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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Test  >  RRB JE Mock Test Series for ECE 2026  >  RRB JE ECE (CBT I) Mock Test- 1 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 1 Free Online Test 2026


Full Mock Test & Solutions: RRB JE ECE (CBT I) Mock Test- 1 (100 Questions)

You can boost your Electronics and Communication Engineering (ECE) 2026 exam preparation with this RRB JE ECE (CBT I) Mock Test- 1 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of Electronics and Communication Engineering (ECE) 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 90 minutes
  • - Total Questions: 100
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Mathematics

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RRB JE ECE (CBT I) Mock Test- 1 - Question 1
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If P = sin θ (sin θ + sin 3θ), then

Detailed Solution: Question 1

We have

P = sin θ (sin θ + sin 3θ)

= sin θ (2sin 2θ. cos θ)

[∵ sin C + sin D = 2sin (C+D/2).cos (C−D/2)]

⇒ P = sin 2θ (2sin θ. cos θ)

= sin 2θ.sin 2θ = sin2 2θ

P = sin2 2θ ≥ 0

for all real θ

RRB JE ECE (CBT I) Mock Test- 1 - Question 2
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The value of the expression

Detailed Solution: Question 2

4(n+m-1+m-n-2m)x5(m-1+m+n-2-2m-n)x3(m-n+m+n-2-2m+2

= 4-1 x 5-3 x 30

= 1/4 x 1/125 x 1 = 1/500

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RRB JE ECE (CBT I) Mock Test- 1 - Question 3
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Perimeter of an equilateral triangle is equal to the circumference of a circle. The ratio of their areas is :

(Use π= 22/7)

Detailed Solution: Question 3

Accordingly to the question,

3a = 2πr

⇒ a = 2/3πr

Area of an equilateral triangle:

RRB JE ECE (CBT I) Mock Test- 1 - Question 4
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AB = 8 cm and CD = 6 cm are parallel chords on the same side of the centre of a circle. The distance between them is 1cm. The radius of the circle is:

Detailed Solution: Question 4

Let OE= x

OF = x+1, OA=OC= r

AE= 4cm, CF= 3cm

From ∆ OAE

r2 = 16+x2

x2 = r2- 16 ------(1)

From ∆ OCF

(x+1)2 = r2 - 9 ------ (2)

By equation (2) - (1)

x = 3cm

r = 5cm

RRB JE ECE (CBT I) Mock Test- 1 - Question 5
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A tank that would normally be filled in 8 hours is now taking 2 hours more because of a leak. In how much time will the leak empty the full tank?

Detailed Solution: Question 5

Time taken by the inlet pipe to fill the tank = 8 h

Time taken to fill the tank due to leak = 8 + 2 = 10 h

Let the leak take x hours to empty the filled tank.

Now, according to the question:

Time taken by leak to empty the tank = X = 40 hours

RRB JE ECE (CBT I) Mock Test- 1 - Question 6
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A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of interest be 10% compounded annually, then the value of each installment is

Detailed Solution: Question 6

Let x be the value of installment.

Principal = Present value of x for 1 year + Present value of x for 2 years

RRB JE ECE (CBT I) Mock Test- 1 - Question 7
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The difference between simple and compound interest for the fourth year is ₹ 7280 at 20% p.a. What is the sum?

Detailed Solution: Question 7

Difference between CI and SI for the nth year

CI for 4th year = 20736 - 17280 = 3456

SI for 4th year = 2000

Difference between CI and SI = 1456 for Rs10000

So, the difference of Rs7280 is for Rs50,000

RRB JE ECE (CBT I) Mock Test- 1 - Question 8
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At present, the sum of ages of R and K is 63 years. The ratio of their ages after 7 years will be 7:4; what is the present age of R?

Detailed Solution: Question 8

Let the present age of R is x

⇒ present age of K is 63 - x

7 years later

R's age = x+7

K's age = 63-x+7 = 70-x

⇒ (x+7)/(70-x) = 7/4

⇒ 4x+28 = 490-7x

⇒ 11x = 462

⇒ x = 42 years

therefore present age of R = 42 years

RRB JE ECE (CBT I) Mock Test- 1 - Question 9
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If cosec θ = x + 1/4x, then the value of cosec θ + cot θ is:

Detailed Solution: Question 9

Let cosec θ + cot θ = k

⇒ cosec θ - cot θ = 1/k

On adding, we get

2 cosec θ = k + 1/k = 2(x+1/4x) = k+1/k

⇒ 2x+1/2x = k+1/k ⇒ k = 2x

RRB JE ECE (CBT I) Mock Test- 1 - Question 10
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Base of a right pyramid is a square of area 324 sqm. If the volume of the pyramid is 1296 cu.m., then the area (in m2) of the slant surface is:

Detailed Solution: Question 10

RRB JE ECE (CBT I) Mock Test- 1 - Question 11
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In the given figure AB || CD || EF || GH and BH = 100 cm. Find the value of DF.

