1.1.1

From 0 to 6 kl: free and from 7 to 8 kl R2/kl

1.1.2

From 0 to 6 kl: free and from 7 to 8 kl R2/kl i.e. cost = R(6 x 0 + 2 x 2) = R4,00

1.2

49,12 kl is rounded up to 50 kl.

Amount due = R(6 x 0 + (29 - 6) x 2 + (49 - 30) x 2,40 + 2 x 3,50)

= R98,60

1.3

Amount due = R(6 x 0 + (29 - 6) x 2 + (49 - 30) x 2,40 + 7 x 3,50)

= R116,10

The amount is therefore in excess by R19.55.

2.1

Total expenses = R8 520,00

No, the expenses are R(8 520 - 7 600) = R920 more than the income.

2.2

Not yet. The expenses are still inexcess by R420 over the income. Jack can spend less on

entertainment and clothes. Spend R200 less on clothes and R420 on entertainment. Then

the budget balances.

2.3

Spend R300 on clothes, decrease entertainment by R200 and motor maintenance by R20.

3.1

Total expenses = R18 300. Peter cannot afford all his expenses, because the expenses

exceed the income by R1 300.

3.2

Not yet. The shortage /deficit is still R900. Spend R450 less on clothes and

R450 on entertainment.

3.3

The shortage is R1 200. Cut down expenses by spending R500 less on clothes, R500 on

entertainment and R200 on motor maintenance - save on fuel, use the car less.

4.1

The phone number is 001 968 0001.

4.2

The postal addres is P. O. Box 34, Hereweare 001908

4.3

The invoice is dated 28 / 03 / 2016.

4.4

Last date for payment is 14/04/2016.

4.5

The account is settled and thus that amount must be deducted from the account balance

so that the outstanding amount is R0,00.

4.6

Water consumed = (1241 - 1206) kl = 35 kl and electricity = (8324 - 80136) = 188 units

4.7

Amount = R(6 x 0 + (35 - 6) x 1,10) = R31,90

VAT = R31,90 x 0,14 = R4,47 so that Total cost = R31,90 + R4,47 = R36,37

The account is completely in error.

4.8

Electricity consumption is 188 units and not 288. The cost is therefore erroneous.

Cost = R188 x 1,45 = R272,60 so that VAT = R272,60 x 0,14 = R38,16 and

Total = R310,76. The amount due is therefore wrong.

The amounts should be : Amount = R424,50; VAT = R59,43 and Amount = R894,05

4.9

No. At 1,6% the levy is R350 000 x 0,016 = R5 600,00.

The levy of R4 921,44 is calculated at (4921,44 ÷ 350000) x 100 = 1,406%

5.1

The date is May 2, 2005 and the number of the statement is 51.

5.2

The commas are used to separate hundreds and thousands and the fullstops are used

to separate the integers from the decimals.

5.3

The interest rate is 1% per annum or 0,083333% per month.

5.4

The interest earned = Pi = R1 986.32 x 0.00083333 = R1.66

5.5

It means that the interest earned is added to the Balance to form the new Capital.

5.6

Balance = R1,985.32 + 1.66 = R1,987.98

5.7

The amounts in the Debit column are the amounts that are withdrawn from the Balance to

pay other accounts or fees. The Balance is therefore decreased by that amount.

5.8

The interest earned = R1,168.78 x 0.00083333 = R0.97 so that the Balance = R1,169.75

6.1

The date is May 2, 2005 and the number of the statement is 251

6.2

The Balance brought forward is an amount that Miss Marais owes the Bank - her account

is overdrawn - she "borrows" money from the Bank.

6.3

The amounts in the Debits column are the amounts that were used to pay certain other debts

or cash withdrawals. Money is withdrawn from the account and therefore the Balance must

be decreased by that amount.

6.4

Balance = R3,814.21- + R7,420.00 = R3,605.79

The sign is positive which means that the account is no longer overdrawn, there is

"money in the Bank", there are funds available.

6.5

Balance = R20.96- The account is once more overdrawn by R20.96.

6.6

Balance = R2,451.21-

6.7

Balance = R2,4763.21- An amount of R800 is deposited into the account which

decreases the amount overdrawn.

