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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- 6 - Civil Engineering (CE) MCQ

RRB JE CE (CBT I) Mock Test- 6 - 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- 6

RRB JE CE (CBT I) Mock Test- 6 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- 6 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The RRB JE CE (CBT I) Mock Test- 6 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- 6 below.
Solutions of RRB JE CE (CBT I) Mock Test- 6 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- 6 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- 6 | 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- 6 - Question 1
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Solve the following expression: 452 - 35 * 68 - 58 + 60% of 730.

Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 1

= 452 - 35 * 68 - 58 + 60% of 730

= 2025 - 2380 - 58 + 438

= 2463 - 2438

= 25

= 52

RRB JE CE (CBT I) Mock Test- 6 - Question 2
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The angles of elevation of the top of a tower 90 m high, from two points on the level ground on its opposite sides are 45° and 60°. What is the distance between the two points (rounded off to the nearest integer)?(√3 = 1.732)

Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 2

Let AB be the tower and C and D the two points.

Now AB/BC=tan⁡45∘=1

AB=BC=90

Also AB/BD=tan⁡60∘=√3

BD=90/√3=30√3

Hence CD = CB = 90 + 30√3 = 90 + 51.96 = 141 = 142m

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RRB JE CE (CBT I) Mock Test- 6 - Question 3
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Two quadratic equations: 16x2 - Ax + 25 and Bx2 + Ax + 4 = 0 have real and equal roots, then what is the value of √B?

Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 3
16x2 - Ax + 25 has real and equal roots

(-A)2 = 4 * 16 * 25

A = 40

Bx2 + Ax + 4 = 0 has real and equal roots

(A)2 = 4 * B * 4

(40)2 = 4 * B * 4

B = 100

√B = √100

√B = 10

RRB JE CE (CBT I) Mock Test- 6 - Question 4
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Roshan deposited Rs 17500 in the bank which offers compound interest at 8% per annum. What is the interest obtained on the sum after 2 years?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 4
Amount after two years = 17500 x (1+8/100) 2

= 17500 x 1.08 x 1.08 = 20412

Interest = 20412 - 17500 = Rs 2912

RRB JE CE (CBT I) Mock Test- 6 - Question 5
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Length of a rectangle is increased by 10%,, and breadth is decreased by 10%. Find the percent effect in the area.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 5
Let length = l

And breadth = b

Area = lb

After increment, length = l x (100 + 10)/100 = l x 110/100 = l x 11/10

After decrement, breadth = b x (100 - 10)/100 = b x 90/100 = b x 9/10

Area = l x 11/10 x b x 9/10

= 99lb/100 lb

% decrease = (lb - 99lb/100)/lb x 100

= (100lb - 99lb)/100 x 1/lb x 100

= lb/100 x 1/lb x 100

= 1%

RRB JE CE (CBT I) Mock Test- 6 - Question 6
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One of the angles of a triangle is half the larger angle of a parallelogram. The respective ratio between the adjacent angles of the parallelogram is 14 : 31 . The smallest angle of the triangle is half the smaller angle of the parallelogram. What is the value of the largest angle of the triangle?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 6
We know that the sum of the adjacent angles of a parallelogram = 180o

Let the adjacent angles of the parallelogram be 14x and 31x

Therefore, 14x + 31x = 180o

45x = 180o

X = 4

RRB JE CE (CBT I) Mock Test- 6 - Question 7
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Aman donated 10% of his Savings to Charity and gave 15% of his savings to his sister. Of the remaining savings his gave 60% to his son and gave 80% of the still remaining saving to his daughter. If the sum of the money given to the sister and the money left with Aman finally is Rs 84,000, what was his savings?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 7
Let his savings be Rs 100x

Money donated to charity = Rs 10x and Money given to sister = Rs 15x

Money left with Aman = 100x - 10x - 15x = Rs 75x

Money left with him after giving to son = (40/100)*75x = Rs 30x

Money left with him after giving to daughter = (20/100)*30x = Rs 6x

Given, 15x + 6x = 84000

=> 21x = 84000

=> x = 4000

His Savings = Rs 4,00,000

RRB JE CE (CBT I) Mock Test- 6 - Question 8
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The ratio of 4th and 2nd term of an A.P of positive terms is 5:3, and the product of the 1st and the 3rd terms is 72. what is the product of the 5th and 6th terms of the AP?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 8
Let the first term be 'a' and the common difference be 'd'.

