Explore Exams
Login Signup
Sign in Open App
Mechanical Engineering Exam  >  Mechanical Engineering Test  >  RRB JE Mock Test Series (ME) 2027  >  RRB JE ME (CBT I) Mock Test- 1 - Mechanical Engineering MCQ

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


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

You can boost your Mechanical Engineering 2026 exam preparation with this RRB JE ME (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 Mechanical Engineering 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 ME (CBT I) Mock Test- 1 - Question 1
Save

Two trains, 165 m and 175 m long run at the speed of 50 km/h and 52 km/h respectively in opposite directions on parallel tracks. What is the time (in seconds) they take to cross each other?

Detailed Solution: Question 1

Relative speed = (50 + 52) km/h = (102 x 5/18) m/s = (255/9) m/s.

Distance covered in crossing each other = (165 + 175) = 340 m.

∴ Required time = (340 x 9/255) = 12 s.

RRB JE ME (CBT I) Mock Test- 1 - Question 2
Save

If two equations: 2x2 - Ax - 20 = 0 and 2x2 - 11x + 12 = 0 have one root in common, then what is the value of 'A' if the value of A is positive?

Detailed Solution: Question 2

2x2 - 11x + 12 = 0

(2x - 3)(x - 4) = 0

x = 1.5 and 4

Since both the equations have one root in common-

Case 1: When x = 1.5 is common-

2(1.5)2 - A(1.5) - 20 = 0

Clear all your doubts with EduRev
RRB JE ME (CBT I) Mock Test- 1 - Question 3
Save

Divya has 3 types of coins having denomination of 20p, 10p and 5p. Their numbers are in the ratio of 1: 2 : 3. If Divya has total money of Rs. 44 then how many 5p coins are there?

Detailed Solution: Question 3

Since the ratio between the numbers of coins of 20p, 10p and 5p is 1: 2 : 3

So let us assume the number of 20p coins = a

Then the number of 10p coins is 2a and the number of 5p coins is 3a.

Total worth of 20p coins = number of coins X worth of each coin = a X (1/5) = Rs (a/5)

Total worth of 10p coins = 2a X (1/10) = Rs (a/5)

Similarly total worth of 5p coins = 3a X (1/20) = Rs (3a/20)

Since total money with the Divya is Rs 44 hence

a/5 + a/5 + 3a/20 = 44 => 2a/5 + 3a/20 = 44

=> 8a/20 + 3a/20 = 44 => 11a/20 = 44 => a = 80

So the number of 5p coins = 3a = 80 x 3 = 240

RRB JE ME (CBT I) Mock Test- 1 - Question 4
Save

The difference between 42% and 23% of a number is 304. What is 72% of that number?

Detailed Solution: Question 4

Let the number be 100x.

∴ 42x - 23x = 304 => 19x = 304 => x = 16

∴ The number is 100 x 16 = 1600.

=> 72% of the number = 0.72 x 1600 = 1152

RRB JE ME (CBT I) Mock Test- 1 - Question 5
Save

What is the value of 60 ÷ [{21 ÷ 7 x √(√625 + √121)} − 6] x 5 - 42 ÷ 3 x 2 = ?

Detailed Solution: Question 5

60 ÷ [{21 ÷ 7 x √(√625 + √121)} − 6] x 5 - 42 ÷ 3 x 2

= 60 ÷ [{3 x √(25 + 11)} − 6] x 5 - 14 x 2

= 60 ÷ [{3 x √(36)} − 6] x 5 - 28

= 60 ÷ [{3 x 6} − 6] x 5 - 28

= 60 ÷ [18 − 6] x 5 - 28

= 60 ÷ 12 x 5 - 28

= 5 x 5 - 28

= -3

RRB JE ME (CBT I) Mock Test- 1 - Question 6
Save

Three pipes x, y and z can fill a water tanker in 20, 25 and 30 minutes respectively. All the three pipes are opened to fill the tanker, and if the pipe y is closed 2 minutes before the water tanker gets filled, then in how much time will the tank get filled?

