CH 10 ADDITIONAL PROBLEM SOLUTIONS

CH 10 ADDITIONAL PROBLEM SOLUTIONS - CH 10 ADDITIONAL...

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CH 10 ADDITIONAL PROBLEM SOLUTIONS Exercise 10-8  (20 minutes) 1. The standard price of a kilogram of white chocolate is determined  as follows: Purchase price, finest grade white chocolate. ........................ £9.00 Less purchase discount, 5% of the purchase price of £9.00. . (0.45) Shipping cost from the supplier in Belgium. ........................... 0.20 Receiving and handling cost. .................................................   0.05     Standard price per kilogram of white chocolate. ..................... £8.80 2. The standard quantity, in kilograms, of white chocolate in a dozen  truffles is computed as follows: Material requirements. .............................. 0.80 Allowance for waste. ................................. 0.02 Allowance for rejects. ................................ 0.03 Standard quantity of white chocolate. ....... 0.85 3. The standard cost of the white chocolate in a dozen truffles is  determined as follows: Standard quantity of white chocolate (a). ...... 0.85 kilogram Standard price of white chocolate (b). ........... £8.80 per kilogram Standard cost of white chocolate (a) × (b). ... £7.48
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Exercise 10-9  (30 minutes) 1. a. Notice in the solution below that the materials price variance is  computed on the entire amount of materials purchased,  whereas the materials quantity variance is computed only on  the amount of materials used in production. Actual Quantity of  Inputs, at  Actual Price Actual Quantity of  Inputs, at  Standard Price Standard Quantity  Allowed for Output, at  Standard Price (AQ × AP) (AQ × SP) (SQ × SP) 70,000 diodes ×  $0.28 per diode 70,000 diodes ×  $0.30 per diode 40,000 diodes* ×  $0.30 per diode = $19,600 = $21,000 = $12,000 Price Variance,  $1,400 F 50,000 diodes × $0.30 per diode = $15,000 Quantity Variance,  $3,000 U *5,000 toys × 8 diodes per toy = 40,000 diodes Alternative Solution: Materials Price Variance = AQ (AP – SP) 70,000 diodes ($0.28 per diode – $0.30 per diode) = $1,400 F Materials Quantity Variance = SP (AQ – SQ) $0.30 per diode (50,000 diodes – 40,000 diodes) = $3,000 U
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Exercise 10-9  (continued) b. Direct labor variances: Actual Hours of  Input, at the Actual  Rate Actual Hours of  Input, at the  Standard Rate Standard Hours  Allowed for Output,  at the Standard Rate (AH × AR) (AH × SR) (SH × SR) $48,000 6,400 hours ×  $7 per hour 6,000 hours* ×  $7 per hour = $44,800 = $42,000 Rate Variance,  $3,200 U Efficiency Variance,  $2,800 U Total Variance, $6,000 U *5,000 toys × 1.2 hours per toy = 6,000 hours Alternative Solution: Labor Rate Variance = AH (AR – SR) 6,400 hours ($7.50* per hour – $7.00 per hour) = $3,200 U *$48,000 ÷ 6,400 hours = $7.50 per hour Labor Efficiency Variance = SR (AH – SH) $7 per hour (6,400 hours – 6,000 hours) = $2,800 U
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Exercise 10-9  (continued) 2. A variance usually has many possible explanations. In particular,  we should always keep in mind that the standards themselves may  be incorrect. Some of the other possible explanations for the  variances observed at Topper Toys appear below: Materials Price Variance
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This note was uploaded on 04/27/2008 for the course ACCT 112 taught by Professor Reiter during the Spring '08 term at Binghamton University.

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CH 10 ADDITIONAL PROBLEM SOLUTIONS - CH 10 ADDITIONAL...

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