Explore Exams
Login Signup
Sign in Open App
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- 7 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 7 Free Online Test 2026


Full Mock Test & Solutions: RRB JE ECE (CBT I) Mock Test- 7 (100 Questions)

You can boost your Electronics and Communication Engineering (ECE) 2026 exam preparation with this RRB JE ECE (CBT I) Mock Test- 7 (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

Sign up on EduRev for free and get access to these mock tests, get your All India Rank, and identify your weak areas to improve your marks & rank in the actual exam.

RRB JE ECE (CBT I) Mock Test- 7 - Question 1
Save

The ratio of the number of Matches played by Rohit and Kohli is 6:5. Kohli scored an average of 52 runs per match, and Rohit's average is 25% less than that. If the total runs scored by Kohli is 546 more than Rohit, what is the number of matches Kohli has played?

Detailed Solution: Question 1

Let the number of matches played by Rohit and Kohli be 6k and 5k respectively.

Average runs scored in a match by Rohit = 52(1 - 25/100) = 39

Given, 52*5k - 39*6k = 546

=> 26k = 546

=> k = 21

Number of Matches played by Kohli = 5*21 = 105

RRB JE ECE (CBT I) Mock Test- 7 - Question 2
Save

What is the value of

Detailed Solution: Question 2

Clear all your doubts with EduRev
RRB JE ECE (CBT I) Mock Test- 7 - Question 3
Save

For what value of 'k', the quadratic equation 4x2 - 3m√2x + k = 0 (where, m>0) has equal roots? (Given: 2m2 + 7m - 60 = 0)

Detailed Solution: Question 3

For 2m2 + 7m - 60 = 0:

2m2 - 8m + 15m - 60 = 0

m = -15/2,

Quadratic equation 4x2 - 3m√2x + k = 0 has equal roots, therefore,

D = 0 = (-3m√2)2 - 4 x 4 x k

For m = 4:

18 x (4)2 = 16 x k

k = 18

RRB JE ECE (CBT I) Mock Test- 7 - Question 4
Save

Study the following table carefully and answer the related question.

The following table represents the marks of the top four students of the class in four subjects.

Who got approximately 71.5% percentage in the class?

Detailed Solution: Question 4

Total marks = 100 + 100 + 150 + 150 = 500

Total marks of Rakhi = 78 + 60 + 96 + 124 = 358

Rakhi's percentage = (358/500) x 100 = 71.6%

Total marks of Jay = 56 + 80 + 88 + 92 = 316

Jay's percentage = (316/500) x 100 = 63.2%

Total marks of Amit = 95 + 90 + 80 + 75 = 340

Amit's percentage = (340/500) x 100 = 68%

Total marks of Urvi = 70 + 70 + 120 + 105 = 365

Urvi's percentage = (365/500) x 100 = 73%

RRB JE ECE (CBT I) Mock Test- 7 - Question 5
Save

A sum of money at simple interest amounts to Rs 2240 in 2 years and to Rs 2600 in 5 years. What is the principal amount?

Detailed Solution: Question 5

Interest for 3 years = Rs 2600 - Rs 2240 = Rs 360

=> Interest for 2 years = 360x2/3 =Rs 240

Hence, principal = (2240 - 240) = Rs 2000

RRB JE ECE (CBT I) Mock Test- 7 - Question 6
Save

Number of kids in 10 families is given below. Find the median number of kids.

3, 4, 2, 1, 0, 2, 4, 0, 1 and 2

Detailed Solution: Question 6

Ascending order of number of kids in the family-

0, 0, 1, 1, 2, 2, 2, 3, 4, 4

Total terms = 10 = Even

Median = [(n/2)th term + {(n/2) + 1} th term]/2

=> [(10/2)th term + {(10/2) + 1}th term]/2

RRB JE ECE (CBT I) Mock Test- 7 - Question 7
Save

There are four persons A, B,C and D, and the ratio of the present age of A to C is 4: 3, C is 7 years younger to E who is 16 years old at present. Present age of B is 75% of the present age of D and B is 6 years older than A. What is the average present age of persons B, C and D?

Given below are the steps involved. Arrange them in the sequential order.

Let the age of A = 4A

(A) The ages of A and C are 12 years and 9 years respectively.

