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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  RRB JE Mock Test Series for ECE 2025  >  RRB JE ECE (CBT I) Mock Test- 7 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 7 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series for ECE 2025 - RRB JE ECE (CBT I) Mock Test- 7

RRB JE ECE (CBT I) Mock Test- 7 for Electronics and Communication Engineering (ECE) 2024 is part of RRB JE Mock Test Series for ECE 2025 preparation. The RRB JE ECE (CBT I) Mock Test- 7 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT I) Mock Test- 7 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ECE (CBT I) Mock Test- 7 below.
Solutions of RRB JE ECE (CBT I) Mock Test- 7 questions in English are available as part of our RRB JE Mock Test Series for ECE 2025 for Electronics and Communication Engineering (ECE) & RRB JE ECE (CBT I) Mock Test- 7 solutions in Hindi for RRB JE Mock Test Series for ECE 2025 course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt RRB JE ECE (CBT I) Mock Test- 7 | 100 questions in 90 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study RRB JE Mock Test Series for ECE 2025 for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
RRB JE ECE (CBT I) Mock Test- 7 - Question 1
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The ratio of the number of Matches played by Rohit and Kohli is 6:5. Kohli scored an average of 52 runs per match, and Rohit's average is 25% less than that. If the total runs scored by Kohli is 546 more than Rohit, what is the number of matches Kohli has played?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 1
Let the number of matches played by Rohit and Kohli be 6k and 5k respectively.

Average runs scored in a match by Rohit = 52(1 - 25/100) = 39

Given, 52*5k - 39*6k = 546

=> 26k = 546

=> k = 21

Number of Matches played by Kohli = 5*21 = 105

RRB JE ECE (CBT I) Mock Test- 7 - Question 2
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What is the value of

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 2

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RRB JE ECE (CBT I) Mock Test- 7 - Question 3
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For what value of 'k', the quadratic equation 4x2 - 3m√2x + k = 0 (where, m>0) has equal roots? (Given: 2m2 + 7m - 60 = 0)

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 3

For 2m2 + 7m - 60 = 0:

2m2 - 8m + 15m - 60 = 0

m = -15/2,

Quadratic equation 4x2 - 3m√2x + k = 0 has equal roots, therefore,

D = 0 = (-3m√2)2 - 4 x 4 x k

For m = 4:

18 x (4)2 = 16 x k

k = 18

RRB JE ECE (CBT I) Mock Test- 7 - Question 4
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Study the following table carefully and answer the related question.

The following table represents the marks of the top four students of the class in four subjects.

Who got approximately 71.5% percentage in the class?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 4
Total marks = 100 + 100 + 150 + 150 = 500

Total marks of Rakhi = 78 + 60 + 96 + 124 = 358

Rakhi's percentage = (358/500) x 100 = 71.6%

Total marks of Jay = 56 + 80 + 88 + 92 = 316

Jay's percentage = (316/500) x 100 = 63.2%

Total marks of Amit = 95 + 90 + 80 + 75 = 340

Amit's percentage = (340/500) x 100 = 68%

Total marks of Urvi = 70 + 70 + 120 + 105 = 365

Urvi's percentage = (365/500) x 100 = 73%

RRB JE ECE (CBT I) Mock Test- 7 - Question 5
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A sum of money at simple interest amounts to Rs 2240 in 2 years and to Rs 2600 in 5 years. What is the principal amount?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 5

Interest for 3 years = Rs 2600 - Rs 2240 = Rs 360

=> Interest for 2 years = 360x2/3 =Rs 240

Hence, principal = (2240 - 240) = Rs 2000

RRB JE ECE (CBT I) Mock Test- 7 - Question 6
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Number of kids in 10 families is given below. Find the median number of kids.