Detailed Solution: Question 6

Correct option: C. The tank is filled in 324/37 minutes.

Let the total filling time be t minutes.

Pipe x fills 1/20 part of the tank per minute, pipe y fills 1/25 part per minute, and pipe z fills 1/30 part per minute.

Since pipe y is closed 2 minutes before the tank is full, pipes x and z run for t minutes each, while pipe y runs for t - 2 minutes.

t × (1/20) + (t - 2) × (1/25) + t × (1/30) = 1

Combine the rates of pipes x and z: 1/20 + 1/30 = 1/12. So the equation becomes t × (1/12) + (t - 2) × (1/25) = 1.

Multiply both sides by 300 (LCM of 12 and 25) to clear denominators: 25t + 12(t - 2) = 300.

Simplify: 25t + 12t - 24 = 300 → 37t = 324.

Therefore t = 324/37 minutes, which verifies option C.

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

If [4x - (1/12x) = 8], then what is the value of [x2 + { /(2304 x 2)}]?

Detailed Solution: Question 7

4x - 1/12x = 8

=> 4(x - 1/48x) = 8

=> x - 1/48x = 2 ------ (i)

Squaring both sides, we get

(x - 1/48x)2 = 4

=> x2 + 1/(2304x2) - 2x/48x = 4 ------ [(a - b)2]

= a2 + b2 - 2ab] x2 + 1/2304 x2

= 4 + 1/24 = 97/24

RRB JE ME (CBT I) Mock Test- 1 - Question 8
Save

A boy throws a ball of mass 4 kg vertically downwards for a height of 20 metres. What is the velocity of the ball which touches the ground? (Take g= 10 m/s2)

Detailed Solution: Question 8

According to the question,

Mass of ball = m = 4 kg, elevation from ground = h = 20 m; v = velocity at touch point

So, mgh = ½ (mv2)

Gives v = 20 m/s

mg = 1/2 (mv2)

2gh = v2

v2 = 2 x 10 x 20 = 400

v = 20 m/sec

RRB JE ME (CBT I) Mock Test- 1 - Question 9
Save

Express 0.78414141.. into (p/q) form.

Detailed Solution: Question 9

Let x = 0.78414141..

=> 100x = 78.414141.. (I)

=> 10000x = 7841.414141.. (II)

(II) - (I) gives, 9900x = 7763

=> x = 7763/9900

RRB JE ME (CBT I) Mock Test- 1 - Question 10
Save

Nidhi borrowed Rs. 10000 from a bank for two years at the rate of 10% per annum. From that amount, she invested Rs. 3500 in another bank which offers simple interest @12.5% per annum, and the remaining amount she invested into a chit fund which offers compound interest of 10% per annum. What is the profit obtained by Nidhi at the end of two years?

Detailed Solution: Question 10

Total interest that Nidhi has to pay after 2 years = Rs. (10000 x 2 x 10)/100 = Rs. 2000

Interest obtained by Nidhi from the bank = Rs. (3500 x 2 x 12.5)/100 = Rs. 875

Interest obtained by Nidhi from Chit fund = 6500(1 + 0.1)2 - 6500 = Rs. 1365

So total interest obtained by Nidhi = Rs. (875 + 1365) = Rs. 2240

Profit = Rs. (2240 - 2000) = Rs. 240

RRB JE ME (CBT I) Mock Test- 1 - Question 11
Save

A book was marked initially at Rs. 500 which A purchased at 8% discount. He marked up the price by 15% on his purchased price and sold after Rs. 20 discount to B. B marked up the price by 20% above his purchased price and sold to C after allowing Rs. 50.9 discount. What is the percent profit gained by B when sold to C?

