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kipiarov [429]
2 years ago
9

Biscayne Bay Water Inc. bottles and distributes spring water. On May 14 of the current year, Biscayne Bay Water Inc. reacquired

24,200 shares of its common stock at $68 per share. On September 6, Biscayne Bay Water Inc. sold 12,500 of the reacquired shares at $74 per share. The remaining 11,700 shares were sold at $65 per share on November 30.
Required:

A. Journalize the transactions of May 14, September 6, and November 30. Refer to the Chart of Accounts for exact wording of account titles.
B. What is the balance in Paid-In Capital from Sale of Treasury Stock on December 31 of the current year?
C. Where will the balance in Paid-In Capital from Sale of Treasury Stock be reported on the balance sheet?
D.
For what reasons might Biscayne Bay Water Inc. have purchased the treasury stock?
Business
1 answer:
nordsb [41]2 years ago
6 0

Answer:

treasury stock   1,645,600 debit

             cash        1,645,600 credit

--to record he purchase of TS--

cash       925,000 debit

    treasury stock         850,000 debit

   additional paid-in TS  75,000 debit

-to record sale of TS above their cost--

cash                         760,500 debit

additional paid-in TS  35,100 debit

           Treasury Stock    795,600 credit

Explanation:

<em><u>treasury stock cost:</u></em>

24,200 shares x $68 each = 1,645,600

sale of TS

proceeds:  12,500 x $74 =925,000

cost:           12,500 x $68 = 850,000

additional paid-in TS 75,000

second sale of TS:

proceeds _ 11,700 x $65   760,500

Cost:           11,700 x $68    795,600

 additional paid-in TS          35,100

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Answer:

The correct answers are:

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i. The marginal gain from 8:00 AM to 9:00 AM is calculated thus:

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Marginal gain = (Work done at 9:00 AM) - (work done at 8:00) = 40 - 0 = 40 problems

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Marginal gain = 90 - 70 = 20 problems

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8:00 AM to 9:00 AM = 40 - 0 = 40 problems (hour 1)

9:00 AM to 10:00 AM = 70 - 40 = 30 problems (hour 2)

10:00 AM to 11:00 AM = 90 - 70 = 20 problems (hour 3)

11:00 AM to Noon = 100 - 90 = 10 problems (hour 4)

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1 hour of reading = 25 problems

Comparing the reading and solving problems, we see that the number of problems solved in hour-1 and hour-2  (40 and 30 respectively) are more than the equivalent number of problems solved in reading for 1 hour, while hour-3 and hour-4 (20 and 10 respectively), have lesser equivalent number of problems each, than reading for one hour. Therefore to make the most out of the 4 hours, Maria will spend the first 2 hours solving problems and the next two hours reading, giving a total of: 40 + 30 + 25 + 25 = 120 problems solved.

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