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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- 2 - Electrical Engineering (EE) MCQ

RRB JE EEE (CBT I) Mock Test- 2 - 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- 2

RRB JE EEE (CBT I) Mock Test- 2 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- 2 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE EEE (CBT I) Mock Test- 2 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- 2 below.
Solutions of RRB JE EEE (CBT I) Mock Test- 2 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- 2 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- 2 | 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- 2 - Question 1
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D and E are the mid-points of AB and AC of ∆ABC, BC is produced to any point P; DE, DP and EP are joined. then, area of:

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 1
(By mid-point theorem)

DE ∥ BC

DE = 1/2 BC

⇒ ar(∆BDE) = 1/4 × ar(∆ABC)

And ∆BDE = ∆PED

[ ∵ both triangles lie on the same base DE and between two parallel lines DE and BP.]

∴ ar(∆PED) = 1/4 × ar(∆ABC)

RRB JE EEE (CBT I) Mock Test- 2 - Question 2
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One side other than the hypotenuse of right angle isosceles triangle is 6 cm. The length of the perpendicular on the hypotenuse from the opposite vertex is:

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

Let √B be right angle in ΔABC

and BD be perpendicular on hypetenuse AC then

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RRB JE EEE (CBT I) Mock Test- 2 - Question 3
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The speeds of two trains are in the ratio 3 : 4. They are going in opposite directions along parallel tracks. If each takes 3 seconds to cross a telegraph post, find the time taken by the trains to cross each other completely?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 3
Speed of both trains = 3 : 4

x 3 ↓ : ↓ x 3 ⟶ time

Length = 9m : 12m.

= 3 sec.

RRB JE EEE (CBT I) Mock Test- 2 - Question 4
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A mother is 3 times faster than her daughter. If the daughter completes a piece of work in 15 days, how long will it take for both mother and daughter to complete the same work?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 4
Daughter’s one day work = 1/15

Mother’s one day work = ⅕

Both can complete their work = 1/5+1/15 = (3+1)/15 = 4/15

Hence, required days= 15/4 days .

RRB JE EEE (CBT I) Mock Test- 2 - Question 5
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Study the following table carefully to answer the questions that follow.

Number of soldiers (in thousands) joining five different forces during six different years.

The total number of soldiers joining BSF in the years 2004, 2005 and 2006 was approximately what percent of the total number of soldiers joining Navy over all the years together?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 5
Total no. of soldiers joining BSF in 2004, 2005, 2006

= 13400

Total no. of soldiers joining in Navy = 10900

Required Percentage= (13400/10900)×100 = 123%

RRB JE EEE (CBT I) Mock Test- 2 - Question 6
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Study the following table carefully to answer the questions that follow.

Number of soldiers (in thousands) joining five different forces during six different years.

What was the ratio of the number of soldiers joining Army in the year 2008 to the number of soldiers joining coast guard in the year 2006?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 6
Required ratio = 6.5/1.3 = 5/1 = 5:1
RRB JE EEE (CBT I) Mock Test- 2 - Question 7
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If tan (5x – 10°) = cot (5y + 20°), then the value of (x+y) is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 7
tan (5x – 10°) = cot (5y + 20°)

tan (5x – 10°) = tan (90° – {5y + 20°})

5x – 10° = 90° – (5y + 20°)

5x + 5y = 90° + 10° – 20°

5x + 5y = 80°

x + y = 16°

RRB JE EEE (CBT I) Mock Test- 2 - Question 8
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If a train, with a speed of 60 km/hr, crossed a pole in 30 second, the length of the train (in meters) is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 8
Speed = 60 km/hr.

S = 60 x 5/18 m./sec. = 50/3 m./sec.

T = 30 sec.

Length= S x T =

l = 500m.

RRB JE EEE (CBT I) Mock Test- 2 - Question 9
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The supplement of an angle is one-fourth of itself. Determine the angle and its supplement.
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 9
Let the measure of the angle be x°

⇒(180°- x) = x

⇒ 720 - 4x = x

⇒ 5x = 720

⇒ x = 144

Thus, the measure of the angle is 144°,the and the measure of its supplement.

= 180°- 144° = 36°.

