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MPPGCL JE Electronics Mock Test - 4 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test MPPGCL JE Electronics Mock Test Series 2025 - MPPGCL JE Electronics Mock Test - 4

MPPGCL JE Electronics Mock Test - 4 for Electronics and Communication Engineering (ECE) 2024 is part of MPPGCL JE Electronics Mock Test Series 2025 preparation. The MPPGCL JE Electronics Mock Test - 4 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The MPPGCL JE Electronics Mock Test - 4 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MPPGCL JE Electronics Mock Test - 4 below.
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MPPGCL JE Electronics Mock Test - 4 - Question 1

Find the value of current density for a compensated p-type material having the mobility of 500 cm2/V-sec and intensity of Electric field as 5 V. The carrier concentrations are Na = 1016/cm3 and Nd = 5 × 1015/cm3

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 1

Concept:

For a compensated semiconductor with an acceptor concentration greater than the donor concentration, the majority carrier concentration is calculated as:

With Na - Nd ≫ ni, the above equation becomes:

n0 ≅ Nd - Na

The conductivity for compensated p-type material is given as:

σp = qμp[Na - Nd]     ---(1)

μp = mobility of holes 

Current density in compensated p-type material is:

J = σpE     ---(2)

E = Electric Field

Calculation:

Given:

Na = 1016/cm3 

Nd = 5 × 1015/cm3

μp = 500 cm2/V-sec

E = 5 V

From equation (1) conductivity can be calculated:

σp = 1.6 × 10-19 × 500 × [10 - 5] × 1015

σp = 0.4 Ω-1cm-1

Now the current density can be calculated from equation (2):

J = 0.4 × 5

J = 2 A/cm2

MPPGCL JE Electronics Mock Test - 4 - Question 2

RMS value of voltage v(t) = 8 + 6 cos(3t) is

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 2

Concept:

f(t) = A0 + A1 sin (ω1t + ϕ1) + Asin (ω2t + ϕ­2) + …

The rms value of a sinusoidal equation of the above form is given by:

Application:

v(t) = 8 + 6 cos(3t)

The RMS value of the voltage will be:

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MPPGCL JE Electronics Mock Test - 4 - Question 3

In the circuit shown below, find out the value of i(0+) and . If switch is closed at t = 0 -

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 3

Concept:

At steady state for DC excitation inductor will be replaced by a short circuit and the capacitor will be replaced by an open circuit.

An inductor does not allow sudden change of current .

Hence, iL(0-) = iL(0+)

A capacitor does not allow sudden change of voltage.

Hence, VC(0-) = VC(0+)

Voltage across inductor → VL = L diL/dt

Calculation:

Before operation of the switch at t = 0- 

 iL(0-) = 0 A

VC(0-) = 0 V

After closing the switch at t = 0+

 iL(0-) = iL(0+) = 0 A (zero current implies replace inductor with open circuit )

VC(0-) = VC(0+) = 0 V (zero voltage implies replace capacitor with close circuit)

i(0+) = 0 A

Apply KVL on the circuit at t = 0+

-10 + 10i(0+) + VL(0+) + VC(0+) = 0

-10 + (10 × 0) + VL(0+) + 0= 0

VL(0+) = 10

VL(0+) = L diL(0+)/dt 

VL(0+) =  di(0+)/dt  [ ∵ iL(t) = i(t)]

di(0+)/dt = 10 A/sec

MPPGCL JE Electronics Mock Test - 4 - Question 4

The forward transfer function   and the feedback gain H(s) = -0.8. Find the closed loop transfer function of the SFG.

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 4

Concept:

The closed-loop transfer function for negative feedback is given by:

G(s) = Forward path transfer function

H(s) = Feedback gain transfer function

Application:

Given:

H(s) = - 0.8

The minus sign indicates that the feedback used is negative.

