## NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

- Class 7 Maths Algebraic Expressions Exercise 12.1
- Class 7 Maths Algebraic Expressions Exercise 12.2
- Class 7 Maths Algebraic Expressions Exercise 12.3
- Class 7 Maths Algebraic Expressions Exercise 12.4

**NCERT Solutions for Class 7 Maths Chapter 12 Algebraic Expressions Exercise 12.3**

Ex 12.3 Class 7 Maths Question 1.

If m = 2, find the value of:

(i) m – 2

(ii) 3m – 5

(iii) 9 – 5m

(iv) 3m^{2} – 2m – 7

(v) \(\frac{5 m}{2}-4\)

Solution:

(i) m – 2

Putting m = 2, we get

2 – 2 = 0

(ii) 3m – 5

Putting m = 2, we get

3 × 2 – 5 = 6 – 5 = 1

(iii) 9 – 5m

Putting m = 2, we get

9 – 5 × 2 = 9 – 10 = -1

(iv) 3m^{2} – 2m – 7 Putting m = 2, we get

3(2)^{2} – 2(2) – 7 = 3 × 4 – 4 – 7

=12 – 4 – 7 = 12 – 11 = 1

(v) \(\frac{5 m}{2}-4\)

Putting m = 2, we get

\(\frac{5 \times 2}{2}-4=5-4=1\)

Ex 12.3 Class 7 Maths Question 2.

If p = -2, find the value of:

(i) 4p + 7

(ii) -3p^{2} + 4p + 7

(iii) -2p^{3} – 3p^{2} + 4p + 7

Solution:

(i) 4p + 7

Putting p = -2, we get 4(-2) + 7 = -8 + 7 = -1

(ii) -3p^{2} + 4p + l

Putting p = -2, we get

-3(-2)^{2} + 4(-2) + 7

= -3 × 4 – 8 + 7 = -12 – 8+ 7 = -13

(iii) -2p^{3} – 3p^{2} + 4p + 7

Putting p = -2, we get

– 2(-2)^{3} – 3(-2)^{2} + 4(-2) + 7

= -2 × (-8) – 3 × 4 – 8 + 7

= 16 – 12 – 8 + 7 = 3

Ex 12.3 Class 7 Maths Question 3.

If a = 2, b = -2, find the value of:

(i) a^{2} + b^{2}

(ii) a^{2} + ab + b^{2}

(iii) a^{2} – b^{2}

Solution:

(i) a^{2} + b^{2}

Putting a = 2 and b = -2, we get

(2)^{2} + (-2)^{2} = 4 + 4 = 8

(ii) a^{2} + ab + b^{2}

Putting a = 2 and b = -2, we get

(2)^{2} + 2(-2) + (-2)^{2} = 4 – 4 + 4 = 4

(iii) a^{2} – b^{2}

Putting a = 2 and b = -2, we get

(2)^{2} – (-2)^{2} = 4 – 4 = 0

Ex 12.3 Class 7 Maths Question 4.

When a = 0, b = -1, find the value of the given expressions:

(i) 2a + 2b

(ii) 2a^{2} + b^{2} + 1

(iii) 2a^{2}b + 2ab^{2} + ab

(iv) a^{2} + ab + 2

Solution:

(i) 2a + 2b = 2(0) + 2(-1)

= 0 – 2 = -2 which is required.

(ii) 2a^{2} + b^{2} + 1

= 2(0)^{2} + (-1)^{2} + 1 =0 + 1 + 1 = 2 which is required.

(iii) 2a^{2}b + 2ab^{2} + ab

= 2(0)^{2} (-1) + 2(0)(-1)^{2} + (0)(-1)

=0 + 0 + 0 = 0 which is required.

(iv) a^{2} + ab + 2

= (0)^{2} + (0)(-1) + 2

= 0 + 0 + 2 = 0 which is required.

Ex 12.3 Class 7 Maths Question 5.

Simplify the expressions and find the value if x is equal to 2.

