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Mechanical Engineering Exam  >  Mechanical Engineering Tests  >  RRB JE Mock Test Series for Mechanical Engineering (ME) 2025  >  RRB JE ME (CBT I) Mock Test- 3 - Mechanical Engineering MCQ

RRB JE ME (CBT I) Mock Test- 3 - Mechanical Engineering MCQ


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30 Questions MCQ Test RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 - RRB JE ME (CBT I) Mock Test- 3

RRB JE ME (CBT I) Mock Test- 3 for Mechanical Engineering 2025 is part of RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 preparation. The RRB JE ME (CBT I) Mock Test- 3 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The RRB JE ME (CBT I) Mock Test- 3 MCQs are made for Mechanical Engineering 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ME (CBT I) Mock Test- 3 below.
Solutions of RRB JE ME (CBT I) Mock Test- 3 questions in English are available as part of our RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 for Mechanical Engineering & RRB JE ME (CBT I) Mock Test- 3 solutions in Hindi for RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 course. Download more important topics, notes, lectures and mock test series for Mechanical Engineering Exam by signing up for free. Attempt RRB JE ME (CBT I) Mock Test- 3 | 100 questions in 90 minutes | Mock test for Mechanical Engineering preparation | Free important questions MCQ to study RRB JE Mock Test Series for Mechanical Engineering (ME) 2025 for Mechanical Engineering Exam | Download free PDF with solutions
RRB JE ME (CBT I) Mock Test- 3 - Question 1
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If length, breadth, and height of a cuboid is increased by a%, b%, and c% respectively, then its volume is increased by-

Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 1
Let us suppose that each side of the cuboid be 100 units. Then its volume 1003 units

Now new sides of the cuboid are: (100 + a), (100 + b) and (100 + c)

Then its new volume = (100 + a)(100 + b)(100 + c)

= 1003+1002(a+b+c)+100(ab+bc+ca)+abc

Then % change in volume

= 1002(a+b+c)+100(ab+bc+ca)+abc/1003) × 100

= (a+b+c + ab+bc+ca/100 + abc/(100)2)%

RRB JE ME (CBT I) Mock Test- 3 - Question 2
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Surendra started a mocktail (soft drink + soda) counter. Initially, he had 140 liters of mocktail, which had 30% soda in it. He sold 20 liters of the mocktail. Then he added an equal amount of soft drink and soda. Now the ratio of soda to soft drink became 2:3. How much soda was added later on?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 2
Initial amount of soft drink = 100% mock tail – 30% soda = 70% soft drink

Surendra sold 20 liter mock tail, so now he has 140 - 20 = 120 liters mocktail

In this 120l mock tail, the ratio will be the same, i.e., 30% soda and 70% soft drink.

So, amount of soft drink = 70% of 120 = 84 liters

So, Amount of soda = 120 – 84 = 36 liters

Now, let Surendra add x liters of soft drink and x liters of soda

36+x/84+x = 23

108 + 3x = 168 + 2x

x = 60 liters = amount of soda added

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RRB JE ME (CBT I) Mock Test- 3 - Question 3
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Find the remainder in the expression 557×653×672/9.

Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 3

557×653×672/9

557 is divided by 9, remainder = 8

653 is divided by 9, remainder = 5

672 is divided by 9, remainder = 6

So, now multiplication of remainders 8×5×6 = 240 is divided by 9, find the required remainder.

Required remainder = 6

RRB JE ME (CBT I) Mock Test- 3 - Question 4
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Read the following graph carefully and answer the question given below.

The bar-graph given above shows the percentage of marks obtained by six students namely P, Q, R, S, T and U in a class test. There were three class tests viz. Hindi, English and Science and each test was of 20 marks. Any student scoring less than 58% marks is considered failed in the exams.

