Securing a job at the Nigerian National Petroleum Corporation (NNPC) is a highly coveted achievement for many individuals. To stand out from the competition and increase your chances of success, it is essential to be adequately prepared for the recruitment process. One valuable resource that can greatly assist you in your preparation is the NNPC recruitment past questions and answers. In this blog post, we will delve into the importance of past questions, their benefits, and how they can help you succeed in your NNPC job application.
1. Understanding the Recruitment Process:
The NNPC recruitment process can be rigorous and challenging. Familiarizing yourself with past questions provides insight into the types of questions asked, the format of the exam, and the specific areas on which to focus your preparation.
2. Getting Accustomed to the Exam Structure:
By studying past questions and answers, you become familiar with the structure, style, and format of the examination. This enables you to navigate the actual exam with ease, saving valuable time and reducing stress.
3. Identifying Commonly Tested Areas:
Past questions give you an understanding of the recurring topics and subject areas that are frequently tested in NNPC recruitment exams. This allows you to tailor your study plan accordingly and prioritize the areas that carry more weight in the assessment process.
4. Practicing Time Management:
Solving past questions helps you develop effective time management skills during the exam. By familiarizing yourself with the time constraints, you learn to pace yourself and allocate the appropriate amount of time to each section or question.
5. Enhancing Knowledge and Skills:
Past questions serve as a comprehensive review of your knowledge and skills in the particular subject matter. They reinforce concepts, refresh your memory, and contribute to a deeper understanding of the topics covered during the exam.
6. Improving Confidence:
Practicing with past questions builds confidence and reduces anxiety. By testing your knowledge and skills in an exam-like setting, you become more assured of your abilities, allowing you to approach the actual exam with a positive mindset.
7. Recognizing Question Patterns:
Studying past questions allows you to identify recurring question types, patterns, and the expected approach to solving them. This familiarity enables you to answer questions more efficiently and accurately during the exam.
8. Assessing Progress and Identifying Weaknesses:
Regularly practicing with past questions assists in assessing your progress and identifying areas where further improvement is needed. This allows you to focus on your weaknesses, ensuring a well-rounded preparation.
9. Gaining Insights and Strategic Tips:
Past questions provide valuable insights into the thinking of the examiners and the specific areas they emphasize. By studying the answers and explanations provided with the past questions, you can gain strategic tips to approach similar questions in the actual exam confidently.
10. Boosting Exam Performance:
Ultimately, utilizing NNPC recruitment past questions and answers significantly enhances your exam performance. The familiarity, confidence, and knowledge gained from studying past questions contribute to better accuracy, a higher score, and a greater chance of success in your job application.
This past questions are focused on three subjects which are Physics, Chemistry and ICT.
A. True
B. False
Answer: B
Explanation: Watt’s law gives relationship between power to current, voltage and resistance.
A. True
B. False
Answer: B
Explanation: Energy = Power*Time (not multiply by voltage).
P = VI.
P = (V^2)/R.
P = (I^2)*R.
A. True
B. False
Answer: B
Explanation: Battery capacity is measured in ampere-hours (amp-hours)
A. True
B. False
Answer: B
A. True
B. False
Answer: A
A. 1.3 kWh
B. 13.3 kWh
C. 1.2 kWh
D. 12 kWh
Answer: D
Explanation: The power is multiplied by time then the energy is: E = PT. E = 40030 = 12*10^3.
= 12KWH.
A. 1.286 kWh B. 12.85 kWh C. 535 kWh D. 252 kWh
Answer: A
Explanation: 18 kwh for 14 days. For one day power consumption is = 18/14.