(i) x + 7 +4(x – 5)

(ii) 3(x + 2) + 5x – 7

(iii) 6x + 5(x – 2)

(iv) 4(2x – 1) + 3x + 11

Solution:

(i) x + 7 + 4(x – 5) = x + 7 + 4x – 20 = 5x – 13

Putting x = 2, we get

= 5 × 2 – 13 = 10 – 13 = -3

which is required.

(ii) 3(x + 2) + 5x – 7 = 3x + 6 + 5x -7 = 8x – 1

Putting x = 2, we get

= 8 × 2 – 1 = 16 – 1 = 15

which is required.

(iii) 6x + 5(x – 2) = 6x + 5x – 10

= 11 × – 10

Putting x = 2, we get

= 11 × 2 – 10 = 22 – 10 = 12

which is required.

(iv) 4(2x – 1) + 3x + 11 = 8x – 4 + 3x + 11

= 11x + 7

Putting x = 2, we get

= 11 × 2 + 7 = 22+ 7 = 29

Ex 12.3 Class 7 Maths Question 6.

Simplify these expressions and find their values if x = 3, a = -1, b = -2.

(i) 3x – 5 – x + 9

(ii) 2 – 8x + 4x + 4

(iii) 3a + 5 – 8a + 1

(iv) 10 – 3b – 4 – 55

(v) 2a – 2b – 4 – 5 + a

Solution:

(i) 3x – 5 – x + 9 = 2x + 4

Putting x = 3, we get

2 × 3 + 4 = 6 + 4 = 10

which is required.

(ii) 2 – 8x + 4x + 4 = -8x + 4x + 2 + 4 = -4x + 6

Putting x = 2, we have

= -4 × 2 + 6 = -8 + 6 =-2

which is required.

(iii) 3a + 5 – 8a +1 = 3a – 8a + 5 + 1 = -5a + 6

Putting a = -1, we get

= -5(-1) + 6 = 5 + 6 = 11

which is required.

(iv) 10 – 3b – 4 – 5b = -3b – 5b + 10 – 4

= -8b + 6

Putting b = -2, we get

= -8(-2) + 6 = 16 + 6 = 22

which is required.

(v) 2a – 2b – 4 – 5 + a = 2a + a – 2b – 4 – 5

= 3a – 26 – 9

Putting a = -1 and b = -2, we get

= 3(-1) – 2(-2) – 9

= -3 + 4 – 9 = 1 – 9 = -8

which is required.

Ex 12.3 Class 7 Maths Question 7.

(i) If z = 10, find the value of z^{2} – 3(z – 10).

(ii) If p = -10, find the value of p^{2} -2p – 100.

Solution:

(i) z^{2} – 3(z – 10)

= z^{2} – 3z + 30

Putting z = 10, we get

= (10)^{2} – 3(10) + 30

= 1000 – 30 + 30 = 1000 which is required.

(ii) p^{2} – 2p – 100

Putting p = -10, we get

(-10)^{2} – 2(-10) – 100

= 100 + 20 – 100 = 20 which is required.

Ex 12.3 Class 7 Maths Question 8.

What should be the value of a if the value of 2x^{2} + x – a equals to 5, when x = 0?

Solution:

2x^{2} + x – a = 5

Putting x = 0, we get

2(0)^{2} + (0) – a = 5

0 + 0 – a = 5

-a = 5

⇒ a = -5 which is required value.

Ex 12.3 Class 7 Maths Question 9.

Simplify the expression and find its value when a = 5 and b = -3.

2(a^{2} + ab) + 3 – ab

Solution:

2(a^{2} + ab) + 3 – ab = 2a2 + 2ab + 3 – ab

= 2a^{2} + 2ab – ab + 3

= 2ab + ab + 3

Putting, a = 5 and b = -3, we get

= 2(5)^{2} + (5)(-3) + 3

= 2 × 25 – 15 + 3

= 50 – 15 + 3

= 53 – 15 = 38

Hence, the required value = 38.

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