By what percentage, the marks scored by P in Hindi more or less than that scored by T in Science?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 4

Marks scored by P in Hindi = 20 × 0.85 = 17

Marks scored by T in science = 20 × 0.40 = 8

Percentage value

= 17−8/8 × 100 = 112.5%

RRB JE ME (CBT I) Mock Test- 3 - Question 5
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Two poles of the height 12 m and 17 m stand vertically upright on a plane ground. If the distance between their feet is 12 m, find the distance between their tops.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 5

In triangle BDE,

BD2 = BE2 + DE2

= 52 + 122 = 144 + 25 = 169

BD = 13 m

RRB JE ME (CBT I) Mock Test- 3 - Question 6
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What is the difference between CI and SI, if the sum is Rs. 8,000 for 4 years at a rate of 4%?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 6

Difference between SI and CI = Sum × r2 × (300+r)/1003

= 8000 × 42 × (300+4)/1003

= 38912000/1000000 = 38.91

RRB JE ME (CBT I) Mock Test- 3 - Question 7
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One glass has wine and soda in the ratio 4:3 while other same quantity of glass has in the ratio 3:2. If both glasses are poured in a vessel, what will be the final ratio of soda to wine in the vessel?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 7

Glass 1 : Wine : soda = 4 : 3

Glass 2 : Wine : soda = 3 : 2

Vessel Wine = Glass 1 + Glass 2 = 4/7+3/5 = 41/35

Vessel soda = Glass 1 + Glass 2 = 3/7+2/5 = 29/35

Soda to wine ratio = 29/35 : 41/35 = 29 : 41

RRB JE ME (CBT I) Mock Test- 3 - Question 8
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Read the following graph carefully and answer the question given below.

3 different products (in Thousands) produced by a company in five different years

What was the total number of all the products produced by the company in the year 2005 and 2007 together?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 8
Required number of all products = (7.5+15+25+20+15+30) thousand

= 112500

RRB JE ME (CBT I) Mock Test- 3 - Question 9
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Read the following graph carefully and answer the question given below.

The bar-graph given above shows the percentage of marks obtained by six students namely P, Q, R, S, T and U in a class test. There were three class tests viz. Hindi, English and Science and each test were of 20 marks. Any student scoring less than 58% marks is considered failed in the exams.

What is the average score of the class in these three tests?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 9
As we don't know the total number of student in the class. Hence, we cannot determine the average of marks of the class.
RRB JE ME (CBT I) Mock Test- 3 - Question 10
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Three-fourth part of a tank is filled with water. 50% of the water is removed from the tank and 60 liters of pure milk is added to it. If now the ratio of milk-water in the tank becomes 5:4, then the capacity of the tank will be:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 10

As the new ratio of milk and water is 5:4

The quantity of milk = 60 liters and hence the quantity of water has to be 48 liters.

Now, 1/2 of 3/4th of the capacity of tank = 48 liters

So, capacity of the tank = 48×2×4/3 = 128 liters

RRB JE ME (CBT I) Mock Test- 3 - Question 11
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Seats for Science, Arts and Commerce in a school are in the ratio 6:9:10. There is a proposal to increase these seats by 30%, 20% and 65% respectively. What will be the ratio of increased seats?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 11
Originally, let the number of seats for Science, Arts and Commerce be 6x, 9x and 10x respectively.

Number of increased seats are (130% of 6x), (120% of 9x) and (165% of 10x).

(130/100 × 6x) : (120/100 × 9x) : (165/100×10x)

= 78x : 108x : 165x

= 26 : 36 : 55

RRB JE ME (CBT I) Mock Test- 3 - Question 12
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50% of a class of 120 students passed in physics and only 20% not passed in chemistry. The maximum possible number of students who can neither passed in physics nor in chemistry is?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 12
Number of students does not pass in physics = 50% of 120 = 60

Number of students does not pass in chemistry = 20% of 120

= 20/100×120 = 24

So, the maximum possible number of students who can neither passed in physics nor in chemistry is = 24

RRB JE ME (CBT I) Mock Test- 3 - Question 13
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What is the positive square root of
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 13
[19 + 4√21]

= (√7)2 + (2√3)2 + 2 x 2√3 x √7

Which is a perfect square,

= (√7 + 2√3)2

Its square root = √7 + 2√3

RRB JE ME (CBT I) Mock Test- 3 - Question 14
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Read the following graph carefully and answer the question given below.

