OpenTuition.com Free resources for accountancy students
Free ACCA lectures and course notes | ACCA AAT FIA resources and forums | ACCA Global Community
ACCA F2 / FIA FMA lectures Download ACCA F2 notes
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?
John Moffat says
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…
You must be logged in to post a comment.
OpenTuition.com is dedicated to providing all accountancy students throughout the world with the resources they need to study for the major … Learn more
Please log in to get the most from OpenTuition, registration is free!