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RRB JE CE (CBT I) Mock Test- 1 - 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- 1

RRB JE CE (CBT I) Mock Test- 1 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- 1 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The RRB JE CE (CBT I) Mock Test- 1 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- 1 below.
Solutions of RRB JE CE (CBT I) Mock Test- 1 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- 1 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- 1 | 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- 1 - Question 1
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Two trains 165 m and 175 m long run at the speed of 50 km/h and 52 km/h respectively in opposite directions on parallel tracks. What is the time (in seconds) they take to cross each other?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 1
Relative speed = (50 + 52) km/h = (102 x 5/18) m/s = (255/9) m/s.

Distance covered in crossing each other = (165 + 175) = 340 m.

∴ Required time = (340 x 9/255) = 12 s.

RRB JE CE (CBT I) Mock Test- 1 - Question 2
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If two equations: 2x2 - Ax - 20 = 0 and 2x2 - 11x + 12 = 0 have one root in common, then what is the value of 'A' if the value of A is positive?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 2
2x2 - 11x + 12 = 0

(2x - 3)(x - 4) = 0

x = 1.5 and 4

Since both the equations have one root in common-

Case 1: When x = 1.5 is common-

2(1.5)2 - A(1.5) - 20 = 0

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RRB JE CE (CBT I) Mock Test- 1 - Question 3
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Divya has 3 types of coins having denomination of 20p, 10p and 5p. Their numbers are in the ratio of 1: 2 : 3. If Divya has total money of Rs. 44 then how many 5p coins are there?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 3
Since the ratio between the numbers of coins of 20p, 10p and 5p is 1 : 2 : 3 So let us assume the number of 20p coins = a

Then number of 10p coins is 2a and number of 5p coins is 3a.

Total worth of 20p coins = number of coins X worth of each coin = a X (1/5) = Rs (a/5)

Total worth of 10p coins = 2a X (1/10) = Rs (a/5)

Similarly total worth of 5p coins = 3a X (1/20) = Rs (3a/20)

Since total money with the Divya is Rs 44 hence

a/5 + a/5 + 3a/20 = 44 => 2a/5 + 3a/20 = 44

=> 8a/20 + 3a/20=44 => 11a/20 =44 => a = 80

So the number of 5p coins = 3a = 80 x 3 = 240

RRB JE CE (CBT I) Mock Test- 1 - Question 4
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The difference between 42% and 23% of a number is 304. What is 72% of that number?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 4
Let the number be 100x.

∴ 42x - 23x = 304

=> 19x = 304

=> x = 16

∴ The number is 100 x 16 = 1600.

=> 72% of the number = 0.72 x 1600 = 1152

RRB JE CE (CBT I) Mock Test- 1 - Question 5
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What is the value of 60 ÷ [{21 ÷ 7 x √(√625 + √121)} − 6] x 5 - 42 ÷ 3 x 2 = ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 5
60 ÷ [{21 ÷ 7 x √(√625 + √121)} − 6] x 5 - 42 ÷ 3 x 2

= 60 ÷ [{3 x √(25 + 11)} − 6] x 5 - 14 x 2

= 60 ÷ [{3 x √(36)} − 6] x 5 - 28

= 60 ÷ [{3 x 6} − 6] x 5 - 28

= 60 ÷ [18 − 6] x 5 - 28

= 60 ÷ 12 x 5 - 28

= 5 x 5 - 28

= -3

RRB JE CE (CBT I) Mock Test- 1 - Question 6
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Three pipes x, y and z can fill a water tanker in 20, 25 and 30 minutes respectively. All the three pipes are opened to fill the tanker and if the pipe y is closed 2 minutes before the water tanker gets filled, then in how much time will the tank get filled?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 6
Let the time after which the water tanker gets filled be 't' minutes.

In one minute pipe x can fill 1/20 part.

In t minutes, pipe x can then fill t/20 part.

Therefore, t/20 + t-2/25 + t/30 = 1

On solving, we get t = 324/37 minutes.

RRB JE CE (CBT I) Mock Test- 1 - Question 7
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If [4x - (1/12x) = 8], then what is the value of [x2 + {1/(2304x2)}] ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 7
4x - 1/12x = 8

=> 4(x - 1/48x) = 8

=> x - 1/48x = 2 ------ (i)

Squaring both sides, we get

(x - 1/48x)2 = 4

=> x2 + 1/(2304x2) - 2x/48x = 4 ------ [(a - b)2 = a2 + b2 - 2ab]

x2 + 1/2304 x2 = 4 + 1/24 = 97/24

RRB JE CE (CBT I) Mock Test- 1 - Question 8
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A boy throws a ball of mass 4 kg vertically downwards for a height of 20 metres. What is the velocity of the ball which touches the ground? (Take g= 10 m/s2)
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 8
According to question,

Mass of ball = m = 4 kg, elevation from ground= h= 20 m; v= velocity at touch point

So, mgh = ½ (mv2)

Gives v= 20 m/s

mg = 1/2 (mv2)

2gh = v2

v2 = 2 x 10 x 20 = 400

v = 20 m/sec

RRB JE CE (CBT I) Mock Test- 1 - Question 9
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Express 0.78414141.. into (p/q) form.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 9
Let x = 0.78414141..

=> 100x = 78.414141.. (I)

=> 10000x = 7841.414141.. (II)

(II) - (I) gives,

9900x = 7763

=> x = 7763/9900

RRB JE CE (CBT I) Mock Test- 1 - Question 10
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Nidhi borrowed Rs. 10000 from a bank for a period of two years at the rate of 10% per annum. From that amount, she invested Rs. 3500 in another bank which offers simple interest @12.5% per annum and remaining amount she invested into a chit fund which offers compound interest of 10% per annum. What is the profit obtained by Nidhi at the end of two years?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 10
Total interest that Nidhi has to pay after 2 years = Rs. (10000 x 2 x 10)/100 = Rs. 2000

Interest obtained by Nidhi from the bank = Rs. (3500 x 2 x 12.5)/100 = Rs. 875

Interest obtained by Nidhi from Chit fund = 6500(1 + 0.1)2 - 6500 = Rs. 1365

So total interest obtained by Nidhi = Rs. (875 + 1365) = Rs. 2240

Profit = Rs. (2240 - 2000) = Rs. 240

RRB JE CE (CBT I) Mock Test- 1 - Question 11
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A book was marked initially at Rs. 500 which A purchased at 8% discount. He marked up the price by 15% on his purchased price and sold after Rs. 20 discount to B. B marked up the price by 20% above his purchased price and sold to C after allowing Rs. 50.9 discount. What is the percent profit gained by B when sold to C?

Given below are the steps involved. Arrange them in the sequential order.

(A) Initial marked price of book = Rs. 500 and purchased price for A = 92% of 500 = Rs. 460

(B) Purchased price of book by B = 529 - 20 = Rs. 509

(C) Per cent profit = [(559.9 - 509)/509] x 100 = 10%

(D) Marked price of book by A = 115% of 460 = Rs. 529

(E) Marked price of book by B = 120% of 509 = Rs. 610.8 and purchased price of book by C = 610.8 - 50.9 = Rs. 559.9
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 11
The correct order is

(A) Initial marked price of book = Rs. 500 and purchased price for A = 92% of 500 = Rs. 460

(D) Marked price of book by A = 115% of 460 = Rs. 529

(B) Purchased price of book by B = 529 - 20 = Rs. 509

(E) Marked price of book by B = 120% of 509 = Rs. 610.8 and purchased price of book by C = 610.8 - 50.9 = Rs. 559.9

(C) Per cent profit = [(559.9 - 509)/509] x 100 = 10%

RRB JE CE (CBT I) Mock Test- 1 - Question 12
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Seema purchased an item for Rs.9,600 and sold it for a loss of 5 percent. From that money she purchased another item and sold it for a gain of 5 percent. What is her overall gain/ loss?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 12

SP = 9600 x 95 / 100 = Rs. 9,120

Second S.P. = 9120 x 100 / 100 = Rs. 9,576

Loss = 9600 - 9576 = 24

RRB JE CE (CBT I) Mock Test- 1 - Question 13
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What is the angle between the hands of a clock at 8:30 a.m.?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 13
At 8:30 a.m. minute hand points at 6 and and hour hand points exactly between 8 and 9.

So these two hands are apart by 2.5 blocks.

We know that the difference of angle of one block is 30°.

Hence the angle between the hands = 2.5 X 30° = 75°

RRB JE CE (CBT I) Mock Test- 1 - Question 14
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Following data given below shows the sex ratio of India in last 10 years. Find the mode of sex ratio in India.

976, 979, 1084, 993, 1084, 993, 935, 979, 991, 979
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 14
The mode is a statistical term that refers to the most frequently occurring number found in a set of numbers.

Since, 979 is occurring frequently and number of occurrence of 979 is 3.

Hence mode of sex ratio in India in last 10 years is 979.

RRB JE CE (CBT I) Mock Test- 1 - Question 15
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If 7x5 - 2x4 + 3x3 - 4x2 + 6x + 2k is divisible by (x + 1), then what is the value of k?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 15
Since, 7x5 - 2x4 + 3x3 - 4x2 + 6x + 2k is divisible by (x + 1)

=> x = -1 must satisfy the given equation

=> 7(-1)5 - 2(-1)4 + 3(-1)3 - 4(-1)2 + 6(-1) + 2k = 0

=> -7 - 2 - 3 - 4 - 6 + 2k = 0

=> 2k = 22

=> k = 11

RRB JE CE (CBT I) Mock Test- 1 - Question 16
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The ratio of two numbers is 7:5 and their H.C.F. is 6. What is their L.C.M.?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 16
Let the numbers be 7x and 5x.

as, 7 and 5 are coprime pairs.

∴ Their H.C.F. = x = 6 (given).

=> The numbers are 42 and 30.

L.C.M. of 42 and 30 = 210.

RRB JE CE (CBT I) Mock Test- 1 - Question 17
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From an external point O, a secant is drawn on a circle which meets the circle at points A and B respectively, if OA = 4 cm and AB = 60 cm. The length of the tangent drawn on circle from point O is (in cm)
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 17

Let the tangent be T

OT2 = OA.OB

=> OA = 4

=> OB = OA + AB

=> 4+60

=> 64

OT2 = 4x64

=> 256

=> OT = 16 cm

RRB JE CE (CBT I) Mock Test- 1 - Question 18
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A copper-wire when bent in the form of a square enclose an area of 121 cm2. If the same wire is bent in the form of a circle, then the area of the circle is
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 18
Side of square = √121 = 11 cm

Length of wire = circumference of circle.

44 = 2 π r or r = (44 x 7)/(2 x 22) = 7 cm

Hence, area of circle = πr2 = 22/7 x 7 x 7 = 154 cm2

RRB JE CE (CBT I) Mock Test- 1 - Question 19
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Table given below shows the total number of employees of four different grades in a department and average age of employees in those difference grades.

Find the average age of the employees of department P ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 19

Sum of ages of all employees in grade A = 25 x 28 = 700 years

Sum of ages of all employees in grade B = 40 x 24 = 960 years

Sum of ages of all employees in grade C = 15 x 32 = 480 years

Sum of ages of all employees in grade D = 20 x 22 = 440 years

Sum of ages of all the employees of the department P = 700 + 960 + 480 + 440 = 2580 years

Total employees in department P = 25 + 40 + 15 + 20 = 100

Average age of the employees of department P = 2580/100 = 25.8 years

RRB JE CE (CBT I) Mock Test- 1 - Question 20
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What will be the percentage profit after selling an article at a certain price if there is a loss of 12.5% when the article is sold at half of the previous selling price?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 20
Suppose the previous selling price = Rs. x

Selling price = Rs. x/2

There is a loss of 12.5% when selling price is Rs. x/2.

Cost Price = x/2 x 100/(100-12.5)= (100 x)/175= 4x/7

Now, when selling price is Rs. x, % profit =[(x-(4x/7))/(4x/7)] x 100 = (3x/4x) x 100 = 75%

RRB JE CE (CBT I) Mock Test- 1 - Question 21
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If α, 3α+6, and 71α are in arithmetic progression, then find the value of α.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 21
α, 3α + 6, and 71α are in arithmetic progression.

=> 3α + 6 - α = 71α - 3α - 6 = common difference of the arithmetic progression.

=> 2α = 68α - 6 - 6

=> 66α = 12

=> α = 12/66 = 2/11

RRB JE CE (CBT I) Mock Test- 1 - Question 22
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One-fourth of Lainey's savings in Kisan Vikas Patra is equal to one-fifth of her savings in Employee Provident Fund. How much has she saved in Employee Provident Fund, if she has a total savings of ₹ 180,000?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 22
Let her savings in Employee Provident Fund be ₹ x.

