Quiz: Master Plumber Review - Financial Arithmetic (Engineering Economy) Part 1 Of 2

25 Questions | By Zuproc66 | Updated: Mar 21, 2022 | Attempts: 513

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Quiz: Master Plumber Review - Financial Arithmetic (Engineering Economy) Part 1 Of 2 - Quiz

Part 1 of Part 2: This is a twenty five (25) item refresher quiz for the Financial Arithmetic (Engineering Economy) for Master Plumber Board Exam.

Questions and Answers

1.

Php 4,000 is borrowed for 75 days at 16% per year simple interest. How much will be due at the end of 75 days?
- A.
  
  Php 4,150.00
- B.
  
  Php 4,133.33
- C.
  
  Php 4,166.67
- D.
  
  Php 4,333.33
2.

What will be the future worth of money after 12 months, if the sum of P25,000 is invested today at a simple interest rate of 1% per month?
- A.
  
  P 30,000
- B.
  
  P 29,000
- C.
  
  P 28,000
- D.
  
  P 27,000
3.

If you borrowed money from your friend with simple interest if 12%, find the present worth of P 50,000 which is due at the end of 7th month.
- A.
  
  P 44, 893
- B.
  
  P 45, 789
- C.
  
  P 46,200
- D.
  
  P 46, 730
4.

A man borrowed P 100,000 at the interest of 12% per year, compounded semi-quarterly, what is the effective rate?
- A.
  
  12 %
- B.
  
  12.55%
- C.
  
  12.64 %
- D.
  
  13.2 %
5.

What is the corresponding effective rate of 18% compounded semi-quarterly?
- A.
  
  19.55%
- B.
  
  19.48%
- C.
  
  18.95%
- D.
  
  18.46%
6.

The effective rate of 14% compounded semi annually is:
- A.
  
  12.36%
- B.
  
  14.49%
- C.
  
  14.88%
- D.
  
  14.94%
7.

An interest rate is quoted as being 7.5% compounded quarterly. What is the effective annual interest rate?
- A.
  
  7.71%
- B.
  
  7.22%
- C.
  
  15.78%
- D.
  
  21.81%
8.

Find the present worth of the future payment of P 10,000 to be made in 10 years with an interest rate of 12% compounded quarterly.
- A.
  
  P 3,044.40
- B.
  
  P 3,054.60
- C.
  
  P 3,065.80
- D.
  
  P 3,300.90
9.

The amount of P 12,800 in 4 years at 5% compounded quarterly is
- A.
  
  P 14,785
- B.
  
  P 15,614
- C.
  
  P 15,837
- D.
  
  P 16,311
10.

At an interest rate of 10 percent compounded annually, how much will a deposit of P 1,500 be in 15 years?
- A.
  
  P 6,165
- B.
  
  P 6,235
- C.
  
  P 6,265
- D.
  
  P 6,435
11.

How long in years will it take the money to quadruple if it earns 7% compounded semi-annually?
- A.
  
  20.15
- B.
  
  26.35
- C.
  
  33.15
- D.
  
  40.35
12.

In how many years is required for P 2,000 to increase by P 3,000 if the interest at 12% compounded semi-annually?
- A.
  
  7
- B.
  
  8
- C.
  
  9
- D.
  
  10
13.

A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after ten years. Find the book value after five years using the straight line method.
- A.
  
  P 12,500
- B.
  
  P 16,400
- C.
  
  P 22,300
- D.
  
  P 30,000
14.

A unit welding machine cost P 45,000 with an estimated life of five years. Its salvage value is P 2,500. Find its depreciation rate using the straight line method.
- A.
  
  19.89%
- B.
  
  18.89%
- C.
  
  17.79%
- D.
  
  15.59%
15.

The annual maintenance cost of a machine is Php 69,994.00. If the cost of making a forging is Php 56.00 per unit and its selling price is Php 135.00 per unit, find the number of units to be forged to break even.
- A.
  
  866
- B.
  
  876
- C.
  
  886
- D.
  
  896
16.

A company invest P10,000 today to be repaid in five years in one lump sum at 12% compounded annually. How much profit in present day is realized?
- A.
  
  P 7,036.00
- B.
  
  P 7,236.00
- C.
  
  P 7,623.00
- D.
  
  P 8,632.00
17.

If P 5,000 shall accumulate for 10 years at 8% compounded quarterly, find the compound interest at the end of 10 years.
- A.
  
  P 6,020.00
- B.
  
  P 6,030.00
- C.
  
  P 6,040.00
- D.
  
  P 6,050.00
18.

A man expect to receive P 25,000.00 in eight years. How much is that money worth now considering interest at 8% compounded quarterly.
- A.
  
  P 13,265
- B.
  
  P 13,675
- C.
  
  P 13,859
- D.
  
  P 13,958
19.

The P 500,000 was deposited 20.15 years ago at an interest rate of 7% compounded semi annually. How much is the sum now?
- A.
  
  P 2,000,015.00
- B.
  
  P 2,000,150.00
- C.
  
  P 2,015,000.00
- D.
  
  P 2,150,000.00
20.

$ 200,000.00 was deposited by Engr. Ali Dela Torre last January 1, 1988 at an interest rate of 24% compounded semi-annually. How much would the sum be on January 1993?
- A.
  
  $ 321,170.00
- B.
  
  $ 421,170.00
- C.
  
  $ 521,170.00
- D.
  
  $ 621,170.00
21.

An asset is purchased for P 500,000. The salvage value in 25 years is P 100,000. What is the total depreciation in the first three years using SLM.
- A.
  
  P 48,000
- B.
  
  P 32,000
- C.
  
  P 24,000
- D.
  
  P 16,000
22.

A machine has an initial cost of P 50,000 and a salvage value of P 10,000 after 10 years. What is the book value after 5 years using SLM.
- A.
  
  P 15,000
- B.
  
  P 25,000
- C.
  
  P 30,000
- D.
  
  P 35,000
23.

An equipment costs P 10,000 with a salvage value of P 500 at the end of 10 years. Calculate the annual depreciation cost by sinking fund method at 4%.
- A.
  
  P 971.12
- B.
  
  P 950.00
- C.
  
  P 845.32
- D.
  
  P 791.26
24.

A company purchased an asset for P 10,000 and plans to keep it for 20 years. If the salvage value is zero at the end of 20th year, what is the depreciation in the third year? Use the SYD method
- A.
  
  P 1,000,000
- B.
  
  P 857.14
- C.
  
  P 837.14
- D.
  
  P 737.14
25.

A bank charges 12% simple interest on a P300,000 loan. Engr. Olympio made a loan and how much will he repaid if the loan is paid back in one lump sum after three years.
- A.
  
  P 336,000
- B.
  
  P 408,000
- C.
  
  P 415,000
- D.
  
  P 418,000

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