Class VII

Simplification of given expression 16a2 b2 ⁄48a3 b gives
  1. b⁄3a
  2. 3a⁄b
  3. 24a
  4. 6a⁄7b
If we make a subject of given formula (p + a)⁄5 = 3p we get
  1. 12p⁄a
  2. a + p⁄5
  3. a = 14p
  4. a = 15p
Given that A = (1⁄3)πr2 h + (4⁄3)πr3 find A when r = 7, h = 15 and π= 3.142
  1. 2207.8
  2. 2100
  3. 564.89
  4. 2206.7
Making a subject of (a⁄b) - (a⁄c) = 1 we get
  1. a = b⁄c
  2. a = c⁄b
  3. a = c + b + 1
  4. a = bc⁄(c - b)
Simplification of (a2 - 4b2)⁄((a2+2ab)⁄ab) yields
  1. b(a - 2b)
  2. a(b - 2a)
  3. a + 2b
  4. a - 2b
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