∴ The closed-loop transfer function will be:

MPPGCL JE Electronics Mock Test - 4 - Question 5

Determine pass band gain and Cut off frequency of filter shown below-
The value of capacitance shown is C1 = 0.01 μF

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 5

Concept:

For an Non inverting type Active filter shown below

Passband Gain (AV) = (1 + RF/R2)

Cut off frequency (fc) = 1/(2π R1C1 )

Calculation:

Given;

RF = 10K

R2 = 10K

R1 = 15K

C1 = 0.01 μF

Passband Gain (AV) = (1 + RF/R2) = (1 + 10/10) = 2

Cut off frequency (fc) = 1/(2π R1C1) = 1/(2 × 3.14 × 15000 × 0.01 × 106) = 1061.57 ≈ 1 KHz

MPPGCL JE Electronics Mock Test - 4 - Question 6
The band gap in a material is 1 eV; the corresponding wavelength emitted is ______.
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 6

Concept:

The freq. and wavelength are related by the relation:

 

Also, the energy of a photon with wavelength (λ) is given by:

 

Where,

h : Planck’s constant = 6.626 × 10-34 Js

c : 3 × 108 m/sec

 

 

      ------(1)

Calculation:

Given:

E = 1 eV

From equation (1)

 

λ = 1.24 μm

MPPGCL JE Electronics Mock Test - 4 - Question 7
The errors that can be pointed out by the compiler are
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 7

The correct answer is Syntax errors.

  • A syntax error is an error in the source code of a program.
  • Since computer programs must follow strict syntax to compile correctly, any aspects of the code that do not conform to the syntax of the programming language will produce a syntax error.
  • Unlike logic errors, which are errors in the flow or logic of a program, syntax errors are small grammatical mistakes, sometimes limited to a single character.
  • For example, a missing semicolon at the end of a line or an extra bracket at the end of a function may produce a syntax error.

Key Points

  • Internal errors are due to faulty logic or coding in the program.
    • Common types of internal errors include Bounds errors, Inserting a null pointer into a collection, Attempting to use a bad date.
  • A semantic error is a violation of the rules of the meaning of a natural language or a programming language.
    • semantic errors are the hardest to debug because the interpreter provides no information about what is wrong.
  • A logic error (or logical error) is a mistake in a program's source code that results in incorrect or unexpected behavior.
    • It is a type of runtime error that may simply produce the wrong output or may cause a program to crash while running.
MPPGCL JE Electronics Mock Test - 4 - Question 8

Given is a voltage divider circuit, such that V = 10 V and R2 = 1 kΩ. For what value of R1 will v2 be 10% of V?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 8

Concept:

By using the voltage division rule,

The voltage applied across the resistor R1 is 

The voltage applied across the resistor Ris 

Calculation:

Given:

V = 10 V

R2 = 1 kΩ

v2 = 10% of 10 = 1 V

According to the voltage division rule:

R1 = 9 kΩ

MPPGCL JE Electronics Mock Test - 4 - Question 9

An FM signal with a deviation 'δ' is passed through a mixer and has its frequency reduced five fold. The deviation in the output of the mixer is -

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 9

Key Points

Mixers are used in a variety of RF/microwave applications, including military radar, cellular base stations, and more. An RF mixer is a three-port passive or active device that can modulate or demodulate a signal. The purpose is to change the frequency of an electromagnetic signal while preserving every other characteristic (such as phase and amplitude) of the initial signal.  

Frequency deviation will remain the same at the mixer output.

Additional Information 

MPPGCL JE Electronics Mock Test - 4 - Question 10
How many byte of instruction XCHG is used to executed in 8085 microcontroller? 
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 10

XCHG command

  • This is an instruction to exchange contents of HL register pair with DE register pair.
  • This instruction uses implied addressing mode.
  • This instruction requires 1-Byte, 4-Machine Cycles and 4 T-States for execution.
  • As it is1-Byte instruction, so it occupies only 1-Byte in the memory.
  • For example: HL and DE register pairs are having ABCDH and 6789H contents respectively.
  • After execution of instruction XCHG the contents of HL and DS register pairs will be 6789 H and ABCD H respectively. 
MPPGCL JE Electronics Mock Test - 4 - Question 11
Two coils having self inductances of 16 mH and 25 mH are mutually coupled. The maximum possible mutual inductance is
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 11

Concept:

Mutual Inductance

When two coils are placed close to each other, a change in current in the first coil produces a change in magnetic flux, which lines not only the coil itself but also the second coil as well. The change in the flux induced voltage in the second coil. The voltage is called induced voltage and the two coils are said to have a mutual inductance.

The coupling coefficient K represents how closely they are coupled.

We have expression 

Where 

M = Mutual inductance

L1 = Inductance of coil one

L2 = Inductance of coil two

Calculation:

Given

L1 = 16 mH

L2 = 25 mH

K = 1

For the maximum value of inductance, the value of K must be equal to 1

MPPGCL JE Electronics Mock Test - 4 - Question 12

What is the value of maximum number of modes that requires for the successful propagation of light in the fibre?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 12

Optical Fibre communication:

The type of communication in which information or data is transmitted in the form infrared wave through optical fibre cable.

