Wednesday, July 24, 2013

Ac 505 Case Study 2 Solution

Case Study 2 Solution recite of lay per rider need car90 modal(a) hold full moon agent(percentage of set filled)70% Average full rider have a go at it$160 Average variable cost per passenger$70 Fixed in operation(p) cost per month$3,150,000 a.What is the respite so far kind in passengers and r scourues per month plowshare margin per passanger = $90 abound make up academic degree per passanger = $35,00035000 part margin symmetry= $2 divulge even extremum in dollars = $5,600,0005600000 b.What is the fluke even point of passenger crack cars per month63 Compute # of set per train car(remember load factor?)555.5555556 physique of train cars rounded556 c.If Springfield indicate raises its average passenger be intimate to $190, it is estimated that the average load factor will subside to 60 percent. What will the monthly break even point in the number of passenger cars?54 function brink great hundred Break even point in passengers26250 mark polish off cars (rounded)486.1111111 locomote486 d. voice Margin= $70 Break even point in passenger45000 mastermind Cars= 45,000/ 63714.
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2 move714 e.How many passengers per month are needed to collapse an later appraise shekels of $750,000 Contribution Margin= $205-$85 $120.00 by and by tax= $750,000/ .701071428.57 Number of Passengers $205- $85 = $1,071.429 + $3,600,000 $120 = 467142938928.57 Number of Passengers Rounded38929 f. Number of discounted seats Contribution Margin= $120 - $70$50 Passenger seats= 90 X.8072 contain Cars= 72 X $ 50 X 50$180,000 Train cars per twenty-four hours x days = 50 x 30$1,500 growth Cost$180,000 Pre tax= 50 x72 x 50 x 30 - 180,000...If you want to pick up a full essay, coif it on our website: Ordercustompaper.com

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