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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  RRB JE Mock Test Series for Civil Engineering (CE) 2025  >  RRB JE CE (CBT I) Mock Test- 10 - Civil Engineering (CE) MCQ

RRB JE CE (CBT I) Mock Test- 10 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series for Civil Engineering (CE) 2025 - RRB JE CE (CBT I) Mock Test- 10

RRB JE CE (CBT I) Mock Test- 10 for Civil Engineering (CE) 2024 is part of RRB JE Mock Test Series for Civil Engineering (CE) 2025 preparation. The RRB JE CE (CBT I) Mock Test- 10 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The RRB JE CE (CBT I) Mock Test- 10 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE CE (CBT I) Mock Test- 10 below.
Solutions of RRB JE CE (CBT I) Mock Test- 10 questions in English are available as part of our RRB JE Mock Test Series for Civil Engineering (CE) 2025 for Civil Engineering (CE) & RRB JE CE (CBT I) Mock Test- 10 solutions in Hindi for RRB JE Mock Test Series for Civil Engineering (CE) 2025 course. Download more important topics, notes, lectures and mock test series for Civil Engineering (CE) Exam by signing up for free. Attempt RRB JE CE (CBT I) Mock Test- 10 | 100 questions in 90 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study RRB JE Mock Test Series for Civil Engineering (CE) 2025 for Civil Engineering (CE) Exam | Download free PDF with solutions
RRB JE CE (CBT I) Mock Test- 10 - Question 1
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

In which year the total number of candidates admitted in both the institutions together was the second lowest?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 1
In 2002 = 2 + 3.5 = 5.5

In 2003 = 3.5 + 4.0 = 7.5

RRB JE CE (CBT I) Mock Test- 10 - Question 2
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

What was the ratio between the number of candidates admitted in institution A in the year 2006 and the total number of candidates admitted in both the institutions together in the year 2004?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 2
A in 2006 = 4

A, B in 2004 = 9

hence required ratio= 4 : 9

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RRB JE CE (CBT I) Mock Test- 10 - Question 3
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Prabhakar deposited Rs. 5,000 in the bank. The bank offered simple interest at the the .rate of 12 percent per year. After 5 years the total amount received by him will be:

Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 3
Here sum = Rs.5000 = P, Rate = 12% = R

Time = 5 years = T

RRB JE CE (CBT I) Mock Test- 10 - Question 4
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Find the value of

Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 4

RRB JE CE (CBT I) Mock Test- 10 - Question 5
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

Total number of candidates admitted in institution A over all the years together was approximately what percentage of the total number of candidates admitted in institution B over all the years together?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 5
Total No. of candidates in institution A = 2 + 3.5 + 5 + 6 + 4+6.5

= 27 (in thousand)

Total no. of candidates in institution B =3.5 + 4 + 4 + 5.5 + 5 + 6 = 28 (in thousand)

Required percentage=2700/28

=96.42≈ 96 %

RRB JE CE (CBT I) Mock Test- 10 - Question 6
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In the figure, ABCD is a parallelogram with AD = a units, DC = 2a units and DE : EC = 1 : 2. CEFG is a rectangle with FE = 3AE. What is the ratio of the area of a parallelogram and the rectangle?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 6

Area of parallelogramABCD

= DC x AE + AE x EC

= AE x DC

Area of rectangle EFGC = FE x EC

= 1/2

RRB JE CE (CBT I) Mock Test- 10 - Question 7
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The ages of Ram and Shyam are in the ratio 2 : 1. After five years, the ratio of their ages will be 3 : 2. Their present ages are:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 7

Suppose present ages of Ram and Shyam are 2x and x year respectively.

Or 4 x + 10 = 3x + 15

Or x = 5

Hence, present age of Ram = 10 years

And present age of Shyam = 5 years

RRB JE CE (CBT I) Mock Test- 10 - Question 8
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

What was the total number of candidates admitted in institution B over all the years together?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 8
In 2002,Number of candidates admitted in institution B = 3500

In 2003,Number of candidates admitted in institution B = 4000

In 2004,Number of candidates admitted in institution B = 4000

In 2005,Number of candidates admitted in institution B = 5500

In 2006,Number of candidates admitted in institution B = 5000

In 2007,Number of candidates admitted in institution B = 6000

The total number of candidates admitted in institution B over all the years = 28000

RRB JE CE (CBT I) Mock Test- 10 - Question 9
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A rectangle ABCD is inscribed in a circle with centre O. If AC is the diagonal and ∠BAC = 30°, then the radius of the circle will be equal to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 9

RRB JE CE (CBT I) Mock Test- 10 - Question 10
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A hollow sphere of external and internal diameters 8 cm and 4 cm respectively, is melted into a cone of base diameter 8 cm, then the height of the cone will be equal to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 10
Volume of the material of the shell =

Volume of the cone formed =

Therefore

Hence, the height of cone formed is 14cm.