Given, (a + 3d)/(a + d) = 5/3

=> 3a + 9d = 5a + 5d

=> 2a = 4d

=> a = 2d

Also, a(a + 2d) = 72

=> 2d(2d + 2d) = 72

=> 8d2 = 72

=> d2 = 9

=> d = 3

∴ a = 2*3 = 6

d ≠ −3 as the 1st term will become a = 2(−3) = − 6 but all terms are positive.

5th term = a + 4d = 6 + 4*3 = 18

6th term = 18 + 3 = 21

Required product = 18*21 = 378

RRB JE CE (CBT I) Mock Test- 6 - Question 9
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Three pipes A, B and C can fill a cistern in 10 hrs. After working together for 4 hours, C is closed and A and B fill the cistern in 9 hrs. Then find the time in which the cistern can be filled by pipe C?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 9
A, B and C can fill in 1 hour = 1/10 of the cistern

A, B, and C can fill in 4 hour= 4/10 =2/5 of the cistern

Remaining part = 1 - 2/5 =3/5 of cistern

3/5 of the cistern is filled by A+B in 9 hours

Therefore, A and B can fill the cistern in 9*5/3 = 15 hrs

Hence in one hour C filled the cistern = 1/10 - 1/15 = 1/30 of the cistern

Hence C can fill the cistern in 30 hrs.

RRB JE CE (CBT I) Mock Test- 6 - Question 10
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If the average of twelve consecutive even numbers be 113, then find the sum of the 3rd and 8th numbers among them?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 10
Let the 1st number = x

According to question,

x + (x+2) + (x+4) + ...... + (x+22) = 113*12

12x + 132 = 1356

12x = 1224

x = 102

Therefore 3rd number = 102 + 4 = 106

8th number = 102 + 14 = 116

Hence the required answer = 106 + 116 = 222

RRB JE CE (CBT I) Mock Test- 6 - Question 11
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What approximate value should come in the place of the question mark in the following question?

(72)2 ÷ ∛46656=?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 11
? = (72)2 ÷ ∛46656

? = (72)2 ÷ ∛(36x36x36) = (72)2 ÷ 36

? = (72 X 72)/36 = 144

RRB JE CE (CBT I) Mock Test- 6 - Question 12
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Ranjan purchases an article at Rs.660 and marks up the price by 25%, and after allowing the discount of Rs.99, he gains __% profit on the sale.

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

A) Per cent profit = (66/660) * 100 = 10%

B) Selling price of the item after discount of Rs.99 = 825 - 99 = Rs.726

C) Profit amount earned on the item = 726 - 660 = Rs.66

D) Cost price of the article is Rs.660 and marked price = 125% of 660 = Rs.825
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 12
The correct arrangement is

D) Cost price of the article is Rs.660 and marked price = 125% of 660 = Rs.825

B) Selling price of the item after discount of Rs.99 = 825 - 99 = Rs.726

C) Profit amount earned on the item = 726 - 660 = Rs.66

A) Per cent profit = (66/660) * 100 = 10%

RRB JE CE (CBT I) Mock Test- 6 - Question 13
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Study the following table carefully and answer the related question.

The following,table represents the three types of products A, B and C manufactured in a company during Five months.

In which of the following month there is a maximum percentage increase/decrease in the manufacturing of all three types of products together?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 13
Total products manufactured

In January = 18 + 22 + 25 = 65

In February = 20 + 12 + 32 = 64

In March = 40 + 10 + 13 = 63

In April = 21 + 22 + 24 = 67

In May = 25 + 21 + 16 = 62

Percentage increase/decrease:

In February = ((65 - 64)/65) x 100 = 1.54%

In March = ((64 - 63)/64) x 100 = 1.56%

In April = ((67 - 63)/63) x 100 = 6.35%

In May = ((67 - 62)/67) x 100 = 7.46%

RRB JE CE (CBT I) Mock Test- 6 - Question 14
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Two trains 140 m and 160 m long run at the speed of 60 kmph and 40 kmph respectively in opposite directions on parallel tracks. The time (in seconds) which they take to cross each other is :
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 14
time = Distance / Velocity

Distance =140+160=300m=0.3km

Relative velocity =60+40=100kmph (opp. direction) Time

=(0.3/100)h

or (0.3/100)×3600=10.8sec

Time =10.8 sec.

