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Naomi says

Hi there on qn 6. where did you get 0.995

John Moffat says

S = 6300 + 423 + 0.005S

If you subtract 0.005S from both sides, then on the left hand side S – 0.005S = 0.995S

Naomi says

Thank you Sir.

Ruksar says

Hello Sir

I tried both ways and they gave me almost the same answers. However i miss out on your lecture in terms of audio after 14.50min as i can not hear what you are explaining. please check.

Thanks.

John Moffat says

Both ways will give the same answer.

ashiq says

sir,what is the answer for 5th question in test

John Moffat says

The answer is C.

(All the answers are at the back of the free Lecture Notes – see the contents page)

David says

I didnt get the 0.15 x 0.20 at first but the penny has just dropped, and now I understand.

May I ask:

Would you be asked in any of the exams to do the algebraic version for recharging dept’s?

John Moffat says

No. In the exam, if reciprocal apportionment is asked then you will always be able to do it either way – whichever way you prefer.

David says

Ok cheers for that John.

Dhruv says

Hello sir John Moffat, I did not understand in algebraic method that how you got 0.03s & 0.97s.

Thanks for giving us free notes & lectures it really helps us

John Moffat says

Have you looked at the answer at the back of the course notes?

Dhruv says

Sir I have looked at the back of the course notes but still cannot understand

happyfeet says

I am also struggling with this part. I don’t understand how 0.03 becomes 0.97 and the value of S stays the same.

happyfeet says

It’s ok, I have it now! So…

S = 20,000 + 2,250 + 0.03S

Take 0.03 away from both sides so..

0.97S = 22,500

Bonang says

I still don’t understand where the 2250 comes from , can anyone please do the algebraic method of example 6 for me step by step,,,I am totally lost, I have watched this lecture so many times but it is still not clear.

John Moffat says

Have you looked at the answer at the back of the Course Noted? It sets out the steps.

Heba says

Hello sir!

Thankyou for your lectures.i have a doubt in algebraic method

For eg: the last question 6

S=6300+0.05M

M=8450+0.10S

S=6300+0.05(8450+0.10S)

S=6300+423+0.005S

0.995S =6723

Where did this figure 0.995S come from??

rest everything is fine but could you please explain that bit.

John Moffat says

S = 6300 + 423 + 0.005S

If you subtract 0.005S from both sides, then on the left hand side S – 0.005S = 0.995S

Heba says

Thanks

dosan says

Thank you so much. I became super clear..

I have last question)))

A company has a budgeted labor cost of 180 000 dollar for the production 30 000 units per month. Each unit is budgeted to take 3 hours of labor. The actual labour cost during the month was 160 000 dollar for 28 000 units and 85 000 hours were worked.

What is the labour efficiency variance?.

John Moffat says

Have you watched all the lectures? This question relates to the chapter and lecture on variance analysis, and so please do not post it under a lecture on accounting for overheads.

You should post the question in the F2 Ask the Tutor forum (after you have watched the lecture on variance analysis).

dosan says

Hi John

Can u help me with this question? The answer is 8 (from mock exam). can not fih=gure out somehow. Thanks

A company produces 2 products P and Q that are both worked in 2 departments of 1 and 2.

Each unit of P spends 1 hour in department 1 and 2 hours in department 2

Each unit of Q spends 2 hours in department 1 and 2 hours in department 2

The total budgeted production of P is 5000 units and Q is 10 000 units.

The total budgeted overheads are 50 000 US dollar for department 1 and 90 000 US dollar for department 2

What is the overhead per unit of P?

John Moffat says

Hi Dosan

Have a look at my answer to the question below – it is exactly the same approach

Imran says

Sir, could please explain the 4th test question of chapter 7,

it’s really confusing..

A company manufactures two products L and M in a factory divided into two cost centres, X and Y. The fol- lowing budgeted data are available: Cost centre X Y Allocated and apportioned fixed overhead

__________________________________X___________Y_______________________

costs____________________________$88,000 $96,000_____________________

Direct labour hours P.u: Product L___ 3·0________ 1·0___________________________

___________________Product M___2·5 _________2·0__________________________

Budgeted output is 8,000 units of each product. Fixed overhead costs are absorbed on a direct labour hour basis.

What is the budgeted fixed overhead cost per unit for Product M?

John Moffat says

First you have to calculate the absorption rate for each department separately.

For department X, the total overheads are $88,000.

The total hours worked are 8,000 x 3.0 (for product L) and 8,000 x 2.5 (for product M). That comes to a total of 44,000 hours.

So for department X, the absorption rate is 88,000/44,0000 = $2 per hour.

You can do the same exercise for department Y, and you should get $4 per hour.

To get the cost for product M, it will be 2.5 hours in X at $2 per hour, plus 2.0 hours in Y at $4 per hour.

Imran says

thank you sir.

John Moffat says

You are welcome

Munazza says

This was the great lecture…but some comments make me confused…

Your explanation is better than live…

Thanks alott…

Satiam says

First of all, ur lectures are really helpful… u explain the concepts so well

My problem is that when u do the algebraic method(for e.g in Q6 in ur lecture), after calculating S and M through algebra and apportioning it to production dept 1(60% and 75% respectively), what do we do with the remaining 5%?

in my head its like 5% of the overheads are ignored and NOT apportioned..is there any way in which the 5% should be apportioned?

John Moffat says

Its not the easiest thing to explain by typing

If you calculate the remaining percentage for each of the two service departments, then you will find that they both come to exactly the same amount, and effectively ‘cancel each other out’.

Also, if you do repeated apportionment (instead of using the algebra) then you end up with exactly the same answer, which in a sense ‘proves’ that the algebra does work

Munazza says

You mean the 10% of stores (6757*.10 = 675) and 5% (9126*.05 = 456) on maintenance will be equal…How ? A bit confused…

John Moffat says

No. If you look at the answer at the back of the Course Notes (the part using the algebraic method) then you will see what I mean about them being equal.

zebbo says

i still dont understand how u got the 0.97

John Moffat says

Have you looked at the answer at the back of the Course Notes?

1 – 0.03 = 0.97

Omari says

How did you subtract from both side sorry I couldn’t get it

John Moffat says

I subtracted the same figure from both sides of the equation.

Its a bit difficult to be able to say more here. All I can suggest is that you either watch the lecture again, or work through the answer at the back of the course notes.

Abdullah says

is algebraic method required in CBE ?

Abdullah says

also i didn’t understand how did you get this value 0.97s @4:50

John Moffat says

I make it clear in the lecture that you can do it either way – they will both give the same answer. Do it whichever way you find easiest and quickest.

For the 0.97S, I subtract 0.03S from both sides. (S – 0.03S = 0.97S)

(Remember that the answers are also at the end of the Course Notes)

michellesp says

thank you

michellesp says

Can someone please explain #5 in the test question for me please.

John Moffat says

If you look below, you will find the answer to question 5 explained (4 or 5 posts further down this page)

Mahrukh says

Salam tutor,

I wanted to know if hours are given instead of percentage as the basis of apportionment, would the algebric method still be applicable??

John Moffat says

Yes, but the algebraic approach is only relevant if two service departments are doing work for each other.

Ihuoma says

Hello

Please I do not know how you got 0.97s= 22250

Please help me

thank you

gabriele says

Can you give me the details for the questions 4-5? Thanks

John Moffat says

4:

The total overheads for centre X are $88,000.

The total hours worked in X are (8,000 x 3.0) + (8,000 x 2.5) = 44,000 hours.

So the absorption rate for X is 88,000/44,000 = $2 per hour.

You can do the same for Y and calculate the absorption rate for centre Y per hour.

To get the cost per unit of M, it is 2.5 hours at $2 per hour, plus 2.0 hours at the absorption rate for centre Y.

5:

Since labour is paid $8 per hour, they are working 400/8 = 50 hours.

So the total cost is the total of materials (300), plus the labour (400) plus production overheads (50 hours at $26 per hour) plus non-production overheads (120% x (300 + 400)).

Yanique says

what is the correct answer for Question 4?

John Moffat says

The answers are at the back of the Course Notes.

Temperance says

@ John Moffat

Help!I’m sooo close to screaming lol ok. I can’t seem to get the correct answer i.e. (D). This is what I did :

Cost Centers

Allocated app o/h = 88,000 (x)

Product M (2.5/5.5 x 88,000) 40,000 (x)

Total o/h in cost center x = 128,000

Allocated app o/h = 96,000 (y)

Product M 2/3 x 96,000 = 64,000

Total o/h in cost center (y) = 160,000

Total No. hours in X = 8000 X 2.5 = 20,000 therefore :

Fixed Cost per unit = 128,000/20,000 = $6.40

Total No. hours in Y = 8000 X 2 = 16,000 therefore :

Fixed Cost per unit = 160,00/160,000 = $10

Total fixed cost = $16.40

Where did I go wrong?

Temperance says

JM this is question 4 in chapter 7 test btw. Thanks dear

Temperance says

Help!I’m sooo close to screaming lol ok. I can’t seem to get the correct answer i.e. (D). This is what I did :

Cost Centers

Allocated app o/h = 88,000 (x)

Product M (2.5/5.5 x 88,000) 40,000 (x)

Total o/h in cost center x = 128,000

Allocated app o/h = 96,000 (y)

Product M 2/3 x 96,000 = 64,000

Total o/h in cost center (y) = 160,000

Total No. hours in X = 8000 X 2.5 = 20,000 therefore :

Fixed Cost per unit = 128,000/20,000 = $6.40

Total No. hours in Y = 8000 X 2 = 16,000 therefore :

Fixed Cost per unit = 160,00/160,000 = $10

Total fixed cost = $16.40

Where did I go wrong?

Temperance says

@JM

Soo im so silly:(

I totally absorbed wrong.

its for product M so its 40000 + 64000/8000 =13

my bad my bad

Temperance says

I love that these lectures don’t drag on for hours upon hours and the lecturer has a sense of humor! THANK U OT:)!

James says

Is it just me or is the sound out of sync with the video?

It just makes it slightly more confusing (or difficult) when trying to follow the equations.

James says

But I managed to follow and worked out the solution to Q6 using algebra. I never studied it before so I’m quite proud of myself!

John Moffat says

I think the sound is OK, but I will check with admin.

Anyway, congratulations on sorting out the algebra

Martin says

Great lecturing. Thanks for the help.

devikaramlugun says

For example 6 chapter 7.

why is 3% added to 20,000+2,250, i.e. 20,000+2,250+0.03s

John Moffat says

We are replacing M in the equation.

0.15M = 0.15(15000 + 0.20S)

0.15 x 15000 = 2250

0.15 x 0.20S = 0.03S

Have you watched my lecture, because I do talk through this in the lecture?

devikaramlugun says

ok thank you so much Sir

John Moffat says

You are welcome

MVS says

Hi Mr Moffat

I was going through the revision notes downloaded from open tuition website. Can you please explain me a point on below question.

OVERHEAD ALLOCATION AND ABSORPTION

Jones Ltd has allocated overheads between departments as follows:

Dept $

A 336,000

B 210,000

Repairs 42,000

Maintenance 28,000

In addition there are general overheads of $308,000 which should be apportioned:

A: 40%; B: 30%; Repairs: 20%; Maintenance: 10%.

A & B are production departments. The repairs and maintenance service production department as follows:

A B Repairs Maintenance

Repairs 60% 40% – –

Maintenance 40% 40% 20% –

Budgeted labour hours:

A: 40,000 hrs; B: 8,000 hrs

Budgeted machine hours:

A: 5,000hrs; B: 60,000 hrs

Why is it A has been calculated on Labour hours and B is calculated on Machine hours. This is how the workings are shown on Answers.

MVS says

Is it simply chosen based on labour and machine intensive? And is this what we might have to do in exam if not clearly asked for?

John Moffat says

Yes – that is the reason.

And yes – although usually you would be told the basis for absorption, you could be asked to choose the most sensible way on the information given.

MVS says

Thanks very much.

bentzion says

Hi,

I’ve worked out the answer to Question 6, ch 7 on pg 44 of the notes and I dont seem to come out with the correct answer. I reapportion Stores and Maintenance a few times to Production Depts 1 and 2 and I come out with a number 27 903 at the end for Dept 1. I’ve done it a few times but not come to the correct figure it gives on the answer. Any ideas how best I can check what I’m doing wrong?

Thanks so much for your help.

John Moffat says

I assume that you mean production department X?

If so, then the answer cannot possibly be 27903, because there is already 70,000 overheads in department X even before we start apportioning extra from the service departments

Without seeing your answer it is impossible for me to tell you where you went wrong. However if you look at the back of the notes there is the full answer (done both ways – repeated distribution and algebra) and so you should be able to check there.

bentzion says

Thanks for reply, but I’m not talking about EXAMPLE 6 I’m referring to QUESTION 6 on pg 44, the multiple choice test.

John Moffat says

Sorry – I got confused

It is a bit difficult to write up the full answer on here, because the tabbing does not work here.

Anyway, this should help you check:

If you recharge stores first, then 3780 goes to dept 1 and 630 goes to Maintenance. ( I am not going to bother typing what goes to dept 2 because it is irrelevant)

That gives a total now on Maint of 9080.

If you recharge this then 6810 goes to dept 1 and 454 to stores

If you recharge this 454 from stores, then 272 goes to 1, and 45 goes to Maint

If you recharge this 45 from Maint, then 34 goes to 1 and 2 goes to Stores

This 2 is recharged to 1.

So the total to dept 1 is 17500 + 3780 + 6810 + 272 + 34 + 2 = 28398

bentzion says

Thank you so much! I made one silly mistake in recharging Stores – so that was definitely a worthwhile exercise to see how careful one needs to be! Thanks so much I really appreciate it.

Kristy says

in question 6 in chapter 7,

S= 6300 + 0.05 M

M=8450 + 0.10 S

S= 6300 + 0.05 (8450 + 0.10 S)

=6300 + 422.5 + 0.005 S

Where did the 422.5 come from??? thanks in advance =)

John Moffat says

It’s 0.05 x 8450.

Kristy says

oh i see…. i can be so silly sometimes…. anyway, awesome and very enlightening lectures and thank you!!!!

John Moffat says

No problem – glad you are now sorted out

bentzion says

Hi

Please can someone help me with the Test at the end of Chapter 7, Question number 5 on Page 44 of the lecture notes. I do not know how the answer is worked out. Thanks alot.

John Moffat says

The material is 300 and the labour is 400, and so the prime cost is 700.

Non-production overheads are absorbed at 120% of prime cost, and so they are 120% x 700 = 840.

Production overheads are 26 per hour. Because the total labour cost is 400, and labour is paid 8 per hour, it means that there must be 400/8 = 50 hours of labour. So…..production overheads are 50 hours x 26 = 1300.

So…..the total cost is 300 (material) + 400 (labour) + 1300 (production overheads) + 840 (non-production overheads) = 2840.

bentzion says

thanks

onica1 says

OMG, question 4 on page 44 really had me. Thanks J

Reena says

Dear Sir, PLease explained below where we got the Reapportion J 9540

A Factory consist two department cost centres (G and H) and two service cost centres (J and K).The total overheads alocated and aportioned to each centre are as follows:

G H J K

$40000 $ 50000 $30000 18000

G H J K

The percentage of service cost centre J to 30% 70% – –

The percentage service cost centre K to 50% 40% 10% –

The company apportions service cost centre costs to production cost centres using method that fully regconises any work by one service cost centre for another.

What are the total overheads for production cost centre G after the reapportionment of all service costre costs?

THe answer is $58540 where got the sum $9540?

John Moffat says

recharging K means that 9000 (50% x 18000) goes to G, and 1800 (10% x 18000) goes to J.

The total on J now is 31800 (30000 + 1800) and 30% of this goes to K. 30% x 31800 = 9540

So the total on K is 40000 (already there) + 9000 (from the first line of this answer) + 9540 (from the second line of this answer).

The total comes to 58540

bentzion says

But surely since K has already been recharged you cant include the 40000 again – otherwise it’s like double-counting? is that not correct?

John Moffat says

Sorry – I meant to type G in the third line of my reply, and not K.

The total for G is the 40,000 that is already there, plus 9,000 (recharged from K) plus 9,540 from J.

Sorry about that

The question asks for the total for G, which is 58540.

Amna Zaman says

Please explain Test , question 1.

I understand proportion of cost centre Y. I also understand proportion of cost centre X that is 46000× 50 % , but where and why did 30000 × 10% come from. Thanks

John Moffat says

Service department Y has overheads of 30,000, and does 10% of its work for department X.

So the total costs in department X are the 46000 per the question, plus 10% of Y’s 30,000.

wang9ackles says

Can someone plz work out question 4 from the Lecture notes of chapter 7 for me, cudnt find any examples related to it and hence unable to solve:

wang9ackles says

oh i got it.lol

anne85 says

can you help me work out test question number 1 coz i didnt quite understand

John Moffat says

There is already 95,000 for centre A.

In addition there is 30% of the work done by Y, so that means an extra 30% x 30000 = 9000.

Also, 10% of Y’s work is for X which means that the total for X is 46000 + (10% x 30000) = 49000

X does 50% of its work for A, so that means we need to give A 50% x 49000 = 24500.

So….the total for A is 95000 + 9000 + 24500 = 128500

Hope that helps

marembon says

Can u help me work out no. 6 in chapter 7

John Moffat says

@marembon, the correct answer is C

If S = stores and M = maintenance, then:

S = 6300 + 0.05M

M = 8450 + 0.10S

Substituting for M in the first equation,

S = 6300 + 0.05 (8450 + 0.10S)

= 6300 + 422.5 + 0.005S

So 0.995S = 6722.5

S = 6756

Substituting for S in the second equation, M = 8450 + 0.10 x 6756 = 9126

So, the total for Department 1 is: 17,500 + (0.60 x 6756) + (0.75 x 9126)

= 28398

raeb says

where can i get past paper question

John Moffat says

@raeb, The ACCA does not publish past papers for F2.

nhs14 says

I need help with the Workings for Test Question 4 Please!

John Moffat says

@nhs14, See my answer below.

denzyboo says

Can you help me work # 4 in chapter 7

John Moffat says

@denzyboo, The total hours in X are (8000 x 3.0) + (8000 x 2.5) = 44.000.

The total overhead in X is $88000, so overheads per hour are $2.

If you do the same for Y, there are 24,000 hours and so the overheads per hour are $96,000 / 24,000 = $4.

Product M used 2.5 hours of X (at $2 per hour) and 2.0 hours of Y (at $4 per hour).

So total for a unit of M is (2.5 x $2) + (2.0 x $4) = $13.

desie86 says

Why is the 0.20S end up in solving for S?

John Moffat says

@desie86, The questions says that 20% of S’s work is for the other centre. 20% = 0.20

desie86 says

how did you arrive at 0.97S?

John Moffat says

@desie86, When you rearrange the equation, S – 0.03S = 0.97S

accakeisha says

Description of OT in algebraic form;

Step 1. confused + lost = visiting OT site

Step 2. free lectures + equation 1 = acca graduate

Step 3. tell a friend.

adejumolu says

Please sir, help me with Question 4. I have been struggling with it. Awaiting your response. Thank you.God bless.

John Moffat says

@adejumolu, Have you looked at the answers at the back of the Course Notes?

Reena says

Dear Sir,

Please note the sound disappears at 14:50.Please do needful I cannot follow example 6 .Thanks

John Moffat says

@Reena, The sound stops because you are supposed to be trying the example yourself – the method has already been explained.

The example is in the notes and you can follow the workings on the screen.

denzyboo says

can u help me with ques 4

hixam says

Could you please tell me how you have got the answer of test Question 5 as c. in this chapter.

John Moffat says

@hixam, Materials and labour total $700 and so the prime cost is $700. Non-production overheads are absorbed at 120% of prime cost which is 120% x 700 = $840.

Because the total labour cost is 400 and labour is paid 8 per hour, it means the the number of labour hours is 400/8 = 50. Production overheads are absorbed at $26 per labour hour and so the total production overheads are 50 x 26 = $1300.

So….total cost is: Materials 300 + Labour 400 + production overheads 1300 + non-production overheads 840 = $2840

merces69 says

thank you very much. god bless you!

wrt130 says

Thanks you Sir!!!

faizanfareed says

thank u sir….!!!!

osmansomali says

hellow idil. well let me try my best to make it easier, 0.03S when you take it to other side of the equation it will be -0.03S (it will take minus sign). therefore deduct this from S ( dont forget that S is associated with number 1, so it is 1S). the equation will be like this S – 0.03= 0.97S.

I hope you understand it.

zinnat10 says

@osmansomali, perfect

idil23 says

Can anyone please explain to me how 0.03s transformed into 0.97s step by step as I am not good at algebra. I would appreciate it greatly.

naveedsulaiman says

@idil23, You need to get all the ‘s’ values on one side of the equation and all known values on the other side.

Here is the equation:

s = 20000 + 0.15(15000 + 0.2s)

s = 20000 + 2250 + 0.03s

So so you now need to take ‘0.03s’ from both sides of the equation, it becomes:

s – 0.03s = 20000 + 2250 + 0.03s – .0.03s

Simplified, it becomes:

0.97s = 22250

naveedsulaiman says

@idil23, Remember that ‘s’ is the same as ‘1s’.

kalamq says

The sound disappears at 14:50, other than that it’s great stuff!

faizanfareed says

i have same problem…