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

(A) Initial marked price of book = Rs. 500 and purchased price for A = 92% of 500 = Rs. 460

(B) Purchased price of book by B = 529 - 20 = Rs. 509

(C) Per cent profit = [(559.9 - 509)/509] x 100 = 10%

(D) Marked price of book by A = 115% of 460 = Rs. 529

(E) Marked price of book by B = 120% of 509 = Rs. 610.8 and purchased price of book by C = 610.8 - 50.9 = Rs. 559.9

Detailed Solution: Question 11

The correct order is:

(A) Initial marked price of book = Rs. 500 and purchased price for A = 92% of 500 = Rs. 460

(D) Marked price of book by A = 115% of 460 = Rs. 529

(B) Purchased price of book by B = 529 - 20 = Rs. 509

(E) Marked price of book by B = 120% of 509 = Rs. 610.8 and purchased price of book by C = 610.8 - 50.9 = Rs. 559.9

(C) Per cent profit = [(559.9 - 509)/509] x 100 = 10%

RRB JE ME (CBT I) Mock Test- 1 - Question 12
Save

Seema purchased an item for Rs.9,600 and sold it for a loss of 5 percent. From that money, she bought another item and sold it for a gain of 5 percent. What is her overall gain/ loss?

Detailed Solution: Question 12

SP = 9600 x 95/100 = Rs.9,120

Second S.P. = 9120 x 105/100 = Rs.9,576

Loss = 9600 - 9576 = 24

RRB JE ME (CBT I) Mock Test- 1 - Question 13
Save

What is the angle between the hands of a clock at 8:30 a.m.?

Detailed Solution: Question 13

At 8:30 a.m., minute hand points at 6 and hour hand points exactly between 8 and 9.

So these two hands are apart by 2.5 blocks.

We know that the difference in angle of one block is 30°.

Hence the angle between the hands = 2.5 X 30° = 75°

RRB JE ME (CBT I) Mock Test- 1 - Question 14
Save

Following data given below shows the sex ratio of India in the last 10 years. Find the mode of sex ratio in India.

976, 979, 1084, 993, 1084, 993, 935, 979, 991, 979

Detailed Solution: Question 14

The mode is a statistical term that refers to the most frequently occurring number found in a set of numbers.

Since 979 is frequently occurring, and the number of occurrence of 979 is 3.

Hence the mode of sex ratio in India in the last 10 years is 979.

RRB JE ME (CBT I) Mock Test- 1 - Question 15
Save

If 7x5 - 2x4 + 3x3 - 4x2 + 6x + 2k is divisible by (x + 1), then what is the value of k?

Detailed Solution: Question 15

Since, 7x5 - 2x4 + 3x3 - 4x2 + 6x + 2k is divisible by (x + 1)

=> x = -1 must satisfy the given equation

=> 7(-1)5 - 2(-1)4 + 3(-1)3 - 4(-1)2 + 6(-1) + 2k = 0

=> -7 - 2 - 3 - 4 - 6 + 2k = 0

=> 2k = 22

=> k = 11

RRB JE ME (CBT I) Mock Test- 1 - Question 16
Save

The ratio of two numbers is 7:5, and their H.C.F. is 6. What is their L.C.M.?

Detailed Solution: Question 16

Let the numbers be 7x and 5x. As 7 and 5 are co-prime pairs.

∴ Their H.C.F. = x = 6 (given).

=> The numbers are 42 and 30.

L.C.M. of 42 and 30 = 210.

RRB JE ME (CBT I) Mock Test- 1 - Question 17
Save

From an external point O, a secant is drawn on a circle that meets the circle at points A and B, respectively, if OA = 4 cm and AB = 60 cm. The length of the tangent drawn on the circle from point O is (in cm)

Detailed Solution: Question 17

Let the tangent be T

OT2 = OA.OB

=> OA = 4

=> OB = OA + AB

=> 4 + 60

=> 64

OT2 = 4 x 64

=> 256

=> OT = 16 cm

RRB JE ME (CBT I) Mock Test- 1 - Question 18
Save

A copper wire, when bent in the form of a square, enclose an area of 121 cm2. If the same wire is bent in the form of a circle, then the area of the circle is

Detailed Solution: Question 18

Side of square = √121 = 11 cm

Length of wire = circumference of circle.