(B) Present age of E = (3A + 7) = 16 => A = 3

(C) Required average = (18 + 9 + 24)/3 = 17 years

(D) Present age of B = 12 + 6 = 18 years and present age of D = 18 * (100/75) = 24 years

(E) Present age of C = 4A * (3/4) = 3A

Detailed Solution: Question 7

The correct order is

(E) Present age of C = 4A * (3/4) = 3A

(B) Present age of E = (3A + 7) = 16 => A = 3

(A) Age of A and C is 12 years and 9 years respectively

(D) Present age of B = 12 + 6 = 18 years and present age of D = 18 * (100/75) = 24 years

(C) Required average = (18 + 9 + 24)/3 = 17 years

RRB JE ECE (CBT I) Mock Test- 7 - Question 8
Save

The average age of a man and his son is 54 years. The ratio of their ages is 23:13. What will be the ratio of their ages after 6 years?

Detailed Solution: Question 8

Let their ages be 23x and 13x respectively.

Average age is 54 years which means the sum of their ages = 108 years

i.e. 36x = 108

or x = 3

Hence their ages are 69 and 39 years respectively.

After 6 years, their ages would be 75 and 45 years

So the required ratio = 75:45 = 5:3

RRB JE ECE (CBT I) Mock Test- 7 - Question 9
Save

Out of 5 numbers, whose average is 36, the first one is one fifth of the sum of the last four. The first number is?

Detailed Solution: Question 9

Let the five numbers be a, b, c, d and e

Given, (a+b+c+d+e)/5=36

a+b+c+d+e = 180

a = (b+c+d+e)/5

a = (180-a)/5

a = 30

RRB JE ECE (CBT I) Mock Test- 7 - Question 10
Save

What value we get when: √(92 - 43 + 3) is multiplied by √(25 * 10-1)?

Detailed Solution: Question 10

= √(92 - 43 + 3) * √(25 * 10-1)

= √(81 - 64 + 3) * √(32 ÷ 10)

= √20 * √3.

= √(20 * 3.2)

= √64

= 8

RRB JE ECE (CBT I) Mock Test- 7 - Question 11
Save

Following graph shows the number of questions attempted and the number of correct questions in preliminary and mains exams by 5 persons.

{[376.PNG}}

What is the ratio of the total number of non-correct questions out of total number of questions attempted by all 5 persons in preliminary exam to the total number of non-correct questions out of total number of questions attempted by all 5 persons in mains exam?

Detailed Solution: Question 11

Total number of non-correct questions out of total number of questions attempted by all 5 persons in preliminary exam

= (10 +5 +5 +10 +5)

= 35.

Total number of non-correct questions out of total number of questions attempted by all 5 persons in mains exam

= (30 +15 +11 +30 +40)

= 126

Required ratio= 35/126

RRB JE ECE (CBT I) Mock Test- 7 - Question 12
Save

If n is the greatest number that will divide the three numbers 1236, 3708 and 4944 leaving the same remainder in each case. What is the product of digits of 'n'?

Detailed Solution: Question 12

n = HCF of (3708 - 1236), (4944 - 3708) and (4944 - 1236)

= HCF of 2472, 1236 and 3708

= 1236

Product of digits of n = 1 x 2 x 3 x 6 = 36

RRB JE ECE (CBT I) Mock Test- 7 - Question 13
Save

LCM of three numbers is 840 and their ratio is 2: 3: 5 respectively. What will be the HCF of these numbers?

Detailed Solution: Question 13

Let the three numbers be 2a, 3a and 5a respectively.

According to the question:

840 = 2a x 3a x 5a

a = 28

Therefore, the numbers are (2 x 28), (3 x 28) and (5 x 28) respectively.

Hence, HCF of these numbers = 28

RRB JE ECE (CBT I) Mock Test- 7 - Question 14
Save

A discount of 30% on one article is the same as a discount of 35% on another article. The marked prices of two articles can be (in ₹.)

Detailed Solution: Question 14

Since discounts are 30 % and 35 % respectively, Therefore, discounts are 0.3x and 0.35y, where x and y are the marked prices respectively.

Further, it is given that these discounts are equal, so 0.3x = 0.35y=> x: y = 7: 6

i.e. marked price ratio is 7:6.

Now, check for all the given options.

RRB JE ECE (CBT I) Mock Test- 7 - Question 15
Save

A boat takes 4 more hours to travel 180 km upstream than to travel the same distance downstream. If the still water speed of the boat is 28 km/hr, what is the stream speed?

Detailed Solution: Question 15

Let the stream speed be 'r' km/hr.

180/(28 - r) = 4 + 180/(28 + r

45/(28 - r) - 45/(28 + r) = 1

90r/(784 - r2) =

r2 + 90r - 784 = 0

(r - 8)(r + 98) = 0

r = 8

RRB JE ECE (CBT I) Mock Test- 7 - Question 16
Save

The ratio of heights of Asif and Bharat is 3:4 and the ratio of heights of Bharat and Chatur is 6:7. If the average of their heights is 175 cm, what is the height of Asif?

