GATE Mock Test Electronics Engineering (ECE)- 10 - Electronics and Communication Engineering (ECE) MCQ

# GATE Mock Test Electronics Engineering (ECE)- 10 - Electronics and Communication Engineering (ECE) MCQ

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## 30 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - GATE Mock Test Electronics Engineering (ECE)- 10

GATE Mock Test Electronics Engineering (ECE)- 10 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The GATE Mock Test Electronics Engineering (ECE)- 10 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 10 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Electronics Engineering (ECE)- 10 below.
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GATE Mock Test Electronics Engineering (ECE)- 10 - Question 1

### Rohit and Ronit start together from a point in opposite direction on bike. Rohit’s speed is 21Km/h and Ronit’s speed is 15km/h. What will be the distance between them after 20 mintues?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 1

Rohit’s speed = 21 km/h

Ronit’s speed = 15 Km/h

As we know that they start from opposite direction.

Thus, the relative speed = 21 Km/h + 15 Km/h = 36 Km/h.

The distance between Rohit and Ronit after 20 minutes = 36/3

= 12 km

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 2

### There are 60 students in a class. The students are divided into three groups A, B and C of 15, 20 and 25 students, respectively. The groups A and C are combined to form group D. What is the average weight of the students in group D?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 2
There is no indication of weights of the students in the question. Therefore, it is not possible to find the relation between the weights of students in groups A, B, C, and D.
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GATE Mock Test Electronics Engineering (ECE)- 10 - Question 3

### Each of the letters in the figure below represents a unique integer from 1 to 9. The letters are positioned in the figure such that each of (A+B+C),(C+D+E),(E+F+G) and (G+H+K) is equal to 13 each. Which integer does E represent?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 3
A + B + C = C + D + E = E + F + G

=G + H + K = 13

If we add all, we will get = 4×13 = 52

But sum of all natural number 1 to 9 =45=(9×10)/2

A + B + C + C + D + E + E + F + G + G + H + K

= 52

A + B + C + D + E + F + G + H + K =45

Hence, C + E + G=7

Also, C + D + E =13

D - G = 6

E = 4

Alternative Method

By checking other equations

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 4

The probability that a man will live 10 more years is 1/4 and the probability that his wife will live 10 more years is 1/3. The probability that neither of them will be alive in 10 years is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 4
Probability that a man will live 10 more years = 1/4

Probability that man will not live 10 more years = 1 - 1/4 = 3/4

Probability that his wife will live 10 more years = 1/3

Probability that his wife will not live 10 more years = 1 - 1/3 = 2/3

Then, probability that neither will be alive in 10 years = (3/4) x (2/3) = 1/2

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 5

Following bar chart provides the percentage of Adult Males, Adult Females and Children out of total population in five colonies A, B, C, D and E

If total number of residents in colonies A, B and C are 1250, 2050 and 1800 respectively, then what is the total number of adult females in colonies A, B and C together?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 5

The total number of adult females in colonies A, B and C together

= 1250 x 36/1000 + 2050 x 30/100 + 1800 x 42/100

= 450 + 615 + 756

= 1821

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 6

Voting is the privilege for which wars have been fought, protests have been organized, and editorials have been written. "No taxation without representation," was a battle cry of the American Revolution. Women struggled for suffrage, as did many minorities. Eighteen year olds clamored for the right to vote, saying that if they were old enough to fight in the war, then they should be allowed to vote. Yet Americans have a deplorable voting history, and many will tell you that they have never voted.

Which of the following words is the best synonym for the word 'privilege' as used in the passage?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 6
The context mentions that wars have been fought for 'voting' and 'privilege' means 'a special right'.
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 7

In the basement of an apartment, the total number of wheels of all cars and bikes is 70 more than twice the number of parked vehicles. The number of cars parked in the basement is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 7
Let the total number of cars and bikes parked in the basement be ‘a’ and ‘b’ respectively.

As we know that a car and a bike has four and two wheels respectively.

Therefore, total number of wheels = 4a + 2b

Also, it is given that the total number of wheels is 70 more than twice the number of parked vehicles.