= 1.2857
A. 938 Wh
B. 0.938 Wh
C. 56.25 Wh
D. 5.6 Wh
Answer: B
Explanation: As per ohm’s law,
R = V / I
I = V / R
= 15 / 12
= 1.25 A
And Power P = V * I
= 15*1.25
= 18.75 w
Given duration is 3 miniute i.e. 3 / 60 hour. Therefore, Watt-hour = 18.75 * (3 / 60) = 0.9375
A. 0.58 Ah
B. 2.1 Ah
C. 21 Ah
D. 58 Ah
Answer: C
Explanation: The capacity of battery generally expressed in AH, so that in this case AH is 6A*3.5H=21AH
A. 8.57%
B. 42.85%
C. 4.28%
D. 85.7%
Answer: D
Explanation: Efficiency=output/input
=. 6.7
=. 857
% efficiency =. 857*100 = 85.7%
A. 20.16 kWh
B. 201.6 kWh
C. 2.01 kWh
D. 8.4 kWh
Answer: B
Explanation: Kwh = (3502424)/1000
= 201.6kWh
A. 25 kW
B. 0.00025 mW
C. 2,500 µW
D. 25 mW
Answer: D
Explanation: 0.025w=25*10^-3=25mw
A. overheated
B. shorted
C. open
D. reversed
Answer: B
Explanation: r=0 when short circuited r=infinity(high) when open circuited.
A. 33
B. 330
C. both resistors
D. neither resistor
Answer: D
A. a fuse is not necessary
B. 10 A
C. 24 A
D. 20 A
Answer: Option C
Explanation: If current flow exceeds 24A then fuse works and breaks up the circuit, till then circuit is safe
A. current
B. voltage
C. resistance
D. wattage
Answer: C
A. current
B. voltage
C. resistance
D. none of the above
Answer: A
18) A circuit breaker is a
A. fuse
B. switch
C. resettable protective device
D. resistor
Answer: C
Explanation: Circuit breaker break the circuit when it sense any abnormal condition in the circuit using relay. Relay is a very important component in a circuit breaker.
B. 5
C. none
D. depends on the type of atom
Answer: B
Explanation: 1st orbit = 2 electrons, 2nd orbit = 3 electron. Atomic no = no.of protons in nucleus. Therefore 2+3=5
A. switch
B. photoconductive cell
C. thermistor
D. potentiometer
Answer: D
Explanation: Wiper is connected on metal rod of potentiometer, we can vary it.
A. 10.5 C
B. 105 C
C. 3.4 C
D. 34 C
Answer: A
Explanation: Charge(Q) = current/time(sec)=i/t so i = Q/t
Q = it=(61.75) Q = 10.5
A. current
B. voltage
C. resistance
D. current, voltage, and resistance
Answer: D
B. conductor
C. semiconductor
D. valence
Answer: A
A. yellow, violet, red, gold
B. yellow, violet, orange, gold
C. yellow, violet, red, silver
D. orange, violet, red, silver
Answer: C
Explanation: BLACK-0, BROWN-1, RED -2, ORANGE-3, YELLOW-4, GREEN-5, BLUE -6, VIOLET-7, GREY -8, WHITE-9
The first band gives the first digit.
The second band gives the second digit.
The third band indicates the number of zeros.
The fourth band is used to shows the tolerance (precision) of the resistor
4700 means,
4 =yellow
7 =violet
00 =red(10^2)
Ten percent tollerance =silver
So, =4700.
A. 1.6 A
B. 16 A
C. 2 A
D. 0.2 A
Answer: D
Explanation: Q = 8/10 given; t = 4 s
Now current i = Q / t i = (8/10)/4
i = 2/10;
i = 0.2 A
A. wear resistance
B. red hardness
C. toughness
D. all of the above
Answer: D
A. electro-chemical machining
B. ultra-sonic machining
C. electro-discharge machining
D. laser machining
Answer C
A. side relief angle
B. end relief angle
C. back rake angle
D. side rake angle
Answer B
A. straight fluted reamer
B. left hand spiral fluted reamer
C. right hand spiral fluted reamer
D. any one of the above
Answer C
A. 5º
B. 10º
C. 15º
D. 20º
Answer B
A. internal taper
B. external taper
C. internal and external taper
D. no taper
Answer A
A. Regulating wheel diameter
B. Speed of the regulating wheel
C. Angle between the axes of grinding and regulating wheels
D. all of the above
Answer D
A. trimming the surface left by sprues and risers on castings
B. grinding the parting line left on castings
C. removing flash on forgings
D. all of the above
Answer D
A. hobbing
B. shaping with pinion cutter
C. shaping with rack cutter
Answer B
A. conical locator
B. cylindrical locator
C. diamond pin lovator
D. vee locator
Answer A
A. mild steel
B. cast iron
C. high speed steel
D. high carbon steel
Answer B
A. halide torch
B. sulphur sticks