Number of modes depend on V number.

V= 

a = Radius of core

NA = Numerical aperture of optical fibre

Hence mode is directly proportional to 

Hence correct option is "2"

MPPGCL JE Electronics Mock Test - 4 - Question 13
If the output of the system at steady state does not agree with the input, then the system is said to have _________ which determines the _________ of the system.
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 13

Steady-state error-

Steady-state error is defined as the difference between the input (command) and the output of a system in the limit as time goes to infinity (i.e. when the response has reached a steady-state). The steady-state error will depend on the type of input (step, ramp, etc.) as well as the system type (0, I, or II).

We can calculate the steady-state error for this system from either the open or closed-loop transfer function using the Final Value Theorem.

The steady-state error is a measure of system accuracy.

These errors arise from the nature of the inputs, system type, and from nonlinearities of system components such as static friction, backlash, etc.

These are generally aggravated by amplifier drifts, aging, or deterioration.

MPPGCL JE Electronics Mock Test - 4 - Question 14
Which condition is created by two wattmeters W1 and W2 for power factor is zero? 
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 14

Measurement of power

According to Blondel's Theorem, for the measurement of power in n-phase, n-wire without neutral, (n - 1) wattmeters are required.

For measurement of power in 3-phase, (3 - 1 = 2) wattmeters are required.

The readings of both the wattmeters are given by:

The power factor is given by:

Power factor = cosϕ 

At zero power, cosϕ = 0

At, ϕ = 90°, the readings of both the wattmeters are:

Thus, W1 = -ve, W2  = +Ve 

Thus at zero power factor, the reading of both wattmeters are equal and opposite.

MPPGCL JE Electronics Mock Test - 4 - Question 15
Which of the following is associated with a delay in transmission line?
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 15

Key Points 

Amplitude Distortion:

  • For linear systems, the transfer function can be written as the function of the angular frequency ω as
  •  

           

  • Amplitude distortion can occur when                  

Phase distortion:

  • For the above linear phase system Phase or delay, distortion will occur when
  • It means that the system does not have a constant delay as a function of frequency.

Non-Linear Distortion:

  • It occurs when the system includes nonlinear elements.
  • The output signal is not exactly proportionate to the input signal and harmonics are generated.

Multipath Fading:

  • It is a type of degradation that occurs in radio communication.
  • It occurs when more than one version of the transmitted signal arrives at the receiver.

Hence, Phase distortion is associated with a delay in the transmission line.

MPPGCL JE Electronics Mock Test - 4 - Question 16

The Ebers moll model is applicable to:

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 16

Ebers-Moll Model:

It is an ideal model for BJT, that can be used in forward or reverse active mode of operation, as well as in both saturation and cut-off mode also.

The Model contains two diodes and two currents source shown below:

Analysis:

  • The two diodes represent Base emitter and Base collector diodes.
  • The current sources quantify the transport of minority carriers through the base region. These current sources depend on the current through the diode.
  • IE,S, and IC,S are the saturation currents of the base-emitter and base-collector diodes.
  • αF & αR are forward and reverse transport factors respectively.

By applying KCL at Node a,

IE = IF - IR ∝ R, IC = -IR + αF IF

The Ebers-Moll Parameters are related by:

IE,S αF = IC,S αR

MPPGCL JE Electronics Mock Test - 4 - Question 17
The two basic components of a CPU are: 
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 17

The correct answer is option 2.

Concept:

The two basic components of a CPU are the Arithmetic Logic Unit and Control Unit.

arithmetic logic unit (ALU):

The arithmetic logic unit (ALU) is responsible for the computer's arithmetic and logical functions. The input data is held in the A and B registers, and the result of the operation is received in the accumulator. The instructions for the ALU are stored in the instruction register.

Control unit:

A control unit, or CU, is circuitry within a computer's CPU that guides operations. It instructs the logic unit, memory, and both input and output devices of the computer on how to respond to program instructions. CPUs and GPUs are examples of devices that use control units.

Hence the correct answer is Arithmetic Logic Unit and Control Unit.