Three different companies of bikes used by persons (in lacs)

Find the ratio of average of hero bikes used in the 2012 and 2017 and average of Honda bikes used in the same years.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 14
Average of hero bikes used in 2012 and 2017 = 2000000+1000000/2 = 1500000

Average of honda bikes using in 2012 and 2017 = 3000000+4000000/2 = 3500000

Required ratio = 1500000/3500000 = 3 : 7

RRB JE ME (CBT I) Mock Test- 3 - Question 15
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Read the following graph carefully and answer the question given below.

Three different companies of bikes used by persons (in lacs)

Find out the percentage increase or decrease in the number of persons using Hero from 2014 to 2015?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 15
Number of persons using Hero bikes in 2014 = 1000000

Number of persons using Hero bikes in 2014 = 3000000

Required percentage increase = 3000000−1000000/1000000×100

= 2000000/1000000×100 = 200%

= 200 %

RRB JE ME (CBT I) Mock Test- 3 - Question 16
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Amit and Bhanu invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and Amit's share is Rs. 8550, then find the profit share of Bhanu.
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 16
Let the total profit be Rs. 100.

After paying to charity, Amit's share = Rs. (95× 3/5 ) = Rs. 57

If Amit’s share is Rs. 57, total profit = Rs. 100

If Amit's share is Rs. 8550 then total profit = 8550 × 100/5

= 15000

Bhanu’s share = (15000×95/100×25) = Rs. 5700

RRB JE ME (CBT I) Mock Test- 3 - Question 17
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In triangle PQR, A is the point of intersection of all the altitudes and B is the point of intersection of all the angle bisectors of the triangle. If ∠PBR = 1050, then what is the value of ∠PAR?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 17

∠PQR = 300

∠PAR = 1800 - 300 = 1500

RRB JE ME (CBT I) Mock Test- 3 - Question 18
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The average monthly income of A and B is Rs 7760. The average monthly income of B and C is Rs. 10990 and that of C and A is Rs. 9070. What is the annual income of B?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 18

Total income of A and B = 2 x 7760

So, A+B = 15520 ... (i)

Similarly,

B +C = 10990 x 2 = 21980... (ii)

and A+C = 9070 x 2 = 18140 ... (iii)

Adding Eqs. (i), (ii) and (iii),

2(A + B + C) = 55640

⇒ A+B + C =27820 ... (iv)

Subtracting Eq. (iii) from Eq. (iv),

B = 27820 – 18140 = 9680

Hence, annual income of B = 9680 x 12 = Rs. 116160

RRB JE ME (CBT I) Mock Test- 3 - Question 19
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If a shopkeeper sells a TV at a discount of 42%, he loses 13%. At what discount should he sell the TV to earn a profit of 5%?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 19

SP = (1−42/100) MP= (1−13/100) CP

MP = 0.87×CP/0.58= 1.5×CP

To earn profit of 5%, let the discount be D%.

New SP= (1+5/100) CP = (1−D/100) MP

1.05CP = (1−D/100) × 1.5 × CP

D= 30%

Short trick:

100 − D1/100 − D2 = 100+P1/100+P2

58/100−D2 = 87/105

⇒ D2 = 100−70 = 30%

RRB JE ME (CBT I) Mock Test- 3 - Question 20
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The profit obtained by selling an article for Rs. 856 is twice the loss incurred when the article is sold for Rs. 760. The cost price of the article is?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 20
Let ‘C’ be the cost price of the article.

⇒ (856 - C) = 2(C - 760)

⇒ 3C = 856 + 1520

⇒ 3C = 2376

⇒ C = 792

∴ The cost price of the article = 792

RRB JE ME (CBT I) Mock Test- 3 - Question 21
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The length of two parallel chords of a circle of radius 5cm are 6 cm and 8 cm in the same side of the centre. The distance between them is-
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 21
Here, O is the centre of the circle of radius 5 cm and AB and CD are two chord of length 8 cm and 6 cm respectively.

