Selina Concise Class 6 Math Chapter 18 Substitution Exercise 18A Solution
EXERCISE 18A
(1) Fill in the following blanks:
(i) x + y = 3 + 6 = 9
(ii) y – x = 6 – 3 = 3
(iii) y/x = 6/3 = 2
(iv) c ÷ b = 32 ÷ 8 = 4
(v) z ÷ x = 18 ÷ 3 = 6
(vi) y × d = 6 × 0 = 0
(vii) d ÷ x = 0 ÷ 3 = 0
(viii) ab + y
= (2 × 8) + 6
= 16 + 6
= 22
(ix) a + b + x
= 2 + 8 + 3
= 13
(x) b + z – d
= (8 + 18) – 0
= 26
(xi) a – b + y
= (2 – 8) + 6
= – 6 + 6
= 0
(xii) z – a – b
= 18 – 2 – 8
= 18 – 10
= 8
(xiii) d – a + x
= 0 – 2 + 3
= 1
(xiv) xy – bd
= (3 × 6) – (8 × 0)
= 18 – 0
= 18
(xv) xz + cd
= (3 × 18) + (32 × 0)
= 54
(2) Find the value of:
(i) p + 2q + 3r
= 1 + (2 × 5) + (3 × 2)
= 1 + 10 + 6
= 17
(ii) 2a + 4b + 5c
= (2 × 5) + (4 × 10) + (5 × 20)
= 10 + 40 + 100
= 150
(iii) 3a – 2b
= (3 × 8) – (2 × 10)
= 24 – 20
= 4
(iv) 5x + 3y – 6z
= (5 × 3) + (3 × 5) – (6 × 4)
= 15 + 15 – 24
= 30 – 24
= 6
(v) 2p – 3q + 4r – 8s
= (2 × 10) – (3 × 8) + (4 × 6) – (8 × 2)
= 20 – 24 + 24 – 16
= (20 + 24) – (24 + 16)
= 44 – 40
= 4
(vi) 6m – 2n – 5p – 3q
= (6 × 20) – (2 × 10) – (5 × 2) – (3 × 9)
= 120 – 20 – 10 – 27
= 120 – (20 + 10 + 27)
= 120 – 57
= 63
(3) Find the value of:
(4) If a = 3, b = 0, c = 2 and d = 1, find the value of:
(i) 3a + 2b – 6c + 4d
= (3 × 3) + (2 × 0) – (6 × 2) + (4 × 1)
= 9 + 0 – 12 + 4
= (9 + 4) – 12
= 13 – 12
= 1
(ii) 6a – 3b – 4c – 2d
= (6 × 3) – (3 × 0) – (4 × 2) – (2 × 1)
= 18 – 0 – 8 – 2
= 18 – (8 + 2)
= 18 – 10
= 8
(iii) ab – bc + cd – da
= (3 × 0) – (0 × 2) + (2 × 1) – (1 × 3)
= 0 – 0 + 2 – 3
= – 1
(iv) abc – bcd + cda
= (3 × 0 × 2) – (0 × 2 × 1) + (2 × 1 × 3)
= 0 – 0 + 6
= 6
(v) a2 + 2b2 – 3c2
= (3)2 + [2 × (0)2] – [3 × (2)2]
= 9 + .0 – (3 × 4)
= 9 – 12
= – 3
(vi) a2 + b2 – c2 + d2
= (3)2 + (0)2 – (2)2 + (1)2
= 9 + 0 – 4 + 1
= (9 +1) – 4
= 10 – 4
= 6
(vii) 2a2 – 3b2 + 4c2 – 5d2
= [2 × (3)2] – [3 × (0)2] + [4 × (2)2] – [5 × (1)2]
= (2 × 9) – 0 + (4 × 4) – 5
= 18 + 16 – 5
= 34 – 5
= 29
(5) 5x2 – 3x + 2
= [5 × (2)2] – (3 × 2) + 2
= (5 × 4) – 6 + 2
= 20 – 4
= 16
(6) 3x3 – 4x2 + 5x – 6
= [3 × (- 1)3] – [4 × (- 1)2] + [5 × (- 1)] – 6
= – 3 – 4 – 5 – 6
= – (3 + 4 + 5 + 6)
= – 18
(7) L. H. S. = x3 – 8x2 + 12x – 5
= (1)3 – [8 × (1)2] + (12 × 1) – 5
= 1 – 8 + 12 – 5
= (1 + 12) – (8 + 5)
= 13 – 13 = 0 = R. H. S. (proved)
(8) State true and false:
(i) L.H.S. = x + 5
= 1 + 5 = 6 = R. H. S. (TRUE)
(ii) L. H. S. = 2x – 3
= (2 × 0) – 3
= 0 – 3
= – 3 ≠ R. H. S. (FALSE)
(9) If x = 2, y = 5 and z = 4, find the value of each of the following:
(10) If a = 3, find the values of a2 and 2a.
(a)2 = (3)2 = 9
(2)a = (2)3 = 8
(11) If m = 2, find the difference between the value of 4m3 and 3m4.
(4m3) = [4 × (2)3] = 4 × 8 = 32
(3m4) = [3 × (2)4] = 3 × 16 = 48
Then, 48 – 32 = 16