MPPGCL JE Electronics Mock Test - 4 - Question 18

Which fiber is preferred for long distance communication?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 18
  • Single-mode step indexed fibers are widely used for wideband communications are preferred for long-distance communication.
  • Single-mode step-index fiber is used to eliminate modal dispersion during optical communication.
  • In this fiber, a light ray can travel on only one path so minimum refraction takes place hence, no pulse spreading permits high pulse repetition rates.

Advantages of single-mode fiber:

1) Low signal loss

2) No modal dispersion

3) Does not suffer from modal dispersion

4) Can be used for higher bandwidth applications

5) Long-distance applications

6) Cable TV ends

7) High speed local and wide area network

Important Point

Single-mode means the fiber enables one type of light mode to be propagated at a time. This is explained with the help of the following diagram:

Single-mode fiber core diameter is much smaller than multimode fiber.

2) For single-mode fiber, the B.W ranges from 50 to 100 GHz/km

Multimode fibers:

Fibers that carry more than one mode are called multimode fibers. There are two types of multimode fibers:

1) Step Index

2) Graded Index

The comparison of the refractive index profile for step and graded fibers are respectively shown as:

The multimode step-index multimode fiber suffers from Modal dispersion.

  • Rays of light enter the fiber with different angles to the fiber axis. The limit is the fiber’s acceptance angle.
  • Rays that enter with a shallow angle travel a more direct path and arrive sooner than those that enter at steeper angles.
  • This arrival of different modes of light at different times is called modal dispersion.
MPPGCL JE Electronics Mock Test - 4 - Question 19
A/An ______ is a special high-speed storage mechanism.
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 19

The correct answer is Cache.

Key Points

  • A/An cache is a special high-speed storage mechanism used in computers.
  • A cache is a hardware or software component that stores data so that future requests for that data can be served faster.
  • With Internet browsers, the cache is a temporary storage area where website data is stored.
  • Cache memory has the shortest access time.
  • The processor can retrieve data or instructions from the cache memory.
  • Cache memory acts between CPU and RAM.
MPPGCL JE Electronics Mock Test - 4 - Question 20
The operation executed on data stored in registers is called ______
Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 20

Micro operation:

  • Performs basic operations on data stored in the registers.
  • Micro operation also transferring data between registers or between register and buses of CPU and performing arithmetic or logical operations on registers.
MPPGCL JE Electronics Mock Test - 4 - Question 21

The minimum change in the measured variable which produces an effective response of the instrument is known as

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 21

The minimum change in the measured variable which produces an effective response of the instrument is known as resolution.

Concept:

The static performance parameters of the instruments are:

  1. Accuracy
  2. Precision
  3. Resolution
  4. Threshold
  5. Static sensitivity
  6. Linearity
  7. Hysteresis
  8. Backlash
  9. Dead band
  10. Drift
  • Accuracy: It is the closeness with which an instrument reading approaches the true value of the quantity being measured. It means conformity to truth.
  • Precision: It is defined as the ability of the instrument to reproduce a certain set of readings within a given accuracy
  • Drift: It is defined as the variation of output for a given input caused due to change in the sensitivity of the instrument to certain interfering input like temperature changes, component instability, etc.
  • Resolution: It is defined as the smallest increment in the measured value that can be detected with certainty by the instrument.
  • Threshold: It is defined as the minimum value of input below which no output can be detected.
  • Static sensitivity: It is defined as the ratio of the magnitude of response to the magnitude of the quantity being measured.
  • Linearity: It represents the output of the instrument which is a linear function of the input.
  • Hysteresis: It is defined as the magnitude of error caused in the output for a given value of input when this value is approached from opposite directions, i.e. from ascending order and then descending order.
  • Dead band: It is defined as the largest change of the measurand to which the instrument does not respond.
  • Backlash: It is defined as the maximum distance or angle through which any part of the mechanical system may be moved in one direction without causing motion of the next part.
MPPGCL JE Electronics Mock Test - 4 - Question 22

Consider the signal,

  

The signal x(t) is:

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 22

Concept:

A signal will be called periodic if it is periodic for the entire range of domain, i.e.

If it is periodic for - to +, the signal is said to be periodic.

Calculations:

The given signal is defined as:

For t < 0,

x(t) = 2 cos t + cos 2t

2.cos t is period with period = 2π

Cos 2t is also periodic with period 

So, 2 cos t + cos (2t) will be periodic with period of 2 π. (LCM of π and 2π)

Similarly,

For, t ≥ 0,

x(t) = 2 sin t + sin (2t)

sin t is periodic with a period of π.

and sin 2t is also periodic with period of π.

