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Elanso [62]
2 years ago
6

he following information was drawn from the accounting records of Chapin Company. On January 1, Year 1, Chapin paid $56,000 cash

to purchase a truck. The truck had a five-year useful life and a $6,000 salvage value. As of December 31, Year 1, Chapin Company had a $68,000 balance in its Accounts Receivable account and a zero balance in its Allowance for Doubtful Accounts account. Sales on account for Year 1 amounted to $320,000. Chapin estimates that 5 percent of credit sales will be uncollectible. Required a. Record the year-end adjusting entry for depreciation expense on the truck in T-accounts. b. Determine the book value of the truck that will appear on the December 31, Year 1, balance sheet. c. Record the year-end adjusting entry of uncollectible accounts expense. d. Determine the net realizable value of receivables that will appear on the December 31, Year 1, balance sheet.
Business
1 answer:
ololo11 [35]2 years ago
8 0

Answer:

a. Record the year-end adjusting entry for depreciation expense on the truck in T-accounts.

December 31, 202x, accrued depreciation expense on truck

Dr Depreciation expense 10,000

    Cr Accumulated depreciation - truck 10,000

b. Determine the book value of the truck that will appear on the December 31, Year 1, balance sheet.

Truck $46,000

c. Record the year-end adjusting entry of uncollectible accounts expense.

December 31, 202x, allowance for doubtful accounts

Dr Bad debt expense 16,000

    Cr Allowance for doubtful accounts 16,000

d. Determine the net realizable value of receivables that will appear on the December 31, Year 1, balance sheet.

Accounts receivable $52,000

Explanation:

truck's depreciation expense straight depreciation = ($56,000 - $6,000) / 5 years = $10,000 per year

accounts receivable balance December 31 = $68,000

allowance for doubtful accounts = $0

total sales on account = $320,000

5% of credit sales are uncollectible

accounts receivable = $68,000 - $16,000 = $52,000

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If it costs $100000 to put on an event for four weeks (28 consecutive nights) how much revenue per night is needed to make $2000
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2 years ago
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Granite State Airlines serves the route between New York and Portsmouth, NH, with a single-flight-daily 100-seat aircraft. The o
TEA [102]

Answer:

Given data: One flight with total seats = 100

Full fare passengers, cost per ticket=$150, mean=56 passengers, SD=23

Discount fare passengers, cost per ticket=$100, mean=88 passengers, SD=44

(a) Here, though there is a hint to use the CDF, since the confidence interval is not given we will make some simplying assumptions that will reduce the complexity of the question, of course keeping the question statistically correct.

this question wants us to maximize total revenue per flight (one way), we can do that by taking only full fare passengers or total revenue will be 150*100=$15,000, but since historical probability shows a mean of 56 with a standard deviation of 23, we can assume in best case scenario total full fare ticket passengers will be 56+23=79, leaving 21 tickets for discount passenger, in this case the total revenues will be 79*150+21*100=$13,950

(b) Now, the new constrained policy is giving a clear cut number of seats to each category of pasengers, 44 for discount (total revenues 44*100) and 56 for full fare (total revenues 56*150) both of which are within the probabilities given earlier (full fare mean=56, discount mean=88). Total revenues in case will be 44*100+56*150=$12,800.

(c) Gain is the difference of the excess revenues in both cases of optimal total revenues and limited seats policy or answer (a) - answer (b) = $13,950- $12,800=$1,150

(d) Realistically speaking, there is no answer for this question without a clear cut confidence interval. Another simplifying assumption we can make here is taking the mean passengers as expected bookings (can be tweaked once confidence interval or degree of significance is given). so total revenues in this case will be 44*100 from discount and 56*150 from full fare passengers. That is still similar to answer (c) due to our assumption/lack of constraints, so our optimal booking will be 54 full fare tickets and 44 discount passenger tickets. You can also take worst case scenario by subtracting SD of each passenger type from the mean or go the best case scenario in which SD of full fare will be added to the mean while the pending seats (left over from 100) will be the total to discount fare for optimal revenue collection.

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Answer:

Please consider the explanation below

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carrying cost = 120 units *$130 = $15600

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c.The number of orders per year

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• d.The time between orders (in working days)

= 364 / 12.5 (considered one leave)

= 29.12 days

=29 days

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