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CERTIFIED INDUSTRIAL ENGINEER (CIE) EXAMINATION REVIEW ASSESSMENT SOLUTIONS
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NAME: _____________________________________________________________________________ 2010 CERTIFIED INDUSTRIAL ENGINEER EXAM REVIEW ASSESSMENT - SOLUTIONS PART I – General Engineering Engineering Math: 1. In what quadrants do the secant and cosecant of an angle have the same algebraic sign?...

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  • October 8, 2024
  • 17
  • 2024/2025
  • Exam (elaborations)
  • Questions & answers

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  • lomoarcpsd certified industrial eng

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  • CERTIFIED INDUSTRIAL ENGINEER
  • CERTIFIED INDUSTRIAL ENGINEER

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lOMoARcPSD|47166518




2019 CERTIFIED INDUSTRIAL
ENGINEER (CIE) EXAMINATION
REVIEW ASSESSMENT
SOLUTIONS




July 2019




Downloaded by Austine Wanjala ()

, lOMoARcPSD|47166518




Certified Industrial Engineer (CIE) Exam Review Assessment 2019




NAME: _____________________________________________________________________________

2010 CERTIFIED INDUSTRIAL ENGINEER EXAM REVIEW ASSESSMENT - SOLUTIONS

PART I – General Engineering

Engineering Math:
1. In what quadrants do the secant and cosecant of an angle have the same algebraic sign?
a. I and II b. I and III c. II and III d. II and IV

Answer: b. I and III

2. The lengths of side AB and side BC of a scalene triangle ABC are 12 cm and 8 cm respectively. The
o
size of angle C is 59 . Find the length of side AC.
a. 12 b. 14 c. 16 d. 18

Answer: b. 14

Let x be the length of side AC. Use the cosine law
122 = 82 + x2 - 2*8*x*cos(59o)
Solve the quadratic equation for x: x = 14.0 and x = -5.7

Elementary Electrical Engineering:
3. Find the equivalent resistance for the circuit shown below.




a. 13.9 ohms b. 12.6 ohms c. 11.1 ohms d. 10.7 ohms

Answer: c. 11.1 ohms

Note that the 4 and 5 are NOT in parallel because the 2 and 8 are in between. The 1 and 2 are NOT in
series because the 4 is in between. The 1 and 7 are NOT in parallel because the 4 is in between. The
only combination you can make (at first) is the 3, 6, and 9 are in series, so:




Now the 5 and 18 are in parallel:




2 IEdeas Review Center



Downloaded by Austine Wanjala ()

, lOMoARcPSD|47166518




Certified Industrial Engineer (CIE) Exam Review Assessment 2019


The 2, 3.9, and 8 are in series:




The 4 and 13.9 are in parallel:




Finally, all three are in series




Req = 1 + (4 || {2 + [5 || (3+6+9)] + 8}) + 7
= 11.1 ohms

4. Find the current through the 5 volt source by simplifying the resistors to a single equivalent resistance.




a. 220 mA b. 210 mA c. 200 mA d. 190 mA

Answer: b. 210 mA

The 10 and 20 are in parallel (because the tops and the bottoms are connected so that they share the
same voltage) and the 30 and 40 are in parallel.




Now the two resistors are in series, so:



3 IEdeas Review Center



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