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Mechanical Engineering Exam  >  Mechanical Engineering Test  >  RRB JE Mock Test Series (ME) 2026  >  RRB JE ME (CBT I) Mock Test- 10 - Mechanical Engineering MCQ

RRB JE ME (CBT I) Mock Test- 10 Free Online Test 2026


Full Mock Test & Solutions: RRB JE ME (CBT I) Mock Test- 10 (100 Questions)

You can boost your Mechanical Engineering 2026 exam preparation with this RRB JE ME (CBT I) Mock Test- 10 (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

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RRB JE ME (CBT I) Mock Test- 10 - Question 1
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Find the HCF of (29 x 36 - 12 x 22) and the average of three numbers 827, 157, and 528.

Detailed Solution: Question 1

= (29 x 36 - 12 x 22)

= 780

Average of 827, 157 and 529 = (827 + 157 + 528)/3 = 504

HCF of 780 and 504 = 12

RRB JE ME (CBT I) Mock Test- 10 - Question 2
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Sum of three numbers is 54, and the first number is 10 more than the third and 10 less than the second number. Find the second number.

Detailed Solution: Question 2

Let first, second and third numbers are a, b and c respectively.

a + b + c = 54

b = a + 10

c = a - 10

a + a - 10 + a + 10 = 54

a = 18

second number = b = a + 10 = 18 + 10 = 28

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RRB JE ME (CBT I) Mock Test- 10 - Question 3
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Which of the following is irrational?

a = (2√2 - 3√3)16

b = √1764 + 2√1089

c = √(14 x 59 - 67 x 3)

d = √(24 x 22 + 16 x 31)

Detailed Solution: Question 3

a = (2√3 - √32)√216 = 32√2 - 48√3 (irrational)

b = √1764 + 2√1089 = 108

c = √(14 x 59 - 67 x 3) = 25

d = √(24 x 22 + 16 x 31) = 32

RRB JE ME (CBT I) Mock Test- 10 - Question 4
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A Faulty watch gains 5 seconds in 5 minutes. It was set right at 8 a.m. If, later that day, it shows 11:15 p.m. in the watch, what is the actual time?

Detailed Solution: Question 4

Duration from 8 a.m. to 11:15 p.m. = 15 hours 15 minutes = 915 minutes = 54900 seconds

305 seconds in defective watch = 300 seconds of actual time.

⇒ 305*(54900/305) seconds in defective watch = 300*(54900/305) seconds of actual time

⇒ 54900 seconds in defective watch = 54000 seconds of actual time

= 15 hours of actual time

Actual time is 8 a.m. + 15 hours = 11:00 p.m.

RRB JE ME (CBT I) Mock Test- 10 - Question 5
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The population of a village in 2016 is 2000 and it increases by 20% every year from the previous year. What is the population gained during 2018-2019?

Detailed Solution: Question 5

Population in 2019 = 2000(1 + 20/100)3 = 3456

Population in 2018 = 2000(1 + 20/100)2 = 2880

Population gained during 2018-2019 = 3456 - 2880 = 576

RRB JE ME (CBT I) Mock Test- 10 - Question 6
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If 40% of a is equal to two times the value of b, find a: b.

Detailed Solution: Question 6

(40/100) x a = 2 x b

a = 5b

a: b = 5: 1

RRB JE ME (CBT I) Mock Test- 10 - Question 7
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Aman invested Rs.16000 for 4 years at a 15% simple rate of interest. After 4 years, total amount received he invested for 2 years at 20% compound rate of interest. With the amount received after 2 years, he purchased an article and sold this article at Rs.40000, then what is the profit received by him after selling the article?

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

(A) Total amount after 4 years when invested at 15% SI = 16000 + [(16000 * 15 * 4)/100] = Rs.25600

(B) Cost price of article = Rs.36864 and selling price = Rs.40000

(C) Profit amount earned = 40000 - 36864 = Rs.3136

(D) Total amount after 2 years when invested at 20% CI = 25600 * (1.2)2 = Rs.36864

Detailed Solution: Question 7

The correct order is

(A) Total amount after 4 years when invested at 15% SI = 16000 + [(16000 * 15 * 4)/100] = Rs.25600

(D) Total amount after 2 years when invested at 20% CI = 25600 * (1.2)2 = Rs.36864

(B) Cost price of article = Rs.36864 and selling price = Rs.40000

(C) Profit amount earned = 40000 - 36864 = Rs.3136

RRB JE ME (CBT I) Mock Test- 10 - Question 8
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A car crosses a man walking at 6 km/hr. The man can see things up to 450 m, only in one direction due to fog. He sees the car, which was going in the same direction for 4.5 min. What is the speed of the car?

