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July 17, 2017 at 2:52 am
Q1 is very confusing, I got 56% as the answer.
Allow me to rewrite the question to illustrate my point.
A company is intending to produce a new product.
They have produced one test unit which took 18 hours. The learning curve is 56%
How many hours will the second product take to product (to the nearest hour)?
A: 5 hours
B: 3 hours
C: 14 hours
D: 10 hours (correct answer)
John Moffat says
July 17, 2017 at 8:31 am
What you have written is wrong!
The doubling rule applies to the average time per unit.
If the first takes 18 hours and the second takes 10 hours, then the average time per unit is (18 + 10)/2 = 14 hours.
Using the doubling rule, if the learning rate is r, then the average time per unit if 2 are made = 18 x r = 14. So r = 0.78 (or 78%)
If the learning rate was 56% then the average time per unit if 2 were made would be 18 x 0.56 = 10 hours. This is NOT the time for the second unit, but the average time per unit is we make 2. So the total time for 2 units is 2 x 10 = 20 hours.
Since the first unit took 18 hours, the time for the second = 20 – 18 = 2 hours.
I do suggest that you watch the free lectures on this.
July 18, 2017 at 1:47 am
I understand it now.
July 18, 2017 at 7:43 am
You are welcome 🙂
June 26, 2017 at 11:43 am
Q2?sir?I don’t understand the0.8^3
why is it 3?
June 26, 2017 at 3:50 pm
Because to make 8 in total means doubling 3 times.
Have you not watched the free lectures on this?
February 18, 2017 at 11:48 am
Activity level 800 units 1200 unit
Total cost $16400 $23600
The fixed cost of the business step up by 40 % at 900 units
What is variable cost per unit
A : $8.00
B : $18
c : $19.67
sir this is the question about high low method from kaplan practice question i calculate variable cost $18 but answer saying its $8.
is there anything i missing ?
February 18, 2017 at 3:57 pm
Kaplan’s answer is correct, but ask in the Ask the Tutor Forum and I will explain (not as a comment on a lecture).
February 7, 2017 at 8:37 pm
sir for q1, the average time per unit for 2 units = (18 + 10) / 2 = 14 hours so what i am getting it that for 2nd unit we require 14 hours that is average time and if there are 3 units so we divide it by 3 for average and the answer would give the average time required specifically for making 3rd unit..is i am right
February 8, 2017 at 7:44 am
No, you are not right.
You need to watch my lectures on learning curves.
May 23, 2017 at 12:31 am
That’s the tabular approach in in learning curve use only when production doubles. To solve the problem of calculating time for the 3rd or 5th unit u need to use the formula – y = axb (y is average time per unit to produce x units, a is time taken for the first unit, x is the total number of units and be is to the power: log the learning curve over log 2) of course u can use the formula on even number of units. I have tried it ?Hope this helps.
May 23, 2017 at 12:37 am
*That’s the tabular approach use in learning curve only when production doubles
May 23, 2017 at 8:31 am
Which is as I explain in my free lectures. It is important to watch the lectures before attempting the tests.
January 31, 2017 at 2:37 am
Referring to Question 3 in this practice questions,
Using the highlow method sir I thought it was supposed to be 960000-883200 not 885120 that you use
January 31, 2017 at 7:34 am
No – you use the highest and lowest levels of production.
May 23, 2017 at 12:36 am
Sir the cost for the lowest production shouldn’t be the lowest cost overall. Yes I have use the lowest and highest production levels but by right automatically the cost suppose to be the lowest and highest for the levels of production selected. However the lowest cost there is not under the lowest units produced. Rather it’s under the level of production of 20,160
May 23, 2017 at 8:32 am
You should watch my free lectures on the high-low method. We use the costs for the highest and lowest production levels (regardless of whether or not they are the highest and lowest costs).
May 24, 2017 at 7:03 pm
Okay sir thank u. U are really doing a good job.
