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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- 4 - Mechanical Engineering MCQ

RRB JE ME (CBT I) Mock Test- 4 - 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- 4

RRB JE ME (CBT I) Mock Test- 4 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- 4 questions and answers have been prepared according to the Mechanical Engineering exam syllabus.The RRB JE ME (CBT I) Mock Test- 4 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- 4 below.
Solutions of RRB JE ME (CBT I) Mock Test- 4 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- 4 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- 4 | 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- 4 - Question 1
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A canal is 5 m wide at the bottom and 8 m wide at the the top. The depth of water is 4 m. Water is flowing in the canal with the speed of 10 m/s. How much cubic m. of water will flow in 1 minute?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 1
Area of the base 1/2 (sum of parallel sides) × Height

= 1/2(8+5)×4 = 26 m2

∵ Speed of water = 10 m/sec.

∴ Volume of water = 26 × 10 × 60 = 15600 m3

RRB JE ME (CBT I) Mock Test- 4 - Question 2
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In how many years will a sum of Rs. 3,500 yield an interest of Rs. 2,100 at 15% per annum?

Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 2
T = 2100×100/3500×15

T = 4 Years

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RRB JE ME (CBT I) Mock Test- 4 - Question 3
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The average of all the prime and composite numbers up to 100 is:

Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 3
Since 1 is neither prime nor composite number. Thus there are only 99 number viz.

2, 3, 4, 5, 6, ... 99, 100.

RRB JE ME (CBT I) Mock Test- 4 - Question 4
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A, B and C can do a work in 6, 8 and 12 days respectively. Doing that work together they get an amount of Rs. 1350, the share of B in that amount is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 4
A’s one day work = 1/6

B’s one day work = 1/8

C’s one day work = 1/12

A’s share: B’s share: C’s share = 1/6:1/8:1/12

= 4 : 3 : 2

∴ B’s share = 1350×3/9 = ₹ 450

RRB JE ME (CBT I) Mock Test- 4 - Question 5
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The cost price of an article is ₹ 3,90 and sold at a profit of 3.12%, how much would be its approximate selling price?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 5
S.P. = 390 + 3.12% of 390 = ₹ 402.168
RRB JE ME (CBT I) Mock Test- 4 - Question 6
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The two chords AB and CD of a circle cut each other perpendicularly at point O. OA = 2 cm, OC = 3 cm and OB = 6 cm, then what will be the diameter of the circle?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 6

Alternate solution

Now, diameter

Diameter

RRB JE ME (CBT I) Mock Test- 4 - Question 7
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17 articles were bought for ₹ 3,910 and sold for ₹ 4,590. How much was the approximate profit percentage per article?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 7
Profit = 4590 - 3910

= Rs. 680

Profit percentage = 680/3910 x 100

= 17.39%

= 17% approximately.

RRB JE ME (CBT I) Mock Test- 4 - Question 8
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If cos A + cos 2A = 1, then sin 2A + sin 4A = ?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 8
cos A = 1 – cos2 A

= sin2 A

Now, since

sin2 A + cos2 A = 1

⇒ sin2 A + (sin2 A)2 = 1

⇒ sin2 A + sin4 A = 1

RRB JE ME (CBT I) Mock Test- 4 - Question 9
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If θ is a positive acute angle and tan 2θ.tan 3θ = 1, then what will be the value of (2 cos2 5θ/2−1)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 9

tan2θ.tan3θ = 1

tan2θ = cot3θ

⇒ θ = 180

RRB JE ME (CBT I) Mock Test- 4 - Question 10
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A mixture contains spirit and water in the ratio 3:2. If it contains 3 liters more spirit than water, the quantity of spirit in the mixture is?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 10
Let the amount of water be x liters

∴ x+3/x = 3/2

or 2x + 6 = 3x or, x= 6

∴ The quantity of spirit in the mixture = x + 3 = 6 + 3

= 9 liters

RRB JE ME (CBT I) Mock Test- 4 - Question 11
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What will be the compound interest on a sum of Rs. 25,000 after three years at the rate of 12% per annum?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 11

RRB JE ME (CBT I) Mock Test- 4 - Question 12
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A round balloon of radius r subtends an angle α at the eye of an observer while the angle of elevation of its center is β. The height of the center of the balloon is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 12
Let O be the center of balloon of radius r. The observer’s eye is at C.

∠ACB = α and ∠OCD = β clearly, CA and CB are tangents to the circle. So

∠ACO = ∠BCO = α/2

In right angled ∆OBC,

In right angled △OCD

∴ Height of the center of the balloon is r sin β. cosec α/2

RRB JE ME (CBT I) Mock Test- 4 - Question 13
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The largest and the second largest angles of a triangle are in the ratio of 3 : 2 respectively. The smallest angle is 20% of the sum of the rest two angles. What is the sum of the smallest and the second largest angles?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 13
If the largest and the second largest angles be 3x and 2x respectively then, third angle = x

∴ x + 2x + 3x = 180°

⇒ x = 30o

∴ Required sum

= x + 2x = 3x = 90°

RRB JE ME (CBT I) Mock Test- 4 - Question 14
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Study the following Pie Chart carefully and answer the questions given below:

Budget estimated by a family for their monthly expenses

What is the budget estimated by the family on Clothing and Grocery together?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 14
Budget estimated on Clothing and Grocery together= (25 + 8)% of ₹ 32,000

= (33/100) × 32000 = ₹ 10,560

RRB JE ME (CBT I) Mock Test- 4 - Question 15
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Study the following Pie Chart carefully and answer the questions given below:

Budget estimated by a family for their monthly expenses

Due to a sudden marriage, the family incurs Miscellaneous expenditure of ₹ 3,040. How much is the increase in the amount under this head from that of the budgeted?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 15
Budget estimated on miscellaneous 7% of ₹ 32,000

= (7/100) × 32000 = ₹ 2240

Required change= 3040-2240 = 800

RRB JE ME (CBT I) Mock Test- 4 - Question 16
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PQ is a direct common tangent of two circles of radii r1 and r2 touching each other externally at A. Then the value of PQ2 is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 16

Since opposite sides are parallel and interior angles are 90° therefore OPQR is a rectangle.