Accordingly,

4a + 2b = 2(a + b) + 70

On simplifying we get,

a = 35.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 8

Two trains start at the same time from two stations and proceed towards each other at speeds of 90 km/h and 70 km/h, respectively. When they meet, it is found that one train has travelled 140 km more than the other. What is the distance (in km) between the two stations?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 8

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 9

Direction: In the given question, an idiom/phrase has been printed in bold in the sentence. Choose the alternative that best expresses the meaning of the idiom/phrase

The family of the victim said that the killer had got his just deserts when he was jailed for life.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 9
To get (or receive) one's just deserts receive what one deserves, especially appropriate punishment.
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 10

A tree of the graph shown below is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 10

Following figure shown that a d f is a tree.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 11

If Eigen values of A2×2Matrix are 2, 3 then Eigen values of A2 – 3A + 4I will be 'g' and 'h'. What will be the sum of 'g' and 'h'.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 11
Eigen values of f(A) = A2−3A+4Iare f(2), f(3)

∴ 2, 4 are Eigen values

g + h = 6

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 12

Directions: Select the most appropriate word/phrase among the choices to complete the sentence.

You can make your quixotic experiments with someone else, I do not wish to be your ________.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 12

'Guinea pig' is term used to refer any person or thing used in an experiment. As first part of sentence refers the experiments as quixotic (impractical or daring), so perfect word suited to this blank is 'guinea pig', which is used in labs for testing on biological experiments.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 13

One hundred identical coins, each with probability P of showing up heads are tossed once. If 0 < P < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins. Then the values of P is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 13
Given100C50q50p50

= 100 C51q49p51

51 q = 50 p

⇒ 51 ( 1 - p )

⇒ 50p, p = 51/101

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 14

If prices reduce by 20% and sales increase by 15%, what is the net effect on gross receipts?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 14
Let original price = p, and original sales = s. Therefore, original gross receipts = ps.

Let new price = 0.80p, and new sales = 1.15s. Therefore, new gross receipts = 0.92ps.

Gross receipts are only 92% of what they were. Hence, gross receipts decrease by 8%.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 15

A specimen of a Si has a resistivity of 200kΩ -cm, magnetic flux B = 0.3Wb / m2 and d = w = 6mm. if the value of hall voltages and hall currents are 60mV and 10μA. Then the value of μp is (cm2 / V - sec )

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 15
μ = σRH

σ = 1 / resistivity

=1 / 200kΩ−cm

= 5 × 10−4 mho / m

RH = VHW / BI

= 360m3 / C

Thereforerμ = σ RH

= 5 х 10 -4 х 360

= 1800 cm2 / V - sec

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 16

Kirchhoff's current law is applicable to:

1. Closed loops in a circuit
2. Junction in a circuit
3. Magnetic circuits

Which of the above is/are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 16

Applying KCL at mode 'N':
I1 + I2 = I3

KCL is applicable to junction in the circuit.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 17

In the circuit given below, what is the operating region of the transistor, if it has a β of 100?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 17

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 18

Most Reality TV shows centre on two common motivators: fame and money. The shows transform waitresses, hairdressers, investment bankers, counsellors and teachers, to name a few, from obscure figures to household names. A lucky few successfully parlay their fifteen minutes of fame into celebrity. The luckiest stars of Reality TV also reap huge financial rewards for acts, including eating large insects, marrying someone they barely know, and revealing their innermost thoughts to millions of people.

Which of the following options best supports the above paragraph?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 18
The correct answer is the first option as it is stated in the passage that most Reality TV shows centre on two common motivators: fame and money.

Choices 2 and 4 are not supported by the passage because passage is about Reality TV stars and not Reality TV.

