Question

Problem 6-6A (Part Level Submission) You are provided with the following information for Marigold Corp. Marigold Corp. uses the periodic system of accounting for its inventory transactions March 1 Beginning inventory 2,050 Iters at a cost of 604 per ter March ) Purchased 2.455 liters at a cost of 65€ per liber. March 5 Sold 2,255 wers for $1.10 per liter. March 10 Purchased 4,130 liters at a cost of 724 per liter March 20 Purchased 2,305 liters at a cost of 80 per liter. March 30 Sold 5,175 ters for $1.30 per liber. (a1) Your answer is incorrect. Try again Calculate the value of ending Inventory that would be reported on the balance sheet, under each of the following cost flow assumptions. (Round answer mal places, e.g. 125.25.) (1) Specific identification method assuming: 09 The March 5 sale consisted of 1.000 liters from the March 1 beginning inventory and 1.255 from the March) purchase and (1) The March 30 se consisted of the following number of units sold from beginning inventory and each purchase: 400 Mers from March 1: 570 liters from March 3: 2.900 sters from March 10, 1,305 Mers from March 20. (2) FIFO (3) UFO Ending Inventory Specite entification PRO LIFO

please answer the ending inventory for the specific identification, FIFO and LIFO

Answer 1

	Date of Purchase	Liters	Cost per liter	Total Cost
	Beg. Inventory	2050	$                    0.60	$      1,230.00
	Mar-03	2455	$                    0.65	$      1,595.75
	Mar-10	4130	$                    0.72	$      2,973.60
	Mar-20	2505	$                    0.80	$      2,004.00
		11140		$      7,803.35
	Liters available for sale =		11140
	Cost of goods available for sale =		$                  7,803
	Liters sold = 2255+5175 =		7430
	Ending inventory in liters =		3710
					Ending Inventory
1]	Specific identification:
	Cost of goods sold = 1000*0.60+1255*0.65+400*0.60+570*0.65+2900*0.72+1305*0.80 =
				$      5,158.25
	Ending inventory = 7803-5187970 =				$           2,645.10
2]	FIFO = 2505*0.80+(3710-2505)*0.72 =				$           2,871.60
3]	LIFO = 2050*0.60+(3710-2050)*0.65 =				$           2,309.00