RRB JE CE (CBT I) Mock Test- 10 - Question 11
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The value of sin θ + sin (θ + 120°) + sin (θ + 240°) is equal to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 11
The given expression is sinθ + sin (θ+120°) + sin (θ+240°)

RRB JE CE (CBT I) Mock Test- 10 - Question 12
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The area of a rhombus is 120 cm2. If one of its diagonals is of length 10 cm, then the length of one of its sides is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 12

RRB JE CE (CBT I) Mock Test- 10 - Question 13
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A mixture of 70 litres contains milk and water in the the ,’,’,’,’ratio of 3 :4. How many litres of milk must be added to mixture so as to make the ratio 5 : 4?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 13
Milk : Water

New ratio 5 : 4

Old ratio 3 : 4

Difference 2 : 0

Sum of old ratios = 3 + 4 = 7

∴ Milk added = (2/7)×70=20 litres

RRB JE CE (CBT I) Mock Test- 10 - Question 14
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A man rows boat to a place covering 72 km distance and back in 15 hours. He finds that he can row 3 km with the stream in the same time as 2 km against the stream. Find the speed of the stream.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 14
Speed downstream : speed upsteam = 3 : 2

∴ Time taken to row downstream : Time taken to row upsteam = 2 : 3

But, total time taken = 15 hours

RRB JE CE (CBT I) Mock Test- 10 - Question 15
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In an examination A gets 10% marks less than B and B gets 25% marks more than C and C gets 20% marks less than D. If A gets 270 marks in 500 then D get-
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 15

RRB JE CE (CBT I) Mock Test- 10 - Question 16
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If from the top of a light house 100 metre high the angle of depression of a boat is tan-1(5/12) then the distance (in metre) between the boat and the light house is equal to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 16

Given that the height of light house = 100m

RRB JE CE (CBT I) Mock Test- 10 - Question 17
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The perimeter of a rhombus is 40 cm and the measure of an angle is 60o, then the area of it is?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 17

RRB JE CE (CBT I) Mock Test- 10 - Question 18
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The average of three numbers is 40. The first number is twice the second and the second one is thrice the third number. The difference between the largest and the smallest numbers is?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 18
Let the third number be x

∴ second number = 3x

RRB JE CE (CBT I) Mock Test- 10 - Question 19
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ABCD is a trapezium in which AB is parallel to DC. If the diagonals intersect at O, then which one of the following is correct?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 19

RRB JE CE (CBT I) Mock Test- 10 - Question 20
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In a right angled triangle, one acute angle is double the other. If the length of the smallest side is a. The length of the hypotenuse is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 20

RRB JE CE (CBT I) Mock Test- 10 - Question 21
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Study the following graph and answer the questions that follow.

No. of candidates admitted in two different Institution(In thousands)

What was the difference between the total number of candidates admitted in institution A in the year 2004 and 2007 together and the total number of candidates admitted in institution B in the year 2003 and 2007 together?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 21
A in 2004 and 2007

= 5 + 6.5 = 11.5

B in 2003 and 2007

4 + 6 = 10.0

Difference = 1500

RRB JE CE (CBT I) Mock Test- 10 - Question 22
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In the given figure, if x = 20°, find y.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 22

RRB JE CE (CBT I) Mock Test- 10 - Question 23
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Find the value of (x3+y3+z3 - 3xyz), if x +y+z = 9 and xy + yz+zx = 26:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 23
Given (x+y+z) = 9

∴ (x + y + z)2 = 92 = 81

∴ x2 + y2 + z2 + 2 × 26 = 81

∴ x2 + y2 + z2 = 81 - 52 = 29

Now, x3 + y3 + z3 - 3xyz

= (x + y + z) [(x2 + y2 + z2)] - (xy + yz + xz)]

= 9 × (29 - 26) = 27

RRB JE CE (CBT I) Mock Test- 10 - Question 24
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If then the volume of is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 24

RRB JE CE (CBT I) Mock Test- 10 - Question 25
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Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 25
Hence, first number is 1 - 1 = 0

we know that, any number multiplied by 0 always gives 0.

∴ 0 is the answer.

RRB JE CE (CBT I) Mock Test- 10 - Question 26
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A carriage driving in a fog passed a man who was walking at the rate of 3 km an hour in the same direction. He could see the carriage for 4 minutes and it was visible to him upto a distance of 100 m. What was the speed of the carriage?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 26
The distance travelled by the man in 4 min = [(3×1000)/60] ×4=200 meters

∴ distance travelled by the carriage in 4 min = (200 + 100) = 300 meters

∴ speed of the carriage = (300/4)×(60/1000) km/hour

=9/2 km/hour

RRB JE CE (CBT I) Mock Test- 10 - Question 27
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Two partners invested Rs. 50,000 and Rs. 70,000 respectively in a business and agreed that 70% of profit should be divided equally between them and the remaining profit in the ratio of investment. If one partner gets Rs. 90 more than the other then the total profit made in the business is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 27
Ratio of profits = 50,000 : 70,000 = 5 : 7

RRB JE CE (CBT I) Mock Test- 10 - Question 28
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The ratio of the first and second train fares between two stations is 3:1 and that of the numbers of passengers travelling between the two stations by first and second classes is 1:50. In on a particular day, ₹1325 are collected from passengers travelling between the two stations, then the amount collected from the second class passengers is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 28

RRB JE CE (CBT I) Mock Test- 10 - Question 29
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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 CE (CBT I) Mock Test- 10 - Question 29
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 CE (CBT I) Mock Test- 10 - Question 30
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If a man’s age is 125% of what it was 10 years ago, but 83(⅓)% of what it will be after 10 years. His present age is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 10 - Question 30
Let the present age be x years

Then 125% of (x-10) = x;and 83(⅓)% of (x+10)= x

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