1.2 Tracy Limited is authorized to issue 2 000 000 ordinary shares. 1.2.1 Calculate the average price of shares before the shares were bought back on 28 September 2019?.

INFORMATION:

Ordinary Share Capital (600 000 shares issued) 3600 000
Retained Income 476 000
Shareholders for dividends215 000
SARS (Income Tax) 120 000 Cr

Additional information:
(2019 Mar 18) Pay the outstanding amounts due to SARS for income tax and to shareholders for dividends.

May 17 Issue 300 000 shares for R9 00 per share.

Aug 31 The directors decided to pay an interim dividend of R130 000.

The company paid their first provisional tax of R225 000
Sep 28 The directors decided to buy 120 000 shares back for R8 each.

2020
Feb 26 A second provisional tax payment of R245 000 was made.

Feb 29 The directors recommended and declared a dividend of 105 cents per share.

The income tax for the year amounted to R475 000​

16 is the absolute value

step-by-step explanation:

i think it's eleven too, don't woosh me here, but according to the "experts" it's 2.

the proof starts from the peano postulates, which define the natural

numbers n. n is the smallest set satisfying these postulates:

  p1.   1 is in n.

  p2.   if x is in n, then its "successor" x' is in n.

  p3.   there is no x such that x' = 1.

  p4.   if x isn't 1, then there is a y in n such that y' = x.

  p5.   if s is a subset of n, 1 is in s, and the implication

      (x in s => x' in s) holds, then s = n.

then you have to define addition recursively:

  def: let a and b be in n. if b = 1, then define a + b = a'

      (using p1 and p2). if b isn't 1, then let c' = b, with c in n

      (using p4), and define a + b = (a + c)'.

then you have to define 2:

  def:   2 = 1'

2 is in n by p1, p2, and the definition of 2.

theorem:   1 + 1 = 2

proof: use the first part of the definition of + with a = b = 1.

      then 1 + 1 = 1' = 2   q.e.d.

note: there is an alternate formulation of the peano postulates which

replaces 1 with 0 in p1, p3, p4, and p5. then you have to change the

definition of addition to this:

  def: let a and b be in n. if b = 0, then define a + b = a.

      if b isn't 0, then let c' = b, with c in n, and define

      a + b = (a + c)'.

you also have to define 1 = 0', and 2 = 1'. then the proof of the

theorem above is a little different:

proof: use the second part of the definition of + first:

      1 + 1 = (1 + 0)'

      now use the first part of the definition of + on the sum in

      parentheses:   1 + 1 = (1)' = 1' = 2   q.e.d.

step-by-step explanation: