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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

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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…