Test Series: March, 2018 MOCK TEST PAPER FINAL (OLD) COURSE: GROUP – II PAPER – 5: ADVANCED MANAGEMENT ACCOUNTING SUGGESTED ANSWERS/HINTS 1. (a)

The following is a possible scorecard for “North Garden” Financial Perspective

Economic Value Added Revenue per villa % repeat customers Number of customer complaints Service rating of spa Staff hours per guest % cost spent for maintenance Travel guide rank for restaurant Employee retention Number of new services offered

Customer Perspective Internal Business

Innovation and Learning (b)

Statement Showing Manufacturing Cost and Buying Cost (` in lakhs) Year

Present Value Factor @ 10%

When the Component is Manufactured

When the Component is Bought

Cash Outflows*

Present Value of Cash Outflows

Cash Outflows (Cost of Buying)

Present Value of Cash Outflows

0

1.000

4

4.000

-

-

1

0.909

6+2

7.272

9

8.181

2

0.826

7+2

7.434

10

8.260

3

0.751

8+2

7.510

11

8.261

4

0.683

10+2

8.196

14

9.562

34.412

34.264

Cash Outflows* means Capital Cost plus Manufacturing Cost plus Opportunity Cost. The above statement shows that there is a saving in buying the component amounting to `0.148 lakh (i.e. ` 34.412 lakhs – 34.264 lakhs). Hence, it is beneficial to buy the component from outside. Note: It may be noted that the loss of ` 2 lakhs of cash inflow for each of the 4 years due to inability of the firm to operate another machine when it manufactures the component has to be treated as an opportunity cost. (c)

Let x1, x2 and x 3 be the number of acres allotted for cultivating radish, mutter and potato respectively.

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Objective function: Since the average yield of radish is 1,500 kg per acre, and the selling price for radish is ` 5/kg hence the selling amount which the agriculturist gets from one acre is` 5 × 1,500 kg Or,

` 7,500

To produce 100 kg of radish, the manure cost is `12.50, so the manure cost per acre will be-

`12.50 × 1,500 kg 100kg Or,

` 187.50

Labour cost per acre for radish` 40 × 6 man days Or,

` 240

Profit per acre for radish` 7,500 − ` 187.50 − ` 240 Or,

` 7,072.50

Similarly, the selling price, manure cost, labour cost and profit per acre of land for mutter and potato are also calculated and presented in the following tablePer Acre

Radish

Selling Price

Mutter

Potato

`7,500

`7,200

`6,000

(`5 × 1,500 kg)

(`4 × 1,800 kg)

(`5 × 1,200 kg)

Less: Manure Cost

` 187.50

` 225

` 187.50

 `12.50 × 1,500 kg    100kg  

 `12.50  1,800kg    100kg  

 `12.50  1,200kg    80kg  

Less: Labour Cost

` 240

` 200

` 240

( ` 40 × 6 man days)

( ` 40 × 5 man days)

( ` 40 × 6 man days)

` 7,072.50

` 6,775

` 5,572.50

Profit

Since, the agriculturist wants to maximise the total profit, hence the objective function of the problem is given byZ

=

7,072.50x 1 + 6,775.00x 2 + 5,572.50x 3

The linear programming model for the problem: Maximise Z = 7,072.50x 1 + 6,775.00x 2 + 5,572.50x 3 Subject to the Constraints:

Where (d)

x1 + x2 + x3

≤ 125

6x1 + 5x2 + 6x3

≤ 500

x 1, x 2, x 3

≥ 0

Journal Entries in Single Plan S. No. Journal Entries

(i)

Debit Amount (`)

Dr. Material Control A/c

√ 2

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Credit Amount (`)

(ii)

(iii)

2. (a)

Dr. or Cr. Material Price Variance A/c Cr. Creditors A/c (Being recording of Price Variance during Purchase of Materials) Dr. WIP Control A/c Dr. or Cr. Material Usage Variance A/c Cr. Material Control A/c (Being recording of Usage Variance at Standard Cost of excess / under utilized Quantity) Dr. Wages Control A/c Dr. or Cr. Labour Rate Variance A/c Cr. Cash (Being recording of Wages at Standard Rate)

√ √

√ √ √

√ √ √

Working for Finding – Missing Figures Cost Variance A

=

0

Cost Variance (A+B)

=

` 1,300 (A)

Yield Variance (A+B)

=

` 270 (A)

Standard Cost and Actual Cost (Incomplete Information) Raw Material

A B Total

Standard Data

Actual Data

Qty. (Kg.) [SQ]

Price (`) [SP]

Amount (`) [SQ x SP]

Qty. (Kg.) [AQ]

Price (`) [AP]

Amount (`) [AQ x AP]

? ?? ? ?? ???

