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Joint and By Products

SFShaheen Fathima7y ago
I found this question in the revision kit (BPP Learning media). Two joint products A and B are produced in a process. Data for the process for the last period are as follows: Product A B Tonnes Tonnes Sales 480 320 Production 600 400 Common production costs in the period were $12,000. There was no opening inventory. Both products had a gross profit margin of 40%. Common production costs were apportioned on a physical basis. What was the gross profit for Product A in the period? i) $2,304 ii) 2,880 iii) 3,840 iv) 4,800 The answer that I got was option (ii) $2,880. I first calculated the total production and since this is on a physical units basis, I got 1,000 tonnes (Product A: 600 tonnes + Product B: 400 tonnes). Then I apportioned the total common production costs of $12,000 to product A and got $7,200 [(600/1,000) x $12,000]. And then I divided the sales by the total production of product A and multiplied it by the cost apportioned to product A [(480/600) x $7,200] = $5,760. Then I calculated the profit margin for Product A, which gave me $2,880 ($5,760 x 40%). However, when I checked said that the correct answer was option (iii) $3,840. The first three steps that I worked out till getting the cost apportioned to Product A were right, however, the twist came about in the last step that required me to calculate the profit margin for Product A. What it showed in the solution was somewhat confusing $5,760 x 40/60 = $3,840. My argument here is that it's a percentage and calculating 40% of $5,760 makes sense to me. What confused me here is the figure 60; I don't understand where that came from. I humbly ask for help to get this confusion sorted.
John MoffatJohn MoffatTutor7y ago#1
$5,760 is the cost. Since the profit margin is 40%, the profit is 40% of the sales and therefore the cost is 60% of the sales. Therefore the sales are 5,760/60% and the profit is 40% x 5,760/60% (which is the same as what is written in the answer: 5,760 x 40/60)
SFShaheen Fathima7y ago#2
Thank you professor!
John MoffatJohn MoffatTutor7y ago#3
You are welcome :-)
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