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Aap Tutorial 3 Answer

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chapter 21 Test your understanding 3 ­ Hanford & Stopple Answer 1 (a) Hanford consolidated statement of financial position at 30 September 20X1 Non­current assets Property, plant and equipment (W8) Goodwill (W3) $000 $000 109,510 6,850 ––––––– 116,360

Current assets Inventory (7,450 + 4,310) Accounts receivable (12,960 + 4,330 – 820 (W7)) Bank

11,760 16,470 520 ––––––– 28,750 ––––––– 145,110 –––––––

Equity and liabilities Equity attributable to the equity holders of the parent: Ordinary shares of $1 each (20,000 +10,000 (W6)) Reserves: Share premium (10,000 + 10,000 (W6)) Retained earnings (W5)


20,000 65,750 –––––––

Non­controlling interests (W4)

85,750 ––––––– 115,750 6,950 ––––––– 122,700 6,000

Non­current liabilities 8% Loan notes 20X4 Current liabilities Trade accounts payable (5,920 + 4,160 – 620 (W7)) Bank overdraft Provision for taxation (3,070 + 2,180) 9,460 1,700 5,250 ––––––– Total equity and liabilities

16,410 ––––––– 145,110

Questions & Answers

Workings (all figures in $000) (W1) Group Structure Hanford | 75% Stopple 6m/8m

Note: the unrealised profit on the sale of the plant was initially $400,000, of this 10% i.e. $40,000 has been realised via Stopple's depreciation charge, giving a net adjustment of $360,000 to both Hanford's profits (W5) and the carrying value of the plant. (W2) Net assets in subsidiary At acquisition $000 8,000 2,000 10,000 4,000 –––––– 24,000 –––––– At reporting date $000 8,000 2,000 14,000 4,000 (200) –––––– 27,800 ––––––

Share capital Share premium Retained earnings (6,000 + 4,000) Fair value adjustment Administration charges

Fair value adjustments The insurance claim is a contingent asset and cannot be recognised by Stopple. IFRS revised requires Hanford to make an assessment of all assets, liabilities and contingent liabilities of Stopple at the date of acquisition. Contingent assets can only be recognised if they are virtually certain, the receipt of the claim is merely ‘highly likely’. Therefore it is not recognised by the group at the date of acquisition. Therefore the fair value adjustment is only for the land.



chapter 21

(W3) Goodwill ­ Proportionate share method $000 24,850 (18,000) ______ 6,850 ______

Cost of investment For 75% Net assets at acq (75% × 24,000) Goodwill

(W4) Non­controlling interest 25% × 27,800 (W2) = 6,950

(W5) Consolidated retained earnings $000 63,260 (360) 2,850

Hanford Profit on sale of plant (see below) Stopple Group share post acquisition profits (27,800 ­ 24,000) × 75%

–––––– 65,750 ––––––

(W6) Share exchange Hanford acquired six million shares in Stopple. On the basis of an exchange of five for three, Hanford would issue 10 million new shares. The total value of the consideration is $24.85 million of which $4.85 million was for cash, therefore the value of the 10 million shares would be $20 million, or $2 each i.e. they were issued at a premium of $1 each.



Questions & Answers

(W7) Elimination of current accounts The difference on the current accounts is due to the invoice for central administration of $200,000. A summary of the intra­group adjustment/cancellation is: Dr Hanford’s overdraft Accounts payable Accounts receivable 200 620 –––– 820 –––– 820 –––– 820 –––– Cr

(W8) Property, plant and equipment Amount from question – Hanford – Stopple Fair value adjustment Unrealised profit in transfer of plant 78,690 27,180 4,000 (360) ––––––– 109,510 –––––––

Note: the unrealised profit on the sale of the plant was initially $400,000, of this 10% i.e. $40,000 has been realised via Stopple's depreciation charge, giving a net adjustment of $360,000 to both Hanford's profits (W5) and the carrying value of the plant. (b) The reasons why a parent company may not wish to consolidate a subsidiary can be broken down into two broad groups; (i) to improve the reported position of the group financial statements; and (ii) where consolidating a subsidiary might not give a fair presentation of the performance and position of the group. Improvement of the financial position The financial statements of a subsidiary could show any of the following: – – –

substantial operating losses a poor liquidity position, or high levels of borrowing (high gearing).

