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March 20, 2014 at 1:26 am
@ 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 :
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? 🙁
March 20, 2014 at 1:37 am
JM this is question 4 in chapter 7 test btw. Thanks dear
March 20, 2014 at 1:24 am
March 20, 2014 at 2:23 am
Soo im so silly:(
I totally absorbed wrong.
its for product M so its 40000 + 64000/8000 =13
my bad my bad
March 20, 2014 at 12:21 am
I love that these lectures don’t drag on for hours upon hours and the lecturer has a sense of humor! THANK U OT:)!
March 10, 2014 at 12:44 pm
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.
March 10, 2014 at 1:24 pm
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
March 10, 2014 at 1:29 pm
I think the sound is OK, but I will check with admin.
Anyway, congratulations on sorting out the algebra 🙂
March 6, 2014 at 12:36 pm
Great lecturing. Thanks for the help.
February 28, 2014 at 10:52 am
For example 6 chapter 7.
why is 3% added to 20,000+2,250, i.e. 20,000+2,250+0.03s
February 28, 2014 at 11:25 am
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?
February 28, 2014 at 11:48 am
ok thank you so much Sir
February 28, 2014 at 1:18 pm
You are welcome 🙂
December 25, 2013 at 11:19 pm
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:
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.
December 25, 2013 at 11:43 pm
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?
December 26, 2013 at 11:24 am
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.
December 26, 2013 at 5:35 pm
Thanks very much.
October 15, 2013 at 8:38 pm
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.
October 16, 2013 at 5:27 pm
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.
October 18, 2013 at 9:35 am
Thanks for reply, but I’m not talking about EXAMPLE 6 I’m referring to QUESTION 6 on pg 44, the multiple choice test.
October 18, 2013 at 5:54 pm
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
October 22, 2013 at 11:33 pm
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.
October 14, 2013 at 6:33 am
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 =)
October 14, 2013 at 6:42 am
It’s 0.05 x 8450.
October 15, 2013 at 2:19 am
oh i see…. 🙂 i can be so silly sometimes…. anyway, awesome and very enlightening lectures and thank you!!!! 🙂
October 15, 2013 at 5:13 pm
No problem – glad you are now sorted out 🙂
October 13, 2013 at 9:09 pm
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.
October 14, 2013 at 4:57 pm
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.
October 14, 2013 at 8:32 pm
December 7, 2013 at 6:13 pm
OMG, question 4 on page 44 really had me. Thanks J
October 5, 2013 at 3:49 pm
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?
October 6, 2013 at 7:51 am
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
October 14, 2013 at 12:34 pm
But surely since K has already been recharged you cant include the 40000 again – otherwise it’s like double-counting? is that not correct?
October 14, 2013 at 4:52 pm
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
September 2, 2013 at 8:52 am
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
September 2, 2013 at 9:39 am
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.
July 7, 2013 at 10:21 pm
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:
July 8, 2013 at 10:57 am
oh i got it.lol
July 3, 2013 at 10:19 pm
can you help me work out test question number 1 coz i didnt quite understand
July 4, 2013 at 4:48 pm
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 🙂
October 14, 2012 at 5:21 am
Can u help me work out no. 6 in chapter 7
October 14, 2012 at 2:11 pm
@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)
October 8, 2012 at 1:43 am
where can i get past paper question
October 8, 2012 at 5:13 am
@raeb, The ACCA does not publish past papers for F2.
October 7, 2012 at 12:10 am
I need help with the Workings for Test Question 4 Please!
@nhs14, See my answer below.
October 3, 2012 at 7:29 pm
Can you help me work # 4 in chapter 7
October 8, 2012 at 5:12 am
@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.
September 24, 2012 at 12:25 am
Why is the 0.20S end up in solving for S?
September 24, 2012 at 12:01 pm
@desie86, The questions says that 20% of S’s work is for the other centre. 20% = 0.20
September 24, 2012 at 12:16 am
how did you arrive at 0.97S?
September 24, 2012 at 12:00 pm
@desie86, When you rearrange the equation, S – 0.03S = 0.97S
September 12, 2012 at 2:52 pm
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.
August 30, 2012 at 11:18 am
Please sir, help me with Question 4. I have been struggling with it. Awaiting your response. Thank you.God bless.
August 30, 2012 at 5:06 pm
@adejumolu, Have you looked at the answers at the back of the Course Notes?
August 20, 2012 at 1:18 pm
Please note the sound disappears at 14:50.Please do needful I cannot follow example 6 .Thanks
August 20, 2012 at 3:09 pm
@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.
October 3, 2012 at 7:46 pm
can u help me with ques 4
August 20, 2012 at 12:52 pm
Could you please tell me how you have got the answer of test Question 5 as c. in this chapter.
August 20, 2012 at 3:06 pm
@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
July 5, 2012 at 12:21 pm
thank you very much. god bless you!
May 17, 2012 at 6:30 pm
Thanks you Sir!!!
April 3, 2012 at 7:56 pm
thank u sir….!!!!
February 15, 2012 at 10:41 pm
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.
May 18, 2012 at 8:00 pm
February 15, 2012 at 9:12 pm
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.
March 20, 2012 at 11:00 pm
@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
March 20, 2012 at 11:03 pm
@idil23, Remember that ‘s’ is the same as ‘1s’.
January 20, 2012 at 3:01 am
The sound disappears at 14:50, other than that it’s great stuff!
April 3, 2012 at 7:58 pm
i have same problem…
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