RRB JE CE (CBT I) Mock Test- 6 - Question 15
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a2+b2=5 and a3+b3=9. If ab = 2, what is the value of a4+b4?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 15
Let (a+b) = c

c2 = (a+b)2 = a2+b2 + 2ab = 5 + 4 = 9

=> a + b = 3

Now, (a+b)(a3+b3) = a4+b4+ab(a2+b2)

=> 3 x 9 = a4+b4 + 2 x 5

=> a4+b4 = 17

RRB JE CE (CBT I) Mock Test- 6 - Question 16
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The average weight of 24 students in a class is 40 kg. if the weight of the teacher is included, the average is increased by 500 gms. Find The weight of the teacher?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 16
25*40.5 - 24*40 = 52.5

Hence, the answer is option (2).

RRB JE CE (CBT I) Mock Test- 6 - Question 17
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A shopkeeper makes a profit of 20% by selling a T.V. What would be the new profit percent if the shopkeeper paid 15% less for the T.V and the customer paid 10% more for the T.V?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 17
Let the price at which T.V was purchased by shopkeeper = Rs 100a

Hence the price at which T.V was sold by shopkeeper = Rs 120a

If shopkeeper paid 15% less than the previous amount hence new cost price of the T.V = Rs (100a X 0.85) = Rs 85a

New selling price of the T.V = Rs (120a X 1.1) = Rs 132a

Profit % = (132a-85a)/85a X 100 = 55.3%

RRB JE CE (CBT I) Mock Test- 6 - Question 18
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Solve the following expression: (13/8) of (15/32) of 45% of 3072.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 18
= (13/8) of (15/32) of 45% of 3072

= (13/8) * (15/32) * (45/100) * 3072

= (13/8) * (3/32) * (9/4) * 307

= 13 * 3 * 9 * 3

= 1053

RRB JE CE (CBT I) Mock Test- 6 - Question 19
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Following Pie Chart shows the percentage distribution of people of different ages in a Building. The table shows the ratio of Male and Female people in each age group. There is a total of 800 people in the building.

What is the difference between the total number of Males of age 20 to 60 and the total number of Females of age 20 to 60?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 19
Calculating the number of Male and Females of each age group

total number of Males of age 20 to 60 = 210 + 120 = 330

total number of Females of age 20 to 60 = 150 + 40 = 190

Required difference = 330 - 190 = 140

RRB JE CE (CBT I) Mock Test- 6 - Question 20
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The population of a village is 6000. The next year, the number of male increases by 10% and the number of female increases by 20% and the population becomes 6840. What is the initial population of Male?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 20
Let the initial population of Males and Females be 100x and 100y respectively.

Given, 100x + 100y = 6000

=> x + y = 60..(i)

And, 110x + 120y = 6840

=> 11x + 12y = 684..(ii)

12*(i) - (ii), gives

x = 36

Initial population of Males = 3600

RRB JE CE (CBT I) Mock Test- 6 - Question 21
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The average age of a Sheela and her sister is 16 years. The respective ratio of their ages is 7:1 respectively. Find Sheela's age?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 21
Sum of their ages = 2 X 16 = 32

Let 7x and x be their respective ages, then, 8x = 32 and x = 4

So, Sheela's age = 7x = 7 X 4 = 28 years . Hence, option 1.

RRB JE CE (CBT I) Mock Test- 6 - Question 22
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20% of x% of 60 is equal to y% of 60% of 40, what is the value of the ratio x:y?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 22
20% of x% of 60 = y% of 60% of 40

x = 2y

x:y = 2:1

RRB JE CE (CBT I) Mock Test- 6 - Question 23
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What is the remainder when x20 + x19 + x18 + x17 + x16 + x15 + x14 + x13 + x12 + x11 is divided by x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 23
Numerator = x11 (x9 + x8 + x7 + x6 + x5 + x4 + x3 + x2 + x + 1) = x11 times the denominator.

The denominator divides the numerator, so the answer is 0.

RRB JE CE (CBT I) Mock Test- 6 - Question 24
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A person invested equal amounts in two scheme A and B at the same rate of interest. Scheme A offers simple interest while scheme B offers compound interest. After two years he got Rs. 1920 from the Scheme A as interest and Rs. 2112 from scheme B. If the rate of interest is increased by 4%, what will be the total interest after two years both schemes ?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 24
C.I−S.I=192

S.I for 1 year =960 Interest on RS960 for 1 year is 192 so rate is =(192×100)/(960×1)=20%

so the principal is =(960×100)/20=4800 when interest is increased is 4% then new rate of interest is 24% So simple interest after two years will be = (4800×2×24)/100=2304 and compound interest rate after two years will be =24+24+(24×24)/100=53.76% Compound interest after two years =(53.76×4800)/100=2580.48 So total interest is = 2304 + 2580.48 = Rs. 4884.48

RRB JE CE (CBT I) Mock Test- 6 - Question 25
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How many litres of a solution containing milk and water in the ratio 2:3 should be added to 30 litres of 70 % milk solution in order to get 60% milk solution?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 25
Let the number of liters of solution added be: x

The initial ratio of milk (m) to water (w) is:

=> m/w = 2/3

=> m = 2/5 of the solution

=> m = 40% of the solution

Acc. to the question, after adding x liters of 40% milk solution to 30 liters of 70% milk solution, we get 60% of (x+30) liters milk solution.

Hence, the equation becomes:

=> 40x + 30 * 70 = (x+30) * 60

=> 40x + 2100= 60x + 1800

=> 20x = 300

=> x = 15

RRB JE CE (CBT I) Mock Test- 6 - Question 26
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If 12% of a is equal to 15% of b. Find the respective ratio of a and b.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 26
12% of a = 15% of b

=> a/b = 15/12

=> a/b = 5/4

=> a:b = 5:4

RRB JE CE (CBT I) Mock Test- 6 - Question 27
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Downstream speed of boat A is 20% more than the downstream speed of boat B whose upstream speed is 6 km/hr less than upstream speed of boat A. What is the speed of boat A in still water? The stream for both boat is moving with 10 km/hr.
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 27
Stream speed = 10 km/hr

Downstream speed of boat A = (6/5) x downstream speed of boat B

Upstream speed of boat A - upstream speed of boat B = downstream speed of boat A - downstream speed of boat B = 6

(6/5) x downstream speed of boat B - downstream speed of boat B = 6

Downstream speed of boat B = 30 km/hr

Downstream speed of boat A = 30 + 6 = 36 km/hr

Speed of boat A in still water = 36 - 10 = 26 km/hr

RRB JE CE (CBT I) Mock Test- 6 - Question 28
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If the the ,monthly income of a person is increased by 20% and his expenditures are increased by 40%, then his savings are decreased by what percent? (ratio of income and expenditure is 4: 3 respectively.)
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 28
Let income and expenditure are 'a' and 'b' respectively.

Savings = a - b = a - 3a/4 = a/4

When income and expenditure are increased by 20%. Then,

New income = 120% of a = 1.2a

New expenditure = 140% of b = 1.4 x 3a/4 = 2.1a/2

New savings = 1.2a - 2.1a/2 = 3a/20

Decrement in savings = (a/4) - (3a/20) = a/10

Percentage = a/10)/(a/4 x 100 = 40%

RRB JE CE (CBT I) Mock Test- 6 - Question 29
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A and B can finish a work in 2 days, B and C in 3 days and A and C in 4 days. Find the ratio of time taken by A alone and C alone to finish the work?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 29
Suppose A, B and C take A, B and C days respectively to finish the work on their own.

1/A + 1/B = 1/2 (1)

1/B + 1/C = 1/3 (2)

1/A + 1/C = 1/4 (3)

(1)+(2)+(3) => 2(1/A+1/B+1/C) = 13/12

=> 1/A + 1/B + 1/C = 13/24 (4)

(4) - (2) => 1/A = 5/24

Thus A takes 24/5 = 4.8 days.

(4) - (1) => 1/C = 1/24

Hence C takes 24 days.

Required ratio = 4.8: 24 = 1:5

RRB JE CE (CBT I) Mock Test- 6 - Question 30
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Which of the following is irrational?
Detailed Solution for RRB JE CE (CBT I) Mock Test- 6 - Question 30
Product of (3 - √27) and (√9 + 3√3)

= (3 - 3√3)(3 + 3√3)

= - 18 (which is not irrational)

Option (2):

= Sum of (5 - 4√2) and 4(6 + √2)

= (5 - 4√2) + 4(6 + √2)

= 5 - 4√2 + 24 + 4√2

= 29 (which is not irrational)

Option (3):

= Sum of √4761 and √5175

= 69 + 15√23 (which is irrational)

Option (4):

= Product of √16875 and √27

= 75√3 x 3√3

= 675 (which is not irrational)

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