January 26, 2017 at 10:45 am
Hi Sir, i don’t think question 4 is covered in the lecture right? i don’t remember learning it.
January 26, 2017 at 10:47 am
ok i get it, it is from chapter 11 instead.
January 26, 2017 at 1:36 pm
I am glad you have got it 🙂
January 27, 2017 at 6:16 am
Thank you Sir
January 27, 2017 at 7:33 am
January 13, 2017 at 7:15 pm
Using the highlow method sir I thought it was supposed to be 960000-883200 not 885120 that you used.
Or am I getting this wrong sir?
January 13, 2017 at 7:36 pm
Which question are you referring to?
March 14, 2016 at 9:44 am
isn’t the question 3 in practical questions of quantitative analysis of budgeting actually from CVP?
March 14, 2016 at 12:33 pm
No – it is testing the high-low method (which is covered in our lectures on quantitative analysis).
March 14, 2016 at 6:40 pm
March 14, 2016 at 6:42 pm
And what about question 4? Multi product profit volume charts?
March 8, 2016 at 12:44 am
sir, why 0.8^3? i didn’t get where is 3 comes from.
March 8, 2016 at 7:28 am
Because 8 involves doubling 3 times.
I do suggest that you watch the free lecture, which explains the doubling rule and works through an almost identical example.
(You should not attempt the tests unless you have watched the relevant lecture first 🙂 )
ali imran says
December 6, 2015 at 8:59 am
In Q.2 by solving Y = axb formula we get :42*7^-.321928=22.448. But this is wrong. plz identify my mistake.
December 6, 2015 at 2:08 pm
What you are calculating is the time for the first 7, which is not what is asked for.
The have already made 1, and so another 7 means a total of 8. The only way you can do it is to calculate the time for the first 8 and subtract the time for the first.
By all means use the formula, but it is faster to use the doubling rule (especially since you are not given a value for b – it is always given in the exam if you need to use the formula), and, of course, you need to be able to use the doubling rule anyway because you can be specifically tested on it.
I do suggest that you watch our free lectures on learning curves where all of the above is explained, with examples.
(Our lectures are a complete course for Paper F5 and cover everything you need to be able to pass the exam well)
December 6, 2015 at 2:53 pm
December 6, 2015 at 3:34 pm
December 6, 2015 at 1:24 am
Question 2 how was the decimal 0.83 arrived at?
December 6, 2015 at 6:56 am
Oops – it is a typing error.
The average time per unit for 8 units = 42 x 0.80^3 = 21.504
Everything else in the solution is fine.
I will have it corrected.
November 24, 2015 at 12:28 pm
Hi Sir, the first took 18 hour and second took 10 hour , how calculate the learning rate?
November 24, 2015 at 3:17 pm
The average time per unit for 2 units = (18 + 10) / 2 = 14 hours
Therefore the learning rate = 14/18 = 0.78 (or 78%)
May 19, 2016 at 7:33 pm
With regard to Q1, the second unit takes 10 hrs which means 18 * r^1. That gives r = 56%. Please correct me if I am wrong. Also, when we deduct 56% from 18, it gives 10 hours which is what it takes for the second unit.
November 22, 2015 at 6:58 am
Sir kindly also update the calculations workings for arriving at a particular solution especially for the numerical
November 22, 2015 at 9:51 am
The current software does not allow for the calculations to be shown (which is why you should ask here if you have a problem and then I will explain).
When we can afford to upgrade the software then the calculations will appear, but given that we provide all this free of charge there is obviously a limit as to what we can afford 🙂 )
November 11, 2015 at 8:58 pm
Hi Sir, not sure I understood the question here – using high low wouldn’t you end up with a vc of 1.81 hence having a formula tc = fc + vc
960000= Fc + 22080(1.81)?
November 20, 2015 at 3:23 pm
I don’t know where you got 1.81 from.
Using high low, the variable cost = (960,000 – 885,120) / (22,080 – 19,200) = 26 per unit.
(The Paper F2 free lectures on high-low will help you)
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