Detailed Solution: Question 8

According to question.

s1−s2=d/t

s1−6=(450/1000)/(4.5/60)

s1=6+6=12 km/hr

RRB JE ME (CBT I) Mock Test- 10 - Question 9
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The ratio of the speeds of 3 Bikers is 6:4:1, and the ratio of the time for which they have travelled is 2:1:4. If the sum of the distance travelled by all of them is 60 km, what is the difference between the distance travelled by the slowest Biker and the fastest Biker?

Detailed Solution: Question 9

The distances they travelled will be in the ratio

(6*2) : (4*1) : (1*4) = 12:4:4 = 3:1:1

Let them be 3x km, x km and x km respectively.

Given, 3x + x + x = 60

⇒ 5x = 6

⇒ x = 12

difference between the distance travelled by the slowest Biker and the fastest Biker = 3x - x = 2x = 24 km

RRB JE ME (CBT I) Mock Test- 10 - Question 10
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Five children are born to Riya each at a gap of 3 years such that the age of eldest child seven years later if doubled becomes 3 less than Riya's present age. 15 years hence, the average of the three youngest children becomes 23 less than Riya's age 3 years ago. Find the age of Riya and the age of the youngest child, respectively.

Detailed Solution: Question 10

Let present age of Riya = 'm' years

Five children in ascending order of their present age be a, (a+3) , (a+6) , (a+9) and (a+12)

Now,

2*(a+12+7) = m - 3

m = 2a + 41 ....(i)

15 years hence,

(a+a+3+a+6+15*3)/3 + 23 = 2a + 41 - 3

a + 18 + 23 = 2a + 38

a = 3

So,

Riya = m = 2a + 41 = 47 years

Youngest = a = 3 years

RRB JE ME (CBT I) Mock Test- 10 - Question 11
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In an election, there are five candidates and, each got the different number of votes. P gets more votes than U, who gets less votes than R. T receives more votes than S, who gets less votes than P. P didn't get more votes than both T and R. U didn't gets the lowest number of votes and R didn't gets the highest number of votes.

How many candidates get more votes than S, but less votes than R?

Detailed Solution: Question 11

Here,

→ P gets more votes than U, who gets less votes than R.

So,

R > P > U

Or

P > R > U

Now,

→ T gets more votes than S, who gets less votes than P.

So,

T > P > S

Or

P > T > S

Since,

→ P didn't get more votes than both T and R.

So, we get,

R > P > U and T > P > S

So, combining both we get,

R, T > P > U, S

Since,

→ U didn't get the lowest number of votes and R didn't gets the highest number of votes.

Final arrangement is,

T > R > P > U > S

Hence,

Two candidates P and U get more votes than S, but less votes than R

RRB JE ME (CBT I) Mock Test- 10 - Question 12
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The difference between CI and SI on a certain sum of money at 20% per annum for 2 years is Rs 800, what would be the difference of CI and SI on that same sum of money after 3 years if the rate of interest was 10%?

Detailed Solution: Question 12

Let the principle be Rs 'P'.

Given, P(20/100)2 = 800

P = 20,000

Difference between CI and SI in 3 years

= P(R/100)2((300 + R)/100) [symbols have their usual meanings]

= 20,000(10/100)2((300 + 10)/100)

= Rs 620

RRB JE ME (CBT I) Mock Test- 10 - Question 13
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2 men and 3 women can do a piece of work in 10 days while 3 men and 2 women can do the same work in 8 days. In how many days can 2 men and 1 woman do the work?

Detailed Solution: Question 13

Let 1 man's 1 day work = x and 1 woman's 1 day work = y.

Then,

2x + 3y = 1/10 and 3x + 2y = 1/8

Solving, we get:

X = 7/200 and y = 1/100

Therefore, (2 men and 1 woman)'s 1 day's work = (2 * (7/200) + 1 * (1/100)) = 16/200 = 2/25

Therefore, (2 men and 1 woman) do the work in 25/2 days.

Hence, option c is correct.

RRB JE ME (CBT I) Mock Test- 10 - Question 14
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In a bag 40% of balls are red, 20% balls are blue and the remaining are green color balls. If the number of green and blue balls is 45, then find the number of red balls in the bag.

Detailed Solution: Question 14

Let the number of balls in a bag be P’ In a bag 40% of balls are red. 20% balls are blue, so green balls will be.

BGreen = 100 - (40 + 20) = 100 - 60 = 40%

According to the given conditions.

Number of red balls in the bag is.

Therefore, the number of red balls in the bag is 30.

RRB JE ME (CBT I) Mock Test- 10 - Question 15
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Prices of 10 items are given below in Rupees. Find the difference between their mean and median.

