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Answers

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Part 1 Examination Paper 1.2

Financial Information for Management December 2005 Answers

Section A

1 D

2 C

3 C

4 C5 B

6 D

7 C

8 C

9 A

10 B

11 A

12 B

13 C

14 D

15 A

16 C

17 D

18 A

19 A

20 C

21 D

22 D

23 D

24 D

25 B

1 D

2 C 1,700 units Breakeven level units (1,200) = 500 units

3 C Contribution per unit = 22 055 045 = 18

Breakeven point = 198,000 18 = 11,000

4 C Variable cost per unit = [(170,000 5,000) 140,000)] (22,000 17,000) = 5

Total fixed cost above 18,000 units = 170,000 (22,000 5) = 60,000

Total cost of 20,000 units = (20,000 5) + 60,000 = 160,000

5 B

6 D

7 C Weighted average after 13th = [(200 9,300 300) + (600 33)] (200 + 600) = 3250

Closing stock valuation = 300 3250 = 9,750

8 C EOQ = [(2 160 9,000) (008 40)]05 = 949

9 A

10 B

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11 A Absorption rate = 247,500 30,000 = 825

Absorbed cost = 28,000 825 = 231,000

Actual cost = 238,000

Under absorption = 7,000

12 B Marginal costing profit = 36,000 (2,000 63,000 14,000) = 27,000

13 C Process F: expected output = 092 65,000 = 59,800

actual output = 58,900

abnormal loss

Process G: expected output = 095 37,500 = 35,625

actual output = 35,700

abnormal gain

14 D

Actual hours at standard rate (27,000 850) 229,500

Standard hours of production at standard rate 253,980

Labour efficiency variance is 24,480 Favourable

15 A Sales price variance:

Actual sales at standard price (4,650 6) 27,900

Actual sales at actual price 30,225

Favourable price variance 2,325

Adverse sales volume contribution variance:

350 units (6 060) 1,260

16 C

17 D

18 A

19 A Coefficient of determination = r2 = 06 06 = 036 = 36%

20 C

21 D 4,000 [(20,000 2,500) 1025] = 32,800

22 D Production (units):

J: (6,000 100 + 300) = 6,200

K: (4,000 400 + 200) = 3,80010,000

Joint costs apportioned to J: (6,200 10,000) 110,000 = 68,200

23 D Material required to meet maximum demand:

6,000 (13 4) + 8,000 (19 4) = 57,500 litres

Material available: 50,000 litres

Material is a limiting factorLabour required to meet maximum demand:

6,000 (35 7) + 8,000 (28 7) = 62,000 hours

Labour available: 60,000 hours

Labour is a limiting factor

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24 D Profits maximised when: marginal revenue (MR) = marginal cost (MC)

MR = 50 005Q

MC = 15

MR = MC 50 005Q = 15

and Q = 700

P = 50 (0025 700) = 3250

25 B When P = 20 then 20 = 50 0025Qand Q = 1,200

Total revenue (P Q) = 1,200 20 = 24,000

Less total costs 2,000 + (15 1,200) = 20,000

Profit 4,000

Section B

1 (a) Using the high-low method:

Units Total cost ()

120,000 (W1) 700,000102,000 619,000

18,000 81,000

Working (W1)

Full capacity = 102,000 085 = 120,000

(i) Variable cost per unit = 81,000 18,000 = 450

(ii) Total fixed costs = 700,000 (120,000 450) = 160,000

(iii) Selling price per unit = variable cost per unit (100 040)

= 450 06 = 750

(iv) Contribution per unit = (750 450) = 300

(b) New business: per unit

Selling price (080 750) 600

Less variable cost (450)

Contribution 150

Contribution from 15,000 units (15,000 150) 22,500

Less opportunity cost (15,000 6) 300 (7,500)

Net increase in contribution (and profit) 15,000

(c) An opportunity cost is the cost of the best alternative forgone in a situation of choice. Opportunity costs are relevant costs.

In the situation of Pointdextre Ltd, if it goes ahead with the new business (that is the decision) then it will lose (forgo) thecontribution from some existing sales. This lost contribution is an opportunity cost relevant to the decision.

2 (a) Process I

Litres Litres

Input 50,000 365,000 Output (W1) 47,000 634,500

Conversion 256,000 Normal loss (008 50,000) 4,000

Abnormal gain (W2) 1,000 13,500 51,000 634,500 51,000 634,500

Workings:

W1 Cost per litre (365,000 + 256,000) (50,000 092) = 1350

Output value = 47,000

1350 = 634,500W2 Abnormal gain = 47,000 (50,000 092) = 1,000

Valuation (1,000 1350) = 13,500

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(b) Workings:

Cost per equivalent litre (EL): Conversion

EL

Completion of opening WIP 3,000

Started and finished within the month (50,000 5,000) 45,000

Work done so far on closing WIP 1,00049,000

Cost per EL = 392,000 49,000 = 8

(i) Output = 80,000 + (45,000 1350) + (48,000 800) = 1,071,500

(ii) Closing WIP = (2,000 1350) + (1,000 800) = 35,000

(c) The disposal costs would be debited to the process account. Alternatively, they could be shown as a negative value on the

credit side of the account.