Detailed Solution: Question 16

Ratio of heights of Asif and Bharat = 3:4 = 9:12

Ratio of heights of Bharat and Chatur = 6:7 = 12:14

Ratio of heights of Asif, Bharat and Chatur = 9:12:14

Sum of parts of the ratio = 9 + 12 + 14 = 35

Asif's height =(9/35)*175*3 = 135 cm

RRB JE ECE (CBT I) Mock Test- 7 - Question 17
Save

A alone can do a piece of work in 20 days, B alone in 40 days and C alone in 60 days. Everyone begins to do the work together, but A leaves after 4 days and B leaves 6 days before the completion of the work. How many days did the work last?

Detailed Solution: Question 17

Total work in 1 day by A, B and C = 1/20 + 1/40 + 1/60 = 11/120

In 4 days = 44/120

By C in last 6 days = 6/60 = 1/10

Total = 44/120 + 1/10 = 56/120

Work left = 1 - 56/120 = 64/120

Work done by B and C in one day = 1/40 + 1/60 = 5/120

Work left is done by B and C. Number of days it takes = (64/120)*(120/5) = 64/5

Total time = 10 + 64/5 = 22 4/5

RRB JE ECE (CBT I) Mock Test- 7 - Question 18
Save

A faulty watch gains 5 minutes every 40 minutes. After how many days will it show the correct time again?

Detailed Solution: Question 18

It will show the correct time again when it gains 24 hours.

It gains 5 minutes in 40 minutes

It gains 1 minute in 8 minutes

It gains 1 hr in 480 minutes

It gains 24 hr in 480*24 minutes or (480*24)/(60*24) or 8 days

RRB JE ECE (CBT I) Mock Test- 7 - Question 19
Save

If x/y + y/x + 1 = 0, what is the value of (x - y)6/9x3y3?

Detailed Solution: Question 19

x/y + y/x + 1 = 0

=> x2 + y2 + xy = 0

=> x2 - 2xy + y2 + 3xy = 0

=> (x - y)2 = −3xy

=> (x - y)6 = −27x3y3

(x - y)6/9x3y3

∴ = (−27x3y3)/9x3y3

= −3

RRB JE ECE (CBT I) Mock Test- 7 - Question 20
Save

A TV costs Rs.20000 and the shopkeeper announces that the price of the TV will reduce at 10% per annum. What will be the cost price of the item after 2 years?

Detailed Solution: Question 20

Whenever the dimensions of the quantity are decreased by 'x' percentage then the overall percentage change = (-x%)+(-x%)+-x%)(- x%/100

Overall change in the price of the TV = -10-10+100/100=-19%

Cost Price of the TV after two years = 20000-(19/100 X 20000)=Rs.16200

RRB JE ECE (CBT I) Mock Test- 7 - Question 21
Save

In the following diagram ABCD is a square inscribed in a circle of centre O. If perimeter of the square ABCD is 40 cm, then what is the area of the blue portion?(take π = 3.14)

Detailed Solution: Question 21

Side of the square = Perimeter/4 = 40/4 =10 cm

Diagonal AC of square = Side x √2 =10√2 cm

AC is also the diameter of a circle.

Radius of circle = 5√2 cm

Area of circle = 3.14 x (5√2)2 = 3.14 x 50 = 157 cm2

Area of square = 100 cm2

Area of blue portion = 157-100 = 57 cm2.

RRB JE ECE (CBT I) Mock Test- 7 - Question 22
Save

Ramesh salary got 2 salary hikes, one of 10%, the next one of 25%. If the total increase in Ramesh salary was 24000, what is the increase in his salary from the first hike to the second Hike?

Detailed Solution: Question 22

Let Ramesh's initial salary be Rs 100x

His salary after 1st Hike = (110/100)*100x = Rs 110x

His salary after 2nd Hike = (125/100)*110x = Rs 137.5x

Given 137.5x - 100x = 24000

=> 37.5x = 24000

=> x = 640

increase in his salary from the first hike to the second Hike

= 137.5x - 110x

= 27.5x

= Rs 17,600

RRB JE ECE (CBT I) Mock Test- 7 - Question 23
Save

Which of the following is irrational?

