RS Aggarwal Class 8 Math Sixth Chapter Operation on Algebric Expressions Exercise 6E Solution
EXERCISE 6E
OBJECTIVE QUESTIONS
Tick (√) the correct answer in each of the following:
(1) The sum of (6a + 4b – c + 3), (2b – 3c + 4), (11b – 7a + 2c – 1) and (2c – 5a – 6) is
Ans: (c) (- 6a + 17b)
Solution: (6a + 4b – c + 3) + (2b – 3c + 4) + (11b – 7a + 2c – 1) + (2c – 5a – 6)
= 6a + 4b – c + 3 + 2b – 3c + 4 + 11b – 7a + 2c – 1 + 2c – 5a – 6
= – 6a + 17b
(2) (3q + 7p2 – 2r3 + 4) – (4p2 – 2q + 7r3 – 3) =?
Ans: (d) (3p2 + 5q – 9r3 + 7)
Solution: 3q + 7p2 – 2r3 + 4 – 4p2 + 2q – 7r3 + 3
= 3p2 + 5q – 9r3 + 7
(3) (x + 5) (x – 3) =?
= x2 + 5x – 3x – 15
= x2 + 2x – 15
Ans: (d)
(4) (2x+ 3) (3x – 1)
= 6x2 + 9x – 2x – 3
= 6x2 + 7x – 3
Ans: (b)
(5) (x + 4) (x + 4)
= x2 + 4x + 4x + 16
= x2 + 8x + 16
Ans: (c)
(6) (x – 6) (x – 6)
= x2 – 6x – 6x + 36
= x2 – 12x + 36
Ans: (d)
(7) (2x + 5) (2x – 5)
= 4x2 + 10x – 10x – 25
= (4x2 – 25)
Ans: (b)
(8) 8a2b3 ÷ (- 2ab)
Ans: (c)
(9) (2x2 + 3x + 1) ÷ (x + 1)
= 2x + 1
Ans: (b)
(10) (x2 – 4x + 4) ÷ (x – 2)
= (x – 2)
Ans: (a)
(11) (a + 1) (a – 1) (a2 + 1)
= [(a)2 – (1)2] (a2 + 1)
= (a2 – 1) (a2 + 1)
= (a2)2 – (1)2
= (a4 – 1)
Ans: (c)
(15) (82)2 – (18)2
= (80 + 2)2 – (20 – 2)2
= [(80)2 + (2 × 80 × 2) + (2)2] – [(20)2 – (2 × 20 × 2) + (2)2]
= (6400 + 320 + 4) – (400 – 80 + 4)
= 6724 – 324 = 6400
Ans: (c)
(16) (197 × 203)
= (200 – 3) × (200 + 3)
= (200)2 – (3)2
= 40000 – 9 = 39991
Ans: (a)
(17) If (a + b) = 12 and ab = 14, then (a2 + b2) =?
⇒ (a + b) = 12
⇒ (a + b)2 = (12)2
⇒ a2 + 2ab + b2 = 144
⇒ (a2 + b2) + (2 × 14) = 144
⇒ (a2 + b2) = 144 – 28 = 116
Ans: (b)
(18) If (a – b) = 7 and ab = 9, then, (a2 + b2) =?
⇒ (a – b) = 7
⇒ (a – b)2 = 72
⇒ a2 – 2ab + b2 = 49
⇒ (a2 + b2) – (2 × 9) = 49
⇒ (a2 + b2) = 49 + 18
⇒ (a2 + b2) = 67
Ans: (a)
(19) If x = 10, then the value of (4x2 + 20x + 25) =?
= 4x2 + 20x + 25
= [4 × (10)2] + (20 × 10) + 25
= (4 × 100) + 200 + 25
= 400 + 200 + 25
= 625
Ans: (c)
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