Detailed Solution: Question 11

AG/BH = CE/DF

120/100 = 30/DF ⇒ DF = 100×30/120 = 25

RRB JE ECE (CBT I) Mock Test- 1 - Question 12
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If sin (10° 6’ 32”) = a, then the value of cos (79° 53’ 28”) + tan (10° 6’ 32”) is:

Detailed Solution: Question 12

cos 790 53’ 28’’ + tan 100 + 6’ 32’’

RRB JE ECE (CBT I) Mock Test- 1 - Question 13
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In the figure AB || CD, ∠DPL= 1/2∠NPO, OP is the perpendicular on MN, the value of xº is:

Detailed Solution: Question 13

∠ONP = 60°

∠OPN = 90° - ∠ONP = 90° -60°

⇒ OPN = 30°

RRB JE ECE (CBT I) Mock Test- 1 - Question 14
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Find the area between two concentric circles of circumference 88 cm and 132 cm.

Detailed Solution: Question 14

Let the radius of the inner circle be r1, and the radius of the outer circle be r2.

A.T.Q.,

RRB JE ECE (CBT I) Mock Test- 1 - Question 15
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A man moving at a speed of 36 kmph covered the shadow of a tower of height 75 m in 3 seconds. If the height of the person is 1.75 m, the length of the shadow cast by the man in meters is:

Detailed Solution: Question 15

Distance travelled in 3 seconds with a speed of 36 kmph is =

(5/18)×(3)×(36) = 30m.

From the figure tan θ = BC/AB = 75/30 = 5/2

The height of the person is 1.75 m

Let the length of the shadow be x m.

tan θ = 1.75/x ⇒ 52 = 1.75/x

⇒ x = 1.75/5(2) = 0.7

RRB JE ECE (CBT I) Mock Test- 1 - Question 16
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A dress is 25% off. If both the cost and selling prices are ₹ 100 less, the profit would be 35%. Find its cost price?

Detailed Solution: Question 16

C.p = (Initial profit + increase in profit %)/Increase in profit%×Amount

C.P = 25+10/10×100

= ₹ 350

RRB JE ECE (CBT I) Mock Test- 1 - Question 17
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The speed of a car during the second hour of its journey is thrice that in the first hour. Also, its third-hour speed is the average speed of the first two hours. Had the car travelled at the second hour’s speed during all the first three hours, then it would have travelled 150 km more. Find the percentage reduction in time if the original distance covered by the second hour’s speed:

Detailed Solution: Question 17

∴ Percentage increase in speed = 3x/6x × 100 = 50%

Since speed is increased by (50%) ½

Therefore, time will reduce by (33.33%) 1/3

RRB JE ECE (CBT I) Mock Test- 1 - Question 18
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If A+B = 45°, then find the value of tan A + tan B + tan A tan B.

Detailed Solution: Question 18

A+B = 45°

tan (A+B) = tan 45°

tan A + tanB + tanAtanB = 1

RRB JE ECE (CBT I) Mock Test- 1 - Question 19
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If a + b +c = 13, what is the maximum value of (a -3)(b -2) (c + 1) ?

Detailed Solution: Question 19

If x+y +z is constant, the product xyz takes maximum value when each of x, y, z takes equal value.

∵ a + b +c = 13

∴ (a -3) + (b -2) + (c + 1) = 13 – 3 – 2 + 1 = 9

For the maximum value of (a -3)(b -2) (c + 1)

= (a -3)= (b -2)= (c + 1)= 9/3 = 3

So, (a -3)(b -2) (c + 1) = 3 × 3 × 3 = 27

RRB JE ECE (CBT I) Mock Test- 1 - Question 20
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The length of three medians of a triangle are 9 cm, 12 cm and 15 cm. The area (in sq. cm) of the triangle is:

Detailed Solution: Question 20

Area of triangle = Area of a triangle formed by taking median as a side of the triangle

A = 9, B = 12, C = 15

RRB JE ECE (CBT I) Mock Test- 1 - Question 21
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The floor of a room is of size 4 m x 3 m, and its height is 3 m. The walls and ceiling of the room require painting. The area to be painted is

Detailed Solution: Question 21

Area of the four walls of the room

= 2 x height (length + breadth)

= 2 x 3(4+3) = 42 sq. metre

Area of ceiling = 4 x 3

= 12 sq. metre

Required area = 42 + 12

= 54 sq. metre

RRB JE ECE (CBT I) Mock Test- 1 - Question 22
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If 5 girls can embroider a dress in 9 days, then the number of days taken by 3 girls will be _____.

Detailed Solution: Question 22

Days taken by 5 girls to embroider a dress = 9 days

Days taken by 1 girl to embroider a dress = 9 x 5 days

Days taken by 3 girls to embroider a dress = 9 × 5/3 days = 15 days

RRB JE ECE (CBT I) Mock Test- 1 - Question 23
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How many metres length of cloth 10 m wide will be required to make a conical tent with base radius of 14 m and height is 48 m?

