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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…