24 30

??? ??? ? ??

??? 70 ? ??

30 ???

??? ??? ???

Material Cost Variance A

= Standard Cost – Actual Cost

0

= (SQA × ` 24 – AQA × ` 30)

SQA

= 1.25 AQA

Material Yield Variance (A+B) = Average Standard Price per unit of Standard Mix × [Total Standard Quantity (units) – Total Actual Quantity (units)] ` 270 (A)

 ` 24 x SQ A + ` 30 x SQ B  =   × SQ A + SQ B   [(SQ A + SQB) – (AQA+70)]

SQA = SQB as Standard Mix is in ratio 1:1 ` 270 (A)

 ` 24 x SQ A + ` 30 x SQ A  =   × SQ A + SQ A   [(SQ A + SQA) – (AQA+70)]

` 270 (A)

= 27 × [2 x SQ A – (AQA+70)] 3

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`270 (A)

= 27 × [2 x 1.25 AQA – (AQA+70)]

AQA

= 40 Kg.

As SQA

= 1.25 AQA = 1.25 × 40 Kg. = 50 Kg.

As SQB

= SQA = 50 Kg.

Cost Variance (A+B)

= Standard Cost – Actual Cost

1,300 (A)

= (50 Kg. × ` 24 + 50 Kg. × ` 30) – (40 Kg. × ` 30 +70 Kg. × APA)

APA

= ` 40

Standard Cost and Actual Cost (Complete Information) Raw Mat.

Standard Data Qty. Price Amount (Kg.) (`) (`) [SQ] [SP] [SQ x SP]

Qty. (Kg.) [AQ]

A B Total

50 50 100

40 70 110

24 30

1,200 1,500 2,700

Actual Data Price Amount (`) (`) [AP] [AQ x AP]

30 40

1,200 2,800 4,000

Std. Cost of Actual Qty. (`) [AQ x SP]

960 2,100 3,060

Computation of Variances Material Cost Variance

= Standard Cost – Actual Cost = SQ × SP – AQ × AP

(A)

= ` 1,200 – ` 1,200 = `0

(B)

= ` 1,500 – ` 2,800 = ` 1,300 (A)

Total

= ` 0 + ` 1,300 (A) = ` 1,300 (A)

Material Price Variance

= Standard Cost of Actual Quantity – Actual Cost = AQ × SP – AQ × AP

Or = AQ × (SP – AP) (A)

= 40 Kg. × (` 24.00 – ` 30.00) = ` 240 (A) 4

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(B)

= 70 Kg. × (` 30.00 – ` 40.00) = ` 700 (A)

Total

= ` 240 (A) + ` 700 (A) = ` 940 (A)

Material Usage Variance

= Standard Cost of Standard Quantity Output – Standard Cost of Actual Quantity

for

Actual

= SQ × SP – AQ × SP Or (A)

= SP × (SQ – AQ) = ` 24 × (50 Kg. – 40 Kg.) = ` 240 (F)

(B)

= ` 30 × (50 Kg. – 70 Kg.) = ` 600 (A)

Total

= ` 240 (F) + ` 600 (A) = ` 360 (A)

Material Mix Variance

= Total Actual Quantity (units) × (Average Standard Price per unit of Standard Mix – Average Standard Price per unit of Actual Mix)

 ` 2,700 ` 3,060   = 110 Kg. ×    100 Kg. 110 Kg.  = ` 90 (A) Material Yield Variance

= Average Standard Price per unit of Standard Mix × [Total Standard Quantity (units) – Total Actual Quantity (units)]

 ` 2,700  =   × (100 Kg. – 110 Kg.)  100 Kg.  = ` 270 (A) Standard Output

= Standard Input – Standard Loss = 100 Kg. – 10 Kg. = 90Kg.

Actual Output

= 90 Kg.

(Actual Output and Standard Output are always equal numerically in any Material Variance Analysis)

(b) The condition for degeneracy is that the number of allocations in a solution is less than m+n -1. The given problem is an unbalanced situation and hence a dummy row is to be added, since the column quantity is greater than that of the row quantity. The total number of rows and columns 5

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will be 9 i.e. (5 rows and 4 columns). Therefore, m+n-1 (= 8), i.e. if the number of allocations is less than 8, then degeneracy would occur. 3.

