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Computer Science Engineering (CSE) Exam  >  Computer Science Engineering (CSE) Tests  >  RRB JE IT (CBT I) Mock Test- 4 - Computer Science Engineering (CSE) MCQ

RRB JE IT (CBT I) Mock Test- 4 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test - RRB JE IT (CBT I) Mock Test- 4

RRB JE IT (CBT I) Mock Test- 4 for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The RRB JE IT (CBT I) Mock Test- 4 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The RRB JE IT (CBT I) Mock Test- 4 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- 4 below.
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RRB JE IT (CBT I) Mock Test- 4 - Question 1
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Walking at 6/7th of his usual speed, a man is 12 minutes late. The usual time taken by him to cover that distance is-

Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 1

S1: S2 = 6 : 7

1 T :T 2 = 7 : 6 (S ⍺ 1 / T)

Difference between

T1 & T2 = 1 (7 - 6 = 1)

If 1 = 12min.

then 6 = 72 min. = 1 hrs 12min.

RRB JE IT (CBT I) Mock Test- 4 - Question 2
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In selling an article for ₹ 76, there is a profit of 52%, if it is sold for ₹ 75, then the profit percent will be.

Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 2

152% = 76

100% = Cost price = 76 / 152 x 100

C.P = Rs. 50

New selling price = 75

% profit = 75 - 50 / 50 x 100%

= 25 / 50 x 100% = 50%

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RRB JE IT (CBT I) Mock Test- 4 - Question 3
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The lines x + 3y = 4 and 3x – y = 6 intersect the y – axis at A and B respectively. AB = ?

Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 3

RRB JE IT (CBT I) Mock Test- 4 - Question 4
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Read the following bar graph which represents a company's manufacturing and selling(in lakh) of product from 2002 to 2008 and answer the following questions.

Approximately what is the average number of items sold for all the years together?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 4
Average = (60+55+65+50+60+80+75)/7 = 445/7 = lakh

≃ 63 lakh(Appx.)

RRB JE IT (CBT I) Mock Test- 4 - Question 5
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Area of a circle is equal to the area of a rectangle having a perimeter of 100 cm and length is more than the breadth by 6 cm. What is the diameter of the circle?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 5

Let breadth = x, then length

= (x + 6)

2(x + x + 6) = 100

2x + 6 = 50

x = 22 cm

Breadth = x = 22 cm & length

= 22 + 6 = 28 cm

Area of circle = Area of rectangle ir

⇒ r2 = 22 x 28

⇒ r2 = 22 x 28 / 22 x 7 = 7 x 4 x 7

⇒ 7 x 2 = 14 cm

So, Diameter = 2r = 28 cm

RRB JE IT (CBT I) Mock Test- 4 - Question 6
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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 IT (CBT I) Mock Test- 4 - Question 6

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

Second S.P. = 9120 x 105 / 100 = Rs. 9,576

Loss = 9600 - 9576 = 24

RRB JE IT (CBT I) Mock Test- 4 - Question 7
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A mixture of 40 litres of milk and water contains 10% water. How much water must be added to make water 20% in the new mixture?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 7

Milk =90/100 × 40 = 36 litres

water = 5 litres

5x + 20 = 40 + x

X = 5

RRB JE IT (CBT I) Mock Test- 4 - Question 8
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The average weight of 8 persons increases by 2.5 kg when one person whose weight is 65 kg is replaced by a new person. What Is the weight of the new person?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 8
Total increase in weight due to addition of new persons = 2.5 × 8 = 20 kg

Weight of new person = 65 + 20 = 85 kg

RRB JE IT (CBT I) Mock Test- 4 - Question 9
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Read the following bar graph which represents a company's manufacturing and selling(in lakh) of product from 2002 to 2008 and answer the following questions.

During which year the percentage of items unsold was the highest?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 9

% in 2002 = {(10 x 100) / 70}% = 14.3%;

% in 2003 = {(10 x 100) / 65}% = 15.38%;

% in 2004 = {(15 x 100) / 80}% = 18.75%;

% in 2005 = {(10 x 100) / 60}% = 16.67%;

% in 2006 = {(15 x 100) / 75}% = 20%;

% in 2007 = {(10 x 100) / 90}% = 11.11%;

and % in 2008 = {(5 x 100) / 70}% = 6.25%;

It was the highest in 2006.

RRB JE IT (CBT I) Mock Test- 4 - Question 10
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Read the following bar graph which represents a company's manufacturing and selling(in lakh) of product from 2002 to 2008 and answer the following questions.

