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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  RRB JE ECE (CBT I) Mock Test- 10 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 10 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT I) Mock Test- 10

RRB JE ECE (CBT I) Mock Test- 10 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT I) Mock Test- 10 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT I) Mock Test- 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 RRB JE ECE (CBT I) Mock Test- 10 below.
Solutions of RRB JE ECE (CBT I) Mock Test- 10 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & RRB JE ECE (CBT I) Mock Test- 10 solutions in Hindi for Electronics and Communication Engineering (ECE) course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt RRB JE ECE (CBT I) Mock Test- 10 | 100 questions in 90 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
RRB JE ECE (CBT I) Mock Test- 10 - Question 1
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

In which year the total number of candidates admitted in both the institutions together was the second lowest?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 1
In 2002 = 2 + 3.5 = 5.5

In 2003 = 3.5 + 4.0 = 7.5

RRB JE ECE (CBT I) Mock Test- 10 - Question 2
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

What was the ratio between the number of candidates admitted in institution A in the year 2006 and the total number of candidates admitted in both the institutions together in the year 2004?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 2

A in 2006 = 4

A, B in 2004 = 9

hence required ratio = 4 : 9

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RRB JE ECE (CBT I) Mock Test- 10 - Question 3
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Prabhakar deposited Rs. 5,000 in the bank. The bank offered simple interest at a rate of 12 percent per year. After 5 years the total amount received by him will be:

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 3
Here sum = Rs.5000 = P, Rate = 12% = R

Time = 5 years = T

∴ Amount = P + I = 5000 + 3000 = Rs. 8000

RRB JE ECE (CBT I) Mock Test- 10 - Question 4
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Find the value of

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 4

4 x 4 x 4

= 64

RRB JE ECE (CBT I) Mock Test- 10 - Question 5
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

The total number of candidates admitted in institution A over all the years together was approximately what percentage of the total number of candidates admitted in institution B over all the years together?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 5
Total No. of candidates in institution A = 2 + 3.5 + 5 + 6 + 4+6.5

= 27 (in thousand)

Total no. of candidates in institution B =3.5 + 4 + 4 + 5.5 + 5 + 6 = 28 (in thousand)

Required percentage=2700/28

= 96.42≈ 96 %

RRB JE ECE (CBT I) Mock Test- 10 - Question 6
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In the figure, ABCD is a parallelogram with AD = a units, DC = 2a units and DE : EC = 1 : 2. CEFG is a rectangle with FE = 3AE. What is the ratio of the area of a parallelogram and the rectangle?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 6

Area of parallelogram ABCD

= 2(Area of △ADE) + (Area of rectangle AECH)

= 1/2

RRB JE ECE (CBT I) Mock Test- 10 - Question 7
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The ages of Ram and Shyam are in the ratio 2 : 1. After five years, the ratio of their ages will be 3 : 2. Their present ages are:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 7
Suppose present ages of Ram and Shyam are 2x and x years respectively.

or 4 x + 10 = 3x + 15

or x = 5

Hence, the present age of Ram = 10 years.

and present age of Shyam - 5 years

RRB JE ECE (CBT I) Mock Test- 10 - Question 8
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

What was the total number of candidates admitted in institution B over all the years together?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 8
In 2002,Number of candidates admitted in institution B = 3500

In 2003, the Number of candidates admitted in institution B = 4000

In 2004, the Number of candidates admitted in institution B = 4000

In 2005, the Number of candidates admitted in institution B = 5500

In 2006, the ,’,’Number of candidates admitted in institution B = 5000

In 2007,Number of candidates admitted in institution B = 6000

The total number of candidates admitted in institution B over all the years = 28000

RRB JE ECE (CBT I) Mock Test- 10 - Question 9
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A rectangle ABCD is inscribed in a circle with centre O. If AC is the diagonal and ∠BAC = 30°, then the radius of the circle will be equal to:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 9

Join B and O

Then √BOC = 2 √BAC = 60°

Draw OM ⊥ from O on BC then BM = 1/2 BC

∴ ∠BOM = 30°

From △ BMO

RRB JE ECE (CBT I) Mock Test- 10 - Question 10
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A hollow sphere of external and internal diameters 8 cm and 4 cm respectively, is melted into a cone of base diameter 8 cm, then the height of the cone will be equal to:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 10

Hence, the height of the cone formed is 14 cm.

