Class 7, Maths, Chapter 12, Exercise 12.3 Solutions
Q.1. If m = 2, find the value of:
(i) m – 2
(ii) 3m – 5
(iii) 9 – 5m
(iv) 3m2 – 2m – 7
(v) $\frac{5m}{2}-4$
Ans:
(i) m-2
Putting m = 2, we get
⟹ m – 2
⟹ 2 – 2 = 0
(ii) 3m-5
Putting m = 2, we get
⟹ 3m – 5
⟹ (3 × 2 ) – 5
⟹ 6 – 5 = 1
(iii) 9-5m
Putting m = 2, we get
⟹ 9 – 5m
⟹ 9 – (5 × 2) = 9 – 10 = -1
(iv) 3m2 – 2m – 7
Putting m = 2, we get
⟹ 3m2 – 2m – 7
⟹ 3(2)2 – 2 (2) – 7
⟹ (3 × 2 × 2) – (2 × 2) – 7
⟹ 12 – 4 – 7
⟹ 12 – 11 = 1
(v) $\frac{5m}{2}-4$
Putting m = 2, we get
⟹ $\frac{5m}{2}-4$
⟹ $\frac{5(2)}{2}-4$
⟹ $\frac{10}{2}-4$
⟹ 5 – 4 =1
Q.2. If p = – 2, find the value of:
(i) 4p + 7
(ii) – 3p2 + 4p + 7
(iii) – 2p3 – 3p2 + 4p + 7
Ans:
(i) 4p + 7
Putting p = -2 ,we get
⟹ 4p + 7
⟹ 4(-2) + 7 = -8 + 7 = -1
(ii) – 3p2 + 4p + 7
Putting p = -2 ,we get
⟹ – 3p2 + 4p + 7
⟹ – 3(-2)2 + 4(-2) + 7
⟹ – 3(-2 x -2) + 4(-2) + 7
⟹ – 3(4) – 8 + 7
⟹ -12 – 8 + 7
⟹ -20 + 7 = – 13
(iii) – 2p3 – 3p2 + 4p + 7
Putting p = -2 ,we get
⟹ – 2p3 – 3p2 + 4p + 7
⟹ – 2(-2)3 – 3(-2)2 + 4(-2) + 7
⟹ – 2(-2 x -2 x -2) – 3(-2 x -2) + 4(-2) + 7
⟹ – 2(-8) – 3(4) + 4(-2) + 7
⟹ 16 – 12 – 8 + 7
⟹ 16 + 7 – 12 – 8
⟹ 23 – 20 = 3
Q.3. Find the value of the following expressions, when x = –1:
(i) 2x – 7
(ii) – x + 2
(iii) x2 + 2x + 1
(iv) 2x2 – x – 2
Ans:
(i) 2x – 7
Putting x = -1 , we get
⟹ 2x – 7
⟹ 2(-1) – 7
⟹ – 2 – 7 = – 9
(ii) – x + 2
Putting x = -1 , we get
⟹ – x + 2
⟹ – (-1) + 2 = 1 + 2 = 3
(iii) x2 + 2x + 1
Putting x = -1 , we get
⟹ x2 + 2x + 1
⟹ (-1)2 + 2(-1) + 1
⟹ (-1) (-1) + 2(-1) + 1
⟹ 1 – 2 + 1 = 0
(iv) 2x2 – x – 2
Putting x = -1 , we get
⟹ 2x2 – x – 2
⟹ 2(-1)2 – (-1) – 2
⟹ 2(-1 x -1) – (-1) – 2
⟹ 2(1) + 1 – 2
⟹ 2 + 1 – 2
⟹ 3 – 2 = 1
Q.4. If a = 2, b = – 2, find the value of:
(i) a2 + b2
(ii) a2 + ab + b2
(iii) a2 – b2
Ans:
(i) a2 + b2
Putting a = 2 and b = -2, we get
⟹ a2 + b2
⟹ (2)2 + (-2)2
⟹ (2 x 2) + (-2 x -2)
⟹ 4 + 4 = 8
(ii) a2 + ab + b2
Putting a = 2 and b = -2, we get
⟹ a2 + ab + b2
⟹ (2)2 + (2)(-2) + (-2)2
⟹ (2×2) + (2)(-2) + (-2 x -2)
⟹ (4) + (-4) + (4)
⟹ 4 – 4 + 4 = 8 – 4 = 4
(iii) a2 – b2
Putting a = 2 and b = -2, we get
⟹ a2 – b2
⟹ (2)2 – (-2)2
⟹ (2 x 2) – (-2 x -2)
⟹ 4 – 4 = 0
Q.5. When a = 0, b = – 1, find the value of the given expressions:
(i) 2a + 2b
(ii) 2a2 + b2 + 1
(iii) 2a2b + 2ab2 + ab
(iv) a2 + ab + 2
Ans:
(i) 2a + 2b
Putting a = 0 and b = -1, we get
⟹ 2a + 2b
⟹ 2(0) + 2(-1)
⟹ 0 – 2 = – 2
(ii) 2a2 + b2 + 1
Putting a = 0 and b = -1, we get
⟹ 2a2 + b2 + 1
⟹ 2(0)2 + (-1)2 + 1
⟹ 0 + 1 + 1 = 2
(iii) 2a2b + 2ab2 + ab
Putting a = 0 and b = -1, we get
⟹ 2a2b + 2ab2 + ab
⟹ 2(0)2(-1) + 2(0)(-1)2 + (0)(-1)
⟹ 0 + 0 + 0 = 0
(iv) a2 + ab + 2
Putting a = 0 and b = -1, we get
⟹ a2 + ab + 2
⟹ (0)2 + (0)(-1) + 2
⟹ 0 + 0 + 2 = 2
Q.6. 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
Ans:
(i) x + 7 + 4 (x – 5)
We have,
= x + 7 + 4 (x – 5)
= x + 7 + 4x – 20
Putting x = 2, we get
= (2) + 7 + 4(2) – 20
= 2 + 7 + 8 – 20
= 17 – 20
= -3
(ii) 3 (x + 2) + 5x – 7
We have,
= 3 (x + 2) + 5x – 7
= 3x + 6 + 5x – 7
= 3x + 5x + 6 – 7