(a)

(i) Calculation of Missing Figures: Activity

1–2 1–3 1–4 2–4 2–5 3–6 4–6 5–7 6–7 6–8 7–8 8–9

Duration

EST

EFT

LST

LFT

Total Float

Dij

Ei

Ei + Dij

Lj − Dij

Lj

LST− EST

4 12 7 8 5 9 11 13 0 5 7 6

0 0 0 4 4 12 12 9 23 23 23 30

4 12 7 12 9 21 23 22 23 28 30 36

0 2 5 4 5 14 12 10 23 25 23 30

4 14 12 12 10 23 23 23 23 30 30 36

0 2 5 0 1 2 0 1 0 2 0 0

(ii) The Network for the given problem:

(iii) The Various Paths in the Network are: 1–2–4–6–7–8–9 with Duration 36 Days 1–2–5–7–8–9 with Duration 35 Days 1–3–6–7–8–9 with Duration 34 Days 1–2–4–6–8–9 with Duration 34 Days 1–3–6–8–9 with Duration 32 Days 1–4–6–7–8–9 with Duration 31 Days 1–4–6–8–9 with Duration 29 Days The Critical Path is 1–2–4–6–7–8–9 with Duration 36 Days.

6

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(b)

Statement Showing Cost and Profit for the Next Year Particulars

Existing Volume, etc.

Volume, Costs, etc. after 10% Increase

(`) 5,00,000 2,50,000 1,00,000 40,000 1,10,000 60,000 50,000

Sale Less: Direct Materials Direct Labour Variable Overheads Contribution Less: Fixed Cost# Profit

Estimated Sale, Cost, Profit, etc.*

(`) 5,50,000 2,75,000 1,10,000 44,000 1,21,000 60,000 61,000

(`) 5,72,000 2,69,500 1,07,800 43,120 1,51,580 58,800 92,780

(*) for the next year after increase in selling price @ 4% and overall cost reduction by 2%. ( #) Fixed Cost

=

Existing Sales – Existing Marginal Cost – 12.5% on `4,00,000

=

`5,00,000 – `3,90,000 – `50,000

=

`60,000

 `92,780



x 100  Percentage Profit on Capital Employed equals to 23.19%   ` 4,00,000 

Since the Profit of `92,780 is more than 23% of capital employed, the proposal of the Sales Manager can be adopted. 4.

(a) Customer Wise Profitability Statement and Overall Profitability Statement SN. Particulars

P

M

W

Total `

A

Sales (net proceeds) –Table 1

241,288

237,500

272,812

751,600

B

Variable Cost of Goods Sold @150 p.u.

1,50,000 1,42,500

1,87,500

4,80,000

C

Assignable- Marketing and Administration Cost - Table 2 •

Order Taking and

•

Processing

1,200

600

4,500

6,300

Sale Return Processing

150

-

1,200

1,350

•

Billing Cost

200

100

750

1,050

•

Customer Visit

800

-

4,000

4,800

2,350

700

10,450

13,500

250

-

1,250

1,500

8,000

4,000

-

12,000

Total Assignable Marketing and Administration Cost D

E

Assignable- Distribution Cost - Table 2 •

Expedited / Rush

Orders

•

Delivery Costs

•

Inventory Carrying Cost

10,000

9,500

12,500

32,000

Total Assignable Distribution Cost

18,250

13,500

13,750

45,500

-

-

-

100,000

Non- Assignable Fixed Cost 7

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F

Total Costs (B+C+D+E)

G

Net Profit (Step A - F)

H

Profit % of Sales (G / A)

170,600

156,700

211,700

639,000

70,688

80,800

61,112

112,600

29%

34%

22%

15%

Workings Table 1: Customer Sales Analysis - Revenue Analysis All figures in ` Particulars

P

M

W

Total `

Sales {Sale Units × Sale Price (gross)}

2,50,000

2,37,500

3,12,500

8,00,000

Less: Sale Return (Step 1 × Return%)

1,250

-

31,250

32,500

2,48,750

2,37,500

2,81,250

7,67,500

7,462

-

8,438

15,900

2,41,288

2,37,500

2,72,812

7,51,600

97%

100%

87%

94%

Net Sales Less: Cash Discount Net Proceeds Final Collections vs Original Sale

Table 2: Assignable Marketing, Administrative and Distribution Costs All figures in ` Particulars Order Taking and Processing

