Processing math: 100%

PSEB 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.4

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.4 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.4

Question 1.
If 7 is added to five times a number, the result is 57. Find the number.
Solution:
Let the required number = x
Five times the number = 5x
7 added to five times the number = 5x + 7
According to the problem
5x + 7 = 57
5x = 57 – 7
5x = 50
x = \frac {50}{5}
So, x = 10
Hence the required number is 10.

Question 2.
9 decreased from four times a number yields 43. Find the number.
Solution:
Let the required number = x
Four times the number = 4x
9 decreased from four times the number = 4x – 9
According to the problem
4x – 9 = 43
4x = 43 + 9
4x = 52
x = \frac {52}{4}
x = 13
Hence, the required number is 13.

Question 3.
If one-fifth of a number minus 4 gives 3, find the number.
Solution:
Let the required number = x
One fifth of the number = \frac {1}{5}x
One fifth of the number minus 4 = \frac {1}{5}x – 4
According to problem
\frac {1}{5}x – 4 = 3
\frac {1}{5}x = 3 + 4
\frac {1}{5}x = 7
x = 35
Hence the required number is 35.

PSEB 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.4

Question 4.
In a class of 35 students, the number of girls is two-fifth the number of boys. Find the number of girls in the class.
Solution:
Let the number of boys = x
∴ number of girls = \frac {2}{5}x
Total number of students = 35
x + \frac {2}{5}x =35
\frac{5 x+2 x}{5} = 35
7x = 5 × 35
x = \frac{5 \times 35}{7}
x = 25
Therefore number of boys = 25
Number of girls = 35 – 25 = 10.

Question 5.
Sham’s father’s age is 5 years more than three times Sham’s age. Find Sham’s age, if his father is 44 years old.
Solution:
Let Sham’s age = x years
Then Sham’s father age = 3x + 5
But Sham’s fathers age = 44
According to question
3x + 5 = 44
3x = 44 – 5
3x = 39
Dividing both sides by 3
\frac{3 x}{3}=\frac{39}{3}
or x = 13
Hence Sham’s age is 13 years.

Question 6.
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°)
Solution:
Let each base angle of an isosceles triangle = x (in degrees)
Vertex angle = 40°
The sum of angles of a triangle = 180°
∴ x + x + 40° = 180°
2x = 180° – 40°
2x = 140°
Divide both sides by 2
\frac{2 x}{2}-\frac{140^{\circ}}{2}
Or x = 70°
Each equal angle is of 70°

PSEB 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.4

Question 7.
Irfan says that he has 7 marbles more than five times the marbles Pannit has. Irfan has 37 marbles. How many marbles does Pannit have ?
Solution:
Let marbles Parmit has = x
Marbles Irfan has = 5x + 7
But Marbles Irfan has = 37
∴ 5x + 7 = 37
5x = 37 – 7
5x = 30
x = \frac {30}{5} = 6
Therefore Parmit has 6 marbles.

Question 8.
The length of a rectangle is 3 units more than its breadth and the perimeter is 22 units. Find the breadth and length of a rectangle.
Solution:
Let breadth of rectangle (l)
= x units
∵ length of rectangle (b) = (x + 3) units
∴ Perimeter of rectangle = 2(l + b)
= 2 (x + x + 3) units
= 2(2x + 3) units
According to the question
Perimeter = 22 units
2 (2x + 3) =22
\frac{2(2 x+3)}{2}=\frac{22}{2}
2x + 3 = 11
2x = 11 – 3
or 2x = 8
Dividing both sides by 2 we get
\frac{2 x}{2}=\frac{8}{2}
x = 4
∴ breadth = 4 units
Length = (4 + 3) units
= 7 units

Leave a Comment