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(Solution) - Multiple Choice Questions 1 In linear breakeven analysis if a

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Multiple Choice Questions
1. In linear breakeven analysis, if a company expects to operate at a point above the breakeven point, it should select the alternative:
(a) With the lower fixed cost
(b) With the higher fixed cost
(c) With the lower variable cost
(d) With the higher variable cost
2. A company is considering two alternatives to automate the pH of process liquids. Alternative A will have fixed costs of \$42,000 per year and will require 2 workers at \$48 per day each. Together, these workers can generate 100 units of product per day. Alternative B will have fixed costs of \$56,000 per year, but with this alternative, 3 workers will generate 200 units of product. If x is the number of units per year, the variable cost (VC) in \$ per year for alternative B is represented by:
(a) [2(48)/100] x
(b) [3(48)/200] x
(c) [3(48)/200] x + 56,000
(d) [2(48)/100] x + 42,000
3. When the variable cost is reduced for linear total cost and revenue lines, the breakeven point decreases. This is an economic advantage because:
(a) The revenue per unit will increase.
(b) The two lines will now cross at zero.
(c) The profit will increase for the same revenue per unit.
(d) The total cost line becomes nonlinear.
4. An assembly process can be completed using either alternative X or Y. Alternative X has fixed costs of \$10,000 per year with a variable cost of \$50 per unit. If the process is automated per alternative Y, its fixed cost will be \$5000 per year and its variable cost will be only \$10 per unit. The number of units that must be produced each year in order for alternative Y to be favored is closest to:
(a) Y will be favored for any level of production
(b) 125
(c) 375
(d) X will be favored for any level of production
5. A tractor has a first cost of \$40,000, a monthly operating cost of \$1500, and a salvage value of \$12,000 in 10 years. The MARR is 12% per year. An identical tractor can be rented for \$3200 per month (operating cost not included). If n is the minimum number of months per year the tractor must be used in order to justify its purchase, the relation to find n is represented by:
(a) ? 40,000(A/P, 1%, 10) ? 1500n + 12,000(A/F, 1%, 10) = 3200 n
(b) ? 40,000(A/P, 12%, 10) ? 1500 n + 12,000(A/F, 12%, 10) = 3200 n
(c) ? 40,000(A/P, 1%, 120) ? 1500 n + 12,000(A/F, 1%, 120) = 3200 n
(d) ? 40,000(A/P, 11.4%, 10) ? 1500 n + 12,000(A/F, 11.4%, 10) = 3200 n

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