P

M

W

Total

1,200

600

4,500

6,300

250

-

1,250

1,500

8,000

4,000

-

12,000

150

-

1,200

1,350

200

100

750

1,050

800

-

4,000

4,800

10,000

9,500

12,500

32,000

(# of orders × cost per order)

Expedited / Rush Orders (# of orders × cost per order)

Delivery Costs (Distance in km. × cost per km)

Sale Return Processing (# of returns × cost per return)

Billing Cost (# of invoices × cost per invoice) Customer Visit (#of customer visits × cost per visit)

Inventory Carrying Cost (# of units × inventory carrying cost p.u.)

(b) Identification of Bottleneck: Installation of cameras is the bottleneck in the operation cycle. The annual capacity for manufacturing and installation are given to be 750 camera units and 500 camera units respectively. Actual capacity utilization is 500 camera units, which is the maximum capacity for the installation process. Although, Z can additionally manufacture 250 camera units, it is constrained by the maximum units that can be installed. Therefore, the number of units manufactured is limited to 500 camera units, subordinating to the bottleneck installation operation. Therefore, Z should focus on improving the installation process.

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5. (a)

The usual learning curve model is y

=

axb

y

=

Average time per part for x parts

a

=

Time required for first part (100 minutes)

x

=

Cumulative number of parts

b

=

Learning coefficient and is equal to –0.322

Where

(learning rate 80%) Calculation of total time for 40 parts: y

=

100 × (40) -0.322

log y =

log100 − 0.322 × log40

log y =

log100 − 0.322 × [3 × log2 + log5]

log y =

2 − 0.322 × [3 × 0.30103 + 0.69897]

log y =

1.484

y

=

antilog of 1.484

y

=

30.48 minutes

=

40 Parts × 30.48 minutes

=

1,219 minutes (A)

Total time for 40 Parts

Calculation of total time for 60 parts: y

=

100 × (60) −0.322

log y =

log100 − 0.322 × log60

log y =

log100 − 0.322 × [2 × log2 + log5 + log3]

log y =

2 − 0.322 × [2 × 0.30103 + 0.69897 + 0.47712]

log y =

1.4274

y

=

antilog of 1.4274

y

=

26.75 minutes

Total Time for 60 Parts = =

60 Parts × 26.75 minutes 1,605 minutes (B)

Calculation of total time for 41 to 60 parts (B) – (A):

(b) (i)

=

1,605 minutes –1,219 minutes

=

386 minutes

STC receives a 10% Commission on each ticket

=

` 900 (10% × ` 9,000)

Commission per ticket Variable Cost per ticket Contribution per ticket

= = = =

` 900 ` 200 ` 900 − ` 200 ` 700

Fixed Costs

=

` 1,40,000 per month

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(a)

Break-even Number of Tickets

= =

(b)

Fixed Costs Contribution per ticket

`1,40,000 ` 700 per ticket

=

200 tickets

When Target Operating Income

=

` 70,000 per month

Quantity of Tickets required to be sold

=

`1,40,000 `70,000 `700 per ticket

=

`2,10,000 `700

=

300 tickets

(ii) Under the New System, STC would receive only `500 on the ` 9,000 per ticket.

Thus,

Commission per ticket

=

` 500

Variable Cost per ticket

=

` 200

Contribution per ticket

=

` 500 − ` 200

=

` 300

Fixed Costs

=

` 1,40,000 per month

Break-even Number of Tickets

=

`1,40,000 `300

=

467 tickets (rounded up)

=

` 2,10,000 ` 300

=

700 tickets

Quantity of Tickets required to be sold

The ` 500 cap on the Commission paid per ticket causes the Break-even Point to more than double (from 200 to 467 tickets) and The Tickets required to be sold to earn ` 70,000 per month to also more than double (from 300 to 700 tickets). As would be expected, travel agents will react very negatively to the Dolphin Airlines decision to change commission payments. 6. (a)

Ranking of Products When Availability of Time is the Key Factor Products

Market Price (`) Less: Variable Cost (`) Contribution per unit (`) Labour Hours per unit Contribution per Labour Hour Ranking Maximum Demand (units) Total No. of Hours Allocation of 20,000 Hours on the Basis of Ranking (*) Balancing Figure 10