44 = 2 πr or r = (44 x 7)/(2 x 22) = 7 cm

Hence, area of circle = πr2 = 22/7 x 7 x 7 = 154 cm2

RRB JE ME (CBT I) Mock Test- 1 - Question 19
Save

Table given below shows the total number of employees of four different grades in a department and the average age of employees in those different grades.


What will be the average age of the employees of department P

Detailed Solution: Question 19

Sum of ages of all employees in grade A = 25 x 28 = 700 years

Sum of ages of all employees in grade B = 40 x 24 = 960 years

Sum of ages of all employees in grade C = 15 x 32 = 480 years

Sum of ages of all employees in grade D = 20 x 22 = 440 years

Sum of ages of all the employees of the department P = 700 + 960 + 480 + 440 = 2580 years

Total employees in department P = 25 + 40 + 15 + 20 = 100

Average age of the employees of department P = 2580/100 = 25.8 years

RRB JE ME (CBT I) Mock Test- 1 - Question 20
Save

What will be the percentage profit after selling an article at a certain price if there is a loss of 12.5% when the article is sold at half of the previous selling price?

Detailed Solution: Question 20

Suppose the previous selling price = Rs. x

Selling price = Rs. x/2

There is a loss of 12.5% when the selling price is Rs. x/2.

Cost Price = x/2 x 100/(100 - 12.5) = (100 x)/175 = 4x/7

Now, when selling price is Rs. x, % profit = [(x - (4x/7))/(4x/7)] x 100 = (3x/4x) x 100 = 75%

RRB JE ME (CBT I) Mock Test- 1 - Question 21
Save

If α, 3α+6, and 71α are in arithmetic progression, then find the value of α.

Detailed Solution: Question 21

α, 3α + 6, and 71α are in arithmetic progression.

=> 3α + 6 - α = 71α - 3α - 6 = common difference of the arithmetic progression.

=> 2α = 68α - 6 - 6

=> 66α = 12

=> α = 12/66 = 2/11

RRB JE ME (CBT I) Mock Test- 1 - Question 22
Save

One-fourth of Lainey's savings in Kisan Vikas Patra is equal to one-fifth of her savings in the Employee Provident Fund. How much has she saved in Employee Provident Fund if she has a total savings of ₹ 180,000?

Detailed Solution: Question 22

Let her savings in Employee Provident Fund be ₹ x.

∴ Her savings in Kisan Vikas Patra = ₹ (180,000 - x)

∴ (180,000 - x)/4 = x/5

=> 900,000 - 5x = 4x

=> x = ₹ 100,000

RRB JE ME (CBT I) Mock Test- 1 - Question 23
Save

When 20% of 40% of half of a number 'm' is added to 3/4th of 20% of itself, then a second number is obtained, which is 21 less than the 40% of m. What is the difference between the first and second number?

Detailed Solution: Question 23

First number = m

Second number = (40% of m ) - 21

According to the question:

20% of 40% of (m/2) + (3/4) x 20% of m = (40% of m) - 21

m = 100

Second number = (40% of m) - 21 = 40% of 100 - 21 = 19

Difference = 100 - 19 = 81

RRB JE ME (CBT I) Mock Test- 1 - Question 24
Save

What will come in place of (?) in this equation?

125% of [{? + 200 ÷ (16 ÷ 2 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (18 ÷ 3 - 50% of 4)] = ?

Detailed Solution: Question 24

Let '?' be denoted by 'a'.

125% of [{a + 200 ÷ (16 ÷ 2 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (18 ÷ 3 - 50% of 4)] = a

=> 125% of [{a + 200 ÷ (8 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (6 - 2)] = a

=> 125% of [{a + 200 ÷ 4} ÷ 7.5 x 8 ÷ 4] = a

=> 125% of [{a + 50} ÷ 7.5 x 2] = a => 125% of 2(a + 50)/7.5 = a

=> (5/4) x 2(a + 50)/7.5 = a

=> a + 50 = 3a

=> a = 25

RRB JE ME (CBT I) Mock Test- 1 - Question 25
Save

If cos (900 - θ)/cos (300 + θ) = 1, then the value of sin θ + cos 2θ

Detailed Solution: Question 25

cos (900 - θ)/cos (300 + θ) = 1

sin θ = cos (300 + θ)

sin θ = sin (900 - 300 - θ)