∵ Perpendicular drawn from the centre of a circle to the chord bisects the chord.

∴ EB = 4 cm and FD = 3 cm

Now, In ΔΔ OEB

OB2 = OE2 + EB2

RRB JE ME (CBT I) Mock Test- 3 - Question 22
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P and Q are standing at a distance of 350 m from each other on a straight road such that P is to the west of Q. Both of them start walking simultaneously towards each other and cover a distance of 50 m each. Thereafter, P turns to his left and walks 290 m while Q turns to his right and walks 80 m. At the end both of them turn to their left such thatP covers a distance of 50 m while Q covers a distance of 100 m. How far and in which direction is Q from P?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 22

Shortest Distance between P and Q = P'Q' =

Hence Q is 290m from P and in the South- East Direction.

RRB JE ME (CBT I) Mock Test- 3 - Question 23
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What percentages of numbers from 1 to 80 have 1 or 9 in the unit's digit?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 23
Numbers from 1 to 80 have 1 or 9 in the unit's digit

1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69, 71, 79

Number of such number = 16

Required percentage = (16/80 × 100) = 20%

RRB JE ME (CBT I) Mock Test- 3 - Question 24
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If tan θ = p/q, then the value of p cos 2θ + qsin 2θ is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 24

RRB JE ME (CBT I) Mock Test- 3 - Question 25
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The median of six numbers 6, 7, x – 5, x – 3, 17 and 19, in ascending order, is 11. The value of x is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 25
The median of six number

= (x−5) + (x−3)/2

= 2x − 8/2

= x − 4

According to Question-

x - 4 = 11

x = 15

RRB JE ME (CBT I) Mock Test- 3 - Question 26
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If tan2 θ + cot2 θ = x , what is the value of 1/cosθ.sinθ = ?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 26

RRB JE ME (CBT I) Mock Test- 3 - Question 27
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The product of two numbers is 108 and the sum of their squares is 625. The sum of the number is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 27
Let the Numbers be P and Q.

PQ= 108, P2+Q2 = 625

(P+Q)2 = P2 + Q2 + 2PQ

(P+Q)2 = 625 + 2 × 108

(P+Q)2 = 841

P+Q = 29

RRB JE ME (CBT I) Mock Test- 3 - Question 28
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Ria was one more than eleven times as old as Sandhya in January, 2000 and seven more than three times as old as her in January, 2009. How old was Ria in January, 2000?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 28
Let Ria’s and Sandhya’s age in January 2000 be X years and Y years resp.

According to Question-

X = 1+11Y---------- (1)

In January 2009, Ria’s age was (X+9) years and Sandhya’s was (Y+9) years.

So,

X+9 = 7+3(Y+9)

X = 25+3Y------------(2)

On Solving(1) and (2)

X = 34 and Y = 3

Hence. In January 2000 Ria was 34 Years old.

RRB JE ME (CBT I) Mock Test- 3 - Question 29
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What is the unit digit in (795 − 357)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 29

Unit digit in 795 = Unit digit in [(74)23 × 73]

= Unit digit in [(Unit digit in (2401))23 × (343)]

= Unit digit in (123 × 343)

= Unit digit in (343)

= 3

Unit digit in 357 = Unit digit in

=[(34)14 × 31] = Unit digit in [Unit digit in (81)14 × 3]

= Unit digit in [(1)14 × 3]

= Unit digit in (1 x 3)

= Unit digit in (3)

= 3

Unit digit in (795 − 357) = Unit digit in (343 - 3) = Unit digit in (340) = 0.

RRB JE ME (CBT I) Mock Test- 3 - Question 30
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A sum of Rs. 4000 becomes Rs. 5800 in 3 years, when invested in a scheme of simple interest. If the same sum is invested in a scheme of compound interest with same yearly interest rate (compounding of interest is done yearly), then what will be the amount (in Rs) after 2 years?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 3 - Question 30

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