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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  RRB JE Mock Test Series Electrical Engineering 2025  >  RRB JE EEE (CBT I) Mock Test- 8 - Electrical Engineering (EE) MCQ

RRB JE EEE (CBT I) Mock Test- 8 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series Electrical Engineering 2025 - RRB JE EEE (CBT I) Mock Test- 8

RRB JE EEE (CBT I) Mock Test- 8 for Electrical Engineering (EE) 2024 is part of RRB JE Mock Test Series Electrical Engineering 2025 preparation. The RRB JE EEE (CBT I) Mock Test- 8 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE EEE (CBT I) Mock Test- 8 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE EEE (CBT I) Mock Test- 8 below.
Solutions of RRB JE EEE (CBT I) Mock Test- 8 questions in English are available as part of our RRB JE Mock Test Series Electrical Engineering 2025 for Electrical Engineering (EE) & RRB JE EEE (CBT I) Mock Test- 8 solutions in Hindi for RRB JE Mock Test Series Electrical Engineering 2025 course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt RRB JE EEE (CBT I) Mock Test- 8 | 100 questions in 90 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study RRB JE Mock Test Series Electrical Engineering 2025 for Electrical Engineering (EE) Exam | Download free PDF with solutions
RRB JE EEE (CBT I) Mock Test- 8 - Question 1
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The height of a square base right prism is 15 cm. If the total surface area of the prism is 608 cm2, then what will be the volume of the prism?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 1

Let side of square base be ‘a’.

Now, total surface area = Perimeter of square base × Height

+ 2 × Area of square base.

⇒ 608 = 4a × 15 + 2 × a2 ⇒ a2 + 30a –304 = 0 ⇒ a = 8

∴Volume of prism = Area of base × Height

= a2 × 15 = 8 × 8 × 15 = 960 cm3

RRB JE EEE (CBT I) Mock Test- 8 - Question 2
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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 EEE (CBT I) Mock Test- 8 - Question 2
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°

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RRB JE EEE (CBT I) Mock Test- 8 - Question 3
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What will be the value of x in the infinite series

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 3

or x2 = 6 + x

or x2 - x - 6 = 0

or (x-3) (x+2) = 0

ஃ x = 3 or x = -2

Since x is positive

ஃ x = 3

RRB JE EEE (CBT I) Mock Test- 8 - Question 4
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If a (tanθ + cotθ) = 1, sinθ + cosθ = b with 0° < θ < 90°, then the relation between a and b is:

(a)2a = b2 + 1
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 4
a (tanθ + cotθ) = 1[Given]

sinθ + cosθ = b[Given]

a = sinθ cosθ.................(i)

Also, sinθ + cosθ = b

Squaring both sides,

sin2θ + cos2θ + 2sinθ cosθ = b2

1 + 2 a = b2 ⇒ 2a = b2 – 1[From ⇒ equ. (i)]

RRB JE EEE (CBT I) Mock Test- 8 - Question 5
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The driver of an ambulance sees a school bus 40 meters ahead of him. After 20 seconds, the school bus is 60 meters behind. If the speed of the ambulance is 30 km/hr, then what is the speed of the school bus?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 5

Relative Speed,

= (Total.distance/Total.time)

=60+40/20

= 5 m/s =5×18/5 = 18 kmph

Relative Speed = (speed of ambulance - speed of school bus)

Speed of school bus = speed of ambulance - relative speed.

= 30-18 = 12 kmph.

RRB JE EEE (CBT I) Mock Test- 8 - Question 6
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At what percent above the cost price must an article be marked so that seller gains 33% after allowing the customer a discount of 5% ?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 6
Let the marked price be Rs. x

Let the Cost price be Rs. y

Selling Price = x - 5% of x

x = 95x/100 = 19x/20

y + 33% of y = 19x/20

7y = 5x

x = 7/5 y = y + 2/5 y = y + 40% of y

RRB JE EEE (CBT I) Mock Test- 8 - Question 7
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If (x11+1) is divided by (x+1), then the remainder is :
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 7

Putting x = -1 in given expression

X11 + 1 = (-1) 11 + 1 = -1 + 1 = 0

Hence dividing expression by (x + 1) the remainder will be zero

Alt sol:

Lctp(x) = x11 + 1, be the given polynomial.