= 8x – 1
Putting x = 2, we get
= 8(2) – 1
= 16 – 1 = 15
(iii) 6x + 5 (x – 2)
We have,
= 6x + 5 (x – 2)
= 6x + 5x – 10
= 11x – 10
Putting x = 2, we get
= 11(2) – 10
= 22 – 10 = 12
(iv) 4(2x – 1) + 3x + 11
We have,
= 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 8x + 3x – 4 + 11
= 11x + 7
Putting x = 2, we get
= 11(2) + 7
= 22 + 7 = 29
Q.7. 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 – 5b
(v) 2a – 2b – 4 – 5 + a
Ans:
(i) 3x – 5 – x + 9
We have,
= 3x – 5 – x + 9
= 3x – x – 5 + 9
= 2x + 4
Putting x = 3, we get
= 2(3) + 4
= 6 + 4 = 10
(ii) 2 – 8x + 4x + 4
We have,
= 2 – 8x + 4x + 4
= – 8x + 4x + 4 + 2
= – 4x + 6
Putting x = 3, we get
= – 4(3) + 6
= – 12 + 6
= – 6
(iii) 3a + 5 – 8a + 1
We have,
= 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= – 5a + 6
Putting a = -1, we get
= – 5(-1) + 6
= 5 + 6 = 11
(iv) 10 – 3b – 4 – 5b
We have,
= 10 – 3b – 4 – 5b
= – 3b – 5b – 4 +10
= – 8b + 6
Putting b = -2, we get
= – 8(-2) + 6
= 16 + 6 = 22
(v) 2a – 2b – 4 – 5 + a
We have,
= 2a – 2b – 4 – 5 + a
= 2a + a – 2b – 4 – 5
= 3a – 2b – 9
Putting a = -1 and b = -2, we get
= 3(-1) – 2(-2) – 9
= – 3 + 4 – 9
= 4 – 9 – 3
= 4 – 12 = – 8
Q.8. (i) If z = 10, find the value of z3 – 3 (z – 10).
(ii) If p = – 10, find the value of p2 – 2p – 100
Ans:
(i) If z = 10, find the value of z3 – 3 (z – 10).
We have, z3 – 3 (z – 10)
= z3 – 3z + 30
Putting z = 10, we get
= (10)3 – 3(10) + 30
= (10 x 10 x 10) – 3(10) + 30
= 1000 – 30 + 30
= 1000
if z = 10, then the value of z3 – 3 (z – 10) is 1000.
(ii) If p = – 10, find the value of p2 – 2p – 100
We have, p2 – 2p – 100
Putting p = -10, we get
= (-10)2 – 2(-10) – 100
= (-10 x -10) – 2(-10) – 100
= 100 + 20 – 100
= 100 –100 + 20 = 20
If p = – 10, then the value of p2 – 2p – 100 is 20.
Q.9. What should be the value of ‘a’ if the value of 2x2 + x – a equals to 5, when x = 0?
Ans: According to question,
⟹ 2x2 + x – a = 5
Putting x = 0, we get
⟹ 2(0)2 + (0) – a = 5
⟹ 0 + 0 – a = 5
⟹ – a = 5 (Multiplying by -1)
⟹ a = -5
Hence, the value of a is -5
Q.10. Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab
Ans: We have, 2(a2 + ab) + 3 – ab
= 2a2 + 2ab + 3 – ab
= 2a2 + 2ab– ab+ 3
= 2a2 + ab + 3
Putting a = 5 and b = -3, we get
= 2(5)2 + (5)(-3) + 3
= 2(5 x 5) + (5)(-3) + 3
= 2(25) + (-15) + 3
= 50 -15 + 3
= 53 – 15
= 38
Q.10. Simplify the expression and find its value when a = 5 and b = – 3.
2(a2 + ab) + 3 – ab
Ans: We have, 2(a2 + ab) + 3 – ab
= 2a2 + 2ab + 3 – ab
= 2a2 + 2ab– ab+ 3
= 2a2 + ab + 3
Putting a = 5 and b = -3, we get
= 2(5)2 + (5)(-3) + 3
= 2(5 x 5) + (5)(-3) + 3
= 2(25) + (-15) + 3
= 50 -15 + 3
= 53 – 15
= 38
NCERT Solutions For Class 7 Maths, Chapter 12 Algebraic Expressions (All Exercises)
Class 7, Maths, Chapter 12, Algebraic Expressions
Class 7, Maths, Chapter 12, Algebraic Expressions, Exercise 12.1
Class 7, Maths, Chapter 12, Algebraic Expressions, Exercise 12.2
Class 7, Maths, Chapter 12, Algebraic Expressions, Exercise 12.3 ← You are here
Class 7, Maths, Chapter 12, Algebraic Expressions, Exercise 12.4