Choice 3 is incorrect because the paragraph states that some Reality TV stars manage to parlay their fifteen minutes of fame into celebrity.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 19

Consider a circuit as shown in fig below

Equivalent resistance across the terminals A & B is_____Ω.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 19

⇒ Req = 1 + 3/4

= 7/4 Ω

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 20

Directions: Insert the missing number.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 20
Middle number = (Difference between numbers of left column) × (Sum of numbers of right column)

i.e. 80 = (15 - 5) × (2 + 6) = (10) × (8) and 65 = (9 - 4) × (7 + 6) = (5) × (13)

∴ ? = (13 - 11) × (16 + 8) = 2 × 24 = 48

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 21

The ROC of z-transform of the discrete time sequence is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 21

ROC of (1/4)n u(n−1) is | Z | > 14, the signal being right sided

ROC of 5nu(−n−1) is | Z | <5, the signal being left sided

Thus the common ROC will be 1/4 < | Z | < 5

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 22

Consider a discrete time periodic signal x [n] = sin(πn/5) . Let ak be the complex Fourier series coefficients of x[n]. The coefficients {ak}, are non-zero when k = Bm±1, where m is any integer. The value of B is _______.(Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 22
Given the discrete time periodic signal,

So, we have the fourier series coefficients

Also, we have the period of the function as

Since, the fourier series coefficients are also periodic, so we have

i.e. the fourier series coefficients are non zero for

k = 10 m ± 1 where m = 0, 1, 2, ……….

Given that the system coefficients are non-zero for

k = Bm ± 1

We get the value

B = 10

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 23

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 23

The transmission line generates capacitive reactive volt-amperes in its shunt capacitance and absorbing reactive volt-amperes in its series inductance. The load at which the inductive and capacitive reactive volt-amperes are equal and opposite, such load is called surge impedance load i.e the receiving end voltage is equal to the sending end voltage.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 24

A discrete random variable X has possible values xi = i2, i = 1, 2, 3, 4 which occur with probabilities 0.4, 0.25, 0.15, 0.1. The mean value = E [X] of X is ___. (Answer up to two decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 24

= 1.0 (0.4) + 4 (0.25) + 9 (0.15) + 16 (0.1) = 4.35

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 25

In a transistor amplifier, change of 0.025V in signal voltage causes the base current to change by 15 µA and collector current by 1.2 mA. If collector and load resistances are of 6kΩ and 12kΩ, determine input resistance in kΩ.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 25
Input resistance = change in input voltage / change in input current

= 0.025/15 x 10 ­-6

= 1.67kΩ

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 26

A source has an alphabet {a1, a2, a3, a4, a5, a6} with corresponding probabilities {0.1, 0.2, 0.3, 0.5, 0.15, 0.2}. The entropy (in bits / symbol) of this source is ______. (Answer up to one decimal place)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 26

= -(0.1 log20.1 + 0.2 log20.2) + 0.3 log20.3 + 0.05 log20.05 + 0.15 log20.15 + 0.2 log20.2

= 2.4087 bits/symbol

or H(x) = 2.4 bits/symbol

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 27

In the figure shown below, voltage V1 and V2 are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 27
KCL at node o: 30 = I2 + 6 + 9

I2 = 15 A

KCL at node b: 10 + 6 = I1

I1 = 16 A

KCL at node d: 9 + I3 = 40

I3 = 31 A

Applying KVL around the loop a o c d a

V2 – 3I2 – 4I3 – 40 × 1 = 0

V2 – 3 (15) – 4(31) – 40 = 0

V2 = 209 V

Applying KVL around the loop o b c o

V1 – 2I1 + 3I2 = 0

V1 = – 13V

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 28

The Fourier transform of signal is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 28

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 29

A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 29
White noise with 2 sided PSD 5*10-6 V2/Hz.

RC low pass filter has 3 dB cutoff 5 kHz.

Mean Squared value of the output power can be found using following way,

E(y2)=Ry(0)

This will give the Mean squared output power to be 39.27*10-3 V2

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 30

Consider the given circuit and waveform for the input voltage. The diode in circuit has cutting voltage Vy = 0.

The waveform of output voltage vo is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 30
For vi < 4 V the diode is ON and output v0 = 4 V.

For vi > 4 V diode is off and output v0 = vi.

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