3, 4, 2, 1, 0, 2, 4, 0, 1 and 2
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 6
Ascending order of number of kids in the family-

0, 0, 1, 1, 2, 2, 2, 3, 4, 4

Total terms = 10 = Even

Median = [(n/2)th term + {(n/2) + 1} th term]/2

=> [(10/2)th term + {(10/2) + 1}th term]/2

RRB JE ECE (CBT I) Mock Test- 7 - Question 7
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There are four persons A, B,C and D, and the ratio of the present age of A to C is 4: 3, C is 7 years younger to E who is 16 years old at present. Present age of B is 75% of the present age of D and B is 6 years older than A. What is the average present age of persons B, C and D?

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

Let the age of A = 4A

(A) The ages of A and C are 12 years and 9 years respectively.

(B) Present age of E = (3A + 7) = 16 => A = 3

(C) Required average = (18 + 9 + 24)/3 = 17 years

(D) Present age of B = 12 + 6 = 18 years and present age of D = 18 * (100/75) = 24 years

(E) Present age of C = 4A * (3/4) = 3A
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 7
The correct order is

(E) Present age of C = 4A * (3/4) = 3A

(B) Present age of E = (3A + 7) = 16 => A = 3

(A) Age of A and C is 12 years and 9 years respectively

(D) Present age of B = 12 + 6 = 18 years and present age of D = 18 * (100/75) = 24 years

(C) Required average = (18 + 9 + 24)/3 = 17 years

RRB JE ECE (CBT I) Mock Test- 7 - Question 8
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The average age of a man and his son is 54 years. The ratio of their ages is 23:13. What will be the ratio of their ages after 6 years?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 8
Let their ages be 23x and 13x respectively.

Average age is 54 years which means the sum of their ages = 108 years

i.e. 36x = 108

or x = 3

Hence their ages are 69 and 39 years respectively.

After 6 years, their ages would be 75 and 45 years

So the required ratio = 75:45 = 5:3

RRB JE ECE (CBT I) Mock Test- 7 - Question 9
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Out of 5 numbers, whose average is 36, the first one is one fifth of the sum of the last four. The first number is?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 9
Let the five numbers be a, b, c, d and e

Given, (a+b+c+d+e)/5=36

a+b+c+d+e = 180

a = (b+c+d+e)/5

a = (180-a)/5

a = 30

RRB JE ECE (CBT I) Mock Test- 7 - Question 10
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What value we get when: √(92 - 43 + 3) is multiplied by √(25 * 10-1)?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 10

= √(92 - 43 + 3) * √(25 * 10-1)

= √(81 - 64 + 3) * √(32 ÷ 10)

= √20 * √3.

= √(20 * 3.2)

= √64

= 8

RRB JE ECE (CBT I) Mock Test- 7 - Question 11
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Following graph shows the number of questions attempted and the number of correct questions in preliminary and mains exams by 5 persons.

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What is the ratio of the total number of non-correct questions out of total number of questions attempted by all 5 persons in preliminary exam to the total number of non-correct questions out of total number of questions attempted by all 5 persons in mains exam?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 11
Total number of non-correct questions out of total number of questions attempted by all 5 persons in preliminary exam

= (10 +5 +5 +10 +5)

= 35.

Total number of non-correct questions out of total number of questions attempted by all 5 persons in mains exam

= (30 +15 +11 +30 +40)

= 126

Required ratio= 35/126

RRB JE ECE (CBT I) Mock Test- 7 - Question 12
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If n is the greatest number that will divide the three numbers 1236, 3708 and 4944 leaving the same remainder in each case. What is the product of digits of 'n'?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 12

n = HCF of (3708 - 1236), (4944 - 3708) and (4944 - 1236)

= HCF of 2472, 1236 and 3708

= 1236

Product of digits of n = 1 x 2 x 3 x 6 = 36

RRB JE ECE (CBT I) Mock Test- 7 - Question 13
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LCM of three numbers is 840 and their ratio is 2: 3: 5 respectively. What will be the HCF of these numbers?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 13
Let the three numbers be 2a, 3a and 5a respectively.