∴ Her savings in Kisan Vikas Patra = ₹ (180,000 - x)

∴ (180,000 - x)/4 = x/5

=> 900,000 - 5x = 4x

=> x = ₹ 100,000

RRB JE CE (CBT I) Mock Test- 1 - Question 23
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When 20% of 40% of half of a number 'm' is added to 3/4th of 20% of itself, then a second number is obtained which is 21 less than the 40% of m. What is the difference between first and second number?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 23
First number = m

Second number = (40% of m ) - 21

According to the question:

20% of 40% of (m/2) + (3/4) x 20% of m = (40% of m) - 21

m = 100

Second number = (40% of m) - 21 = 40% of 100 - 21 = 19

Difference = 100 - 19 = 81

RRB JE CE (CBT I) Mock Test- 1 - Question 24
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What will come in place of (?) in this equation?

125% of [{? + 200 ÷ (16 ÷ 2 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (18 ÷ 3 - 50% of 4)] = ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 24
Let '?' be denoted by 'a'.

125% of [{a + 200 ÷ (16 ÷ 2 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (18 ÷ 3 - 50% of 4)] = a

=> 125% of [{a + 200 ÷ (8 x 4 x (1/8))} ÷ 7.5 x 8 ÷ (6 - 2)] = a

=> 125% of [{a + 200 ÷ 4} ÷ 7.5 x 8 ÷ 4] = a

=> 125% of [{a + 50} ÷ 7.5 x 2] = a

=> 125% of 2(a + 50)/7.5 = a

=> (5/4) x 2(a + 50)/7.5 = a

=> a + 50 = 3a

=> a = 25

RRB JE CE (CBT I) Mock Test- 1 - Question 25
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If cos (900 - θ)/cos (300 + θ) = 1, then the value of sin θ + cos 2θ
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 25
cos (900 - θ)/cos (300 + θ) = 1

sin θ = cos (300 + θ)

sin θ = sin (900 - 300 - θ)

θ = 900 - 300 - θ

θ = 300

= (sin θ + cos 2θ)

= sin 300 + cos 600

= 1/2 + 1/2

= 1

RRB JE CE (CBT I) Mock Test- 1 - Question 26
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A fraction is such that if it is squared and then the numerator is divided by 2, while the denominator is increased by 25%, the new fraction thus obtained is 4 times the original fraction. What is the new fraction?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 26

 

RRB JE CE (CBT I) Mock Test- 1 - Question 27
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What should come in place of question mark (?) in the following series?

15, 17, 23, 35, 55, (?)
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 27
15 + (1 x 2) = 17

17 + (2 x 3) = 23

23 + (3 x 4) = 35

35 + (4 x 5) = 55

55 + (5 x 6) = 85

RRB JE CE (CBT I) Mock Test- 1 - Question 28
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How much does a watch lose per day, if its hands coincide every 65 minutes?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 28
For a regular clock, we know that the hands of a watch meet after every 65(5/11) minutes.

Hence loss in 65 minutes = 65(5/11) - 65 = (5/11) minutes

Hence loss in 24 hours = (5 x 24 x 60) / (11 x 65) = 10(10/143) minutes

RRB JE CE (CBT I) Mock Test- 1 - Question 29
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Fifteen years hence, the ratio of the ages of a man and his son will be 2:1. If the present age of the man is 45, then what is his son's present age?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 29
Fifteen years hence the ration of age = 2:1.

Let the present age of the son be 'x'.

(45+15)/(x+15) = 2/1

2x + 30 = 60

x = 15

So, the present age of son is 15 years

RRB JE CE (CBT I) Mock Test- 1 - Question 30
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Ramesh can do 20% more work as Mohan can do in same time. If Mohan can finish the work in 30 hours then Ramesh with the help of Mohan, can finish the same work in how many hours ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 1 - Question 30
Mohan can complete a work in 30 hours.

Ramesh can do 6/5 work in 30 hours.

So Ramesh can complete the work in 30 x (5/6) = 25 hours.

Together they can finish 1/30 + 1/25 of work in 1 hour, i.e., 11/150.

So in 150/11 hours, they will finish the work together.

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