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A

B

C

D

150 130 20 3 hrs. 6.67 IV 2,800 8,400 600*

146 100 46 4 hrs. 11.50 III 2,500 10,000 10,000

140 90 50 2 hrs. 25.00 I 2,300 4,600 4,600

130 85 45 3 hrs. 15.00 II 1,600 4,800 4,800

Note: Time required to meeting the demand of 2,500 units of Product D for Division Y is 7,500 hrs. This requirement of time viz. 7,500 hrs for providing 2,500 units of Product D for Division Y can be met by sacrificing 600 hours of Product A (200 units) and 6,900 hours of Product B (1,725 units). Transfer Price

(b) (i)

=

Variable Cost + Opportunity Cost

=

`85 

(6,900 hrs. `11.5  600 hrs. `6.66) 2,500 units

=

` 85 

`79,350  ` 4,000 2,500units

=

`85 + `33.34

=

`118.34

The current cost and profit per unit are calculated as below: Cost Component Revenue

Units

Actual Cost p.a. for 10,000 racks (`)

Actual Cost per rack (`)

10,000 racks

75,00,000

750

5,20,000 sq. ft.

20,00,000

200

1,00,000 hrs.

10,00,000

100

15,000 hrs.

1,50,000

15

200,000 hrs.

30,00,000

300

Total Cost

61,50,000

615

Profit

13,50,000

135

Direct Material Direct Labour Machine Setup Mechanical Assembly

Therefore, the current cost is ` 615 p.u. while the profit is ` 135 p.u. Machine setup is the time required to get the machines and the assembly line ready for production. In this case, 15,000 hours spent on setting up does not add value to the storage racks directly. Hence, it is a non-value add activity. (ii) New sale price per rack is ` 675 per unit. The profit per unit needs to be maintained at `135 per unit. Hence, the new target cost per unit = new selling price per unit – required profit per unit = ` 675 - ` 135 = ` 540 per unit. 7. (a)

Committed Fixed Cost / Discretionary Fixed Cost Committed Fixed Cost

Discretionary Fixed Cost

(i) Salary and Wage increase.

(ii) New Advertisement Cost.

(iii) Rents payable for the next 6 months. (iv) Legal Fees for filing for patent rights. (b)

The product life cycle span the time from the initial R & D on a product to when customer service and support is no longer offered for that product. Life Cycle Costing technique is particularly important when: (i)

High percentage of total life-cycle costs are incurred before production begins and revenue are earned over several years and 11

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(ii) High fraction of the life cycle costs are locked in at the R & D and design stages. F should identify those industries and then companies belonging to those industries where above mentioned feature are prevalent. For example, Automobile and Pharmaceutical Industries companies like Tata Motors Ltd., Sun Pharmaceutical Industries Ltd., and Dabur India Ltd. will be good candidates for study on product life cycle costing. (c) Statement Showing Selling Price Perfect Competition

Monopoly

6,000

1,200

1,80,000

1,80,000

30

150

---

200

Variable Cost per unit

200

---

Selling Price per unit

230

350

Units Contribution (` 1,06,000 + ` 74,000) Contribution per unit 

4





Variable Cost per unit  `150   3

(d) (i)

Invalid Kaizen costing is the system of cost reduction procedures which involves making small and continuous improvements to the production processes rather than innovations or large -scale investment.

(ii) Valid The training of employees is very much a long-term and ongoing process in the Kaizen costing approach. Training enhances the abilities of employees. (iii) Invalid Kaizen Costing approach involves everyone from top management level to the shop floor employees. Every employee’s active participation is a must requirement. (iv) Invalid Though the aim of Kaizen Costing is to reduce the cost but at the same time it also aims to maintain the quality. Kaizen costing also aims to bring the clarity in roles and responsibilities for all employees. (e) The given problem is a balanced minimization assignment problem. The minimum time elements in row 1, 2 and 3 are 70, 50 and 110 respectively. Subtract these elements from all elements in their respective row. The reduced matrix is shown belowA

B

C

1

30

15

0

2

0

20

60

3

0

10

20

The minimum time elements in columns A, B and C are 0, 10, and 0 respectively. Subtract these elements from all the elements in their respective columns to get the reduced time matrix as shown below12

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A

B

C

1

30

5

0

2

0

10

60

3

0

0

20

The minimum number of horizontal and vertical lines to cover all zeros is 3, which is equal to the order of the matrix. The Pattern of assignments among software professionals and programs with their respective time (in minutes) is given belowProgram

Software Professionals

Time (in Minutes)

1

C

70

2

A

50

3

B

120

Total

240

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