RRB JE EEE (CBT I) Mock Test- 2 - Question 10
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3 years ago the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 10
Total age of 5 members, three years ago

= 17 × 5 = 85 years

Three years hence,

Total age of 5 members = 85+3x5 = 85+15 = 100 years

Sum of present ages of 6 members = 17 × 6 = 102 years

Present age of baby= 102 – 100 = 2 years

RRB JE EEE (CBT I) Mock Test- 2 - Question 11
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A ladder leans against a vertical wall. The top of the ladder is 8 meter above the ground. When the bottom of the ladder is moved 2 meter farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 11

Let the length of the ladder be x meter. We have

82 + y2 = x2 and (y + 2) = x

Hence, 64 + (x - 2)2 = x2

⇒ 64 + x2 — 4x + 4 = x2

⇒ 68 = 4x ⇒ x = 17 meter

RRB JE EEE (CBT I) Mock Test- 2 - Question 12
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The length of a Rectangular plot is decreased by 33.33%. By how much % the breadth of the plot will be increased so that the area remains constant?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 12

RRB JE EEE (CBT I) Mock Test- 2 - Question 13
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A man deposited a certain amount in a fixed deposit at r % p.α., interest being compounded annually. If the interest accrued for the fourth and fifth years are Rs 13310 and Rs 14641. what is the total interest accrued for the first three years?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 13
The interests accrued each year on compound interest form a geometric progression with I, as the first year interest and common ratio of

Therefore, interests or 1st, 2nd,3rd,4th and 5th years will be:

respectively.

Given, ------------(1)

And ------------(2)

(2) divided by (1), we get

∴ Sum of the first three years interests

= 10000 + 11000 + 12100

= Rs. 33100

RRB JE EEE (CBT I) Mock Test- 2 - Question 14
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A person earns Rs. 5000 as an interest in 5/2 years on a certain sum invested with a company at the rate of 10% per annum. Find the sum invested by a person in the company?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 14

RRB JE EEE (CBT I) Mock Test- 2 - Question 15
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The Value of is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 15
cot θ = tan (90 - θ) and cos θ = sin (90 - θ)

Hence

RRB JE EEE (CBT I) Mock Test- 2 - Question 16
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A is twice as fast as B and B is one third as fast as C. If they together can complete work in 30 days. In how many days, A, B and C individually can do the same work?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 16
Condition 1- A is twice as fast as

B, means their ratio is 2: 1.

Condition 2- B is one third as fast

as C, means their ratio is 1: 3. Then we can say, the ration of A, B and C is as follows

According to condition 1,

A : B = 2 : 1

According to condition 2;

B : C = 1 : 3

Thus, Ratio of A : B : C = 2 : 1 : 3

This ratio denotes the working efficiency of A, B and C which means;

A, B and C can do = (2 + 1 + 3 = 6) work/day

Now we can calculate total work i.e.

Total work = Total days × Per day work

Total work = 30 × 6 = 180

A can do the same work in;

= (Total work / Efficiency of A) = 180 / 2 = 90 days

B can do the same work in;

= (Total work / Efficiency of B)

180 / 1 = 180 days

C can do the same work in;

= (Total work / Efficiency of C) = 180 / 3 = 60 days

Thus, A, B and C can individually complete the same work in 90, 180 and 60 days, respectively.

RRB JE EEE (CBT I) Mock Test- 2 - Question 17
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If ab= 25, then the minimum value of a+ b is:
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 17
Minimum value of (a+ b) can be obtained when a= b

Then a + b = 5 + 5 (∴ ab= 25 and a=b)

=10

RRB JE EEE (CBT I) Mock Test- 2 - Question 18
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If a + b + c = 1, then Find (1+a) (1+b) (1+c)?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 18
Given, a + b + c = 1

Let observe the value of a,b,c

a + b + c = 1

1 = 1

put the value of a, b, c

(1 + a) (1 + b) (1 + c)

RRB JE EEE (CBT I) Mock Test- 2 - Question 19
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The price of a shirt is ₹260 but the shopkeeper successively discount 15% & 20%. The net sales price is subject to a sales tax of 5%. What does the buyer pay?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 19
1st discount

S.P= Rs.260 - 15% Rs.260

= 260 - 39 = 221

2nd discount

S.P= Rs.221 - 20% Rs.221

= 221 - 44.2 = 176.8

= Rs.8.84

Amount paid by buyer

= Rs. 176.8 + Rs. 8.84

= Rs. 185.64.

