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A mix of three materials into a process produces 0.9 litres of a finished product, with 10% loss of in put in the process. Standard material cost are as follows per 0.9 litres of output.
Material $
X 0.5 litres at $2 per litre. 1.00
Y. 0.4. Litres at $ 1.5 per litre. 0.60
Z 0.1 Litres at $ 4 per litre. 0.40During a control period, 4000 litres of output were produced. These used 2,810 litres of Material X, 1910 litres of Material Y, and 380 litres of Material Z
Yield variance ?
Answer
Litres of input
4000 litres of output require (* 10/9) 4444.44
They did require 5100.00
Yield variance in litres of input material. 666.56 (A)Standard weighted average price per litre of input $2
Yield. Variance. $ 1311(A)THIS ANSWER ARE CALCULATED THE YIELD IN TERM OF LITRES INPUT , COULD YOU PLEASE HELP ME HOW TO CALCULATE THIS IN TERM OF LITRES OF OUT PUT
But why? The answer is correct!
Have you watched my free lectures on mix and yield variances?
Yes sir, I have watched all of your videos. I have got the answer following the method demonstrated by you. However, I was wondering why I am not getting the Same correct answer using the other method. I shall be grateful if you could kindly guide me where I am making the error in the following solution.
Thanks5001 should Yield (5001/(10/9)- 4500.9
They did yield. 4000
500.9
*$2
1001.8I think you should use Std. Material cost per unit if you calculate material yield variance based on input..
Here it is,
5100kg of materials should have yielded (5100*90/100) = 4590
But did yield = 4000
Variance in units = 590
* Std. Material cost per unit (2/0.9) = 2.22
Variance in $ = $1311I think we should arrive at unit cost dividing total cost of input by expected output which is 0.9kg in this example. By doing so we absorbed the cost of 0.1kg- which is the normal loss – into production cost. That’s what we did in F2 when we do ‘Process Costing’..
I think this is the reason. Sir please correct me if I’m wrong 🙂
There are two ways of getting the same answer.
The alternative is as follows:
The standard cost of the actual total input (5,100) at standard mix is:
X: 0.5 x 5,100 = 2,550 x $2 = 5,100
Y: 0.4 x 5,100 = 2,040 x $1.5 = 3,060
Z: 0.1 x 5,100 = 510 x $4 = 2,040Total: $10,200
The standard total input for the actual output of 4,000 = 1/0.9 x 4,000 = 4,444.4
X: 0.5 x 4,444.4 = 2,222.2 x $2 = 4,444.4
Y: 0.4 x 4,444.4 = 1,777.8 x $1.5 = 2,666.7
Z: 0.1 x 4,444.4 = 444.4 x $4 = 1,777.6Total: $8,888.7
Yield variance = the difference = 10,200 – 8,888.7 = $1,311.3
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