15, 5, 8, 10, 22, 25, 12, 30, 20, 18

Detailed Solution: Question 15

Arrangement of price in ascending order:

5, 8, 10, 12, 15, 18, 20, 22, 25, 30

Total number of observations = 10 (even number)

Median = ((n/2)th term + ((n/2) + 1)th term)/2

= ((5)th term + ((6)th term)/2

= (15 + 18)/2 = Rs.16.5

Mean = (5 + 8 + 10 + 12 + 15 + 18 + 20 + 22 + 25 + 30)/10 = Rs.16.5

Difference = 16.5 - 16.5 = Rs.0

RRB JE ME (CBT I) Mock Test- 10 - Question 16
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If 16m2 = 9n2, find m3 % of n3.

Detailed Solution: Question 16

We have,

16m2 = 9n2

⇒ m2/n2 = 9/16

⇒ m/n = √(9/16)

⇒ m/n = 3/4

Now, we want, m3% of n3 = (m3/100)*n3

we cannot calculate the value of (mn)3.

So, answer cannot be determined.

RRB JE ME (CBT I) Mock Test- 10 - Question 17
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A cloth trader gets a profit equal to selling price of 12 metres of cloth when he sells 36 metres of cloth. Find the profit percentage.

Detailed Solution: Question 17

⇒ CP of 36 metres + profit of 36 metres = SP of 36 metres

⇒ But profit of 36 metres= SP of 12 metres

CP of 36 metres + SP of 12 metres= SP of 36 metres

CP of 36 metres = SP of 24 metres

⇒ Profit = (36-24)/24*100

⇒ Profit = 12*100/24

⇒ Profit = 50%

RRB JE ME (CBT I) Mock Test- 10 - Question 18
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A number when divided by 3, 5 and 7 leaves remainder 2 in each case, what will be the remainder when the same number is divided by 9?

Detailed Solution: Question 18

LCM of 3, 5 and 7 = 105

Number when divided by 3, 5 and 7 leaves remainder 2 in each case.

So, the number = 105 + 2 = 107

Remainder when '107' is divided by '9' = 107/9 = (99 + 8)/9 = 8

RRB JE ME (CBT I) Mock Test- 10 - Question 19
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The number of pupils of a class is 1342. The ratio of the number of boy pupils to the number of girl pupils is 5: 6. If 120 boys pupils leave the class and 80 new girl pupils joined the class, then new ratio becomes:

Detailed Solution: Question 19

Number of boy pupils initially = 1342 x = 610

Number of girl pupils initially = 1342 x = 732

Required ratio = (610 -120): (732 + 80) = 490: 812 = 35:58

RRB JE ME (CBT I) Mock Test- 10 - Question 20
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In a Village, there are 10800 people. The %literacy is 55% and the % Female literacy is 48%. If there are 1620 more literate Males than literate Females, what is the Gender ratio(Male : Female) of the Village?

Detailed Solution: Question 20

Number of literate people = (55/100)*10800 = 5940

Let the number of Females be 'x'

Number of Males = 10800 - x

Number of Female literate people = (48/100)*x = 12x/25

Number of Male literate people = 1620 + 12x/25

Given, 1620 + 12x/25 + 12x/25 = 5940

⇒ 24x/25 = 4320

⇒ x = 4500

Number of Males = 10800 - 4500 = 6300

Gender Ratio = 6300:4500 = 7:5

RRB JE ME (CBT I) Mock Test- 10 - Question 21
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The ratio of three numbers is 3:4:6 and their LCM is 36. What is the difference between the largest and the smallest number?

Detailed Solution: Question 21

Let the numbers be 3k, 4k and 6k respectively.

LCM(3k, 4k, 6k) = 36

⇒ k*LCM(3, 4, 6) = 36

⇒ 12k = 36

⇒ k = 3

Difference between the largest and the smallest number = 6k - 3k = 3k = 9

RRB JE ME (CBT I) Mock Test- 10 - Question 22
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Following table represents the ages of some people.

Calculate the variance.

Detailed Solution: Question 22

Mean = 954/70 = 13.63

Variance = 24696/70 - 13.632 = 167.02

RRB JE ME (CBT I) Mock Test- 10 - Question 23
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What is the value of (1/9) of [√169 X 3 + {62 + (3 ÷ 6 X 12)}] X (1/8) of [40 + {20 + (4 ÷ 16 X 16)}]?

Detailed Solution: Question 23

(1/9) of [√169 X 3 + {62 + (3 ÷ 6 X 12)}] X (1/8) of [40 + {20 + (4 ÷ 16 X 16)}]

= (1/9) of [13 X 3 + {36 + (0.5 X 12)}] X (1/8) of [40 + {20 + (0.25 X 16)}]

= (1/9) of [39 + {36 + 6}] X (1/8) of [40 + {20 + 4}]

= (1/9) of [39 + 42] X (1/8) of [40 + 24]

= (1/9) of 81 X (1/8) of 64

= 9 X 8

= 72

RRB JE ME (CBT I) Mock Test- 10 - Question 24
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What is the equation of line passing through midpoint of Points A(4, 3) and B(0, 7) having slope 3?