3 Let X = the number of units of product X

and Y = the number of units of product Y

Contribution per unit:

Product X Product Y per unit per unit

Selling price 60 25

Less variable cost (45) (13)

Contribution 15 12

Objective function:

Total contribution = 15X + 12Y

Constraints:

Material (5 per kg) 3X + Y 4,200

Labour (6 per hour) 4X + 05Y 3,000

Non negative X, Y 0

Using a graphical approach, the constraints (solid lines) and the objective function (dotted line) can be shown as follows:

Note: the objective function line has been shown on the above graph for a total contribution of 9,000 (assumed). Thus 15X +

12Y = 9,000.

Therefore when X = 0, Y = (9,000 12) = 750

and when Y = 0, X = (9,000 15) = 600

The feasible region is the area OABC shown on the graph. If the objective function line is moved away from the origin (at the

same gradient) the last point it reaches in the feasible region is point A which must therefore be the optimal point.

Therefore the optimal production is to produce and sell 4,200 units of product Y and no units of product X.

20

B

A

C0

750

600

4,200

6,000

Yunits

750 1,400

X units

Material

Labour

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An alternative approach would be to calculate the total contributions at points A, B and C shown on the graph and select the point

giving the highest total contribution, as follows:

Point A

Total contribution from 4,200 units of Y is (4,200 12) = 50,400

Point B

To find the units at this point, solve the following equations simultaneously:

3X + Y = 4,200 (1)4X + 05Y = 3,000 (2)

From (1) Y = 4,200 3X

Substituting into (2) 4X + 05(4,200 3X) = 3,000

4X + 2,100 15X = 3,000

25X = 900

X = 360

Substituting into (1) (3 360) + Y = 4,200

Y = 3,120

Total contribution from 360 units of X and 3,120 units of Y is (360 15) + (3,120 12) = 42,840

Point C

Total contribution from 750 units of X is (750 15) = 11,250

Point A gives the highest contribution (50,400 from producing 4,200 units of Y and no units of X) and is therefore the optimal

solution (as before).

4 (a)

Standard cost of actual production [12,500 (11 + 24 + 18)] 662,500

Total variances: Adverse Favourable

Materials (W1) 5,200

Labour (W2) 8,700

Fixed overhead (W3) 5,800 11,000 8,700 2,300 A

Actual cost (142,700 + 291,300 + 230,800) 664,800

Workings:

W1 Variance

Actual cost 142,700

5,200 A

Standard cost of actual production 137,500

W2

Actual cost 291,300

8,700 F


W3

Actual cost 230,800

5,800 AStandard cost of actual production 225,000

(b)

Expenditure variance:

Actual cost 230,800

14,800 A

Budgeted cost (12,000 18) 216,000

Volume variance:

Budgeted cost 216,000

9,000 F


(c) The total direct materials and labour variances would be the same under absorption and marginal costing. The total fixedoverhead variance under marginal costing would be different and would be the same as the expenditure variance under

absorption costing (14,800 A). There is no volume variance under marginal costing as fixed production costs are treated

as period costs and not treated as product costs.

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5 (a) Absorption rates:

Cost centre T: (780,000 16,250) = 48 per machine hour

Cost centre W: (173,400 14,450) = 12 per direct labour hour

(b) Prime costs:

Direct materials 10

Direct labour:

Cost centre T 14Cost centre W 21

45

Production overheads:

Cost centre T: (35 60) 48 28

Cost centre W: (21 6) 12 42115

(c) Products do not pass through service cost centres so the costs of such centres cannot be absorbed directly into products.

Products only pass through production cost centres. Therefore in order to calculate a total production cost per unit, service

cost centre costs have to be reapportioned to production cost centres for absorption.

The method of reapportionment that fully recognises any work that service cost centres do for each is called the reciprocal

method. There are two techniques for applying the reciprocal method a repeated distribution approach or the use ofsimultaneous equations.

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Part 1 Examination Paper 1.2

Financial Information for Management December 2005 Marking Scheme

Marks

Section A

Each of the 25 questions in this section is worth 2 marks 50

Section B

1 (a) (i) Variable cost per unit 2

(ii) Total monthly fixed costs 2

(iii) Selling price per unit 1

(iv) Contribution per unit 1

6

(b) Contribution from new business 2

Opportunity cost 11/2

Net increase in profit 1/2

4

(c) Explanation of opportunity cost 1Reference to Pointdextre Ltd 1

2

12

2 (a) Input and conversion 1

Normal loss 11/2

Abnormal gain 11/2

Output 1

5

(b) Equivalent units for conversion 11/2

Cost per equivalent unit for conversion 1/2

Valuation of output 2

Valuation of closing work in progress 1

5

(c) Debit entry 212

3 (a) Contributions per unit 1

Objective function 1

Constraints 3

5

(b) Graph (or total contributions at feasible points) 3

Optimal plan 1

4

9

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Marks

4 (a) Total materials variance 1

Total labour variance 1

Total fixed overhead variance 1

Reconciliation statement 1

4

(b) Expenditure variance 1

Volume variance 1

2

(c) Direct materials and labour variances the same 1

Total variance = expenditure variance 1

No volume variance with reason 1

3

9

5 (a) Cost centre T absorption rate 1

Cost centre W absorption rate 12

(b) Prime cost 1/2

Production overheads (T) 1

Production overheads (W) 1

Total unit cost 1/2

3

(c) Reapportionment explanation 2

Reapportionment method 1

3

8

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f2 past paper_ans05-2005

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