Mark Scheme
January 2021
Final
Pearson LCCI Certificate in Advanced
Business Calculations Level 3
(ASE3003)
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© Pearson Education Ltd 2021
General Marking Guidance
•
All candidates must receive the same treatment. Examiners must mark the first
candidate in exactly the same way as they mark the last.
•
Mark schemes should be applied positively. Candidates must be rewarded for
what they have shown they can do rather than penalised for omissions.
•
Examiners should mark according to the mark scheme not according to their
perception of where the grade boundaries may lie.
•
There is no ceiling on achievement. All marks on the mark scheme should be
used appropriately.
•
All the marks on the mark scheme are designed to be awarded. Examiners should
always award full marks if deserved, i.e. if the answer matches the mark scheme.
Examiners should also be prepared to award zero marks if the candidate’s
response is not worthy of credit according to the mark scheme.
•
Where some judgement is required, mark schemes will provide the principles by
which marks will be awarded and exemplification may be limited.
•
When examiners are in doubt regarding the application of the mark scheme to a
candidate’s response, the team leader must be consulted.
•
Crossed out work should be marked UNLESS the candidate has replaced it with
an alternative response.
Question
Number
1 (a)
Answer
Mark
Allthat Glitters Ltd:
AU$
Cost
Discount
Factor
NPV (AU$)
8,000,000
-1.000
-8,000,000
Year 1
800,000
0.917
733,600
Year 2
2,700,000
0.842
2,273,400
Year 3
4,500,000
0.772
3,474,000
Year 4
5,000,000
0.708
3,540,000
A1 any
--------------
M1
2,021,000
A1
First M1 is for setting about the NPV calculation
correctly, including the investment cost at par and
shown as negative. A1 for any one of the yearly
contributions to NPV. Second M1 is for adding the
figures. Final A1 is for achieving the correct answer.
Question
Number
1 (b)
M1
Answer
(4)
Mark
Chief Engineer's proposal:
AU$
Cost
Discount
Factor
NPV (AU$)
9,600,000
-1.000
-9,600,000
Year 1
800,000
0.917
733,600
Year 2
2,700,000
0.842
2,273,400
Year 3
5,400,000
0.772
4,168,800
Year 4
6,000,000
0.708
4,248,000
A1
------------1,823,800
A1
A1 for fully correct row for both of Year 3 and Year 4,
and A1 for new NPV.
As a percentage,
=(
1,823,800
) 100 = 90.24245423%
2,021,000
Accept 90% or more accurate
A1
(3)
Question
Number
1 (c)
Answer
Mark
The NPV after 4 years is lower than for the original,
less expensive, proposal
Therefore, I would not agree.
A1ft
A1ft
or
Although the NPV after 4 years is slightly lower than
for the original proposal, if the project continues after
this, it may have a higher output, and thus has
potential to generate higher profits for the
company.(A1)
Therefore, I would agree. (A1)
Question
Number
1 (d)
Reward any other explanation that accords with the
data. One mark for a valid reason. Two marks for a
valid reason with an accompanying
agreement/disagreement.
No marks for "agree" or "disagree" without a valid
justification.
(2)
Answer
Mark
IRR
= 7% +
2,604,300
(9% − 7%)
(2,604,300 − 2,021,000)
= 7% + 8.92954%
= 15.92954%
Accept 16% or more accurate
M1M1
A1
M1 for numerator in formula, and M1 for
denominator.
Alternatively, IRR
=
2,604,300  9% − 2,021,000  7% 9, 291,700
(M1M1)
=
2,604,300 − 2,012,000
583,300
= 15.92954% as before.(A1)
(3)
(Total for Question 1 = 12 marks)
Question
Number
2 (a)
Answer
Year
Mark
Annual
Depreciation
K
Cumulative
Depreciation
K
0
Question
Number
2 (b)(i)
Question
Number
2 (b)(ii)
Question
Number
2 (b)(iii)
Book Value at
end of Year
K
27,500,000
1
3,300,000
3,300,000
24,200,000
2
2,904,000
6,204,000
21,296,000
3
2,555,520
8,759,520
18,740,480
M1 for 12% of K27,500,000 as K3,300,000
M1 for cumulative column
M1 for book value column
A1 for fully correct table
M1
M1
M1
A1(4)
Answer
Mark
Depreciation in Year 5:
K18,740,480 x 0.88 x 0.12
= K1,978,994.688
Accept K1,978,995 or more accurate
M1
A1
(2)
Answer
Mark
Book value at end year 6:
K27,500,000 x (0.88)6
= K12,771,112.39
Accept K12,771,112 or more accurate
M1
A1
Alternatively
K18,740,480 x (0.88)3 (M1)
= K12,771,112.39 (A1)
Accept K12,771,112 or more accurate
(2)
Answer
Mark
Depreciation in year 8:
K27,500,000 x (0.88)7 x 0.12
= K1,348,629.468
Accept K1,348,629
M1
A1
Alternatively, using their answer to (b)(ii),
K12,771,112 x 0.88 x 0.12 (M1)
= K1,348,629.427 (A1)
Accept K1,348,629
(2)
Question
Number
2 (c)
Answer
Mark
Original cost is (annual depreciation x years)
+ scrap value
Cost = (4 x K62,375) + K90,000
= K339,500
M1
A1 (2)
(Total for Question 2 = 12 marks)
Question
Number
3 (a)(i)
Answer
Mark
Liabilities = 145,250€ + 124,750€
= 270,000€
Assets as a percentage
= 159,610€ / 270,000€
= 0.5911481481 = 59.1% to 3sf
M14
Answer
Mark
Paid to unsecured creditors
= 159,610€ - 124,750€
= 34,860€
M1
A1(2)
Answer
Mark
Dividend calculation: 34,860€ / 145,250€
= 0.24
Dividend rate payable is 24c in the euro
Also accept 0.24€ in the euro
M1
Question
Number
3 (b)
Answer
Mark
Question
Number
3 (c)(i)
Answer
Mark
Pierre is the unsecured creditor.
His rate is 7,200€ / 18,000€ = 0.40
= 0.40€ (or 40c) in the €
M1
A1(2)
Answer
Mark
Nicole is both a secured creditor and an
unsecured creditor.
Let the amount owed to Nicole as a secured
creditor be S.
Then, S + 0.4(25,000 - S) = 12,700
Thus 0.6S = 12,700 – 10,000 = 2,700
Amount of secured debt = 2,700 / 0.6
= 4,500€
M1
M1
Question
Number
3 (a)(ii)
Question
Number
3 (a)(iii)
Question
Number
3 (c)(ii)
Nicole received proportionately more than
Pierre.
Therefore, Nicole is the secured creditor.
The second mark is only available to
candidates scoring the first mark.
M1
A1r(3)
A1
(2)
A1
A1(2)
M1
A1(4)
(Total for Question 3 = 15 marks)
Question
Number
4 (a)
Question
Number
4 (b)(i)
Answer
Total costs of Hot Wire Method
= RM3,280,000 + (6,000 x RM1,100)
= RM9,880,000
Total variable costs of Cold Conductor Method
= 6,000 x 550 = RM3,300,000
Fixed costs for Cold Conductor Method
= RM9,880,000 – RM3,300,000
= RM6,580,000
Mark
M1
M1
M1
A1(4)
Answer
Mark
Marks for correct axes and labelling
Position of intersection
Intercept for each line on the y axis
A1
A1
A1A1(4)
Question
Number
4 (b)(ii)
Answer
Mark
Correct region shaded to left of x=6,000
A1 (1)
Question
Number
4 (b)(iii)
Question
Number
4 (c)
Answer
Mark
Correct calculated value is RM9,055,000
Therefore, accept answers from the chart in (b)(i)
between RM8,800,000 and RM9,200,000
A1 (1)
Answer
Mark
Overhead expenses = 0.25 x RM1,100 = RM275
Labour = 0.3 x RM1,100 = RM330
Materials = RM1,100 - RM275 - RM330 = RM495
M1
M1
A1
Alternatively,
Overhead expenses are 25% of the total overhead
expenses
M1
Labour is 30%
Therefore Materials are 45% M1
45% x RM1,100 = RM495 A1
(3)
(Total for Question 4 = 13 marks)
Question
Number
5 (a)(i)
Answer
Mark
2x + 3y = 120 (Eq 1)
x + 2y = 70 (Eq 2)
M1
M1
Multiply Eq 2 by 2: 2x + 4y = 140
M1
Subtract Eq 1: y = 20
M1
Thus, it takes 20 minutes to bake one batch of A1
bagels.
Question
Number
5 (a)(ii)
Alternatively, the equations can be used
to solve first for x, by multiplication of
Eq. 1 by 2 and Eq. 2 by 3, for example.
Accept any other algebraic symbols in
place of x and y.
(5)
Answer
Mark
Substitute in Eq 2: x + (2 x 20) = 70
Thus, x = 30
It takes 30 minutes to bake one batch of
brioches
M1
A1
Alternatively, if x was found in (i), then
substitution into Eq 1 will derive a value for y
in this part.