2 sin t + sin 2t is therefore periodic as well with period = LCM (π, 2π) = 2 π only

If, x(t) is to be period with period 2π

x (-2π) = x(0) = x(2π)

Checking,

x (-2π) = 2 cos (-2π) +cos (-4π)

= 2 + 1 = 3

x(0) = 2 sin 0 + sin 0 = 0

And x (2π) = 2 sin 2π + sin (4π) = 0

Clearly, x (-2π) ≠ x (0) = x (2π)

Since the signal is not periodic for the complete range of = ∞ to ∞, the given signal is not a periodic signal.

MPPGCL JE Electronics Mock Test - 4 - Question 23

The respective ratio between the present ages of Roshan and Rakesh is 6: 11. Four years ago, the ratio of the ages was 1: 2 respectively. What will be Rakesh’s age after five years? 

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 23

MPPGCL JE Electronics Mock Test - 4 - Question 24

Into how many constituencies is the country divided for Lok Sabha elections?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 24
Answer:
Constituencies for Lok Sabha elections:
- The country is divided into constituencies for Lok Sabha elections.
- A constituency is a specific geographic area that elects a representative to the Lok Sabha, the lower house of the Parliament of India.
Number of constituencies:
- The country is divided into a total of 543 constituencies for Lok Sabha elections.
Explanation:
- The total number of constituencies in India for Lok Sabha elections is fixed at 543.
- Each constituency represents a specific area within the country, and the voters of that constituency elect one Member of Parliament (MP) to represent them in the Lok Sabha.
- The number of constituencies is determined based on various factors, including the population size and demographics of each region.
- The delimitation commission is responsible for the division of constituencies and redrawing boundaries after every census to ensure fair representation and equal distribution of seats.
Therefore, the correct answer is B: 543 constituencies.
MPPGCL JE Electronics Mock Test - 4 - Question 25

A’s salary is 25% more than that of B. Then B’s salary is less than that of A by 

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 25

MPPGCL JE Electronics Mock Test - 4 - Question 26

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

Race : Fatigue :: Fast : ?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 26

As the result of Race is Fatigue similarly the result of Fast is Hunger.

MPPGCL JE Electronics Mock Test - 4 - Question 27

A man bought 17 articles for Rs. 480 each, 15 articles for Rs. 600 each and 18 articles for Rs. 900 each. What is the average cost price per article?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 27


Hence, Option A is correct.

MPPGCL JE Electronics Mock Test - 4 - Question 28

Without stoppage, a train travels a certain distance with an average speed of 60 km/h, and withstoppage, it covers the same distance with an average speed of 40 km/h. On an average, howmany minutes per hour does the train stop during the journey?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 28

Since the train travels at 60 kmph, it’s speed per minute is 1 km per minute. Hence, if it’s speed
with stoppages is 40 kmph, it will travel 40 minutes per hour.

MPPGCL JE Electronics Mock Test - 4 - Question 29

Direction: Study the following information carefully and answer the given questions besides.

Some boys are sitting in a row facing the south. Only three boys sit between O and H. Only two boys sit to the left of J. O sits third to the left of M. More than seven boys sit between P and J. V sits fourth to the right of O. Eight boys sit between R and J. More than 12 boys sit between P and H. Only three boys sit between R and M. T sits third to the right of M. P sits fourth to the right of R.
Q. Who among the following sits 3rd to the right of the one who sits immediate right of M?

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 29

From the common explanation, we can say that R sits 3rd to the right of the one who sits immediate right of M. 
Hence, Option D is correct.
Final Arrangement:

Common Explanation:
References:

Only two boys sit to the left of J.
Eight boys sit between R and J.
Only three boys sit between R and M.
Inferences:
From the above references, we get two different cases:

References:
P sits fourth to the right of R.
O sits third to the left of M.
Inferences:
So, from this case 2 will be eliminated as P sits fourth to the right of R.

References:
Only three boys sit between O and H.
T sits third to the right of M.
V sits fourth to the right of O.
More than 12 boys sit between P and H.
More than seven boys sit between P and J.
Inferences:
From the above reference, we get the final arrangement:

MPPGCL JE Electronics Mock Test - 4 - Question 30

The LCM of two numbers is 1920 and their HCF is 16. If one of the number is 128, find the other number.

Detailed Solution for MPPGCL JE Electronics Mock Test - 4 - Question 30

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
We have,
First number × second number = LCM × HCF

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