# Ac 505 Case Study

423 Words2 Pages
Chad Martin AC 505 Case Study 2 Number of seats per passenger train car 90 Average load factor(% of seats filled) 0.7 Average full passenger fare \$160 Average variable cost per passenger \$70 Fixed operating cost per month \$3,150,000 a. Contribution margin per passenger = \$90 Contribution margin ratio = 0.5625 Break-even point in passengers = Fixed costs/Contribution Margin = 35000 Break-even point in dollars = Fixed Costs/Contribution Margin Ratio = \$5,600,000 b. Compute # of seats per train car (remember load factor?) 63 If you know # of BE passengers for one train car and the grand total of passengers, you can compute # of train cars (rounded) =? 556 c. Contribution margin =? \$120 Break-even point in passengers = fixed costs/ contribution margin 26250 Passengers =? 54 train cars (rounded) =? 486 d. Contribution margin =? \$70 Break-even point in passengers = fixed costs/contribution margin 45000 Passengers =? 45000 train cars ( rounded) = ? 918 e. Before tax profit less the tax rate times the before tax profit = after-tax income = \$ ? Profit = (P*Q - V*Q) - Fixed Expense Then, proceed to compute # of passengers -=? 975000=(205*Q-85*Q)-3600000 38,125 Passengers per month needed f. # of discounted seats = ? 9 Contribution margin for discounted fares X # discounted seats = \$ each train X\$ ? train cars per day X ? days per month= \$? minus \$ additional fixed costs = \$? pretax income. CM 50 * 9 = 450 * 50 * 30 - 180,000 = \$495,000