Approximately what is the average number of items unsold for all the years together?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 10

Average = (10+10+15+10+15+10+5)/7 = 75/7 lakh

≈ 1,07,000 lakh(Aprx)

RRB JE IT (CBT I) Mock Test- 4 - Question 11
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A, B and C can finish a work in 8 days, 10 days and 12 days respectively. They worked together and get Rs. 740. How much will each of them will get?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 11

Given,

1 day work of A, B & C is 1 / 8, 1 / 10, 1 / 12 respectively

therefore , Their shares in money = 1/8 : 1/10 : 1/12 = 15 :12 : 10

Sum of their share = ₹740 x 15/37 = ₹300

B’s share = ₹740 x 12/37 = ₹240

C’s share = ₹740 x 10/37 = ₹200

RRB JE IT (CBT I) Mock Test- 4 - Question 12
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A room 8 m long, 6 m broad and 3 m height has two windows 1 1⁄2 m × 1 m and a door 2 m × 1 1⁄2 m. Find the cost of papering the walls with paper 50 cm wide at 25 paise per meter.
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 12

Area of walls = 2(Length + Breadth) x height

= 2(8 + 6) x 3 = 84m2

Area of two windows and a door = (1 ½ x 1) + 2 x 1 ½ ) = 6m2

Area to be covered = 84 - 6

= 78m2

Area of paper = Area to be covered = 78

⇒ (length x breadth) of paper is 78

⇒ length of paper = 78/50 x 100m

= 156 m

So, Cost = 156x25 / 100 = Rs.39

RRB JE IT (CBT I) Mock Test- 4 - Question 13
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Two circles with radii R and r touch each other externally. A direct common tangent with length l is drawn to the circles. Which is true?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 13

Length of direct common tangent l2 = d2 – (R – r)2

Here, d = R + r

Therefore, l2 = (R + r)2 – (R – r)2 = 4Rr

= 2√Rr

RRB JE IT (CBT I) Mock Test- 4 - Question 14
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A certain sum, invested at 4% per annum at compound interest, compounded half-yearly, amounts to Rs. 7,803 at the end of one year. Find the sum.
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 14

If interest is compounded half-yearly rate will be 2 %

C.I. for 2 years = 4.04%

Principal = 7803 / 104.04 x 100

= Rs.7500

RRB JE IT (CBT I) Mock Test- 4 - Question 15
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Read the following bar graph which represents a company's manufacturing and selling(in lakh) of product from 2002 to 2008 and answer the following questions.

The number of items manufactured in 2007 is what percent of the total number of items manufactured in all the years together? (Rounded off to two digits after decimal)?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 15

Reqd. % = {(90 x 100) / 520}% = 17.307% = 17.31 % (Appx.)

RRB JE IT (CBT I) Mock Test- 4 - Question 16
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Find the measures of an angle that is the complement of 50°.
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 16

Let the measure of angle = x°

measure of its complement = 50°

∴ x° + 50°= 90° ⇒ x° = 40°

RRB JE IT (CBT I) Mock Test- 4 - Question 17
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If sin (10° 6’ 32" ) = a, then the value of cos (79° 53’ 28”) + tan (10° 6’ 32”) is:
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 17

cos 79o 53’ 25” + tan 10o 6’ 32”

= sin 10o 6’ 32” + sin 10o 6’ 32” / cos 10o 6’ 32”

RRB JE IT (CBT I) Mock Test- 4 - Question 18
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If cos T = 3/5 and if sin R = 8/17, where T is in the fourth quadrant and R is in the second quadrant, then cos (T-R) is equal to:
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 18

We have

cos T = 3/5

sin T =

=

= √16/25 = -⅘ (Since T is in IV quadrant)

sin R = 8 /17

cos R =

=

√225 / 289 = -15/17 (Since r is in II quadrant)

Now, cos (T-R) = cos T cos R + sin T sinR

=

=

= 77/85

RRB JE IT (CBT I) Mock Test- 4 - Question 19
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There are 13 white and 7 black balls in a bag. Two balls are drawn at random. What is the probability that they are of the same colour?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 19
Probability that both balls are of the same colour =

RRB JE IT (CBT I) Mock Test- 4 - Question 20
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The ratio between the angles of a quadrilateral is 3:4:6:7. Half of the smallest angle of a quadrilateral is the angle of a cyclic quadrilateral. Find another angle of the cyclic quadrilateral?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 20