C. soap and water D. all of the above Answer B
A. humidification
B. dehumidification
C. heating and humidification
D. cooling and dehumidification
Answer D
A. First law of thermodynamics
B. Newton’s law of cooling
C. Newton’s law of heating
D. Stefan’s law
Answer B
A. thermal coefficient
B. thermal resistance
C. thermal conductivity
D. none of the above
Answer B
A. both the fluids at inlet are in their hottest state B. both the fluids at inlet are in their coldest state C. both the fluids at exit are in their hottest state
D. one fluid is coldest and the other is hottest at inlet
Answer A
A. velocity reduction method
B. equal friction method
C. static regain method
D. dual or double method
Answer C
A. improve heat transfer
B. provide support for tubes
C. prevent stagnation of shell side fluid
D. all of the above
Answer D
A. R-11
B. R-12
C. R-22
D. 4
Answer D
A. absorptive power
B. emissive power
C. emissivity
D. none of the above
Answer B
A. wet bulb temperature
B. dry bulb temperature
C. dew point temperature
D. none of the above
Answer A
A. it has low operating pressures
B. it gives higher coefficient of performance
C. it is miscible with oil over large range of temperatures
D. all of the above
Answer C
A. frosting evaporator
B. non-frosting evaporator
C. defrosting evaporator
D. none of the above
Answer A
A. suction pressure
B. discharge pressure
C. critical pressure
D. back pressure
Answer B
A. vertical and uniformly spaced
B. horizontal and uniformly spaced
C. horizontal and non-uniformly spaced
D. curved lines
Answer A
A. ammonia is absorbed in hydrogen
B. ammonia is absorbed in water
C. ammonia evaporates in hydrogen
D. hydrogen evaporates in ammonia
Answer C
A. The constant enthalpy lines are also constant wet bulb temperature lines.
B. The wet bulb and dry bulb temperature are equal at saturation condition.
C. The wet bulb temperature is a measure of enthalpy of moist air.
D. all of the above
Answer D
A. Ammonia
B. Carbon dioxide
C. Sulphur dioxide
D. Flourine
Answer D
B. 2 kW
C. 3 kW
D. 4 kW
Answer D
A. parallel flow type
B. counter flow type
C. cross flow type
D. regenerator type
Answer C
A. small displacements and low condensing pressures
B. large displacements and high condensing pressures
C. small displacements and high condensing pressures
D. large displacements and low condensing pressures
Answer D
A. 1 m³ of wet air
B. 1 m³ of dry air
C. 1 kg of wet air
D. 1 kg of dry air
Answer D
A. Reynold’s number
B. Grashoff’s number
C. Reynold’s number, Grashoff’s number
D. Prandtl number, Grashoff’s number
Answer D
A. The heat transfer in liquid and gases takes place according to convection
B. The amount of heat flow through a body is dependent upon the material of the body
C. The thermal conductivity of solid metals increases with rise in temperature
D. Logarithmic mean temperature difference is not equal to the arithmetic mean temperature difference.
Answer C
A. Backward curved blades has poor efficiency
B. Backward curved blades lead to stable performance
C. Forward curved blades has higher efficiency
D. Forward curved blades produce lower pressure ratio
Answer D
A. True
B. False.
Answer: A
Explanation: In a series RLC circuit, the net reactance= inductive reactance Xl-capacitive reactance Xc. therefore if any one reactances is large enough then other can be neglected and that primarily determines almost the net reactance of the circuit.
A. True
B. False
Answer: A
Explanation: Because at resonance XL=XC.
A. increase
B. remain the same
C. decrease
D. be less selective
Answer: C
Explanation: We know, Bw = F/Q.
When Q increase, the bandwidth decreases.
A. 125 V
B. 250 V
C. 290 V
D. 40 V
Answer: D
Explanation: Vs ^2 = (Vr^2 + (Vl-Vc)^2 ), For resonance , Vl = Vc ,
So , Vs = Vr.
A. increased
B. decreased
C. left alone
D. replaced with inductance
Answer: B
Explanation: Because fr=1/(2piroot(LC)).
