 (y = 5)

L.H.S. = 4y – 3,

= 4(5) – 3 ….(Substituting y = 5)

= 20 – 3

= 17

R.H.S. = 17

∴ L.H.S. = R.H.S.

Hence, the equality of the equation is maintained.

**Solution 3(1):**

a. x = (-5)

For x = (-5),

L.H.S. = x + 5 = (-5) + 5 = 0

R.H.S. = 0

∴ L.H.S. = R.H.S. for x = (-5)

For x = 0,

L.H.S. = x + 5 = 0 + 5 = 5

R.H.S. = 0

∴ L.H.S. ≠ R.H.S. for x = 0

For x = 5,

L.H.S. = x + 5 = 5 + 5 = 10

R.H.S. = 0

∴ L.H.S ≠ R.H.S. for x = 5

**Solution 3(2):**

c. y = 5

For y = 2,

L.H.S. = y – 3 = 2 – 3 = -1

R.H.S. = 2

∴ L.H.S. ≠ R.H.S. for y = 2

For y = 1,

L.H.S. = y – 3 = 1 – 3 = -2

R.H.S. = 2

∴ L.H.S. ≠ R.H.S. for y = 1

For y = 5,

L.H.S. = y – 3 = 5 – 3 = 2

R.H.S. = 2

∴ L.H.S. = R.H.S. for y = 2

**Solution 3(3):**

b. m = 5

For m = 0,

L.H.S. = 6m = 6 × 0 = 0

R.H.S. = 30

∴ L.H.S. ≠ R.H.S. for m = 0

For m = 5,

L.H.S. = 6m = 6 × 5 = 30

R.H.S. = 30

∴ L.H.S. = R.H.S. for m = 5

**Solution 3(4):**

**Solution 4(1):**

x + 8 = 8

Hence, x + 8 – 8 = 8 – 8

(Adding opposite value of 8 on both the sides of the equation)

∴ x = 0

**Solution 4(2):**

a – 7 = 3

∴ a – 7 + 7 = 3 + 7

[Adding opposite value of (-7) on both the sides of the equation]

∴ a = 10

**Solution 4(3):**

**Solution 4(4):**

z – 9 = 0

∴ z = 0 + 9

∴ z = 9

**Solution 4(5):**

y + 5 = (-3)

∴ y = -3 – 5

∴ y = -8

**Solution 4(6):**

**Solution 4(7):**

**Solution 4(8):**

**Solution 4(9):**

**Solution 4(10):**

**Solution 4(11):**

**Solution 4(12):**

**Practice – 2**

**Solution 1(1):**

Let the number of oranges in the smaller box be a.

∴ The number of oranges in seven smaller boxes = 7a

5 oranges more than those present in the seven smaller boxes = 7a + 5

Hence, number of oranges in the bigger box = 7a + 5

But number of oranges in the bigger box = 75 …(Given)

Thus, the equation to find the number of oranges in the smaller box is 7a + 5 = 75

*The question has been rectified.

**Solution 1(2):**

Suppose Kirti has m seeds.

Five times the number of seeds which Kirti has = 5m

Two seeds less than that of five times of what Kirti has = 5m – 2

∴ Number of seeds with Yash = 5m – 2

But number of seeds with Yash = 28 ….(Given)

Hence, the equation to find the number of seeds with Kirti is 5m – 2 = 28

*The question has been rectified.

**Solution 1(3):**

Let Sonal’s age be y years.

Three times Sonal’s age = 3y years

Two years more than three times Sonal’s age = (3y + 2) years

Hence, the age of Sonal’s father = (3y + 2) years

But age of Sonals’ father = 47 years …..(Given)

Thus, the equation to find out Sonal’s age is 3y + 2 = 47

**Solution 1(4):**

Let Vinod’s weight be x kg.

Double the weight of Vinod = 2x kg

5 kg less than double the weight of Vinod = 2x – 5

Hence, Raju’s weight = (2x – 5) years

∴Total weight of Raju and Vinod = x + 2x – 5

But the total weight of Raju and Vinod = 40 kg ….(Given)

∴ x + 2x – 5 = 40

∴ 3x – 5 = 40

Thus, the required equation is 3x – 5 = 40

**Solution 1(5):**

Let the length of the silk cloth be y metres.

∴Three times the length of the silk cloth = 3y metres

∴ 7 metres less than three times the length of the silk cloth = (3y – 7) metres

∴ Total length of cloth = y + 3y – 7 = 4y – 7

But total length of the cloth = 193 metres (Given)

Hence, 4y – 7 = 193

Thus, the required equation is 4y – 7 = 193

**Solution 2:**

- 7x – 2 = 348 (Rs.) Rs. 2 subtracted from seven times a amount is equal to Rs. 348.
- By adding 11 metres to four times the length of a cloth, we get 291 metres.
- When 4 kg is added to three times the weight of rice, the result is 40 kg.
- Adding 4 to Rajesh’s age equals 15 years.

**Solution 3(1):**

**Solution 3(2):**

**Solution 3(3):**

**Solution 3(4):**

**Solution 3(5):**

**Solution 4:**

Let the number of triangles be x.

Then, according to the given information, we have

Number of squares = (x + 1)

Number of rectangles = Number of squares + 1

= (x + 1) + 1

= x + 2

And number of circles = Number of rectangles + 1

= (x + 2) + 1

= x + 3

Thus, total number of shapes

= x + (x + 1) + (x + 2) + (x + 3)

= x + x + 1 + x + 2 + x + 3

= 4x + 6

But total number of shapes = 10 ….(Given)

∴ 4x + 6 = 10

∴ 4x = 10 – 6

∴ 4x = 4

∴ x = 1

∴ Number of triangles = 1

∴ Number of squares = x + 1 = 1 + 1 = 2

∴ Number of rectangles = x + 2 = 1 + 2 = 3

∴ Number of circles = x + 3 = 1 + 3 = 4

The design is as follows: