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RRB JE EEE (CBT I) Mock Test- 10 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series Electrical Engineering 2025 - RRB JE EEE (CBT I) Mock Test- 10

RRB JE EEE (CBT I) Mock Test- 10 for Electrical Engineering (EE) 2024 is part of RRB JE Mock Test Series Electrical Engineering 2025 preparation. The RRB JE EEE (CBT I) Mock Test- 10 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The RRB JE EEE (CBT I) Mock Test- 10 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE EEE (CBT I) Mock Test- 10 below.
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RRB JE EEE (CBT I) Mock Test- 10 - Question 1

Find the minimum number of straight lines to make the given figure.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 1
We get,

Minimum required straight lines are - AB, AC, BC, AH, BD, CD, EF, EG, FH, GH, FG, FM, MG, IK, KL and LJ.

Total = 16

RRB JE EEE (CBT I) Mock Test- 10 - Question 2

If 1.96 of 1.2 of a = 3.43 of 1.6 of b. Find b: a.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 2

1.96 of 1.2 of a = 3.43 of 1.6 of b

1.96 x 1.2 x a = 3.43 x 1.6 x b

3a = 7b

b: a = 3: 7

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RRB JE EEE (CBT I) Mock Test- 10 - Question 3

Following bar graph shows the sales of clothes of 3 Brands: Levi's, Lee Cooper and Spykar in Central Mall in Bangalore for the months of October, November and December.

For all 3 months, what is the difference between the sum of the number of Levi's and Spykar clothes sold and the number of Lee Cooper clothes sold?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 3
sum of the number of Levi's and Spykar clothes sold

= 1500 + 1000 + 1000 + 1800 + 2000 + 1500 = 8800

number of Lee Cooper clothes sold

1200 + 1500 + 1800 = 4500

Difference = 8800 - 4500 = 4300

RRB JE EEE (CBT I) Mock Test- 10 - Question 4

What will come in place of (?) in this equation?

? ÷ [(5/12) of (12/25) of 125 ÷ {√(21 + √16)/2}] = ? + 3

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 4
Let ? be denoted by 'x'.

x ÷ [(5/12) of (12/25) of 125 ÷ {√(21 + √16)/2}] = x + 3

⇒ x ÷ [(5/12) of (12/25) of 125 ÷ {√(21 + 4)/2}] = x + 3

⇒ x ÷ [(5/12) of (12/25) of 125 ÷ {√(25)/2}] = x + 3

⇒ x ÷ [(5/12) of (12/25) of 125 ÷ {5/2}] = x + 3

⇒ x ÷ [25 ÷ 2.5] = x + 3

⇒ x/10 = x + 3

⇒ x = 10x + 30

⇒ x = −10/3

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

How many triangles are present in the given figure?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 5

There are 9 triangles in the given figure.

Triangles are - ABE, ADE, BLM, BCM, DEJ, DJI, FHM, FGM, HLM.

RRB JE EEE (CBT I) Mock Test- 10 - Question 6

Calculate the difference between mean deviation and variance of given numbers.

3, 6, 8, 1, 4, 6, 4, 2, 2, 4

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 6

Mean = (3 + 6 + 8 + 1 + 4 + 6 + 4 + 2 + 2 + 4)/10 = 4

Now, distance of each numbers from their mean are:

4 - 3 = 1

6 - 4 = 2

8 - 4 = 4

4 - 1 = 3

4 - 4 = 0

6 - 4 = 2

4 - 4 = 0

4 - 2 = 2

4 - 2 = 2

4 - 4 = 0

Mean deviation = (1 + 2 + 4 + 3 + 0 + 2 + 0 + 2 + 2 + 0)/10 = 1.6

Variance = (12 + 22 + 42 + 32 + 02 + 22 + 02 + 22 + 22 + 02)/10 = 4.2

Difference = 4.2 - 1.6 = 2.6

RRB JE EEE (CBT I) Mock Test- 10 - Question 7

What is the unit digit of (254+145)42?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 7
The concept of cyclicity is used to identify the last digit of the number.

Cyclicity chart:

(Note: when remainder is 0, power = cyclicity)

= (254+145)42

= 39942

Unit digit of 399 is 9.

From the cyclicity chart, the cyclicity of 9 is 2.

So, we have to divide the power of 399 i.e.42, by 2.

After dividing 42 by 2, remainder = 0. So, power = cyclicity = 2

Unit digit of 92=1

Hence, the unit of (254+145)42 is 1.

RRB JE EEE (CBT I) Mock Test- 10 - Question 8

What is the value of

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 8

RRB JE EEE (CBT I) Mock Test- 10 - Question 9

What value we get if divide (11/5) + (9/4) + (7/2) by (132 - 10)?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 9

(11/5) + (9/4) + (7/2)

= (44 + 45 + 70)/20

= 159/20

(132 - 10)

= 169 - 10

= 159

According to the question-

(159/20) ÷ 159

= (159/20) * (1/159)

= 1/20

= 0.05

RRB JE EEE (CBT I) Mock Test- 10 - Question 10

A Faulty watch gains 1 hour in 6 hours. After how many days will it show the correct time again?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 10

It will show the correct time again when it gains 24 hours.

It gains 1 hour in 6 hours

It gains 24 hours in 6*24 hours or 6 days.

RRB JE EEE (CBT I) Mock Test- 10 - Question 11

If, it is given that, 82 - 28 = 32 and 65 - 42 = 38, then, 76 - 18 = ?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 11
(8*2) + (2*8) = 32

(6*5) + (4*2) = 38

Similarly,

(7*6) + (1*8) = 50

RRB JE EEE (CBT I) Mock Test- 10 - Question 12

If a quadratic equation 3x2 - 4√3x + m = 0 has equal roots, then find the value of 3m2 - 5m + 16.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 12
Quadratic equation 3x2 - 4√3x + m = 0 has equal roots. So,

D = 0 = (-4√3)2 - 4 x 3 x m

m = 4

Therefore,

= 3m2 - 5m + 16

= 3 x 4 x 4 - 5 x 4 + 16

= 44

RRB JE EEE (CBT I) Mock Test- 10 - Question 13

How many numbers are irrational among the numbers given below?

6th root of 2187, 5th root of 4096, √(28 x 23 - 283), √(34 x 35 - 23 x 22), √4802

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 13

1. 6th root of 2187 = 3 x 6th root of 3 (irrational)

2. 5th root of 4096 = 4 x 5th root of 4 (irrational)

3. √(28 x 23 - 283) = 19

4. √(34 x 35 - 23 x 22) = 6√19 (irrational)

5. √4802 = 49√2 (irrational)

RRB JE EEE (CBT I) Mock Test- 10 - Question 14

Select the odd one from the following options.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 14
Option (1)

= Product of (0.003 x 25 + 4.65) and (0.05 x 24 - 0.2)

= 4.725 x 1

= 4.725

Option (2):

= Sum of (23 x 0.005 + 4 x 0.003) and 3.598

= 0.127 + 3.598

= 3.725

Option (4):

= 2 x 3 x 5.6 x 0.005 + 45.57 x 0.1

= 4.725

Hence, option (2) is odd among all four options.

RRB JE EEE (CBT I) Mock Test- 10 - Question 15

A contractor undertakes to complete a work in 19 weeks. He employs 180 men for 7 weeks and they complete 1/2 of the work. He then reduces the number of men to 140, who work for 5 weeks. How many men must be employed for the remaining period to finish the work?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 15
Work done by 180 men in 7 weeks = 1260 men weeks.

This is half the work. It means the remaining half is 1260 men weeks.

Work done by 140 men in 5 weeks = 700 men weeks.

So remaining work = 1260 - 700 = 560 men weeks which has to be completed over 7 weeks.

Required number of men = 560/7 = 80

RRB JE EEE (CBT I) Mock Test- 10 - Question 16

A man bought six different fruits, and, each fruit's quantity bought is different. Quantity of Papaya bought is less than that of Oranges as well as Mango. Quantity of Apples bought is less than both Kiwi and Banana. Quantity of Mango is more than only one fruits and that of oranges is more than at least two fruits. Apple's quantity is more than that of oranges. Which fruit's quantity is less than that of three fruits?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 16

Quantity of Papaya bought is less than that of Oranges as well as Mango.

We get,

Case I:

Papaya < Oranges < Mango

Case II:

Papaya < Mango < Oranges

Now,

Quantity of Apples bought is less than both Kiwi and Banana.

Apple's quantity is more than that of oranges.

We get,

Oranges < Apples < Kiwi/banana < Kiwi/Banana

Since,

Quantity of Mango is more than only one fruits.

And, that of oranges is more than at least two fruits.

Thus, case I becomes invalid.

We get,

Papaya < Mango < Oranges < Apples < Kiwi/banana < Kiwi/Banana

So, Orange's quantity is less than that of three fruits.

RRB JE EEE (CBT I) Mock Test- 10 - Question 17

In ∆ABC, P and Q are points on AB and AC respectively such that AP:PB = 2:1 and AQ:QC = 2:1. If BC = 18 cm, what is the length of PQ?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 17

Since, P and Q divide AB and AC respectively in the same ratio,

PQ is parallel to BC

In ∆APQ and ∆ABC,

∠APQ = ∠ABC and ∠AQP = ∠ACB [Corresponding angles]

So, ∆APQ and ∆ABC are similar,

Corresponding sides are proportional,

BC/PQ = AB/AP

⇒ 18/PQ = (AP + PB)/AP

⇒ 18/PQ = 1 + PB/AP

⇒ 18/PQ = 1 + ½

⇒ PQ = 18*(2/3) = 12 cm

RRB JE EEE (CBT I) Mock Test- 10 - Question 18

The Pie-Chart shows the percentage distribution of the number of employees that work in a particular department within a Company.

If the number of employees that work in R&D is 588, what is the number of employees who work in Senior Management?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 18
14% of total employees = 588

Total employees = 588*(100/14) = 4200

Number of employees who work in Senior Management

= 6% of 4200 = 252

RRB JE EEE (CBT I) Mock Test- 10 - Question 19

Area of a circle is 706.5 cm2 and its radius is increased by a certain amount so that its circumference is increased by 31.4 cm. What is the new area of the circle? (Use π = 3.14)

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 19
Let the original radius of the circle be R

Area of circle = πR2

R2 =706.5/3.14 = 225

R = 15 cm.

Let radius is increased by S cm.

Increase in circumference = (2π x (15 +S)) - (2π x15)

= 2πS

2πS = 31.4

S = 5 cm

New area of circle = π x (20)2 = 3.14 x 400 = 1256 cm2.

RRB JE EEE (CBT I) Mock Test- 10 - Question 20

What is the angle between the initial and final position of the second hand of o watch when it has travelled from 02:37:42 p.m. to 02:39:54 p.m.?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 20
In 15 seconds, the seconds hand travels 90°

In 1 second, the second hand travels 6°

Time duration betweenthe 02:37:42 p.m. and 02:39:54 p.m.

= 2 minutes 12 seconds.

In every minute, the second hand comes back to its original position.

Angle between initial and final position of second hand

= 12*6 = 72°

RRB JE EEE (CBT I) Mock Test- 10 - Question 21

A person has total of Rs.60000 with him, out which 60% he invested in scheme A at 15% SI for 4 years and remaining in scheme B at 25% SI for 4 years. Now, total interest after 4 years he invested in scheme B for 2 years and total principle amount he invested in scheme A for 2 years. What will be the difference between total amount of interest from both the schemes after 6 years?

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

(A) Amount invested in scheme A and scheme B for the second time is Rs.60000 and [21600 + 24000 = Rs.45600] respectively.

(B) Required difference = (21600 + 18000) ~ (24000 + 22800) = Rs.7200

(C) Initial amount invested in scheme A and B is 60% of 60000 = Rs.36000 and 40% of 60000 = Rs.24000 respectively.

(D) Interest from scheme A and scheme B after 2 years = [(60000 * 15 * 2)/100 = Rs.18000] and [(45600 * 25 * 2)/100 = Rs.22800] respectively.

(E) Interest from scheme A and scheme B after 4 years = [(36000 * 15 * 4)/100 = Rs.21600] and [(24000 * 25 * 4)/100 = Rs.24000] respectively.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 21

(C) Initial amount invested in scheme A and B is 60% of 60000 = Rs.36000 and 40% of 60000 = Rs.24000 respectively.

(E) Interest from scheme A and scheme B after 4 years = [(36000 * 15 * 4)/100 = Rs.21600] and [(24000 * 25 * 4)/100 = Rs.24000] respectively.

(A) Amount invested in scheme A and scheme B for the second time is Rs.60000 and [21600 + 24000 = Rs.45600] respectively.

(D) Interest from scheme A and scheme B after 2 years = [(60000 * 15 * 2)/100 = Rs.18000] and [(45600 * 25 * 2)/100 = Rs.22800] respectively.

(B) Required difference = (21600 + 18000) ~ (24000 + 22800) = Rs.7200

RRB JE EEE (CBT I) Mock Test- 10 - Question 22

Express 0.112112112.. into (p/q) form.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 22

Let X = 0.112112112.. (I)

⇒ 1000X = 112.112112112.. (II)

(II) - (I) gives,

999X = 112

⇒ X = 112/999

RRB JE EEE (CBT I) Mock Test- 10 - Question 23

If 70% of (a - b) = 40% of (a + b), find a:b

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 23
70% of (a - b) = 40% of (a + b)

⇒ 70/100 x (a - b) = 40/100 x (a + b)

⇒ 7a - 7b = 4a + 4b

⇒ 7a - 4a = 4b + 7b

⇒ 3a = 11b

⇒ a/b = 11/3

⇒ a:b = 11:3

RRB JE EEE (CBT I) Mock Test- 10 - Question 24

A is as much younger than B as he is older than C. If the sum of the ages of B and C is 15 years, what is definitely the difference between B and A's age?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 24
B - A = B - C => A = C.

Also, B + C = 15 => B + A = 15

So, (B - A) cannot be determined.

RRB JE EEE (CBT I) Mock Test- 10 - Question 25

What is the value of (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(100 ÷ 8 X 2))}]?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 25

(2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(100 ÷ 8 X 2))}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(12.5 X 2))}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ √(25))}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √(80 ÷ 5)}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {√625 − √16}]

= (2/3) of (6/7) of (14/15) of [63 ÷ {25 − 4}]

= (2/3) of (6/7) of (14/15) of [63 ÷ 21]

= (2/3) X (6/7) X (14/15) X 3

= 8/5

RRB JE EEE (CBT I) Mock Test- 10 - Question 26

The product of three numbers a, b and 21 is 10290 and their HCF is 7. If a is greater than b and b is greater than 10, then find the LCM of greatest and smallest number.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 26

let a = 7m , b = 7n

According to the question:

10290 = 7m x 7n x 21

10 = m x n

There are two possible values for m and n.

Case 1: when, m = 10, n = 1

a = 7m = 7 x 10 = 70

b = 7n = 7 x 1 = 7 < 10

So, this case is not valid.

Case 2: when, m = 5, n = 2

a = 7m = 7 x 5 = 35

b = 7n = 7 x 2 = 14

LCM of 35 and 14 = 70

RRB JE EEE (CBT I) Mock Test- 10 - Question 27

Find the value of 13 + (1/3) of 81 - (1/4) of 192 + 23 of 16 ÷ 92.

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 27
= 13 + (1/3) of 81 - (1/4) of 192 + 23 of 16 ÷ 92

= 13 + 27 - 48 + 4

= - 4

RRB JE EEE (CBT I) Mock Test- 10 - Question 28

By selling 45 sweets for Rs 40 a man losses 20%. How much should he sell for Rs 24 to gain 20%?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 28

SP of 45 sweets = Rs 40

SP of 1 sweets = 40/45

L = 20%

CP = 100/80 * 40/45 = 10/9

SP2 = (100 + 20) / 100 * CP

SP2 = 120/100 * 10/9 = 4/3

SP of Rs 4/3 sweets for 1 sweets

SP of Rs 24 for = 24/ (4/3) = (24 * 3) / 4 = 18 sweets

RRB JE EEE (CBT I) Mock Test- 10 - Question 29

A shopkeeper has 5 items such that their mean weight is 30 kg. If weight of these items are consecutive multiples of 5, what will be the sum of the weight of the lightest and heaviest item?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 29
Let 'a' kg is the weight of the lightest item.

Then, the arrangement of weights in ascending order are:

a, a + 5, a + 10, a + 15, a + 20

Mean = 30

5 x 30 = a + a + 5 + a + 10 + a + 15 + a + 20

a = 20 kg

Weight of heaviest item = 20 + 20 = 40 kg

Sum of weight of lightest and heaviest item = 20 + 40 = 60 kg

RRB JE EEE (CBT I) Mock Test- 10 - Question 30

Ram had Rs 2000, part of which he lent at 15% for two years at simple interest and the rest at 10% at compound interest for two years. The total interest was Rs 528. Find the sum lent at simple interest (in Rs)?

Detailed Solution for RRB JE EEE (CBT I) Mock Test- 10 - Question 30
Rate of simple interest = 15%

Let the amount lent at SI be Rs x.

For 2 years, SI = x*15*2/100 = 30*x/100

Amount lent at CI = 2000 - x

For 2 years, CI = (2000 - x)*(1 + 10/100)2 - (2000 - x) = (2000 - x)(21/100)

30x/100 + (2000-x)*21/100 = 528

After solving we get x = Rs 1200

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