chapter 21

If a parent were to consolidate such a subsidiary, it would proportionately worsen the group position in the above areas. Thus a parent may prefer not to consolidate poorly performing subsidiaries. Fair presentation There is a case for excluding subsidiaries from a parent’s consolidated financial statements for the following reasons: – The subsidiary operates under severe long­term restrictions. In effect the parent does not have full control (particularly over the ability to transfer funds to the parent) over the subsidiary, Control is intended to be temporary because the investment is held exclusively with a view to its subsequent resale.

It is apparent that the first group of reasons for non­consolidation is not permitted by International Accounting Standards, whereas in theory the latter group might be. In addition, subsidiaries may sometimes be excluded on the basis of differing activities. Companies that have adopted this approach argue that to add together the assets and liabilities of companies whose activities differ greatly could lead to consolidated financial statements that give a misleading impression (or not provide fair presentation). IAS 27 has never permitted exclusion on these grounds because it feels that ‘differing activity’ problems are overcome by the provision of segmental information. The revised version of IAS 27 Consolidated and Separate Financial Statements does not allow any exclusions from consolidation; all subsidiaries must be consolidated. However, IAS 27 requires disclosure of the nature and extent of any significant restrictions on the ability of a subsidiary to transfer funds to the parent. Where a subsidiary is held exclusively with a view to subsequent resale (provided it has not previously been consolidated) IFRS 5 Non­current Assets Held for Sale and Discontinued Operations requires that it is presented separately in the statement of financial position and that other information is disclosed, so that users of the financial statements are made aware that control is only intended to be temporary.



Questions & Answers Test your understanding 4 ­ Hepburn and Salter Answer 1 Hepburn Consolidated income statement year to 31 March 20X0 $000 Sale revenue (1,200 + 500 – 100 intra­group sales) 1,600 Cost of sales (W7) (890) ––––– Gross profit 710 Operating expenses (120 + 44) (164) Finance costs (12 x 6/12) (6) ––––– Profit before tax 540 Income tax expense (100 + 20) (120) ––––– Profit for the period 420 ––––– Attributable to: Equity holders of the parent 400 Non­controlling interests (200 x 20% × 6/12) 20 ––––– 420 ––––– Consolidated statement of financial position at 31 March 20X0 Non­current assets Property, plant and equipment (620 + 660 + 125) Intangible: Goodwill (W3) Investments (20 + 10) $000 $000

1,405 255 30 ––––– 1,690 510 324 80 –––––

Current assets Inventory (240 + 280 – 10) [W1] Accounts receivable (170 + 210 – 56 (W6)) Bank (20 + 40 + 20 (W5)) Total assets

914 ––––– 2,604 –––––



chapter 21
Equity and liabilities: Equity shares of $1 each (400 + 300 (W3)) Reserves: Share premium (W3) Retained earnings (W6) 700 600 480 –––––

Non­controlling interest (W4)

1,080 ––––– 1,780 250 ––––– 2,030 150

Non­current liabilities 8% Debentures Current liabilities Trade payables (210 + 155 – 36 (W6)) Taxation (50 + 45)

329 95 –––––

424 ––––– 2,604 –––––

Workings (W1) Group Structure Hepburn 80% Salter

The unrealised profit (URP) in inventory is calculated as: Intra­group sales of $100,000 of which one half is in inventory at the year end = $50,000. This has been sold at a mark­up of 25% on cost, therefore the URP in inventory is $50,000 × 25%/125% = $10,000.



Questions & Answers

(W2) Net Assets ­ Salter Share Capital retained earnings (500 + (200x6/12)) Book Value Fair Value adjustment Land Fair value at date of acquisition $000 150 600 ____ 750 125 ____ 875 at reporting date $000 150 700 ____ 850 125 ____ 975

(W3) Goodwill ­ Fair Value (full goodwill method) Hepburn issued five shares for every two shares it acquired in Salter. Therefore Hepburn issued ((150,000/2 x 5) x 80%) = 300,000 shares at a value of $3 each for a total consideration of $900,000. This would be recorded in Hepburn’s books as equity share capital of $300,000 and share premium of $600,000. $000 230 175 —— 55 –––– 255 –––– $000 900 (700) ____ 200

Investment at cost (300 × $3) For 80% (875) Parent shares of goodwill FV of NCI NCI in net assets at date of acq (875 x 20%)

Total Goodwill



chapter 21

(W4) Non­controlling interest (Minority Interest) $000 195 55 ____ 250 ––––

20% 975 NCI in goodwill

(W5) Consolidated reserves $000 410 80 (10) –––– 480 ––––

Hepburn’s reserves Share of Salter’s post acquisition profits (200 x 6/12 x 80%) URP in inventory

(W6) Elimination of current accounts The difference on the current accounts is due to cash in transit of $20,000. A summary of the intra­group cancellations is: $000 20 36 –––– 56 –––– $000

Cash/bank Accounts payable Accounts receivable

56 –––– 56 ––––



Questions & Answers

(W7) Cost of sales $000 650 330 (100) 10 –––– 890 ––––

Hepburn Salter (660 × 6/12) Intra­group sales URP in inventory (below)

Test your understanding 5 ­ Holding Answer 2 Holding has acquired 18 million of Subside’s 24 million equity shares which represents 75% ownership. (a) Goodwill $m Cost of investment Non­controlling interest Share capital Revaluation reserve (W1) Profit and loss reserve (W2) $m 174

24 64 88 –––– 25%x 176

44 ___ 218 (176) –––– 42 ––––

Net assets of subsidiary at acquisition date Goodwill



chapter 21

Workings (W1) Fair value gains/revaluation reserve Property – at 30 June 20X8 – increase up to 1 January 20X9 $m 20 4 –––– 24 –––– 60 (10) –––– 50 (90) –––– 40 –––– 64 ––––

Plant: Net book value 30 June 20X8 Depreciation 6 months to date of acquisition (20 × 6/12) Net book value at date of acquisition Fair value at date of acquisition Fair value increase Total revaluation gains (24 + 40)

(W2) Pre­acquisition profits Retained earnings at 30 June 20X8 Retained profit for period (6/15m × $60m) Total pre­acquisition profits 64 24 –––– 88 ––––



Questions & Answers
(b) Consolidated income statement of Holding for the year to 30 September 20X9 Sales revenue (W1) Cost of sales (W2) Gross profit Operating expenses (W3) Interest payable (10 + (9/12m × 5)) Profit before tax Income tax expense (22 + (9/12m × 10)) Profit for the period Attributable to: Equity holders of the parent Non­controlling interest (W4) $ million 488 (286) –––– 202 (93) (13) –––– 96 (28) –––– 68 –––– 62 6 –––– 68 ––––

Workings: ($ million) (W1) Most of the figures in the consolidated income statement are based on the whole of the holding company’s figures plus the post acquisition figures of the subsidiary. The results of the subsidiary are for a 15 month period, of which nine months is post acquisition. Thus the post acquisition results would be 9/15 or 60% of Subside’s relevant figures. Sales revenue $m 350 168 (30) –––– 488 ––––

Holding Subside (9/12 × 280) Intra­group sales



chapter 21
(W2) Cost of sales $m 200 102 (30) 2 12 –––– 286 ––––

Holding Subside (9/12 × 170) Intra­group sales Unrealised profit in inventory (see below) Additional depreciation (see below)

Unrealised profit A mark­up of 25% on cost is equivalent to 20% of the selling price. Holding has $10 million ($30 m x 1/3) of inventories at the transfer price, thus the unrealised profit is ($10 m x 20%) $2 million. Additional depreciation (plant of Subside) At the date of the acquisition (1 January 20X9) the plant is two and a half years old and has a remaining life also of two and a half years. Therefore the revaluation gain of $40 million will be amortised at $16 million per annum ($40 m/2.5). The post acquisition period is 9 months and would thus require additional depreciation of $12 million (9/12 × $16m). (W3) Operating expenses $m Holding Subside (9/12% × 35) 72 21 –––– 93 ––––

(W4) Non­controlling interest The profit after tax of Subside is $60 million of which $36 million (9/15) is post acquisition. The depreciation adjustment of $12 million (see W4 above) is deducted from this to give an adjusted figure of $24 million. The NCI has a 25% interest in this profit = $6 million.