Now, by remainder theorem, when p(x) is divided by (x + 1), then remainder is given Ity, p(1)

Now, p(x) = x11 + 1

=>p(-1) = (-1)11 + 1= -1 + 1 = 0

So, required remainder = 0

RRB JE EEE (CBT I) Mock Test- 8 - Question 8
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Evaluate : cos80o/sin10o + cos 59o cosec 31o
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 8

RRB JE EEE (CBT I) Mock Test- 8 - Question 9
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The measurement of each angles of a polygon is 160°. The number of its sides is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 9
Let n be the sides of polygon

or 16n = (2n - 4) x 9

or 16n = 18n - 36

2n = 36 ⇒ n = 18

ஃ the numbers of sides of polygons are 18

RRB JE EEE (CBT I) Mock Test- 8 - Question 10
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Two towers of the same height stand on opposite sides of a road 100 m wide. At a point on the road between the towers, the elevations of the towers are 30° and 45°. Find the height of the towers and the position of the point from the farthest tower.
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 10

Let AB, CD are two towers.

E is the point between road BC.

Then in ΔDCE

CD = EC = x = AB

In ΔABE

BE = √3x then

Position of the farthest tower from point = 63.4 m tr.

RRB JE EEE (CBT I) Mock Test- 8 - Question 11
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If (a/b)x−1 = (b/a)x−3 ,then the value of x is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 11

(a/b)x−1 = (b/a)x−3 = (a/b)−(x−3) = (a/b)3−x

∴ x - 1 = 3 - x or 2x = 4

∴ x = 2

RRB JE EEE (CBT I) Mock Test- 8 - Question 12
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If the perimeter of an equilateral triangle is 36.9 metre and the length of an altitude is 8 metre, then the area of the triangle is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 12
Perimeter of the equilateral triangle of side x = 36.9 m

∴ side of equilateral triangle x = 36.9/3 = 12.3 m

height = 8

hence area = 1/2×12.3×8 = 49.2 m2

RRB JE EEE (CBT I) Mock Test- 8 - Question 13
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In what ratio Martina should mix two varieties of juice worth Rs. 52 per liter and Rs. 74 per liter, so that by selling the mixture at Rs.71.28 per liter she can earn a profit of 8%?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 13

RRB JE EEE (CBT I) Mock Test- 8 - Question 14
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40% of the students in a school are girls. If the number of boys is 1680, then what is the total number of students?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 14
% of boys = 100 – 40 = 60%

60% of x = 1680

x = (1680 × 100)/60

x = 2800

RRB JE EEE (CBT I) Mock Test- 8 - Question 15
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Prem invested a certain sum of money in a simple interest. It amounts to Rs 300 at the end of 3 year and to Rs 400 at the end of 8 year. What was the rate of interest in which he invested his sum?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 15

Let principal be P and rate of interest be r%

Then

At the end of another 5 years means 3 + 5 = 8 years.

On Subtracting Eq . (i) from Eq. (ii)

ஃ P x r = 2,000

From Eq (i),

240 x r = 2000

r = 8.33%

RRB JE EEE (CBT I) Mock Test- 8 - Question 16
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10 men can complete a work in 8 days. 20 women can complete the same work in 6 days. In how many days 16 men and 18 women can complete the same work, working together ?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 16

10m → 8 days ⇒ 20m → 4 days ⇒ 1m → 80 days

20w → 6 days ⇒ 20w → 6 days ⇒ 1w → 120 days

ஃ 2m = 3w

Thus. 16m + 18w = 24\v + 18w = 42w

20w x 6 = 42w w x x

(Where x is the number of days taken by 42 women)

ஃ

RRB JE EEE (CBT I) Mock Test- 8 - Question 17
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461 + 462 + 463 + 464 + 465 is divisible by:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 17
461 [1+4+42+43+44] = 461 [1+4+ 16 + 64 +256] = 461 x 341

Since 341 is divisible by 11, then the given expression is also divisible by 11.

RRB JE EEE (CBT I) Mock Test- 8 - Question 18
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A cane contains a mixture of two liquids 'A' and 'B' in the ratio 7 : 5. When 9 liters of a the , the ixture are drawn off and the cane is filled with 'B', the ratio of 'A' and 'B' becomes 7 : 9 liters. Liter of Liquid 'A' contained by the cane initially was?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 18

Initially,

A = 7x litre, B = 5x litre (let)

In 9 litres of mixture ,

In new situation,

252x - 189 = 140x + 147

⇒l 112x = 336

⇒l x = 3

∴ Initial quantity of liquid A

= 7x = 7 x 3 = 21 litre

RRB JE EEE (CBT I) Mock Test- 8 - Question 19
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

