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Civil Engineering (CE) Exam  >  Civil Engineering (CE) Tests  >  RRB JE CE (CBT I) Mock Test- 5 - Civil Engineering (CE) MCQ

RRB JE CE (CBT I) Mock Test- 5 - Civil Engineering (CE) MCQ


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30 Questions MCQ Test - RRB JE CE (CBT I) Mock Test- 5

RRB JE CE (CBT I) Mock Test- 5 for Civil Engineering (CE) 2024 is part of Civil Engineering (CE) preparation. The RRB JE CE (CBT I) Mock Test- 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The RRB JE CE (CBT I) Mock Test- 5 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- 5 below.
Solutions of RRB JE CE (CBT I) Mock Test- 5 questions in English are available as part of our course for Civil Engineering (CE) & RRB JE CE (CBT I) Mock Test- 5 solutions in Hindi for Civil Engineering (CE) 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- 5 | 100 questions in 90 minutes | Mock test for Civil Engineering (CE) preparation | Free important questions MCQ to study for Civil Engineering (CE) Exam | Download free PDF with solutions
RRB JE CE (CBT I) Mock Test- 5 - Question 1
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What should come in place of the question mark (?) in the following question?

36(½) x 64(⅓) ÷ 8(⅓) = ?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 1

RRB JE CE (CBT I) Mock Test- 5 - Question 2
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An equilateral triangle circumscribes all the six circles, each with radius 1 cm. What is the perimeter of the equilateral triangle?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 2

DE = GH = 4, ∠ADG = 60°

Then from 30 - 60 - 90 theorem

AG = 3√, BH =3√, then

AB = AG+GH+BH

Then perimeter of △ABC = 3×(3√+4+3√)=3×(23√+4)=6(3√+2)

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RRB JE CE (CBT I) Mock Test- 5 - Question 3
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Divide 800 into two parts such that the ratio of one may be to the other in the ratio 5:3?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 3
5x + 3x = 800

8x = 800

x = 100

100 × 5 = 500

100 × 3 = 300

= 500, 300

RRB JE CE (CBT I) Mock Test- 5 - Question 4
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A horse takes 2(½) seconds to complete a round around a circular field. If the speed of the horse was 66 m/s, then the radius of the field is :[Given π=22/7]
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 4
Distance covered in one round

= Circumference of circle

= 2πr

Speed = 66 m/s

Time = 5/2

S=D/T

D=S×T

=66×(5/2)

D=165m

2π=165

=2×(22/7)×t=165

R = (165x7)/44 = 26.25m

RRB JE CE (CBT I) Mock Test- 5 - Question 5
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In a mixture of 60 litres, the ratio of milk and water is 2 : 1. What amount of water must be added to make the ratio of milk and water as 1 : 2?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 5
Milk in the original mixture is (2/3)×60= 40 litres and water= 20 litres.

In the new mixture, the ratio of milk to water is 1 : 2.

if milk is 40 litres, water should be 80 litres;

Thus 60 litres of water must be added.

RRB JE CE (CBT I) Mock Test- 5 - Question 6
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Heena and Meena start walking from a fixed point in the opposite direction at an average speed of 2 Kmph. And 4 Kmph. Respectively. How many Km. will they be apart from each other in 5 hours?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 6
Distance covered by Heena = 2 x 5 = 10 Km

Distance Covered by Meena = 4 x 5 = 20 Km

Total distance = (10 + 20) = 30Km

RRB JE CE (CBT I) Mock Test- 5 - Question 7
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A man can row 15 kmph in still water and he finds that it takes him twice as much time to row up than as to row down the same distance in the river. The speed of the current is :
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 7
Let the speed of current be x kmph

∴ Down stream speed = 15 + x kmph

Up stream speed = 15 - x kmph

Let the distance be d km.

∴ Down stream time = d/(15+x)

Upstream time = d/(15-x)

According to question,

2d/(15+x) = d/(15-x)

⇒ 2(15 - x) = (15 + x)

⇒ 30 - 2x = 15 + x

⇒ 15 = 3x

⇒ x = 15/3 = 5

∴ speed of current = 5 kmph

RRB JE CE (CBT I) Mock Test- 5 - Question 8
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For what value of m, the system of equations mx + 2y =2 and 3x + y = 1 will be coincident?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 8
Two linear equations will coincide if there are infinite number of solutions,

I.e. a1/a2 = b1/b2 = c1/c2

for a1 x +b1 y = c1

and a2x +b2y = c2

∴ m/3 = 2/1 = 2/1

⇒ m = 6

RRB JE CE (CBT I) Mock Test- 5 - Question 9
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If the numerator of a fraction is increased by 200% and the denominator is increased by 350%, the resultant fraction is 5/12 . What was the original fraction?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 9
Let the original fraction be x/y

RRB JE CE (CBT I) Mock Test- 5 - Question 10
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if x+y = 25 and x2y3 +y2x3 =25, what is the value of xy?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 10

x2y3 +y2x3 =25

⇒ x2 y2 (x + y) = 25

⇒ (xy)2 (x + y) = 25

⇒ (xy)2 = 1

(∵ x + y = 25)

⇒ xy = ± 1

RRB JE CE (CBT I) Mock Test- 5 - Question 11
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The angle of elevation of the top of a tower at a point on the ground is 30o. On walking 16 meters towards the tower, the angle of elevation becomes 60o. Find the height of the tower.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 11

RRB JE CE (CBT I) Mock Test- 5 - Question 12
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Mohan purchased a bike for Rs 800/- including sales tax of 20%. Find the actual price of the bike?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 12
(800 x 100)/120 = 2000/3 = 666.67
RRB JE CE (CBT I) Mock Test- 5 - Question 13
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If 5 girls can embroider a dress in 9 days, then the number of days taken by 3 girls will be _____.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 13
Days taken by 5 girls to embroider a dress = 9 days

Days taken by 1 girl to embroider a dress = 9 x 5 days

Days taken by 3 girls to embroider a dress = (9 x 5)/3 days = 15 days

RRB JE CE (CBT I) Mock Test- 5 - Question 14
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If p = cosec θ + cot θ, then the value of p + (1/p) =
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 14
Given p=cosec⁡θ+cot⁡θ

⇒ (1/p) =cosec⁡θ−cot⁡θ

P+(1/p)=cosec⁡θ+cot⁡θ+cosec⁡θ−cot⁡θ=2cosec⁡θ = 2/sinθ

RRB JE CE (CBT I) Mock Test- 5 - Question 15
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One man or two women or three boys can do a piece of work in 88 days. One man, one woman and one boy will do it in _____.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 15
1 man+ 1 woman + 1 boy

RRB JE CE (CBT I) Mock Test- 5 - Question 16
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We have to divide a sum of Rs.13,950 among three persons A, B and C. B must get the double of A’s share and C must get Rs.50 less than the double of B’s share. The share of A will be:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 16
Let Rs. x be the part of A, then

x+ 2x + 4x - 50 = 13950

7x = 14000

x = 2000

RRB JE CE (CBT I) Mock Test- 5 - Question 17
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Find the measures of an angle which is the complement of itself.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 17
Let the measure of angle = x°

measure of its complement = x°

∴ x°+x°= 90° ⇒ x° = 45°

RRB JE CE (CBT I) Mock Test- 5 - Question 18
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sin6A+cos6A is equal to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 18
a3 + b3 = (a+b)3 – 3ab (a+b)

Let a = sin2A, b = cos2A, so that

a + b = sin2A + cos2A = 1

⇒ sin6A+cos6A = 1- 3 sin2 A cos2 A

RRB JE CE (CBT I) Mock Test- 5 - Question 19
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The sum of the internal angles of a regular polygon is 1440°. The number of sides is:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 19
Let the number of sides of regular polygon be n.

(n – 2) × 180 = 1440° ⇒ n = 10

RRB JE CE (CBT I) Mock Test- 5 - Question 20
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The difference in selling price of radio at gains of 10% and 15% is Rs.30. Find the price of the radio?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 20
15 - 10 = 5% =30

Then 100% = (100x30)/5 = 600

RRB JE CE (CBT I) Mock Test- 5 - Question 21
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△ABC is a right angled at A, and AD is the altitude to BC. If AB= 7cm and AC= 24cm. Find the ratio of AD is to AM if M is the mid-point of BC:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 21

"AM is the median of a right angled triangle.

And

RRB JE CE (CBT I) Mock Test- 5 - Question 22
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A car travels 120 km from A to B at 30 kmph. but returns the same distance at 20 kmph. The average speed for the round tip is closest to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 22
Average Speed

RRB JE CE (CBT I) Mock Test- 5 - Question 23
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If the average weight of 6 students is 50 kg, the a the average weight of 2 students is 57 kg, and the average weight of 2 students is 55 kg, then the average weight of all students is?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 23
Required average weight

RRB JE CE (CBT I) Mock Test- 5 - Question 24
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In what time period Rs.100, 000,000 will amount to Rs.104,060,401 at 2% per annum at compound interest, interest being compounded half yearly?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 24
104,060,401 = 100,000,000 (1 + (1/100))2t

104,060,401/100,000,000 = (101/100) 2t

(101)4/(100)4 = (101/100) 2t

(101/100)4 = (101/100) 2t

2t = 4

t = 2 years

RRB JE CE (CBT I) Mock Test- 5 - Question 25
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ABC is a right-angled triangle, right angled at B. The external bisector of ∠A meets CB produced at D. If AC = 25 cm, BC = 7 cm, find the length, of CD.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 25

RRB JE CE (CBT I) Mock Test- 5 - Question 26
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DIRECTION: Study the table carefully to answer the questions that follow.

What was the respective ratio between the number of items of type-2 produced by Company M in the month of May and the number of items of type-1 produced by Company-L in the month of March?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 26
Required ratio = 666:864 = 37:48.
RRB JE CE (CBT I) Mock Test- 5 - Question 27
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DIRECTION: Study the table carefully to answer the questions that follow.

What was the average number of items of type-2 produced by all the companies together in the month of January?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 27
Required average= (452 + 654 + 324 + 564)/4

= 1994/4 = 498.5.

RRB JE CE (CBT I) Mock Test- 5 - Question 28
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A sum of Rs. 210 was taken as a loan. This is to be paid back in two equal installments. If the rate of interest be 10% compounded annually, then the value of each installment is
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 28
Let x be the value of installment.

Principal = Present value of x for 1 year + Present value of x for 2 years

RRB JE CE (CBT I) Mock Test- 5 - Question 29
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A child reshapes a cone made up of clay of height 24 cm and radius 6 cm into a sphere. The radius (in cm) of the sphere is
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 29
⇒ Volume of cone = Volume of sphere

⇒1/3×π×r2h=4/3×π×R3

⇒r2×h=4×R3

⇒6×6×24=4×R3

⇒6×6×6=R×R×R

SO WHERE R IS EQUAL TO 6 CM

⇒ Therefore radius of the sphere = 6 CM

RRB JE CE (CBT I) Mock Test- 5 - Question 30
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If θ lies in the second quadrant, then is equal to:
Detailed Solution for RRB JE CE (CBT I) Mock Test- 5 - Question 30

= 2secθ

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