O are in series across a 60 V source. The circuit is at resonance. The voltage across the inductor is
A. 60 V
B. 660 V
C. 30 V
D. 300 V
Answer: D
Explanation: As in resonace impedance equal to resistance
Current = 60/24 = 2.5
Vtg across inductor = 120*2.5 = 300V
A. is not affected
B. increases
C. is reduced to zero
D. decreases
Answer: B
A. 7 kHz
B. 13 kHz
C. 20 kHz
D. 6 kHz
Answer: D
Explanation: B.W = f2-f1, B.W = 13-7 = 6.
A. 9 mA
B. 90 mA
C. 13 mA
D. 130 mA
Answer: D
Explanation: V=IR I=V/R
So.. R is nothing but Z (R=Z) Z=sqrrt(R^2+( xl-xc)) OR (R^2+(xc-xl)) z=sqrrt_(90^2+(50-30^2) Z=SQRRT(8100+400)
z=92.195
I=V/Z=12/92.195=0.13015=130mA.
A. The operating system
B. The motherboard
C. The platform
D. Application software
Answer: A
Internet and those without this access is known as the:
A. digital divide.
B. Internet divide.
C. Web divide.
D. cyberway divide.
Answer: A
A. mainframe
B. supercomputer
C. network
D. client
Answer: C
A. application software. B. system software.
C. operating system software.
D. platform software.
Answer: A
A. bit
B. kilobyte C. gigabyte D. megabyte
Answer: C
A. scanner
B. mouse C. printer
D. keyboard
Answer: C
A. relational
B. megabyte
C. binary
D. processing
Answer: C
A. eight bytes.
B. eight characters
C. eight bits.
D. eight kilobytes.
Answer: C
A. A process
B. Information
C. Software
D. Storage
Answer: B
EXCEPT:
A. Viruses
B. Identity theft. C. Hackers
D. Spam
Answer: D
A. Mouse
B. Keyboard
C. Scanner
D. All the above
Answer: D
A. Softcopy
B. Software
C. Hardware
D. Hardcopy
Answer: B
A. DVD
B. CD ROM
C. Floppy Disk
D. CD RW
Answer: A
A. Keyboard
B. Monitor
C. Central Processing Unit
D. Printer
Answer: C
A. 8 bit
B. 16 bit
C. 32 bit
D. 64 bit
Answer: A
A. CTRL+TAB
B. CTRL+SHIFT+TAB
C. SHIFT+TAB
D. None of these
Answer: A
A. server
B. network
C. supercomputer
D. Enterprise
Answer: B
A. kilobyte
B. megabyte
C. gigabyte
D. terabyte
Answer: A
A. instructions
B. operating system
C. application software
D. system unit
Answer: A
A. system
B. communication
C. application
D. word-processing
Answer: A
A. simulation
B. animation
C. robotics
D. computer forensics.
Answer: D
A. 2
B. 8
C. 10
D. 16
Answer: D
A. Packaged programs
B. Application programs
C. Operating system programs
D. All of these
Answer: D
A. Processes
B. Memory
C. Disks and I/O devices
D. All of the above
Answer: D
A. multi user
B. multi tasking
C. time sharing
D. None of these.
Answer: C
A. Group processing
B. Batch Processing
C. Time sharing
D. None of these.
Answer: B
A. User
B. Process
C. Hardware
D. None of these.
Answer: B
A. Follows a set of rules
B. Can be either a hardware or software device
C. Filters network traffic
D. All the above
Answer: D
A. Monitor, keyboard, mouse, modem
B. Telephone line, PDA, modem and computer
C. Telephone line, modem, computer, and an ISP
D. Modem, computer, PDA and ISP
Answer: C
A. 1 KB
B. 1 MB
C. 1 GB
D. 1 TB
Answer: A
The importance of adequately preparing for the NNPC recruitment process cannot be overstated. Utilizing past questions and answers as a study resource plays a vital role in boosting your understanding, confidence, and exam performance. By studying and practicing with these past questions, you equip yourself with the necessary tools and insights to excel in the exam and stand out among other applicants. So, seize the opportunity to access NNPC recruitment past questions and steer your job application towards success. Good luck!