PSEB 7th Class Maths Solutions Chapter 12 Algebraic Expressions Ex 12.3

Punjab State Board PSEB 7th Class Maths Book Solutions Chapter 12 Algebraic Expressions Ex 12.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

1. Fill in the Table by substituting the values in the given expressions.

Solution:
(i) 10, 1, 16, 37
(ii) 2, 11, 6, 83
(iii) 5, – 76, 189, 7700
(iv) 10, – 80, – 30, – 800.

2. If a = 1, b = – 2 find the value of given expressions

(i) a² – b²
Solution:
a² – b²
Putting a = 1, b = – 2 in a² – b², we get
a² – b² = (1)² – (- 2)²
= 1 – 4
= -3

(ii) a + 2ab – b²
Solution:
a + 2ab – b² = 1 + 2 × 1 × – 2 – (- 2)²
= 1 – 4 – 4
= – 7

(iii) a²b + 2ab² + 5
Solution:
a²b + 2ab² + 5 = 1² × – 2 + 2 × 1 × (- 2)² + 5
= -2 + 2 × 4 + 5
= – 2 + 8 + 5
= 11

3. Simplify the following expressions and find their values for. m = 1, n = 2, p = – 1.

(i) 2m + 3n – p + 7m – 2n
Solution:
2m + 3n – p + 7m – 2n
= 2m + 7m + 3n – 2n – p
= 9m + n – p
Putting m = 1, n = 2, p = – 1, we get
9m + n – p = 9 × 1 + 2- (-1)
= 9 + 2 + 1
= 12.

(ii) 3p + n – m + 2n
Solution:
3p + n- m + 2n = 3p + n + 2n – m
= 3p + 3n – m
Putting m = 1, n = 2, p = – 1
3p + 3n – m = 3 × -1 + 3 × 2 -1
= -3 + 6 – 1
= 2.

(iii) m + p – 2p + 3m
Solution:
m + p – 2p + 3m = m + 3m + p – 2p
= 4m – p
Putting m =1, n = 2, p = -1
4m – p = 4 (1) – (- 1)
= 4 + 1
= 5.

(iv) 3n + 2m – 5p – 3m – 2n + p
Solution:
3n + 2m – 5p – 3m – 2n + p
= 3n – 2n + 2m – 3m – 5p + p
= n – m – 4p
Putting m =1, n = 2, p = – 1
n – m – 4p = 2 – 1 – 4 (-1)
= 2 – 1 + 4
=5

4. What should be the value of a if the value of 2a² + b² = 10 when b = 2 ?
Solution:
2a² + b² = 10
Putting b = 2, we get
2a + (2)² = 10
2a + 4 = 10
2a = 10 – 4 = 6
a = \(\frac {6}{2}\) = 3
a = 3

5. Find the value of x if – 3x + 7y2 = 1 when y = 1.
Solution:
-3x + 7y² = 1
Putting y = 1
-3x + 7y² = 1
-3x + 7 (1)² = 1
-3x + 7 = 1
-3x = 1 – 7
-3x = – 6
x = \(\frac {-6}{-3}\) = 2
x = 2.

6. Observe the pattern of shapes of letters formed from line segment of equal lengths.

If n shapes of letters are formed, then write the algebraic expression for the number of line segment required for making these n shapes in each case.
Solution:
(i) 2n + 1
(ii) 4n + 2

7. Observe the following pattern of squares made using dots.

If n is taken as the number of dots in each row then find the algebraic expression for number of dots in nth figure. Also find number of dots if.
(i) n = 3
(ii) n = 7
(iii) n = 10
Solution:
n² (i) 9, (ii) 49, (iii) 100.

8. Observe the pattern of shapes of digits formed from line segment of equal lengths.

If n shapes of digits are formed then write the algebric expression for the numbers of line segment required to make n shapes.
Solution:
(i) 3n + 1
(ii) 4n + 2
(iii) 5n + 1

9. Multiple Choice Questions :

Question (i).
If l is the length of the side of the regular pentagon, perimeter of a regular Pentagon is.
(a) 3 l
(b) 4 l
(c) 5 l
(d) 8 l.
Answer:
(c) 5 l

Question (ii).
The value of the expression 5n – 2 when n = 2 is.
(a) 12
(b) -12
(c) 8
(d) 3
Answer:
(c) 8

Question (iii).
The value of 3x² – 5x + 6 when x = 1.
(a) 3
(b) 4
(c) – 8
(d) 14.
Answer:
(b) 4