Question
Number
5 (b)
Where candidates have not manipulated
the equations correctly in (a)(i), but
carry out correct manipulation in this
part in order to solve one of the unknown
variables, method marks normally
attributable to (a)(i) may be awarded in
this part.
(2)
Answer
Mark
If Qdemand = Qsupply, then
480 - 200P = -150 + 150P
Hence 350P = 480 + 150 = 630
Thus Price = S$1.80
M1
M1
A1 (3)
Question
Number
5 (c)
Answer
Mark
Substituting back in the supply equation.
Qsupply = -150 + (150 x 1.80)
= 120 batches of brioches
M1
A1ft
Alternatively, substituting in the demand
equation.
Qdemand = 480 – (200 x 1.80) = 120
= 120 batches of brioches
(2)
(Total for Question 5 = 12 marks)
Question
Number
6 (a)
Question
Number
6 (b)
Question
Number
6 (c)
Question
Number
6 (d)
Answer
Mark
Percentage return on capital employed:
RM111,840 / RM745,600 = 0.15 = 15%
M1A1(2)
Answer
Mark
Gross profit: RM111,840 + RM84,070
= RM195,910
RM195,910 / RM753,500 = 0.26 = 26%
M1
Answer
Mark
Average stock: (RM54,230 + RM64,770)/ 2
= RM59,500
COGS = Net sales - Gross profit
= RM753,500 - RM195,910 = RM557,590
Rate of stock turn: RM557,590 / RM59,500
= 9.37126 times per annum
Accept 9.4 times or more accurate
M1
Answer
Mark
Average days in stock:
365 x RM59,500 / RM557,590
= 38.94886924 days
Accept 39 days or more accurate
M1
A1
(2)
M1A1
(3)
M1
M1
A1 (4)
(Total for Question 6 = 11 marks)
Question
Number
7 (a)(i)
Question
Number
7 (a)(ii)
Answer
Mark
Value of the house at start
£560,000
=
= £469,594.35
(1 + 0.045)4
= £469,600 to nearest hundred
M1A1
Answer
Mark
Value of house after one year
= £469,594.35 x 1.045
= £490,726.10
= £490,700 to nearest hundred
M1
A1r(3)
A1r
Also accept use of £469,600 as initial price.
Thus, £469,600 x 1.045 = £490,732 (M1)
= £490,700 to nearest hundred (A1r)
Question
Number
7 (b)
Question
Number
7 (c)(i)
Alternatively, value in 2015
£560,000
=
= £490,726.10 (M1)
(1 + 0.045)3
= £490,700 to nearest hundred (A1r)
(2)
Answer
Mark
Proportional increase = A/P
= £620,882 / £560,000
= 1.108717857
Proportional increase per annum
= 3√1.108717857 = 1.035
Percentage increase per annum = 3.5%
Answer
M1
M1
A1(3)
Mark
Proportion earned
=
£1,000,000
= 1.038000147  1.038
£963,391
Percentage earned per annum
= 2  (1.038 − 1) 100 = 7.6%
M1
M1A1
Alternatively,
1,000,000 − 963,391 = 36,609 M1
Percentage earned per annum
£36,602  2
=
= 7.6% M1A1
£963,391
(3)
Question
Number
7 (c)(ii)
Answer
Mark
Value of the bill
= £963,391 (1 +
0.076
) = £975,593.95
6
Accept £975,594 as fully accurate
M1 A1
(2)
(Total for Question 7 = 13 marks)
Question
Number
8 (a)
Answer
Mark
Index of sales for 2018= 100 x 95,000 / 50,000
= 190
Index of sales for 2019 = 100 x 132,000 / 50,000
= 264
M1
A1
Answer
Mark
Chain index of prices for 2017 = 100
Index for 2018 = 100 x 295 / 320 = 92.1875
Index of prices for 2019
= 100 x 289 / 295 = 97.966
Accept whole numbers or more accurate values.
A1
A1
Question
Number
8 (c)
Answer
Mark
Sales (units) in 2020 = 1.07 x 132,000
= 141,240
M1
A1 (2)
Question
Number
8 (d)
Answer
Mark
Sales value for 2017 = 50,000 x RM320 = RM16,000,000
M1
Year
Sales (units)
Price
Value
Index
A1A1
Question
Number
8 (b)
2017
50,000
RM320
RM16,000,000
100
2018
95,000
RM295
RM28,025,000
175.156
2019
132,000
RM289
RM38,148,000
238.425
A1 for whole number or more accurate values for each of 2018
and 2019.
Correct method for either the index for 2018 or for the index for
2019 earns the second M1
A1(3)
A1(3)
M1 (4)
(Total for Question 8 = 12 marks)