θ = 900 - 300 - θ

θ = 300

= (sin θ + cos 2θ)

= sin 300 + cos 600

= 1/2 + 1/2

= 1

RRB JE ME (CBT I) Mock Test- 1 - Question 26
Save

A fraction is such that if it is squared and then the numerator is divided by 2, while the denominator is increased by 25%, the new fraction thus obtained is 4 times the original fraction. What is the new fraction?

Detailed Solution: Question 26

Correct Option is d) 40.

RRB JE ME (CBT I) Mock Test- 1 - Question 27
Save

What should come in place of question mark (?) in the following series?

15, 17, 23, 35, 55, (?)

Detailed Solution: Question 27

15 + (1 x 2) = 17

17 + (2 x 3) = 23

23 + (3 x 4) = 35

35 + (4 x 5) = 55

55 + (5 x 6) = 85

RRB JE ME (CBT I) Mock Test- 1 - Question 28
Save

How much does a watch lose per day if its hands coincide every 65 minutes?

Detailed Solution: Question 28

For a regular clock, we know that the hands of a watch meet after every 65(5/11) minutes.

Hence loss in 65 minutes = 65(5/11) - 65 = (5/11) minutes

Hence loss in 24 hours = (5 x 24 x 60) / (11 x 65) = 10(10/143) minutes

RRB JE ME (CBT I) Mock Test- 1 - Question 29
Save

After fifteen years, the ratio of the ages of a man and his son will be 2: 1. If the present age of the man is 45, then what is his son's present age?

Detailed Solution: Question 29

After fifteen years, the ratio of age = 2: 1.

Let the present age of the son be 'x'.

(45 + 15)/(x + 15) = 2/1

2x + 30 = 60 x = 15

So, the present age of son is 15 years

RRB JE ME (CBT I) Mock Test- 1 - Question 30
Save

Ramesh can do 20% more work as Mohan can do at the same time. If Mohan can finish the work in 30 hours, then Ramesh, with the help of Mohan, can finish the same work in how many hours?

Detailed Solution: Question 30

Mohan can complete work in 30 hours.

Ramesh can do 6/5 work in 30 hours.

So Ramesh can complete the work in 30 x (5/6) = 25 hours.

Together they can finish 1/30 + 1/25 of work in 1 hour, i.e., 11/150.

So in 150/11 hours, they will finish the work together.

View more questions
158 tests
Join Course
Information about RRB JE ME (CBT I) Mock Test- 1 Page
In this test you can find the Exam questions for RRB JE ME (CBT I) Mock Test- 1 solved & explained in the simplest way possible. Besides giving Questions and answers for RRB JE ME (CBT I) Mock Test- 1, EduRev gives you an ample number of Online tests for practice
Download as PDF

About this Mechanical Engineering Mock Test

This mock test contains 100 questions designed to replicate the pattern, difficulty level and time pressure of the actual Mechanical Engineering exam. It is part of RRB JE Mock Test Series for Mechanical Engineering (ME) 2027 on EduRev, covering all the topics that are being asked as per the current Mechanical Engineering 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 Mechanical Engineering Resources

If you found this RRB JE ME (CBT I) Mock Test- 1 helpful, you can further enhance your Mechanical Engineering preparation with complete mock tests, previous year questions & practice tests available in RRB JE Mock Test Series for Mechanical Engineering (ME) 2027 course on EduRev. The course offers:
  • Video Lectures: covering all the topics of RRB JE Mock Test Series for Mechanical Engineering (ME) 2027
  • Notes for Revision: offering revision notes, important questions, summaries, flashcards, cheatsheets & more for Mechanical Engineering preparation
  • MCQ Test: variety of questions as per the pattern of Mechanical Engineering 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