Punjab State Board PSEB 6th Class Maths Book Solutions Chapter 7 Algebra Ex 7.4 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 6 Maths Chapter 7 Algebra Ex 7.4
1. Write the following statements as algebraic equations:
Question (i)
The sum of x and 3 gives 10.
Solution:
The sum of x and 3 = x + 3
It gives 10.
∴ The algebraic equation is x + 3 = 10
Question (ii)
5 less than a number ‘a’ is 12.
Solution:
5 less than a number ‘a’ = a – 5
It is 12
∴ The algebraic equation is a – 5 = 12
Question (iii)
2 more than 5 times of p gives 32.
Solution:
2 more than 5 times of p = 5p + 2
It gives 32
∴ The algebraic equation is 5p + 2 = 32
Question (iv)
Half of a number is 10.
Solution:
Let Half of a number x = \(\frac {x}{2}\)
It is 10
∴ The algebraic equations is
\(\frac {x}{2}\) = 10
Question (v)
Twice of a number added to 3 gives 17.
Solution:
Let the number be x
Twice of a number added to 3 = 2x + 3
It gives 17
∴ The algebraic equation is 2x + 3 = 17
2. Write the L.H.S. and R.H.S. for the following equations:
Question (i)
l + 5 = 8
Solution:
L.H.S. = l + 5, R.H.S. = 8
Question (ii)
13 = 2m + 3
Solution:
L.H.S. = 13, R.H.S. = 2m + 3
Question (iii)
\(\frac {t}{4}\) = 6
Solution:
L.H.S. = \(\frac {t}{4}\), R.H.S. = 6
Question (iv)
2h – 5 = 13
Solution:
L.H.S. = 2h – 5, R.H.S. = 13
Question (v)
\(\frac {5x}{7}\) = 15.
Solution:
L.H.S. = \(\frac {5x}{7}\), R.H.S. = 15.
3. Solve the following equations by trial and error method:
Question (i)
x + 2 = 7
Solution:
x + 2= l
We try different values of x to make L.H.S. = R.H.S.
| Value of JC | L.H.S. = x + 2 | R.H.S. = 7 | L.H.S. = R.H.S. |
| 1 | 1+ 2 = 3 | 7 | No |
| 2 | 2 + 2 = 4 | 7 | No |
| 3 | 3 + 2 = 5 | 7 | No |
| 4 | 4 + 2 = 6 | 7 | No |
| 5 | 5 + 2 = 7 | 7 | Yes |
From the above table we find that L.H.S. = R.H.S. When x = 5
Question (ii)
5p = 20
Solution:
5p = 20
We try different values of p to make L.H.S. = R.H.S.
| Value of p | L.H.S. = 5p | R.H.S. = 20 | L.H.S. = R.H.S. |
| 1 | 5 × 1 = 5 | 20 | No |
| 2 | 5 × 2 = 10 | 20 | No |
| 3 | 5 × 3 = 15 | 20 | No |
| 4 | 5 × 4 = 20 | 20 | Yes |
From the above table we find that L.H.S. = R.H.S. When p = 4
Question (iii)
\(\frac {a}{5}\) = 2
Solution:
We try different values of a to make L.H.S. = R.H.S.
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From the above table we find that L.H.S. = R.H.S. When a = 10
Question (iv)
2l – 4 = 8
Solution:
21-4 = 8
We try different values of l to make L.H.S. = R.H.S.
| Value of a | L.H.S. | R.H.S. = 8 | L.H.S. ff R.H.S. |
| 1 | 2 × 1 – 4 = 2 – 4 = – 2 | 8 | No |
| 2 | 2 × 2 – 4 = 4 – 4 = 0 | 8 | No |
| 3 | 2 × 3 – 4 = 6 – 4 = 2 | 8 | No |
| 4 | 2 × 4 – 4 = 8 – 4 = 4 | 8 | No |
| 5 | 2 × 5 – 4 = 10 – 4 = 6 | 8 | No |
| 6 | 2 × 6 – 4 = 12 – 4 = 8 | 8 | Yes |
From the above table we find that L.H.S. = R.H.S. When l = 6
Question (v)
3x + 2 = 11.
Solution:
3x + 2 = 11
We try different values of x to make L.H.S. = R.H.S.
| Value of p | L.H.S. = 3x + 2 | R.H.S. = 11 | L.H.S. = R.H.S. |
| 1 | 3 × 1 + 2 = 3 + 2 = 5 | 11 | No |
| 2 | 3 × 2 + 2 = 6 + 2 = 8 | 11 | No |
| 3 | 3 × 3 + 2 = 9 + 2 = 11 | 11 | Yes |
From the above table we find thatL.H.S. = R.H.S. When x = 3
4. Solve the following equations by systematic method.
Question (i)
z – 4 = 10
Solution:
Given Equation is z – 4 = 10 Adding 4 on both sides, we get
2 – 4 + 4 = 10 + 4
⇒ z = 14 is the required solution.
Question (ii)
a + 3 = 15
Solution:
Given equation is a + 3 = 15
Subtracting 3 from both sides, we get
a + 3- 3 = 15 – 3
⇒ a = 12 is the required solution.
Question (iii)
4m = 20
Solution:
Given equation is 4m = 20
Dividing both sides by 4, we get
\(\frac{4 m}{4}=\frac{20}{4}\)
⇒ m = 5 is the required solution.
Question (iv)
3x – 3 = 15
Solution:
Given equation is 3x – 3 = 15
Adding 3 on both sides, we get
3x – 3 + 3 = 15 + 3
⇒ 3x = 18
Dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{18}{3}\)
⇒ x = 6 is the required solution.
Question (v)
4x + 5 = 13.
Solution:
Given equation is 4x + 5 = 13
Subtracting 5 from both sides, we get
4x + 5 – 5 = 13 – 5
⇒ 4x = 8
Dividing both sides by 4, we get
\(\frac{4 x}{4}=\frac{8}{4}\)
⇒ x = 2 is the required solution.
5. Solve the following equation by transposition:
Question (i)
x – 5 = 6
Solution:
Given equation : x – 5 = 6
∴ x = 6 + 5
(Transposing – 5 to other side, it becomes + 5)
⇒ x = 11 is the required solution.
Question (ii)
y + 2 = 3
Solution:
Given equation : y + 2 = 3
⇒ y = 3 – 2
(Transposing + 2 to other side, it becomes – 2)
∴ y = 1 is the required solution.
Question (iii)
5x = 10
Solution:
Given equation : 5x = 10
⇒ x = \(\frac {10}{5}\)
(Transposing ‘multiplication’, it becomes ‘division’)
∴ x = 2 is the required solution.
Question (iv)
\(\frac {a}{6}\) = 4
Solution:
Given equation : \(\frac {a}{6}\) = 4
⇒ a = 4 × 6
(Transposing ‘division’, it becomes ‘multiplication’)
∴ a = 24 is the required solution.
Question (v)
4y – 2 = 30.
Solution:
Given equation : 4y – 2 = 30
⇒ 4y = 30 + 2
(Transposing – 2, it becomes + 2)
⇒ 4y = 32
⇒ y = \(\frac {32}{4}\)
(Transposing ‘multiplication’, it becomes division)
∴ y = 8 is the required solution.
6. Solve the following equations:
Question (i)
x + 7 = 11
Solution:
Given equation :
x + 7 = 11
⇒ x = 11 – 7
(Transposing 7 to R.H.S.)
⇒ x = 4 is the required solution.
Question (ii)
x – 3 = 15
Solution:
Given equation : x – 3 = 15
⇒ x = 15 + 3
(Transposing – 3 to L.H.S. it becomes + 3)
∴ x = 18 is the required solution
Question (iii)
x – 2 = 13
Solution:
Given equation : x – 2 = 13
⇒ x = 13 + 2
(Transposing – 2 to L.H.S.)
∴ x = 15 is the required solution.
Question (iv)
6x = 18
Solution:
Given equation is 6x = 18
Dividing both sides by 6 we get
\(\frac{6x}{6}=\frac{18}{6}\)
∴ x = 3 is the required solution.
Question (v)
3x = 24
Solution:
Given equation 3x = 24
Dividing both sides by 3, we get
\(\frac{3x}{3}=\frac{24}{3}\)
∴ x = 8 is the required solution.
Question (vi)
\(\frac {x}{4}\) = 7
Solution:
Given equation :
\(\frac {x}{4}\) = 7
Multiplying both sides by 4, we get
4 × \(\frac {x}{4}\) = 4 × 7
∴ x = 28 is the required solution.
Question (vii)
\(\frac {x}{8}\) = 5
Solution:
Given equation : \(\frac {x}{8}\) = 5
Multiplying both sides by 8, we get
8 × \(\frac {x}{8}\) = 8 × 5
∴ x = 40 is the required solution.
Question (viii)
2x – 5 = 17
Solution:
Given equation : 2x – 5 = 17
⇒ 2x = 17 – 5
(Transposing – 5 to R.H.S.)
⇒ 2x = 22
⇒ x = \(\frac {22}{2}\)
(Dividing both sides by 2)
∴ x = 11 is the required solution.
Question (ix)
4x + 5 = 21
Solution:
Given equation : 4x + 5 = 21
⇒ 4x = 21 + 5
(Transposing 5 to R.H.S.)
⇒ 4x = 16
⇒ x = \(\frac {16}{4}\)
(Dividing both sides by 4)
∴ x = 4 is the required solution.
Question (x)
5x – 2 = 13.
Solution:
Given equation : 5x – 2 = 13
⇒ 5x = 13 + 2
(Transposing – 2 to R.H.S.)
⇒ 5x = 15
⇒ x = \(\frac {15}{5}\)
(Dividing both sides by 5)
∴ x = 3 is the required solution.