Detailed Solution: Question 23

Option (1):

= Product of 3rd root of 2197 and √6889

= 13 x 83

= 1079

Option (2):

= Product of 4th root of 4096 and √1377

= 8 x 9√17

= 72√17 (irrational)

Option (3):

= Product of 3rd root of 3375 and √5929

= 15 x 77

= 1155

Option (4):

= Product of 4th root of 2401 and √1444

= 7 x 38

= 266

RRB JE ECE (CBT I) Mock Test- 7 - Question 24
Save

A car wiper can clean 693 sq. inch area in its one sweep. If its length is decreased by 3.5 inch, then how much area will it clean in one sweep? (Car wiper covers a semi-circle in one sweep)

Detailed Solution: Question 24

Let original length a car wiper be L inch.

Area cleaned by the car wiper in 1 sweep

= (1/2) x (22/7) x (L)2

L2 = 693 x (7/11) = 63 x 7

L = 21 inch.

Area cleaned by car wiper in a sweep if its length decreased by 3.5 inch.

= (1/2) x (22/7) x (17.5) x (17.5)

= 481.25 sq. inch.

RRB JE ECE (CBT I) Mock Test- 7 - Question 25
Save

A bus reached Nagpur to Amravati in 2:30 hr with an average speed of 70 km/hr. If the average speed increased by 30 km/hr. How much time it will take to cover the same distance.

Detailed Solution: Question 25

Distance = speed *time

= (70 * 150)/60 min

= 175 km

New speed = (70 + 30) km/hr = 100 km/hr

Time = (Distance )/speed

= (175 * 60)/100

= 105 min

RRB JE ECE (CBT I) Mock Test- 7 - Question 26
Save

Two poles of equal height are on either side of a 100 m wide road. From a point on the road, the angles of elevation of their tops are 30° and 60°. Find the height of each pole.

Detailed Solution: Question 26

Let the height of poles be h m.

Tan 60 = h/x

=> √3 = h/x

Or h = √3 x

Similarly,

1/√3 = h/(100-x)

=> 100-x = √3 h = 3x

=> x = 25

Therefore, h = 25√3 m

RRB JE ECE (CBT I) Mock Test- 7 - Question 27
Save

The sales for the month of January was 500 units. Sales increased by 50 units every month for the next 11 months. Find the average monthly sales for the year?

Detailed Solution: Question 27

Total sales for the year = 500 + 550 + 600 + 650 +700 +750 + 800 +850 +900+ 950 + 1000 +1050 = 9300

Average monthly sales for the year = 9300/12 = 775

RRB JE ECE (CBT I) Mock Test- 7 - Question 28
Save

Select the odd one from the following options.

a = 2.5 x 2.4 - 8 x 0.5

b = 12 x 4 x 0.5 - 16

c = 21 - 30 x 0.5

Detailed Solution: Question 28

a = 2.5 x 2.4 - 8 x 0.5 = 2

b = 12 x 4 x 0.5 - 16 = 8

c = 21 - 30 x 0.5 = 6

RRB JE ECE (CBT I) Mock Test- 7 - Question 29
Save

A grocer bought two types of rice A and B, at Rs.20 per kg and Rs.25 per kg. If no profit and no loss occurs on selling both the items together at Rs.24 per kg, then find the ratio of quantity of rice A and B.

Detailed Solution: Question 29

Let quantity of rice A and B are a and b respectively.

Total selling price = 20a + 25b

Total cost price = 24a + 24b

No profit, no loss occurred. So,

Total selling price = total cost price

20a + 25b = 24a + 24b

4a = b

a: b = 1: 4

RRB JE ECE (CBT I) Mock Test- 7 - Question 30
Save

In an election contested by two candidates, a candidate who got 54% of the total votes won by a margin of 288 votes. Total votes are.

Detailed Solution: Question 30

Let total number of votes cast = x

54% of x = 46% of x + 288

8% of x = 288

x = 288 x 100/8 = 3600. Hence, the answer option is (2).

View more questions
161 tests
Join Course
Information about RRB JE ECE (CBT I) Mock Test- 7 Page
In this test you can find the Exam questions for RRB JE ECE (CBT I) Mock Test- 7 solved & explained in the simplest way possible. Besides giving Questions and answers for RRB JE ECE (CBT I) Mock Test- 7, EduRev gives you an ample number of Online tests for practice
Download as PDF

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.

Explore more Electronics and Communication Engineering (ECE) Resources

If you found this RRB JE ECE (CBT I) Mock Test- 7 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:
  • Video Lectures: covering all the topics of RRB JE Mock Test Series for ECE 2026
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for Electronics and Communication Engineering (ECE) preparation
  • MCQ Test: variety of questions as per the pattern of Electronics and Communication Engineering (ECE) exam
Signup to see your scores go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup