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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- 8 - Civil Engineering (CE) MCQ

RRB JE CE (CBT I) Mock Test- 8 - 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- 8

RRB JE CE (CBT I) Mock Test- 8 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- 8 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The RRB JE CE (CBT I) Mock Test- 8 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- 8 below.
Solutions of RRB JE CE (CBT I) Mock Test- 8 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- 8 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- 8 | 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- 8 - Question 1
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Ashok remembers his birthday is after 21st December. While his mother remembers his birthday is before 23rd December. On which date of December is his birthday?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 1
As Ashok remembers, his birthday is after the 21st and his mother remembers his birthday is before 23rd, therefore, his birthday is on 22nd December.
RRB JE CE (CBT I) Mock Test- 8 - Question 2
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If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately-

Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 2
Mode + 2 Mean = 3 Median

Mode = 3 × 22 - 2 × 21 = 66 - 42 = 24.

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RRB JE CE (CBT I) Mock Test- 8 - Question 3
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O is the center of the circle. AC and BD are two chords of the circle intersecting each other at P. If ∠APB=30∘,∠APB=300, and ∠AOB=45∘∠AOB=450 then ∠DOC∠DOC is equal to?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 3

RRB JE CE (CBT I) Mock Test- 8 - Question 4
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A does 60% of a work in 45 days. He then calls B, and they together finish the remaining work in 18 days. How long B alone would take to do the whole work?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 4
Time of A to do complete work =(45/60)×100=75

Time of A+B to do complete work =(18/40)×100=45

Let the time of B to do complete work alone is T =1/45−1/75=1/T

T=112.5 days

RRB JE CE (CBT I) Mock Test- 8 - Question 5
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Abhinash gets 32% marks out of 300 in computer Science. How much marks out of 200 he should get in java language paper so that out of total marks his percentage becomes 46% -
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 5
Abhinash get marks in computer science = 300×32/100=96

Total marks in both subjects = 500×46/100=230

In java = 230−96=134

Required percentage = =134/200×100=67%

RRB JE CE (CBT I) Mock Test- 8 - Question 6
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There are two taps P and Q that can fill a tank in 12 and 24 minutes respectively. Both of them are opened together and after 4 minutes Q is closed. How long more it take for P alone to fill the tank?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 6
Part of the tank filled in 1 minute by tank P=1/12 Part of the tank filled in 1 minute by tan⁡kQ=1/24 Part of the tank filled by P and Q in 4 minutes

Equating Time and part filled, =1/12×2/1=(1/x) x=6 minutes

RRB JE CE (CBT I) Mock Test- 8 - Question 7
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Find the compound interest on a sum of Rs 425 compounded half yearly for 2 years at the rate of 4%.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 7
The compound interest for the given condition can be given by,

Therefore, the required compound interest is,

Therefore, the compound interest is Rs 35.03.

RRB JE CE (CBT I) Mock Test- 8 - Question 8
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Average age of 6 boys is 14 years. The average age of 11 girls is 12 years. What is the average age (in years) of all boys and girls?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 8
Average of 6 boys =14 years And the average of 11 girls =12 years So, Overall average

RRB JE CE (CBT I) Mock Test- 8 - Question 9
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The value of tan 40. tan 430 .tan 470 .tan 860 is-
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 9

We are tasked with finding the value of:

tan⁡(4∘)×tan⁡(43∘)×tan⁡(47∘)×tan⁡(86∘)\tan(4^\circ) \times \tan(43^\circ) \times \tan(47^\circ) \times \tan(86^\circ)

Step 1: Simplify the expression using trigonometric identities

We can simplify the expression by noticing that:

tan⁡(86∘)=cot⁡(4∘)\tan(86^\circ) = \cot(4^\circ)

This is because tan⁡(90∘−x)=cot⁡(x)\tan(90^\circ - x) = \cot(x), and 90∘−4∘=86∘90^\circ - 4^\circ = 86^\circ.

So, we can rewrite the expression as:

tan⁡(4∘)×tan⁡(43∘)×tan⁡(47∘)×cot⁡(4∘)\tan(4^\circ) \times \tan(43^\circ) \times \tan(47^\circ) \times \cot(4^\circ)

Now, tan⁡(4∘)×cot⁡(4∘)=1\tan(4^\circ) \times \cot(4^\circ) = 1, so the expression simplifies to:

1×tan⁡(43∘)×tan⁡(47∘)1 \times \tan(43^\circ) \times \tan(47^\circ)

Step 2: Use the identity for tan⁡(43∘)×tan⁡(47∘)\tan(43^\circ) \times \tan(47^\circ)

We can use the identity:

tan⁡(x)×tan⁡(90∘−x)=1\tan(x) \times \tan(90^\circ - x) = 1

Since 43∘+47∘=90∘43^\circ + 47^\circ = 90^\circ, we have:

tan⁡(43∘)×tan⁡(47∘)=1\tan(43^\circ) \times \tan(47^\circ) = 1

Step 3: Final Answer

Thus, the entire expression simplifies to:

1×1=11 \times 1 = 1

So, the value of tan⁡(4∘)×tan⁡(43∘)×tan⁡(47∘)×tan⁡(86∘)\tan(4^\circ) \times \tan(43^\circ) \times \tan(47^\circ) \times \tan(86^\circ) is:

1\boxed{1}

RRB JE CE (CBT I) Mock Test- 8 - Question 10
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Profit obtained on selling an article for Rs 310 is equal to the loss incurred on selling that article for Rs 230. What will be the loss percentage when the selling price is Rs 180?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 10
Let the profit \& loss be x ATQ. 310−x=230+x

2x=80

x=40

CP of article =Rs.270 when the selling price is Rs 180 loss%

RRB JE CE (CBT I) Mock Test- 8 - Question 11
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Diameter of the cylindrical vessel is 14cm having some water. When a sphere of iron is totally submerged in this water level of water increased by 9(1/3)cm. What is the radii of the sphere?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 11
d=14 then r=7 let the radius of sphere =R

∴ the volume of sphere = volume of the cylinder

∴ the volume of sphere = volume of a ,cylinder (4/3)π×R3=π×7×7×28/3

R3=343

R = 7cm

RRB JE CE (CBT I) Mock Test- 8 - Question 12
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The given figure, AB = 16 cm, AC = 12 cm, BC = 14 cm, then the length of median AD will be -
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 12
(16)2+(12)2=2(AD2+CD2)

256+144=2AD2+98

AD = √151

RRB JE CE (CBT I) Mock Test- 8 - Question 13
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Directions

Each question is followed by 2 statements. Go through these questions and answer the questions as per the instructions : choose your answer as: (A) If the question can be answered by using only statement (I), but not by using statement (II) alone. (B) If the question can be answered by using only statement (II), but not by using statement (I) alone. (C) If the question can be answered by using both the statements together, but cannot be answered by using either statement alone. (D) If the question cannot be answered by using both the statements together.

What is the number?

A. The product of two digits of a number is 27.

B. The sum of the two digits is 12. The ratio of the ten's digit to the unit digit is 3:1.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 13
Let the ten's and unit's digit be x and y respectively. From A,xy=27

From B,x+y=12 and Now, 3y+y=12

4y=12=>y=3

x=3y=9

Number =93

Thus, B alone gives the answer and A alone does not give the answer

RRB JE CE (CBT I) Mock Test- 8 - Question 14
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Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 km/h and 60 km/h respectively. B overtakes A at M, which is on the way from R to S. What is the distance from R, where B overtakes A?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 14
speed of A=40km/hr

speed of B=60km/hr

In two hour A cover distance =2×40=80km relative speed of A and B=60−40=20km/hr the A and B meet at point M in hours =80/20=4 In 4 hours, B cover distance = 4 x 60 = 240km.

RRB JE CE (CBT I) Mock Test- 8 - Question 15
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The area of a square is 72.25 cm2hoursthe the . Find its perimeter (in cm).
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 15
Area of square = side side =72.25

Side =8.5

Perimeter = 4 side = 4 8.5=34cm

RRB JE CE (CBT I) Mock Test- 8 - Question 16
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Arrange the fractions 3/4, 5/12, 13/16, 16/29, 3/8 in their ascending order of magnitude.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 16

RRB JE CE (CBT I) Mock Test- 8 - Question 17
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The concentrations of acid in two solution A and B is 19 percent and 9 percent respectively.In what ratio solution A and B should be mixed to get a solution having 17 percent acid?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 17
Using Allegation method-

The required ratio=8:2=4:1

RRB JE CE (CBT I) Mock Test- 8 - Question 18
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Find the greatest number which on dividing 1580 and 3800 leaves remainders 8 and 1 respectively
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 18
HCF of (1580-8) and (3800-1) will be the greatest number

So HCF of 1572 and 3799

So, 131 Is the number.

RRB JE CE (CBT I) Mock Test- 8 - Question 19
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What must be added to each term of the ratio 2 : 5 so that it may equal to 5 : 6 ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 19
Suppose x must be added.

Then, 2+x/5+x=5/6

12 + 6x = 25 + 5x

x = 13

RRB JE CE (CBT I) Mock Test- 8 - Question 20
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The angle of elevation of the top of a tower at a distance of 25 m from its foot is 600 . The approximate height of the tower is -
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 20
We know that

tan⁡θ=p/b

Therefore:

RRB JE CE (CBT I) Mock Test- 8 - Question 21
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How many whole numbers are there between 52 and 356 which are exactly divisible by 6?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 21
First whole number which is divisible by 6 between 52 and 356 is 54

Last whole number which is divisible by 6 between 52 and 356 is 354

So, by using AP formula:-

Last term = first term + (number of terms-1) × Common difference

354 = 54 + (n-1) × 6

300 = (n-1) × 6

50= (n-1)

So, n =51

Therefore,

There are 51 whole numbers between 52 and 356 which are exactly divisible by 6.

RRB JE CE (CBT I) Mock Test- 8 - Question 22
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Directions

In the following problems, there is one question and some statements given below the question. You have to decide whether the data given in the statements is sufficient to answer the question. Read all the statements carefully and find out which of the statements is/are sufficient to answer the given question. Choose the correct alternative for each question.

Question:

What will be the total weight of 10 poles, each of the same weight ?

Statements:

I. One-fourth of the weight of each pole is 5 kg.

II. The total weight of three poles is 20 kilograms more than the total weight of two poles.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 22
From I, we conclude that weight of each pole = (4 x 5) kg = 20 kg.

So, total weight of 10 poles = (20 x 10) kg = 200 kg.

From II, we conclude that:

Weight of each pole = (weight of 3 poles) - (weight of 2 poles) = 20 kg.

So, total weight of 10 poles = (20 x 10) kg = 200 kg.

RRB JE CE (CBT I) Mock Test- 8 - Question 23
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B starts some business by investing Rs 90000. After 4 months, D joins business by investing Rs 80000. At the end of the year, in what ratio will they share the profit?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 23

RRB JE CE (CBT I) Mock Test- 8 - Question 24
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Calculate the total numbers of prime factors in the expression (9)11 × (5)7 × (3)2 × (17)2
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 24
(9)11×(5)7×(7)5×(3)2×(17)2

⇒(3)22×(5)7×(7)5×(3)2×(17)2

Total no. of prime factors, n= Sum of powers ∴n=22+7+5+2+2

∴ n = 38

RRB JE CE (CBT I) Mock Test- 8 - Question 25
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What is the net discount (in %) for successive discounts of 30% and 50%?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 25
Cumulated Discount = Discount 1+ Discount 2 -
RRB JE CE (CBT I) Mock Test- 8 - Question 26
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In a mixture of 45 litres, the ratio of liquid A and liquid B is 7 : 2. If 11 litres of liquid B is added to the mixture, then what will be the ratio of liquid A and liquid B in the new mixture?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 26
Let Liquid A and Liquid B be 7x and 2x respectively.

ATQ,

7x+2x = 45

9x = 45

So, x = 5

And, Liquid A is 35 and Liquid B is 10 litres respectively.

Liquid B becomes 21 litres.

Therefore,

Liquid A/Liquid B = 35/21 = 5/3

RRB JE CE (CBT I) Mock Test- 8 - Question 27
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If then the value of x is :
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 27

RRB JE CE (CBT I) Mock Test- 8 - Question 28
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A shopkeeper offer Rs. 19% discount on a mobile phone with a marked price of Rs. 7400. If he still earns a profit of Rs. 500, what is the cost price of the phone?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 28
Selling Price = Marked Price × (100-Discount) %

Therefore,

SP = 7400 × (100-19) %

SP=7400×81/100

SP = 5994

So,

Cost Price = Selling Price - Profit

CP = 5994-600

CP = 5494

RRB JE CE (CBT I) Mock Test- 8 - Question 29
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The following equation is incorrect. Which two signs should be interchanged to correct the equation?

12 + 8 - 25 ÷ 10 x 18 = 14
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 29
12+8−25÷10×18

After interchanging signs, 12+8×25÷10−18

⇒12+8×25/10−18

⇒12+20−18

⇒ 14

RRB JE CE (CBT I) Mock Test- 8 - Question 30
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Two pipes can fill a tank in 1hr. and 72 min respectively both the pipes are open together but due to some dust in pipe, the efficiency pipes are decreased to 5/6 and 9/10 of their actually efficiency of respectively. After some time the dust in both pipes get clear together and both pipes run with their full efficiency, and then reaming the tank can be filled in 31 min. Find in how much time it takes to clear the dust.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 8 - Question 30
L.C.M. ⋯=360

A=60−…−6

B=72−⋯−5

An=6×5/6=5,Bn=5×9/10=4.5

(6+5)x 31+(5+4.5)x x= 360, x = 2

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