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NCERT solution class 7 chapter 12 Algebraic expressions 12.3 mathematics

EXERCISE 12.3


Page No 242:

Question 1:

If m = 2, find the value of:

(i) m − 2 (ii) 3m − 5 (iii) 9 − 5m

(iv) 3m2 − 2m − 7 (v) 

Answer:

(i) m − 2 = 2 − 2 = 0

(ii) 3m − 5 = (3 × 2) − 5 = 6 − 5 = 1

(iii) 9 − 5m = 9 − (5 × 2) = 9 −10 = −1

(iv) 3m2 − 2m − 7 = 3 × (2 × 2) − (2 × 2) − 7

= 12 − 4 − 7 = 1

(v) 

Question 2:

If p = −2, find the value of:

(i) 4p + 7

(ii) −3p2 + 4p + 7

(iii) −2p3 − 3p2 + 4p + 7

Answer:

(i) 4p + 7 = 4 × (−2) + 7 = − 8 + 7 = −1

(ii) − 3p2 + 4p + 7 = −3 (−2) × (−2) + 4 × (−2) + 7

= − 12 − 8 + 7 = −13

(iii) −2p3 − 3p2 + 4p + 7

= −2 (−2) × (−2) × (−2) − 3 (−2) × (−2) + 4 × (−2) + 7

= 16 − 12 − 8 + 7 = 3

Question 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

Answer:

(i) 2− 7

= 2 × (−1) − 7 = −9

(ii) − + 2 = − (−1) + 2 = 1 + 2 = 3

(iii) x2 + 2x + 1 = (−1) × (−1) + 2 × (−1) + 1

= 1 − 2 + 1 = 0

(iv) 2x2 − x − 2 = 2 (−1) × (−1) − (−1) − 2

= 2 + 1 − 2 = 1

Question 4:

If a = 2, b = − 2, find the value of:

(i) ab2 (ii) a2 + ab + b2 (iii) a2 − b2

Answer:

(i) a2 + b2

= (2)2 + (−2)2 = 4 + 4 = 8

(ii) a2 + ab + b2

= (2 × 2) + 2 × (−2) + (−2) × (−2)

= 4 − 4 + 4 = 4

(iii) a2 − b2

= (2)2 − (−2)2 = 4 − 4 = 0

Question 5:

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

(i) 2a + 2b (ii) 2ab2 + 1

(iii) 2ab + 2ab2 + ab (iv) a2 + ab + 2

Answer:

(i) 2a + 2= 2 × (0) + 2 × (−1) = 0 − 2 = −2

(ii) 2a2 + b2 + 1

= 2 × (0)2 + (−1) × (−1) + 1

= 0 + 1 + 1 = 2

(iii) 2a2+ 2ab2 + ab

= 2 × (0)2 × (−1) + 2 × (0) × (−1) × (−1) + 0 × (−1)

= 0 + 0 + 0 = 0

(iv) a2 + ab + 2

= (0)2 + 0 × (−1) + 2

= 0 + 0 + 2 = 2

Question 6:

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

(i) x + 7 + 4 (x − 5) (ii) 3 (x + 2) + 5− 7

(iii) 6x + 5 (− 2) (iv) 4 (2x −1) + 3+ 11

Answer:

(i) x + 7 + 4 (x − 5) = x + 7 + 4x − 20

x + 4x + 7 − 20

= 5x − 13

= (5 × 2) − 13

= 10 − 13 = −3

(ii) 3 (x + 2) + 5x − 7 = 3x + 6 + 5x − 7

= 3x + 5x + 6 − 7 = 8x − 1

= (8 × 2) − 1 = 16 − 1 =15

(iii) 6x + 5 (x − 2) = 6x + 5x − 10

= 11x − 10

= (11 × 2) − 10 = 22 − 10 = 12

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

= 11x + 7

= (11 × 2) + 7

= 22 + 7 = 29

Question 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

Answer:

(i) 3x − 5 − x + 9 = 3x − x − 5 + 9

= 2x + 4 = (2 × 3) + 4 = 10

(ii) 2 − 8+ 4x + 4 = 2 + 4 − 8+ 4x

= 6 − 4= 6 − (4 × 3) = 6 − 12 = −6

(iii) 3a + 5 − 8+ 1 = 3a − 8a + 5 + 1

= − 5+ 6 = −5 × (−1) + 6

= 5 + 6 = 11

(iv) 10 − 3b − 4 − 5b = 10 − 4− 3b − 5b

= 6 − 8b = 6 − 8 × (−2)

= 6 + 16 = 22

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

= 3a − 2b − 9s

= 3 × (−1) − 2 (−2) − 9

= − 3 + 4 − 9 = −8

Question 8:

(i) If z = 10, find the value of z3 − 3 (z − 10).

(ii) If p = − 10, find the value of p2 − 2p − 100

Answer:

(i) z3 − 3 (z − 10) = z3 − 3z + 30

= (10 × 10 × 10) − (3 × 10) + 30

= 1000 − 30 + 30 = 1000

(ii) p2 − 2p − 100

= (−10) × (−10) − 2 (−10) − 100

= 100 + 20 − 100 = 20

Question 9:

What should be the value of a if the value of 2x2 + x − a equals to 5, when x = 0?

Answer:

2x2 + x − a = 5, when x = 0

(2 × 0) + 0 − a = 5

0 − a = 5

a = −5

Question 10:

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

2 (aab) + 3 − ab

Answer:

2 (a2 + ab) + 3 − ab = 2a2 + 2ab + 3 − ab

= 2a2 + 2ab − ab + 3

= 2a2 + ab + 3

= 2 × (5 × 5) + 5 × (−3) + 3

= 50 − 15 + 3 = 38

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