What is the average number of students appearing for aptitude test from all the cities together?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 19
(30 + 22.5 + 37.5 + 17.5 + 32.5)/5 = 28
RRB JE EEE (CBT I) Mock Test- 8 - Question 20
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

What is the ratio of the number of students appearing for the aptitude test for city B to that of city A?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 20
22.5 : 30

3 : 4

RRB JE EEE (CBT I) Mock Test- 8 - Question 21
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

The number of students appearing for the aptitude test from city D is approximately what percentage of the number of students appearing for the aptitude test from city C?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 21
17.5×100/37.5 = 47%
RRB JE EEE (CBT I) Mock Test- 8 - Question 22
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

What is the respective ratio of the number of student appearing for the aptitude test from cities C and D together to the number of students appearing for the aptitude test from cities A, D and E together.
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 22

55 : 80

= 11 : 16

RRB JE EEE (CBT I) Mock Test- 8 - Question 23
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The batting average of a batsman in 57 innings is 58 runs. He was out for a duck in 7 innings. His batting average for remaining innings is :
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 23

Total runs of 57 innings = 57 x 58

∴ Average of remaining (57-7) or 50 innings

=57×58/50 = 66.12

RRB JE EEE (CBT I) Mock Test- 8 - Question 24
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If sin(x+y)/sin(x−y) = a+b/a−b , then the value of tan x/tan y is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 24

RRB JE EEE (CBT I) Mock Test- 8 - Question 25
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A and B can do a piece of work in 15 days. B and C can do the same work in 12 days and C and A in 10 days. How many days will A take to complete the work alone?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 25

A + B + C will complete 15/2 work in a day .

A + B + C → 15/2, B + C → 5

Thus , A completes 2.5 unit work in a day .

Thus, total time required by A = 60/25 = 24 days.

RRB JE EEE (CBT I) Mock Test- 8 - Question 26
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If 2 men and 3 women can do a piece of work in 8 days and 3 men and 2 women in 7 days. In how many days can the work be done by 5 men and 4 women working together?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 26

|2 x 8 - 3 x 7| men’s work = |3x 8 - 2 x 7| women’s work.

∴ 5 men’s work = 10 women’s work

∴ 1 men’s work = 2 women’s work

2 men’s work = 4 women’s work

∴ Days taken by 5 men + 4 women = 14 women are 7x8/14 = 4 days

RRB JE EEE (CBT I) Mock Test- 8 - Question 27
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If a and b are two distinct positive numbers, then which of the following is true?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 27
Let a = 16 & b = 9

Option (a) : LHS =

RHS =

Hence , option (a ) is wrong .

Option (b) : LHS =

RHS =

Hence , option (b) is wrong .

Option (c) : LHS =

RHS =

Hence, option (c) is the correct option .

RRB JE EEE (CBT I) Mock Test- 8 - Question 28
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5 year ago, the father’s age was 3 times of his son’s age. 5 years hence the ratio betwwn the ages of the father and the son become 11: 5. Find the father’s present age.
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 28

Father Son

Old ratio 3 : 1

New ratio 11 : 5

Let, 5 years age father’s age was 3x years, where

x=(5+5)×(11−5)/3×5−1×11 =10×6/4 =15

∴ Father’s age (5 years ago) = 3 × 15 = 45 years

∴ His present age = 45 + 5 = 50 years

RRB JE EEE (CBT I) Mock Test- 8 - Question 29
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P and Q starting simultaneously from two different places proceed towards each other at a speed of 20 km/hour and 30 km/hour respectively. By the time they meet each other, Q has covered 36 km more than P. The distance (in k.m.) between the two places is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 29

Let the distance = x km

Let they meet when P travels y km.

then, Q travels in the same time x – y km

from question,

(x – y) – y = 36

x – 2y = 36 .......(i)

Also, 20 × y = 30 (x–y) [∵Time is same]

∴y/x−y−=3/2

2y = 3x – 3y

5y = 3x ........(ii)

Putting the value of y from (ii) to (i),

x−6x/5=36⇒−x/5=36⇒x=180 km

RRB JE EEE (CBT I) Mock Test- 8 - Question 30
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A man walks from A to B at the rate of 5 km per hour and returns back at the rate of 3 kmph. What is the average speed for the whole journey ?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 8 - Question 30

We know that

Average speed = 2xy / x+y

Here, x= 5 km per hour

y= 3 km per hour

therefore, Average speed = 2*5*3 / 5+3 = 30/8 = 15 kmph.

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