RRB JE ECE (CBT I) Mock Test- 10 - Question 11
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The value of sin θ + sin (θ + 120°) + sin (θ + 240°) is equal to:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 11
The given expression is sinθ + sin (θ+120°) + sin (θ+240°)

= sinθ + sin (π + θ)

= sinθ - sinθ

= 0

RRB JE ECE (CBT I) Mock Test- 10 - Question 12
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The area of a rhombus is 120 cm2. If one of its diagonals is of length 10 cm, then the length of one of its sides is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 12

We know that area of a rhombus = 1/2 product of its diagonals

∴ 120 = 1/2 x 10 x 2nd diagonals

or 2nd diagonals = 24 cm

RRB JE ECE (CBT I) Mock Test- 10 - Question 13
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A mixture of 70 litres contains milk and water in a ratio of 3 :4. How many litres of milk must be added to the mixture so as to make the ratio 5 : 4?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 13

Milk : Water

New ratio 5 : 4

Old ratio 3 : 4

Difference 2 : 0

Sum of old ratios = 3 + 4 = 7

∴ Milk added = (2/7)×70 = 20 litres

RRB JE ECE (CBT I) Mock Test- 10 - Question 14
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A man rowed a boat to a place covering 72 km distance and back in 15 hours. He finds that he can row 3 km with the stream in the same time as 2 km against the stream. Find the speed of the stream.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 14
Speed downstream : speed upstream = 3 : 2

∴ Time taken to row downstream : Time taken to row upstream = 2 : 3

But, total time taken = 15 hours

∴ Time taken to row downstream = 2/5 x 15 hours = 6 hours

And time taken to row upstream = 15 - 6 = 9 hours.

∴ Speed downstream = 72/6 = 12 km/h

And speed upstream = 72/9 = 8 km/h

∴ Speed of the stream = (12-8)/2 km/h

RRB JE ECE (CBT I) Mock Test- 10 - Question 15
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In an examination A gets 10% marks less than B and B gets 25% marks more than C and C gets 20% marks less than D. If A gets 270 marks in 500 then D get-
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 15

A : B = 9 : 10

B : C = 5 : 4

C : D = 4 : 5

A : B : C : D = 9 : 10 : 8 : 10

9 = 270

10 = 300

% = 300/500 x 100= 60%

RRB JE ECE (CBT I) Mock Test- 10 - Question 16
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If from the top of a lighthouse 100 metre high the angle of depression of a boat is tan-1(5/12) then the distance (in metre) between the boat and the lighthouse is equal to:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 16
Given that the height of lighthouse = 100m

and the angle of depression of the boat is tan-1 5/12

X = 240

∴ The distance between the boat and the lighthouse is 240 m.

RRB JE ECE (CBT I) Mock Test- 10 - Question 17
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The perimeter of a rhombus is 40 cm and the measure of an angle is 60o, then the area of it is?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 17

RRB JE ECE (CBT I) Mock Test- 10 - Question 18
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The average of three numbers is 40. The first number is twice the second and the second one is thrice the third number. The difference between the largest and the smallest numbers is?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 18
Let the third number be x

∴ second number = 3x

First number = 6x

∴

⇒ 10x = 120 ⇒ x = 12

∴ required difference

= 6x - x = 5x = 5 x 12 = 60

RRB JE ECE (CBT I) Mock Test- 10 - Question 19
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ABCD is a trapezium in which AB is parallel to DC. If the diagonals intersect at O, then which one of the following is correct?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 19

Consider Triangle OAB and Triangle OCD

∠DOC = ∠BOA

∠OCD = ∠OAB

∠ODC = ∠OBA

∴ ΔOAB and Δocd are similar

RRB JE ECE (CBT I) Mock Test- 10 - Question 20
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In a right angled triangle, one acute angle is double the other. If the length of the smallest side is a. The length of the hypotenuse is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 20

Clearly

xo + 2xo = 90o

ie., xo = 30o

RRB JE ECE (CBT I) Mock Test- 10 - Question 21
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

