PSEB 7th Class Maths Solutions Chapter 4 Simple Equations Ex 4.1

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 4 Simple Equations Ex 4.1 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 4 Simple Equations Ex 4.1

1. Complete the following :

Solution:
(i) No
Reason : For x = 5
L.H.S. = x + 5 = 5 + 5 = 10
RHS = 0
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = 5

(ii) Yes
Reason : For x = -5
L.H.S. = x + 5
= -5 + 5 = 0
R.H.S. = 0
Since L.H.S. = R.H.S.
Therefore, given equation is satisfied for
x = – 5

(iii) NO
Reason : For x = 3
L.H.S. = x – 3
= 3 – 3 = 0
R.H.S. = 1
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = 3

(iv) No
Reason : x = – 3
L.H.S. = x – 3
= – 3 – 3 = -6
R.H.S. = 1
Since L.H.S. ≠ R.H.S.
Therefore, given equation is not satisfied for x = – 3

(v) Yes
Reason : For x = 5
L.H.S. = 2x
= 2 × 5 = 10
R.H.S. = 10
Since L.H.S. = R.H.S.
Therefore given equation is satisfied for
x = 5

(vi) No
Reason : For x = – 6
L.H.S. = $$\frac{x}{3}$$
= $$\frac {-6}{3}$$
= -2
R.H.S. = 2
Since L.H.S. ≠ R.H.S.
Therefore given equation is not satisfied for x = – 6

(vii) No,
Reason : For x = 0
L.H.S. = $$\frac{x}{2}$$
= $$\frac {0}{2}$$ = 0
R.H.S. = 2
Since L.H.S. ≠ R.H.S.
Therefore given equation is not satisfied for x = 0

2. Check whether the value given in the brackets is a solution to the given equation or not.

Question (i).
x + 4 = 11 (x = 7)
Yes
Check : For x = 1
L.H.S. = x + 4
= 7 + 4 = 11
R.H.S. = 11
Since L.H.S. = R.H.S.
Therefore x = 7 is the solution to the given equation.

Question (ii).
8x + 4 = 28 (x = 4)
No
Check: Forx-4
L.H.S. = 8x + 4
= 8 × 4 + 4
= 32 + 4
= 36
R.H.S. = 28
Since L.H.S. ≠ R.H.S.
Therefore x = 4 is not solution to the given equation.

Question (iii).
3m – 3 = 0 (m = 1)
Yes
Check : For m = 1
L.H.S. = 3m – 3
= 3 × 1 – 3 = 3 – 3 = 0
R.H.S.= 0
Since L.H.S. = R.H.S.
Therefore, m = 1 is the solution to the given equation.

Question (iv).
$$\frac{x}{5}$$ – 4 = -1 (x = 15)
Yes
Check : For x = 15
L.H.S. = $$\frac{x}{5}$$ – 4
= $$\frac {15}{5}$$ – 4 = 3 – 4 = – 1
R.H.S. = – 1
Since L.H.S. = R.H.S.
Therefore, x = 15 is the solution to the given equation.

Question (v).
4x – 3 = 13 (x = 0)
No
Check : For x = 0
L.H.S. = 4x – 3
= 4 × 0 – 3
= 0 – 3
= -3
R.H.S. = 13
Since L.H.S. ≠ R.H.S.
Therefore x = 0 is not the solution to the given equation.

3. Solve the following equations by trial and error method

Question (i).
5x + 2 = 17

 Value of x L.H.S R.H.S 1 5 × 1 + 2 = 5 + 2 = 7 17 2 5 × 2 + 2 = 10 + 2 = 12 17 3 5 × 3 + 2 = 15 + 2 = 17 17

We observe that for x = 3, L.H.S. = R.H.S.
Hence x = 3 is the solution of the given equation.

Question (ii).
3p – 14 = 4

 Value of p L.H.S. R.H.S. 1 3 × 1 – 14 = 3 – 14 = -11 4 2 3 × 2 – 14 = 6 – 14 = -8 4 3 3 × 3- 14 = 9 – 14 = -5 4 4 3 × 4 – 14 = 12 – 14 = -2 4 5 3 × 5 – 14 = 15 – 14 = 1 4 6 3 × 6 – 14 = 18 – 14 = 4 4

We observe that for p = 6, L.H.S. = R.H.S.
Hence p = 6 is the solution of the given equation

4. Write equations for the following statements.

Question (i).
The sum of numbers x and 4 is 9
x + 4 = 9

Question (ii).
3 subtracted from y gives 9
y – 3 = 9

Question (iii).
Ten times x is 50
10x = 50

Question (iv).
Nine times x plus 6 is 87
9x + 6 = 87

Question (v).
One fifth of a number y minus 6 gives 3.
$$\frac {1}{5}$$x – 6 = 3

5. Write the following equations in statement form :

Question (i).
x – 2 = 6
2 substracted from x is 6

Question (ii).
3y – 2 = 10
2 subtracted from 3 times a number y is 10.

Question (iii).
$$\frac{x}{6}$$ = 6
One sixth of a number x is 6

Question (iv).
7x – 15 = 34
15 subtracted 7 times a number x is 34.

Question (v).
$$\frac{x}{2}$$ + 2 = 8
add 2 to half of a number x to get 8.

6. Write an equation for the following statements :

Question (i).
Raju’s father’s age is 4 years more than five times Raju’s age. Raju’s father is 54 years old.
Let x years be Raju’s age
Five times Raju’s age is 5x years
His father’s age will be 4 years more than five times
Raju’s age = 5x + 4
But 4 years more han five times Raju’s age = Raju’s father’s age
Therefore 5x + 4 = 54

Question (ii).
A teacher tells that the highest marks obtained by a student in his class is twice the lowest marks plus 6. The highest score is 86. (Take the lowest score to be x).
Let the lowest score to be x
Twice the lowest score plus 6 = 2x + 6
The highest score obtained by a student is twice the lowest score plus 6 = 2x + 6
But highest score = 86
Hence 2x + 6 = 86

Question (iii).
In an isosceles triangle, the vertex angle is twice either base angle (Let the base angle be x in degrees. Remember that the sum of angles of a triangle is 180 degrees).
Let the base angle be x (in degrees)
Therefore vertex angle is twice the base angle = 2x (in degrees)
Sum of three angles of a triangle = 180°
∴ x + x + 2x = 180° = 4x = 180°

Question (iv).
A shopkeeper sells mangoes in two types of boxes. One small and one large. The large box contains as many as 8 small boxes plus 4 loose mangoes. The number of mangoes in a large box is given to be 100.