Answer:
c. capitalized as part of the cost of the land.
Explanation:
These are the options for the question
a. depreciated over the period from acquisition to the date the hotel is scheduled to be torn down.
b. written off as an extraordinary loss in the year the hotel is torn down.
c. capitalized as part of the cost of the land.
d. capitalized as part of the cost of the new hotel.
From the question, we are informed about Cotton Hotel Corporation
which recentlyy purchased Emporia Hotel and the land on which it is located with the plan to tear down the Emporia Hotel and build a new luxury hotel on the site. The cost of the Emporia Hotel should be capitalized as part of the cost of the land. In financial accounting, cost of land can be regarded as asset valuation method which can be used to land that shows on the balance sheet of a company. This cost would encompass all amount spent when acquiring the property and other expenses.
,
No Decision have been made
Answer:
A. $8, 167.50
Explanation:
The fact Juniper company returned $1,500 worth of merchandise, means that it is only obliged to pay the amount of $8,250($9,750-$1,500).
However, the payment was made on 16th August, which is the discount period of 10 days, hence, the cash paid on August 16 is computed thus:
cash paid=amount of merchandise owed*(1-discount rate)
discount rate=1%(1% discount if payment is made within 10 days of the purchase date)
cash paid=$8,250*(1-1%)
cash paid=$ 8,167.50
Answer:
$6.00
Explanation:
Given data
quantity demanded ( x ) ∝ 1 / p^3 for p > 1
when p = $10/unit , x = 64
initial cost = $140, cost per unit = $4
<u>Determine the price that will yield a maximum profit </u>
x = k/p^3 ----- ( 1 ). when x = 64 , p = $10 , k = constant
64 = k/10^3
k = 64 * ( 10^3 )
= 64000
back to equation 1
x = 64000 / p^3
∴ p = 40 / ∛x
next calculate the value of revenue generated
Revenue(Rx) = P(price ) * x ( quantity )
= 40 / ∛x * x = 40 x^2/3
next calculate Total cost of product
C(x) = 140 + 4x
Maximum Profit generated = R(x) - C(x) = 0
= 40x^2/3 - 140 + 4x = 0
= 40(2/3) x^(2/3 -1) - 0 - 4 = 0
∴ ∛x = 20/3 ∴ x = (20/3 ) ^3 = 296
profit is maximum at x(quantity demanded ) = 296 units
hence the price that will yield a maximum profit
P = 40 / ∛x
= ( 40 / (20/3) ) = $6