OP = QR = r1

In △OO'R,

∠ORO' = 90°

By Pythagoras theorem,

OR2 + O'R2 = OO'2

OR2 + (r1 - r2)2 = (r1 + r2)2

OR2 = 4r1r2

∴ PQ2 = OR2 = 4r1r2

RRB JE ME (CBT I) Mock Test- 4 - Question 17
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20% of the 4/5th of a number is 60, find the number?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 17

RRB JE ME (CBT I) Mock Test- 4 - Question 18
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If a number x is 10% less than another number y and y is 10% more than 125, then x is equal to
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 18
x = y - 10% of y

x = 9/10 y

New, y = 125 + 10% of 125

= 110/100 × 125

∴ x = 9/10 × 110/100 × 125 = 123.75

RRB JE ME (CBT I) Mock Test- 4 - Question 19
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If A, B, C, D are four points in a straight line such that distance from A to B is 10, B to C is 5, C to D is 4 and A to D is 1, then their correct sequence is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 19
Given that the distance from A to B = 10 and B to C = 5 and C to D = 4, D to A = 1

∴ The correct sequence is A - D - C - B which is in increasing order.

RRB JE ME (CBT I) Mock Test- 4 - Question 20
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The ratio between the speeds of a truck, car and train is 3 : 8: 12. The car moved uniformly and covered a distance of 1040 km in 13 hours. What is the average speed of the truck and the train together?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 20
The ratio between the speeds of the truck, car, and train is 3:8:12

Given that the car covers 1040 km in 13 hours, the speed of the car 80 km/h

Also, if we assume that the speeds of the truck, car, and train are 3x, 8x and 12x

respectively, then

x = 10

Therefore, speed of truck 30 km/h

Speed of train = 120 km/h

The average speed of truck and train is calculated as:

2×30×120/30+120 = 48 km/h

RRB JE ME (CBT I) Mock Test- 4 - Question 21
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A, B, and C contract a work for Rs. 550 together. A and B are supposed to do 7/11 of the work. The amount that C gets is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 21

RRB JE ME (CBT I) Mock Test- 4 - Question 22
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The radius of a cylinder is the same as that of a sphere. There volumes are equal. The height of the cylinder is how many times of its radius?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 22
Radius of cylinder = Radius of sphere and volume of cylinder = volume of sphere

RRB JE ME (CBT I) Mock Test- 4 - Question 23
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A train traveling at 65 kmph leaves Mumbai at 10 A.M. and another train travelling at 75 kmph starts at 12 Noon in the same direction. How many k.m. away from Mumbai will they be together?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 23
In two hours, the first train is 2 x 65 kms i.e, 130 kms ahead of the second train The second train moves 10 kms faster so it will meet the first train in (130 ÷ 10)

i.e.. 13 hours

∴ Distance travelled by the second train = 13 x 75 km = 975 km

Therefore after 975 km both train will meet.

RRB JE ME (CBT I) Mock Test- 4 - Question 24
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If the angles of a pentagon are in the ratio 1 : 3 : 6 : 7 : 10, then the smallest angle is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 24
Suppose the angles of a pentagon are

x°, 3x°, 6x°, 7x°, 10x°

But x° + 3x° + 6x° + 7x° + 10x° = (2x5 -4) x 90°

or 27x° = 6 x 90°

or x = 20°

RRB JE ME (CBT I) Mock Test- 4 - Question 25
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A bob of length 50 cm makes an arc of 16 cm during its motion. What will be the measure of the angle thus formed (approx.)?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 25
Radius = 50 cm

Arc = 16 cm

Let the angle be θ.

16 = θ/360×2×π×50 ⇒ θ = 8×36×7/5×22 ≈ 18o33 ≈ 18°35'

RRB JE ME (CBT I) Mock Test- 4 - Question 26
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If the radius of a cylinder is tripled and the height is halved, what is the ratio between the new volume and the previous volume?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 26
Let the initial radius and height of the cylinder be r cm and h cm respectively.

Then, V1 = πr2 h and V2 = π.(3r).2.h/2 = 9/2 πr2h

New Volume/Previous Volume = 9/2πr2h/πr2h = 9/2 = 9:2

RRB JE ME (CBT I) Mock Test- 4 - Question 27
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Find the value of
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 27

RRB JE ME (CBT I) Mock Test- 4 - Question 28
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The sides of a triangle are 24 cm, 32 cm and 40 cm. What is the height of the perpendicular drawn on the longest side?
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 28
The triangle thus formed is a right-angled triangle.

Thus, the length of perpendicular on the largest side

= 32×24/40 = 19.2 meter.

RRB JE ME (CBT I) Mock Test- 4 - Question 29
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If x sin3 θ + ycos3 θ = sin θ.cos θ and xsin θ = ycos θ, sin θ≠ 0, cos θ≠ 0, then x2 + y2 is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 29

RRB JE ME (CBT I) Mock Test- 4 - Question 30
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If 42x = 1/32 then x is:
Detailed Solution for RRB JE ME (CBT I) Mock Test- 4 - Question 30
42x = 1/32

or (22)2x = 1/25

or 24x = 2-5

∴ 4x = -5

or x = −5/4

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