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June 29, 2015 at 1:41 pm
Hi there on qn 6. where did you get 0.995
John Moffat says
June 29, 2015 at 2:03 pm
S = 6300 + 423 + 0.005S
If you subtract 0.005S from both sides, then on the left hand side S – 0.005S = 0.995S
June 30, 2015 at 8:12 am
Thank you Sir.
April 17, 2015 at 11:30 pm
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.
April 18, 2015 at 10:04 am
Both ways will give the same answer.
April 10, 2015 at 6:08 pm
sir,what is the answer for 5th question in test
April 11, 2015 at 8:53 am
The answer is C.
(All the answers are at the back of the free Lecture Notes – see the contents page)
February 9, 2015 at 1:00 pm
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?
February 9, 2015 at 1:02 pm
No. In the exam, if reciprocal apportionment is asked then you will always be able to do it either way – whichever way you prefer.
February 11, 2015 at 9:45 am
Ok cheers for that John.
December 3, 2014 at 10:23 am
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
December 3, 2014 at 10:48 am
Have you looked at the answer at the back of the course notes?
December 3, 2014 at 10:55 am
Sir I have looked at the back of the course notes but still cannot understand
February 10, 2015 at 9:40 pm
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.
February 10, 2015 at 9:46 pm
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
November 8, 2014 at 11:56 am
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.
November 8, 2014 at 1:27 pm
Have you looked at the answer at the back of the Course Noted? It sets out the steps.
October 19, 2014 at 3:28 pm
Thankyou for your lectures.i have a doubt in algebraic method
For eg: the last question 6
Where did this figure 0.995S come from??
rest everything is fine but could you please explain that bit.
October 19, 2014 at 3:37 pm
October 19, 2014 at 3:45 pm
September 28, 2014 at 4:23 pm
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?.
September 28, 2014 at 8:05 pm
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).
September 28, 2014 at 4:04 pm
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?
September 28, 2014 at 4:05 pm
Have a look at my answer to the question below – it is exactly the same approach
September 28, 2014 at 7:17 am
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
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?
September 28, 2014 at 8:41 am
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.
September 28, 2014 at 9:21 am
thank you sir.
September 28, 2014 at 9:40 am
You are welcome
September 21, 2014 at 8:45 pm
This was the great lecture…but some comments make me confused…
Your explanation is better than live…
July 29, 2014 at 4:35 pm
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?
July 29, 2014 at 5:19 pm
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
September 21, 2014 at 8:43 pm
You mean the 10% of stores (6757*.10 = 675) and 5% (9126*.05 = 456) on maintenance will be equal…How ? A bit confused…
September 22, 2014 at 6:10 am
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.
July 21, 2014 at 5:49 pm
i still dont understand how u got the 0.97
July 21, 2014 at 7:48 pm
Have you looked at the answer at the back of the Course Notes?
1 – 0.03 = 0.97
July 1, 2014 at 6:12 pm
How did you subtract from both side sorry I couldn’t get it
July 1, 2014 at 6:44 pm
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.
May 27, 2014 at 7:04 am
is algebraic method required in CBE ?
May 27, 2014 at 7:16 am
also i didn’t understand how did you get this value 0.97s @4:50
May 27, 2014 at 5:35 pm
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)
May 25, 2014 at 2:59 am
May 24, 2014 at 9:09 pm
Can someone please explain #5 in the test question for me please.
May 24, 2014 at 10:11 pm
If you look below, you will find the answer to question 5 explained (4 or 5 posts further down this page)
May 6, 2014 at 4:20 pm
I wanted to know if hours are given instead of percentage as the basis of apportionment, would the algebric method still be applicable??
May 6, 2014 at 7:25 pm
Yes, but the algebraic approach is only relevant if two service departments are doing work for each other.
May 3, 2014 at 10:51 pm
Please I do not know how you got 0.97s= 22250
Please help me
May 3, 2014 at 2:32 pm
Can you give me the details for the questions 4-5? Thanks
May 3, 2014 at 3:45 pm
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.
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)).
April 29, 2014 at 5:56 am
what is the correct answer for Question 4?
April 29, 2014 at 9:22 am
The answers are at the back of the Course Notes.
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!
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
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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