6.8

Balance = R4,423.76-

6.9

P = R4,423.76- + 99.00- = R4,522.76- and i = 13,5 ÷ (100 x 12) = 0.01125

I = Pi = R4,522.76 x 0.01125 = R50.88

6.10

Balance = R4,522.76- + (R50.88-) = R4,573.64-

7.1

20 items were bought.

7.2

4 plastic carry bags were bought.

7.3

24 liter

7.4

6 items are VAT free.

7.5

R109.70

7.6

R146.45

7.7

18 eggs cost R29,99 and 6 eggs cost R9,99

R29,99

R9,99

1 egg costs ———— = R1,6661111 and 1 egg costs ———— = R1,665

18

6

The 6 egg package is thus a little cheaper.

R34,56

7.8

0,96 kg costs R24,56 and thus 1 kg costs ———— = R36,00

0,96

8.1

There is no cost if no calls are made.

koste

8.2

Yes, because the rate ———— remains constant.

tyd

8.3.1 R75,00 8.3.2 125 minutes
8.3.3 R120,00
8.3.4 225 minutes

R200

8.4

Cost per minute = ————— = R0,80

250 minutes

8.5

8.3.1 Cost = R0,80 x 100 = R80,00

R100

8.3.2 Time = ———— minutes = 125 minutes

R0,80

8.3.3 Cost = R0,80 x 150 = R120,00

R180

8.3.4 Time = ———— minutes = 225 minutes

R0,80

9.1

Time (minutes) |
0 | 10 | 50 | 200 | 250 | 300 | 400 |

Cost (R) |
0 | 5 | 25 | 100 | 125 | 150 | 200 |

cost

9.2

Yes, because the rate ————

time

remains constant.

9.3

See accompanying graph.

9.4.1

A; R50

9.4.2

B; 200 minutes

9.4.3

C; R75

9.4.4

D; 250 minutes

9.5

The line will move upward by half of the

original y-value

(Cost-value), e.g. at 10 minutes the

point will move with half of R5,

i.e. R2,50, opward so that the

new Cost will now be R7,50.

See the accompanying graph.

10.1

The graph starts at the point (0 ; 50). There is a fixed fee of R50 which does not depend

on the time.

10.2.1 R90,00 10.2.2 120 minutes
10.2.3 R120,00
10.2.4 310 minutes

10.3

The fixed fee of R50 must be subtracted from the R150 to find the cost for the

250 minutes.

R100

Cost per minute = ————— = R0,40

250 minutes

10.4

10.2.1 Cost = R50 + (0,40 x 100) = R(50 + 40) = R90,00

10.2.2 Cost for calls made = R(100 - 50) = R50

R50

Time = ———— = 125 minutes

R0,40

10.2.3 Cost = R50 + (0,40 x 150) = R(50 + 60) = R110,00

10.2.4 Cost for calls made = R(178 - 50) = R128

R128

Time = ———— = 320 minutes

R0,40

11.1

The graph starts at point (0 ; 50). There is a fixed cost of R50 which does not

depend on the time.

11.2

From A to B the graph is parallel to the horizontal time-axis so that there is no change

in the value on the vertical cost-axis although the time increases, i.e. from A to B

the cost remains constant on R50 although the time increases from 0 to 100 minutes.

11.3

From B to C the graph slopes upward so that the values on the vertical cost-axis increase

as the values on the horizontal time-axis increase causing an increase in the cost

as the time increases.

11.4.1 R50,00 11.4.2 150 minutes
11.4.3 R125,00
11.4.4 350 minutes

R150

11.5

Cost for minutes = R200 - R50 = R150

Cost per minute = ————— = R0,50

300 minutes

11.6

11.4.1 Cost = R50 + (0,0 x 80) = R50,00

11.4.2 Cost for minutes used = R(80 - 50) = R30

R30

Time = ———— = 60 minutes

R0,50

Time = 60 + 100 minutes = 160 minutes

11.4.3 Cost = R50 + (0,50 x (250 - 100)) = R(50 + 75) = R125,00

11.4.4 Cost for minutes used = R(175 - 50) = R125

R125

Time = ———— = 250 minutes

R0,50

Time = 250 + 100 minutes = 350 minutes