Detailed Solution: Question 24

Midpoint of points A(a1, a2) and B(b1, b2) is C(c1, c2)-

(c1, c2) = (a1 + b1)/2, (a2 + b2)/2

(x, y) = (4 - 0)/2, (3 + 7)/2

(x, y) = (2, 5)

Line of equation = y = mx + c [where m = 3 and line passes through (2, 5)]

5 = 3 * 2 + c

c = -1

Required equation of line = y = 3x - 1

3x - y = 1

RRB JE ME (CBT I) Mock Test- 10 - Question 25
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If 'a' and 'b' are roots of quadratic equation x2 + 4x - 21 = 0 such that b>a, then find the new quadratic equations whose roots will be (2a + 3b) and (2b - 3a).

Detailed Solution: Question 25

x2 + 4x - 21 = 0

x2 +7x - 3x - 21 = 0

x = -7, 3

a = -7, b = 3

Now, (2a + 3b) = 2 x (-7) + 3 x 3 = -5

(2b - 3a) = 2 x 3 - 3 x (-7) = 27

New quadratic equation will be:

x2 - (-5 + 27)x + (-5) x 27 = 0

x2 - 22x - 135 = 0

RRB JE ME (CBT I) Mock Test- 10 - Question 26
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A Metro train of 120 m length is running on a bridge at the rate of 40 km/hr. The train crosses the bridge in 20 second. What is the length of the bridge?

Detailed Solution: Question 26

Speed of Metro

Let the length of the bridge be x m.

Therefore

9(120 + x) = 200

9x = 2000 - 1080

9x = 920

x = 102.22

RRB JE ME (CBT I) Mock Test- 10 - Question 27
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What is the value of 'a' in the following equation?

1/a of [{6.5 X (√100 − √4)} ÷ 13 X {2.5 X (√625 + √9)} ÷ 7] + 1/3 of √(15 ÷ 3 X 8 ÷ 4 - 1) = 6

Detailed Solution: Question 27

1/a of [{6.5 X (√100 − √4)} ÷ 13 X {2.5 X (√625 + √9)} ÷ 7] + 1/3 of √(15 ÷ 3 X 8 ÷ 4 - 1) = 6

⇒ 1/a of [{6.5 X (10 − 2)} ÷ 13 X {2.5 X (25 + 3)} ÷ 7] + 1/3 of √(5 X 2 - 1) = 6

⇒ 1/a of [{6.5 X 8} ÷ 13 X {2.5 X 28} ÷ 7] + 1/3 of √9 = 6

⇒ 1/a of [52 ÷ 13 X 70 ÷ 7] + 1/3 of 3 = 6

⇒ 1/a of [4 X 10] + 1 = 6

⇒ 1/a of 40 = 5

⇒ a = 40/5 = 8

RRB JE ME (CBT I) Mock Test- 10 - Question 28
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If the sum of squares of length(l) and breadth(b) of a rectangle is 89 cm2 and its area is 40 cm2, then what is the area of a square having side equal to the breadth of rectangle(l>b)?

Detailed Solution: Question 28

Let length and breadth of rectangle are l and b respectively.

As per given, l2 + b2 = 89 cm2 and lb = 40 cm2

(l +b)2 = l2 + b2 +2lb = 89 + 80 = 169

l +b =13 cm

l = 13-b

As lb = 40

(13-b) x b =40

13b -b2 =40

b2 - 13b +40 = 0

b = 8 or 5

If b = 8, then l = 5 and if b = 5, then b = 8

as length is greater than breadth. So, b = 5.

Area of square having side equal to breadth of rectangle= 25 cm2.

RRB JE ME (CBT I) Mock Test- 10 - Question 29
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If a + 1/2a2 = 5, what is the value of (a - 5)49 + 1/249a98?

Detailed Solution: Question 29

a + 1/2a2 = 5

⇒ a - 5 = − 1/2a2

∴ (a - 5)49 + 1/249a98

= −1/249a98 + 1/249a98 = 0

RRB JE ME (CBT I) Mock Test- 10 - Question 30
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What is the value of 1 +3 + 5.......... +159?

Detailed Solution: Question 30

Above given series is the sum of the first 80 odd numbers.

Sum of first n odd numbers = n2

Required sum = 80 x 80 = 6400.

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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) 2026 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.

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If you found this RRB JE ME (CBT I) Mock Test- 10 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) 2026 course on EduRev. The course offers:
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