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CCP Braindumps Pdf - Latest CCP Exam Questions Vce
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AACE International Certified Cost Professional (CCP) Exam Sample Questions (Q13-Q18):
NEW QUESTION # 13
An agricultural corporation that paid 53% in income tax wanted to build a grain elevator designed to last twenty-five (25) years at a cost of $80,000 with no salvage value. Annual income generated would be $22,500 and annual expenditures were to be $12,000.
Answer the question using a straight line depreciation and a 10% interest rate.
Which of the following should be included in the life-cycle cost analysis of a power plant?
- A. Capacity factor, end product units, physical dimensions
- B. Factory expenses, distribution expenses, mark-up
- C. Resources, work activities, final cost objects
- D. Construction cost, operating cost, maintenance cost
Answer: D
NEW QUESTION # 14
The following question requires your selection of CCC/CCE Scenario 4 (2.7.50.1.1) from the right side of your split screen, using the drop down menu, to reference during your response/choice of responses.
At the end of Year 3, steel prices will have increased by what percentage over today's price? (round to 1 decimal)
- A. 8.0%
- B. 8.2%
- C. 4.6%
- D. 8.7%
Answer: B
Explanation:
To calculate the total percentage increase in steel prices over three years, we compound the inflation rates for each year:
Year 1: 2.5%
Year 2: 2.5%
Year 3: 3.0%
The formula to calculate the total percentage increase is:
Total Increase=(1+Rate Year 1)×(1+Rate Year 2)×(1+Rate Year 3)-1 ext{Total Increase} = (1 + ext{Rate Year 1}) imes (1 + ext{Rate Year 2}) imes (1 + ext{Rate Year 3}) - 1Total Increase=(1+Rate Year 1)×(1+Rate Year 2)×(1+Rate Year 3)-1 Total Increase=(1+0.025)×(1+0.025)×(1+0.03)-1 ext{Total Increase} = (1 + 0.025) imes (1 + 0.025) imes (1 + 0.03) - 1Total Increase=(1+0.025)×(1+0.025)×(1+0.03)-1 ext{Total Increase} = 1.025 imes 1.025 imes 1.03 - 1 approx 1.082 - 1 = 0.082 ext{ or 8.2%} So, the correct answer is B. 8.2%.
NEW QUESTION # 15
An agricultural corporation that paid 53% in income tax wanted to build a grain elevator
designed to last twenty-five (25) years at a cost of $80,000 with no salvage value. Annual income generated would be $22,500 and annual expenditures were to be $12,000.
Answer the question using a straight line depreciation and a 10% interest rate.
If $50 was invested at 6.0% on January 1, year 1, what would be the value of year-end withdrawals made in equal amounts each year for 10 years and leaving nothing in the fund after the tenth withdrawal?
- A. $2.22
- B. $5.35
- C. $6.80
- D. $3.10
Answer: C
NEW QUESTION # 16
Money is value. Having money when you need it is very important. Money can also be valuable when used wisely by knowing when to spend and when to conserve Also, planning now for future expenses can be a plus to the company rather than a debit.
There are several ways to capitalize money and spending. Basically there is the single payment method that has a compound amount factor and a present worth factor. There is the uniform annual series that has a sinking fund factor, capital recovery factor and also the compound amount factor and present worth factor. At this point, we can assure money is worth 10%.
The following question requires your selection of CCC/CCE Scenario 7 (4.8.50.1.1) from the right side of your split screen, using the drop down menu, to reference during your response/choice of responses.
If $10,000 is scheduled to be paid out 5 years from now, what is the minimum amount we can invest today?
- A. $8,129
- B. $6,209
- C. $3,855
- D. $3,791
Answer: B
Explanation:
Given Scenario:
You need to determine the minimum amount to invest today for a $10,000 payout in 5 years with a 10% interest rate.
The problem requires calculating the present value of a future sum. The present value (PV) is calculated using the formula:
PV=FV(1+r)nPV = rac{FV}{(1 + r)
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