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  1. avatar says

    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?

    • Profile photo of John Moffat says

      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

      • avatar says

        But surely since K has already been recharged you cant include the 40000 again – otherwise it’s like double-counting? is that not correct?

      • Profile photo of John Moffat says

        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.

    • Profile photo of John Moffat says

      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 :-)

    • Profile photo of John Moffat says

      @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)
      = 28398

    • Profile photo of John Moffat says

      @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.

    • Profile photo of John Moffat says

      @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

  2. avatar says

    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.

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