What was the difference between the total number of candidates admitted in institution A in the year 2004 and 2007 together and the total number of candidates admitted in institution B in the year 2003 and 2007 together?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 21
A in 2004 and 2007

= 5 + 6.5 = 11.5

B in 2003 and 2007

4 + 6 = 10.0

Difference = 1500

RRB JE ECE (CBT I) Mock Test- 10 - Question 22
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In the given figure, if x = 20°, find y.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 22

∠AOC + ∠BOC 180°

⇒ 5y + 10 + 2x = 180°

⇒ 5y + 10 + (2 x 20) = 180°

⇒ 5y = 180° - 10 - 40 = 130°

⇒ y = 26°

RRB JE ECE (CBT I) Mock Test- 10 - Question 23
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Find the value of (x3+y3+z3 - 3xyz), if x + y + z = 9 and xy + yz + zx = 26:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 23
Given (x+y+z) = 9

∴ (x + y + z)2 = 92 = 81

∴ x2 + y2 + z2 + 2 × 26 = 81

∴ x2 + y2 + z2 = 81 - 52 = 29

Now, x3 + y3 + z3 - 3xyz

= (x + y + z) [(x2 + y2 + z2)] - (xy + yz + xz)]

= 9 × (29 - 26) = 27

RRB JE ECE (CBT I) Mock Test- 10 - Question 24
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If then the value of tan x/tan y is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 24

RRB JE ECE (CBT I) Mock Test- 10 - Question 25
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Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 25
Hence, first number is 1 - 1 = 0

we know that, any number multiplied by 0 always gives 0.

∴ 0 is the answer.

RRB JE ECE (CBT I) Mock Test- 10 - Question 26
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A carriage driving in a fog passed a man who was walking at the rate of 3 km an hour in the same direction. He could see the carriage for 4 minutes and it was visible to him upto a distance of 100 m. What was the speed of the carriage?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 26
The distance travelled by the man in 4 min = [(3×1000)/60] ×4=200 meters

∴ distance travelled by the carriage in 4 min = (200 + 100) = 300 meters

∴ speed of the carriage = (300/4)×(60/1000) km/hour

= 9/2 km/hour

RRB JE ECE (CBT I) Mock Test- 10 - Question 27
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Two partners invested Rs. 50,000 and Rs. 70,000 respectively in a business and agreed that 70% of profit should be divided equally between them and the remaining profit in the ratio of investment. If one partner gets Rs. 90 more than the other then the total profit made in the business is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 27
Ratio of profits = 50,000 : 70,000 = 5 : 7

∴ Total Profit = Rs. 1800

RRB JE ECE (CBT I) Mock Test- 10 - Question 28
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The ratio of the first and second train fares between two stations is 3:1 and that of the numbers of passengers travelling between the two stations by first and second classes is 1:50. In on a particular day, ₹1325 are collected from passengers travelling between the two stations, then the amount collected from the second class passengers is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 28

53x = 1325

x = 1325/53 = 25

∴ Amount collected from second Class = 50x

25 x 50 = Rs. 1250

RRB JE ECE (CBT I) Mock Test- 10 - Question 29
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If the angles of a pentagon are in the ratio 1 : 3 : 6 : 7 : 10, then the smallest angle is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 29
Suppose the angles of a pentagon are

x°, 3x°, 6x°, 7x°, 10x°

But x° + 3x° + 6x° + 7x° + 10x° = (2x5 -4) x 90°

or 27x° = 6 x 90°

or x = 20°

RRB JE ECE (CBT I) Mock Test- 10 - Question 30
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A man's age is 125% of what it was 10 years ago, but 83 1/3 % of what it will be after ten 10 years. What is his present age?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 10 - Question 30
At present

man age = x years

10 years ago :

Man age = ( x - 10 ) years

10 years after :

Man age = ( x + 10 ) years

According to the problem given ,

125% of (x - 10) = 83⅓% of (x + 10)

(125/100) (x - 10) = (250/300)(x+10)

(5/4) x - 10) = (5/6)(x + 10)

(x - 10)/4 = (x + 10)/6

6(x - 10) = 4(x + 10)

6x - 60 = 4x + 40

6x - 4x = 40 + 60

2x = 100

x = 100/2

x = 50

Therefore ,

Present age of the man = x = 50yrs

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