Detailed Solution: Question 23

Slant height of a cone: ( l ) = √(r2 + h2)

= √(142 + 482)

= √196 + 2304

= √2500

l = 50 m

Lateral surface area of a cone = πrl

Hence length of cloth =

= 220 m

RRB JE ECE (CBT I) Mock Test- 1 - Question 24
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Two trains are travelling in the same direction at 40 kmph, and 22 kmph completely pass off another in 1 minute. If the length of the first train is 125 metres, what is the length of the second train?

Detailed Solution: Question 24

Distance traveled by the relative speed of (40-22) kmph i.e. 18 kmph

In 1 minute= 18×1000/60 = 300 m

∴ Length of second train = (300-125) = 175 m

RRB JE ECE (CBT I) Mock Test- 1 - Question 25
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The ratio between the ages of a father and a son at present is 5: 2, respectively. Four years hence the ratio between the ages of the son and his mother will be 1: 2 respectively. What is the ratio between the present ages of the father and the mother, respectively?

Detailed Solution: Question 25

Let the present ages of father and son be ‘5x’ and ‘2x’ years, respectively.

After 4 years, Son’s age = (2x + 4) years

According to question

After 4 years

The ratio between the present age of the father and mother = 5x: (4x + 4)

∴ We cannot find the exact value of the ratio.

Age of mother = 2 x (2x+ 4) = 4x+ 8 years

Present age of Mother = 4x + 8 - 4

= 4x +4

The ratio between the present age of the father and mother = 5x: (4x + 4)

∴ We cannot find the exact value of the ratio.

RRB JE ECE (CBT I) Mock Test- 1 - Question 26
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If x4 + 1/x4 =119 and x > 1, then the value of x3 − 1/x3 is:

Detailed Solution: Question 26

RRB JE ECE (CBT I) Mock Test- 1 - Question 27
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The total annual Co2 emissions from various sectors are 5 MMT. In the pie chart given below, the percentage distribution to Co2 emission from various sectors are indicated.

Q. Emission of Domestic sector is how much % of transport and commercial sector combined?

Detailed Solution: Question 27

15/50 × 100 = 30%

RRB JE ECE (CBT I) Mock Test- 1 - Question 28
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The total annual Co2 emissions from various sectors are 5 MMT. In the pie chart given below, the percentage distribution to Co2 emission from various sectors are indicated.

Q. Which of the following sectors have an emission difference of 1 MMT between them?

Detailed Solution: Question 28

1 mmt = 20% = Thermal – Domestic

RRB JE ECE (CBT I) Mock Test- 1 - Question 29
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Twenty percent of Anuj’s annual salary is equal to seventy-five per cent of Raj’s annual salary. Raj’s monthly salary is 60% of Ravi’s monthly salary. If Ravi’s annual salary is Rs. 1.44 lacs, what is Anuj’s monthly salary?

Detailed Solution: Question 29

Given:

20% of Anuj's annual salary = 75% of Raj's annual salary

Raj's annual salary = 60% of Ravi's monthly salary

Ravi's annual salary = Rs. 1.44 lakhs = Rs. 144000

Concept:

20% = 1/5, 75% = 3/4, 60% = 3/5

Calculation:

Ravi's monthly salary = 144000/12 = 12000

60% of Ravi's monthly salary = Raj's annual salary

⇒ (3/5) × 12000 = Rs. 7200 = Raj's annual salary

Now, 1/5 of Anuj's annual salary = 3/4 of 7200 = 5400

∴ Anuj's annual salary = 5 × 5400 = Rs. 27000

RRB JE ECE (CBT I) Mock Test- 1 - Question 30
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The product of two successive numbers is 4032. Which is the greater of the two numbers?

Detailed Solution: Question 30

Let the smaller integer be n. Then the next consecutive integer is n + 1. The product equation is:

n(n + 1) = 4032. This simplifies to the quadratic equation:

n2 + n - 4032 = 0.

Using the quadratic formula:

n = (-b ± √(b2 - 4ac)) / (2a)

  • a = 1
  • b = 1
  • c = -4032

Substituting these values into the formula:

n = (-1 ± √(1 + 16128)) / 2 = (-1 ± 127) / 2.

Taking the positive root:

n = 126 / 2 = 63.

Thus, the greater number is 63 + 1 = 64.

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About this Electronics and Communication Engineering (ECE) Mock Test

This mock test contains 100 questions designed to replicate the pattern, difficulty level and time pressure of the actual Electronics and Communication Engineering (ECE) exam. It is part of RRB JE Mock Test Series for ECE 2026 on EduRev, covering all the topics that are being asked as per the current Electronics and Communication Engineering (ECE) syllabus. Attempting this test under timed conditions will help you assess your exam readiness, identify subject-wise weak areas, and improve your speed and accuracy before the actual exam.

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If you found this RRB JE ECE (CBT I) Mock Test- 1 helpful, you can further enhance your Electronics and Communication Engineering (ECE) preparation with complete mock tests, previous year questions & practice tests available in RRB JE Mock Test Series for ECE 2026 course on EduRev. The course offers:
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