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:
