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RRB JE IT (CBT I) Mock Test- 5 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series for Computer Science Engineering 2025 - RRB JE IT (CBT I) Mock Test- 5

RRB JE IT (CBT I) Mock Test- 5 for Computer Science Engineering (CSE) 2024 is part of RRB JE Mock Test Series for Computer Science Engineering 2025 preparation. The RRB JE IT (CBT I) Mock Test- 5 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The RRB JE IT (CBT I) Mock Test- 5 MCQs are made for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE IT (CBT I) Mock Test- 5 below.
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RRB JE IT (CBT I) Mock Test- 5 - Question 1

DIRECTIONS: The pie chart given here, shows various expenses of a publisher in the production and sale of a book. Study the chart and answer questions based on it.

The measure of central angle for the sector on 'Printing cost' is:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 1
According to question 360×35/100=126o
RRB JE IT (CBT I) Mock Test- 5 - Question 2

DIRECTIONS: The pie chart given here shows various expenses of a publisher in the production and sale of a book. Study the chart and answer questions based on it.

The difference between the measures of central angles of sectors for binding charges and advertisement charges is:

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RRB JE IT (CBT I) Mock Test- 5 - Question 3

Find the angle of elevation of the sun if the length of the shadow of a pole is 1/√3 times its height?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 3

tanθ = h /h/√3 = √3

⇒ θ = 60o

= π/3

RRB JE IT (CBT I) Mock Test- 5 - Question 4

If the point (x, y) (1, 2) and (-3, 4) are collinear, then :

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 4

If (x, y) (1, 2) and (-3, 4) are collinear then the triangle formed by points should be zero

∴1/2 [2x + 4 - 3y - 4x + 6 -y] = 0

-2x- 4y + 10 = 0

x + 2y - 5 = 0

RRB JE IT (CBT I) Mock Test- 5 - Question 5

The dimensions of a piece of iron in the shape of a cuboid are 270 cm x 100 cm x 64 cm. If it is melted and recast into a cube, then the surface area of the cube will be :

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

Volume of cube = volume of cuboid

= 270 x 100 x 64 = 1728000 cm3

∴ Length of one side of the cube = 120 cm

∴ Surface area of cube = 6 x 120 x 120 = 86400 cm2

RRB JE IT (CBT I) Mock Test- 5 - Question 6

The average age of a family of five persons is 20 years. If the youngest member is 8 years old then, what was the average age of the family at the birth time of the youngest member?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 6

Sum of the age of family = 5 x 20 = 100 years

8 years before, sum of Ages = 100 - 8 x 5 = 60 years

Average age = Sum / Total person = 60/4 = 15 years

RRB JE IT (CBT I) Mock Test- 5 - Question 7

If a train takes 1.75 sec to cross a telegraph pole and 1.5 sec to a cyclist travelling in the opposite direction at 10 m per second, then the length of the train is:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 7

Suppose speed of train = x m/s

or length of train = l m

Time taken by trains to cross a telegraph cost

= length of train / speed of train

∴ 1.75 = l/x

or x = l/1.75 ……………………(i)

∴ 1.5 = l /x +10

or x + 10 = l / 1.5……………………(ii)

Using (i) and (ii)

l/1.75 + 10 = l/1.5

or l/1.5 - l/1.75 = 10

or 2.5 l = 10 x 1.5 x 1.75

= 105m

RRB JE IT (CBT I) Mock Test- 5 - Question 8

The semi-perimeter of a right-angled triangle ABC is 12 cm, and the shortest median is 5 cm. What is the area of the triangle which has the largest median of triangle ABC as its longest side?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 8

Given: S = 12 cm & BP = 5 cm

∴ AP = PC = BP = 5 cm

∴ AC = 10 cm

We know the ratio of Right triangle like (3:4:5)

then we assume the ratio, (6:8:10)

So, the other two sides = 8 cm, 6 cm

∴ Area of △QBC = ½ x QB x BC = 1/2 x 4 x 6 = 12cm2

RRB JE IT (CBT I) Mock Test- 5 - Question 9

If 4 sin2 x - 3 = 0, 0 < x < 2π, then the values of x are:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 9

4 sin2x - 3 = 0

or sin2x = 3/4

∴ sinx = ± √3 / 2

since 0 < z < 2?

∴ sinx = ± √3 / 2 = sin 60o, sin120o, sin 240o, sin 300o

RRB JE IT (CBT I) Mock Test- 5 - Question 10

Sum of the lengths of any two sides of a triangle is always greater than

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 10
The sum of the other two sides of a triangle is always greater than the third side.
RRB JE IT (CBT I) Mock Test- 5 - Question 11

If 391 bananas were distributed among three monkeys in the ratio of 1/2:2/3:3/4, how many bananas

Did the first monkey get it?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 11

The ratio: 1/2:2/3: 3/4

Converts to 6: 8: 9 (on multiplying by 12)

Thus, the first monkey would get (391/23) × 6 = 102 bananas.

RRB JE IT (CBT I) Mock Test- 5 - Question 12

A is twice as good as B and is, therefore, able to finish a piece of work in 30 days less than B. In how many days can they complete the whole work working together?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 12

Let A can complete the work in x days.

So, B can complete the work in 2x days.

A.T.Q. 2x -x = 30 ⇒ x = 30 days

They both can complete in = 60/2 + 1 = 20 days

RRB JE IT (CBT I) Mock Test- 5 - Question 13

A dishonest fruit seller sells fruits at a 5% loss. If he uses 850 gm weight in place of 1 kg weight, then what is his profit percent?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 13

Let C.P. of 1000 gms = 100

S.P. of 850g gms

= 100 - L

= 100 - 5 = 95 (as 5% on 100 = 5)

S.P. of 1000 gms = 1000 x 95 / 850

= 1900 / 17

P% = P / C.P x 100

=

=

= 200/17

=

RRB JE IT (CBT I) Mock Test- 5 - Question 14

A man calculates his loss% as 16 (2/3)% on S.P. What is the actual loss percent?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 14

Loss% = 16x 2/3% = 50 / 3x100 = 1/6 →L

S.P. = 6

Loss = 1

then C.P. = 7 (as C.P. = S.P. + L)

= 1/7 x 100 = 14 2/7%

RRB JE IT (CBT I) Mock Test- 5 - Question 15

In the figure, OP ⊥ OA and OQ ⊥ OB.

Find ∠POQ if ∠AOB = 200

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 15

∠BOP = 900 - ∠AOB

= 900 - 700

= 200

∴ ∠POQ = 900 - ∠BOP

= 900 - 700

=200

RRB JE IT (CBT I) Mock Test- 5 - Question 16

If 2 men and 3 women can do a piece of work in 8 days and 3 men and 2 women in 7 days. In how many days can the work be done by 5 men and 4 women working together?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 16

|2 x 8 - 3 x 7| men’s work = |3x 8 - 2 x 7| women’s work.

∴ 5 men’s work = 10 women’s work

∴ 1 men’s work = 2 women’s work

2 men’s work = 4 women’s work

∴ Days taken by 5 men + 4 women = 14 women are =7×8/14 = 4 days

Alternate method-

⇒ (2 M + 3W) 8 = (3M + 2W)7

⇒ 16M + 24W = 21M + 14 W

⇒ 10W = 5M

⇒ 2W = M

⇒ 14W × (x) = 7W × 8

= (x) = 4 days

RRB JE IT (CBT I) Mock Test- 5 - Question 17

If the volume of a cube is 729 cm3, then the total surface area of the cube will be :

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 17

Volume of the cube = 729 cm3

∴ Length of one side of the cube = 9 cm

∴ surface area of the cube = 6 x 9 x 9 = 486 cm2

RRB JE IT (CBT I) Mock Test- 5 - Question 18

Two pipes A and B can fill a tank in 60 minutes and 75 minutes, respectively. There is also an outlet C. If A, B, and C are opened together, the tank is full in 50 minutes. How much time will be taken by C to empty the full tank?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 18

(A + B) eff = 9

(A + B - C) eff = 6

Sp, C eff = 3 units

C will empty the tank = 300/3 = 100 min.

RRB JE IT (CBT I) Mock Test- 5 - Question 19

If 10 people can do a job in 20 days. Then 20 persons with the twice efficiency can do the same job in:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 19

Let the number of days be x.

Here efficiency is 2 : 1

Person 20 : 10

Number of days 20 : x

∴ x = 1 x 10 x 20 / 20 x 2 = 5 days

RRB JE IT (CBT I) Mock Test- 5 - Question 20

A sum of Rs.1,075 is divided among Aman, Anu, and Aradhya such that Aman receives 25% more than Aradhya and Aradhya receives 25% less than Anu. What is Aman’s share in the amount?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 20

Let Anu get x, then Aaradhya get 75% of x and Aman gets 125% amount of 75% of x

75% of x + 125% of 75% of x + x = 1075

x = 400

125/100 × 75/100 × 400

= Rs.375.

RRB JE IT (CBT I) Mock Test- 5 - Question 21

The horizontal distance between two towers is 90 m and the angular depression of the top of the first, as seen from the top of the second, is 30o. Find the difference in the heights of the two towers.

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 21

Difference between the height of tower = DE

Again AB = DC

From △CDE

tan 30o = DE/DC

1/√3 = DE /90

DE = 90 / √3

= 30√3 meters

RRB JE IT (CBT I) Mock Test- 5 - Question 22

If the ratio of the areas of two squares is 16: 1, then the ratio of their perimeters is?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 22

Area of square = (side)2

Ratio of sides of two square = 4 : 1

Perimeter of square = 4 x side

Ratio of perimeter of two squares = 16 : 4 = 4 : 1

RRB JE IT (CBT I) Mock Test- 5 - Question 23

If the ratio of areas of two squares is 225: 256, then the ratio of their perimeters is?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 23

Ratio of their perimeters-

15 : 16

RRB JE IT (CBT I) Mock Test- 5 - Question 24

The time duration of 1 hour 45 minutes is what percent of a day?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 24

1 hours 45 minutes = 7/2 hours

= [7/(4×24)]×100 = 7.291%

RRB JE IT (CBT I) Mock Test- 5 - Question 25

Two circles of radii 4 cm each cut each other such that each circle passes through the centre of each other. Thus, what is the length of the common chord?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 25

Let the centers be O & O’

In △OPM,

OP2 = OM2 + PM2

16 = 4 + PM2

PM = √12

∴ Length of the common chord = 2√12 = 4√3 cm

RRB JE IT (CBT I) Mock Test- 5 - Question 26

In ΔABC, medians BE and CF intersect at G. If the straight line AGD meets BC at D in such a way that GD = 1.5 cm, then, length of AD is:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 26

GD = 1/3 AD

or AD = 3 GD

AD = 3 × 1.5 = 4.5 cm

RRB JE IT (CBT I) Mock Test- 5 - Question 27

If two person A and B start at the same time in an opposite direction from two points and after passing each other they complete the journey in a and b hours respectively, then the ratio of A’s and B’s speed will be:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 27

Let total distance be D km and A’s speed be x km/hr, and B’s speed be y km/hr.

RRB JE IT (CBT I) Mock Test- 5 - Question 28

The value of x is:

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 28

⇒ √x = 0.08 x 2/10

⇒ √x = 0.016

⇒ √x = (0.016)2 = 0.000256

RRB JE IT (CBT I) Mock Test- 5 - Question 29

Students of a class stand in a queue. If Ramesh is 19th in order from both ends. How many students are there in the queue?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 29

Ramesh is 19th from both ends.

∴ There are 18 students on his each side

∴ Total students in the queue = 18 × 2 + 1 = 37

RRB JE IT (CBT I) Mock Test- 5 - Question 30

if

then the value of x is :

Detailed Solution for RRB JE IT (CBT I) Mock Test- 5 - Question 30

= = =

∴ x - 1 = x - 3 or 2x = 4

∴ x = 2

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