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...TUTORIAL 1- BUDGETING 1 Question 1 (a) Overhead costs for the 2010 budget: Property cost = $120,000 x 1·05 = $126,000 Central wages = ($150,000 x 1·03) + $12,000 = $166,500 Stationery = $25,000 x 0·6 = $15,000 (b) The road repair budget will be based on 2,200 metres of road repairs; it is common to include a contingency in case roads unexpectedly need repair (see part (c)). The weather conditions could add an extra cost to the budget if poor or bad conditions exist. The adjustment needed is based on an expected value calculation: (0·7 x 0%) + (0·1 x 10%) + (0·2 x 25%) = 6% Hence the budget (after allowing for a 5% inflation adjustment) will be: 2,200 x $15,000 x 1.06 x 1.05 = $36,729,000 This could be shown as: (2,200 x 15,000 x 1·0 x 0·7) + (2,200 x 15,000 x 1·1 x 0·1) + (2,200 x 15,000 x 1·25 x 0·2) = $34,980,000 The $34,980,000 could then be adjusted for inflation at 5% to give $36,729,000 as above. Question 3 (c) Beyond Budgeting (BB) is a responsibility culture in which managers are given goals that have been derived from benchmarks linked to competitors and world class performance. This culture requires an adaptive approach whereby authority is devolved to managers and the organisation’s structure will be a ‘network’ rather than ‘hierarchical’. The principles of BB are: * The organisation structure should have clear principles and boundaries. Everyone should have defined areas of responsibility. ...

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Additional Tutorial 3

...| | Course Title | : | Physics I | Year/ Trimester Session | :: | Year 1 / Trimester 12016/05 | | Lecturer | : | | Additional Questions 3: Kinematics 1. A balloon is 30.0 m above the ground and is rising vertically with a uniform speed when a coin is dropped from it. If the coin reaches the ground in 4.00 s, what is the speed of the balloon? Solution:- Initial velocity of coin = speed of balloon, v. by using the equation [Answer: 12.1 ms–1] 2. A car and train moves together along two parallel paths at 25.0 m s–1. The car then undergoes a uniform acceleration of -2.5 m s–2 because of a red light and comes to rest. It remains at rest for 45.0 s, then accelerates back to a speed of 25 m s–1 at a rate of +2.5 m s–2. How far behind the train is the car when it reaches the speed of 25 m s–1, assuming that the train’s speed has remained constant at 25 m s–1. Solution:- For the car to stop we used the equation v2=v02 + 2as and v = v0 + at m and s For the car to speed up again, m and time taken, s Total distance moved by car in that time = 125 m + 125 m = 250 m. Total distance travelled by the train = 25 × (10+45+10) = 1625 m Therefore the car is (1625 – 250) = 1375 m behind the train. [Answer: 1375 m] *3. A fugitive tries to hop on a freight train traveling at a constant speed of 6.0 m/s. Just as an empty box car passes him, the fugitive starts......

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Mis Tutorial Answer Sheet

...table also contains a key field to uniquely identify each record for retrieval or manipulation. List and describe the three operations of a relational DBMS. In a relational database, three basic operations are used to develop useful sets of data: select, project, and join. • Select operation creates a subset consisting of all records in the file that meet stated criteria. In other words, select creates a subset of rows that meet certain criteria. • Join operation combines relational tables to provide the user with more information that is available in individual tables. • Project operation creates a subset consisting of columns in a table, permitting the user to create new tables that contain only the information required. 3. What are some important database design principles? Define and describe normalization and referential integrity and explain how they contribute to a well-designed relational database. Normalization is the process of creating small stable data structures from complex groups of data when designing a relational database. Normalization streamlines relational database design by removing redundant data such as repeating data groups. A well-designed relational database will be organized around the information needs of the business and will probably be in some normalized form. A database that is not normalized will have problems with insertion, deletion, and modification. Referential integrity rules ensure that relationships between......

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