RRB JE EEE (CBT I) Mock Test- 2 - Question 20
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A cistern from inside is 12.5 m long, 8.5 m broad and 4 m high and is open at the the ,,.top. Find the cost of cementing the inside of a cistern at Rs. 24 per sq. m
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 20
Area of surface to be cemented = 2 x (l+b) x h + (lxb)

i.e, area of four walls + area of the floor

= 2 x (21) x 4 + (106.25)

= 274.25 m2

∴ cost of cementing = 24 x 274.25 = Rs. 6582

RRB JE EEE (CBT I) Mock Test- 2 - Question 21
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A goat is tied to a pole fixed at a corner outside a room with a square base in a grass field. It is tied using a 14 m long rope. The side of the base of the room is 21 m. Find the area of the field over which the goat can graze (in sq. m).
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 21

Let the goat be tied at P.

The area over which the goat could graze = sum of the areas of the regions A and B = Area of a

sector of radius 14 m and central angle 270°

RRB JE EEE (CBT I) Mock Test- 2 - Question 22
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An article is sold at a loss of 10%. Had it been sold for Rs. 9 more, there would have been a gain of on it. The cost price of the article is :
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 22
Let the cost price of the article = Rs. x

S.P. at 10% loss = Rs 0.9x

According to question if it is sold by Rs 9 more there will be gain of 25/2 %

⇒ 180x + 1800 = 225x

⇒ 225x - 180x = 1800

⇒ 45x = 1800

∴ x = Rs. 40

Alternate solution.

Let the CP of the article = 100

x x 22.5 = 9

Required,

⇒ CP x x

⇒

⇒ 40

RRB JE EEE (CBT I) Mock Test- 2 - Question 23
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A water tank is 6 m long, 5 m broad and 3.4 m high. Find the capacity of the tank in litres?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 23

Volume = 6 x 5 x 3.4

Volume = 102 cu. Cm

1 cu. m = 100 cu. Cm

102 cu. m = 102 x 100

= 10200 cu. Cm

1000 cu. cm = 1 litre

1 cu. Cm =

10200 cu. cm = x 10200

= 10.2 liters

RRB JE EEE (CBT I) Mock Test- 2 - Question 24
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The ratio of the area of a square to that of the square drawn on its diagonal is :
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 24
Let the side of the square be ‘α’

Its area = a2

Area of square on the diagonal = (√2a)2 = 2a2

Required ratio

RRB JE EEE (CBT I) Mock Test- 2 - Question 25
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If sec θ + tan θ = p, the

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 25

RRB JE EEE (CBT I) Mock Test- 2 - Question 26
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In the given figure, if ∠ABC = 90°, and ∠A = 30°, then ∠ACD=
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 26
∠ACD = ∠B + ∠A

= 90° + 30°

= 120° (exterior angle)

RRB JE EEE (CBT I) Mock Test- 2 - Question 27
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A certain sum of money yields Rs. 1261 as compound interest for 3 years at 5% per annum. The sum is
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 27
Let the principal , compound interest rate and time be x, R and T respectively

RRB JE EEE (CBT I) Mock Test- 2 - Question 28
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If 2x-1 + 2x+1 = 320, then x =?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 28

2x-1 + 2x+1 = 320

2x -1 (1 + 22) = 320

2x-1 = 320/5 = 64

2x-1 = 26

x - 1 = 6

x = 7

RRB JE EEE (CBT I) Mock Test- 2 - Question 29
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The value of (sin2 25° + sin2 65°) is
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 29
Sin2 25° + sin2 65 = sin2 25° + sin2 (90° - 25°)

= sin2 25° + cos2 25° = 1

RRB JE EEE (CBT I) Mock Test- 2 - Question 30
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A is twice as good as workman as B. Together, they finish the work in 14 days. In how many days can it be done by each separately?
Detailed Solution for RRB JE EEE (CBT I) Mock Test- 2 - Question 30
As per the question, A do twice the work as done by B.

So A:B = 2:1

Also (A+B)'s one-day work = 1/14

To get days in which B will finish the work,

let's calculate work done by B in 1 day

=(1/14)x(1/3) = 1/42

So B will finish the work in 42 days and A will finish the work in 21 days.

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