Let smallest angle is 3x

So,

3x + 4x + 6x + 7x = 360°

Smallest angle = 54°

According to given condition one angle of Cyclic quadrilateral = 27°

Another = 180 – 27 = 153°

RRB JE IT (CBT I) Mock Test- 4 - Question 21
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A, B and C divide an amount of Rs. 9,915 amongst themselves in the ratio 3:5:7 respectively. What is the C’s share in the amount?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 21

A : B : C

3 : 5 : 7

Sum of their ratio = 3 + 5 + 7

= 15 units

Share of C = 7/15 × 9915 = 4,627

RRB JE IT (CBT I) Mock Test- 4 - Question 22
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60 boys working 7 hours a day can do work in 18 days. In how many days will 28 boys working 9 hours a day do the same work?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 22

Using the formula M1D1H1W2 = M2D2H2W1

Since work is same for the two cases

M1D1H1 = M2D2H2

⇒ D2 = M1D1H1 / M2H2 = 60 x 7 x 18 / 28 x 9

= 30 days

RRB JE IT (CBT I) Mock Test- 4 - Question 23
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What should come in place of the question mark (?)?

(833.25 – 384.45) ÷ 24 = ?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 23

(d) ?= (833.25–384.45)/24 = 448.8/24 = 18.7

RRB JE IT (CBT I) Mock Test- 4 - Question 24
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The average of three numbers is 135. The largest number is 195 and the difference between the other two is 20. The smallest number is:
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 24

According to the question,

195 + x + x + 20 = 135 x 3

⇒ 2x + 215 = 405

⇒ 2x = 405 - 215 = 190

∴ x = 190 / 2

= 95

= smallest number

RRB JE IT (CBT I) Mock Test- 4 - Question 25
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Distance between the parallel lines 3x + 4y – 15 = 0 and 9x + 12y – 20 = 0 is
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 25

3x + 4y - 15 = 0

Or 9x + 12y -45 = 0………………..(1)

And 9x + 12y - 20 = 0………………(2)

The distance between two parallel lines

ax + by +c1 = 0 and ax + by +c2 = 0 is

=

= 25/15 = 5/3 units

RRB JE IT (CBT I) Mock Test- 4 - Question 26
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In a container of 56 litres the ratio of petrol to oil is 6:1. How much oil must be added to the mixture to make this ratio 2:1?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 26

Quantity of petrol = (6/7) x 56 = 48 Litre

Quantity of oil = (1/7) x 56 = 8 litre

Given, New Ratio = 48/8+x=12

48 = 16 + 2x

2x = 48 – 16

2x = 32

x = 16

16 litre is to be added

RRB JE IT (CBT I) Mock Test- 4 - Question 27
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'A' starts a business with ₹ 3,500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2:3 what is B's contribution to the capital?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 27

Let B’s investment be Rs x then

3500 x 12/x X 7 = ⅔

⇒ 6000 x 3 = 2x

⇒ x = ₹9,000

RRB JE IT (CBT I) Mock Test- 4 - Question 28
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If a = 12, b = 13 and c = 14, then the value of b² c² + c² a² + a² b² – a⁴ – b⁴ – c⁴ is.
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 28

b² c² + c² a² + a² b² – a⁴ – b⁴ – c⁴

= −12(2a⁴ + 2b⁴ + 2c⁴ – 2a² b² – 2b² c² – 2c² a²)

= −12[(a² − b²)² + (b² – c²)² + (c² – a²)²]

= −12[(144 − 169)² + (169 – 196)² + (196 – 144)²]

= −12[( −25)² + (27)² + (52)²] = –2029

RRB JE IT (CBT I) Mock Test- 4 - Question 29
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The midpoint of the line segment joining (3,-8) and (x + 2, -2) is (4, -5). What is the value of X?
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 29

Midpoint of (x1, y1) and (x2, y2)

X = (x1 + x2)/2

4 = (3 + x + 2)/2

⇒ (x+5)/2 = 4 ⇒ x = 3

RRB JE IT (CBT I) Mock Test- 4 - Question 30
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A man saves 25% of his salary. If due to price rise he increases his monthly expenses by 25% and he is able to save only Rs. 25 per month. Find his monthly salary.
Detailed Solution for RRB JE IT (CBT I) Mock Test- 4 - Question 30

Expenses before price rise = 100 - 25 = 75

25% increase in expenses = 100 + 25

= Rs. 125

Monthly Expenses = (75 x 125) / 100

= 375/4

Saving = 100 - 375/4 = 25/4

If saving in 25/4 then salary = 100

IF saving in 25 then salary

= (100 x 25 x 4)/25

= Rs. 400

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