According to the question:

840 = 2a x 3a x 5a

a = 28

Therefore, the numbers are (2 x 28), (3 x 28) and (5 x 28) respectively.

Hence, HCF of these numbers = 28

RRB JE ECE (CBT I) Mock Test- 7 - Question 14
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A discount of 30% on one article is the same as a discount of 35% on another article. The marked prices of two articles can be (in ₹.)
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 14
Since discounts are 30 % and 35 % respectively, Therefore, discounts are 0.3x and 0.35y, where x and y are the marked prices respectively.

Further, it is given that these discounts are equal, so 0.3x = 0.35y=> x: y = 7: 6

i.e. marked price ratio is 7:6.

Now, check for all the given options.

RRB JE ECE (CBT I) Mock Test- 7 - Question 15
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A boat takes 4 more hours to travel 180 km upstream than to travel the same distance downstream. If the still water speed of the boat is 28 km/hr, what is the stream speed?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 15
Let the stream speed be 'r' km/hr.

180/(28 - r) = 4 + 180/(28 + r

45/(28 - r) - 45/(28 + r) = 1

90r/(784 - r2) =

r2 + 90r - 784 = 0

(r - 8)(r + 98) = 0

r = 8

RRB JE ECE (CBT I) Mock Test- 7 - Question 16
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The ratio of heights of Asif and Bharat is 3:4 and the ratio of heights of Bharat and Chatur is 6:7. If the average of their heights is 175 cm, what is the height of Asif?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 16

Ratio of heights of Asif and Bharat = 3:4 = 9:12

Ratio of heights of Bharat and Chatur = 6:7 = 12:14

Ratio of heights of Asif, Bharat and Chatur = 9:12:14

Sum of parts of the ratio = 9 + 12 + 14 = 35

Asif's height =(9/35)*175*3 = 135 cm

RRB JE ECE (CBT I) Mock Test- 7 - Question 17
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A alone can do a piece of work in 20 days, B alone in 40 days and C alone in 60 days. Everyone begins to do the work together, but A leaves after 4 days and B leaves 6 days before the completion of the work. How many days did the work last?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 17
Total work in 1 day by A, B and C = 1/20 + 1/40 + 1/60 = 11/120

In 4 days = 44/120

By C in last 6 days = 6/60 = 1/10

Total = 44/120 + 1/10 = 56/120

Work left = 1 - 56/120 = 64/120

Work done by B and C in one day = 1/40 + 1/60 = 5/120

Work left is done by B and C. Number of days it takes = (64/120)*(120/5) = 64/5

Total time = 10 + 64/5 = 22 4/5

RRB JE ECE (CBT I) Mock Test- 7 - Question 18
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A faulty watch gains 5 minutes every 40 minutes. After how many days will it show the correct time again?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 18
It will show the correct time again when it gains 24 hours.

It gains 5 minutes in 40 minutes

It gains 1 minute in 8 minutes

It gains 1 hr in 480 minutes

It gains 24 hr in 480*24 minutes or (480*24)/(60*24) or 8 days

RRB JE ECE (CBT I) Mock Test- 7 - Question 19
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If x/y + y/x + 1 = 0, what is the value of (x - y)6/9x3y3?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 19

x/y + y/x + 1 = 0

=> x2 + y2 + xy = 0

=> x2 - 2xy + y2 + 3xy = 0

=> (x - y)2 = −3xy

=> (x - y)6 = −27x3y3

(x - y)6/9x3y3

∴ = (−27x3y3)/9x3y3

= −3

RRB JE ECE (CBT I) Mock Test- 7 - Question 20
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A TV costs Rs.20000 and the shopkeeper announces that the price of the TV will reduce at 10% per annum. What will be the cost price of the item after 2 years?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 20
Whenever the dimensions of the quantity are decreased by 'x' percentage then the overall percentage change = (-x%)+(-x%)+-x%)(- x%/100

Overall change in the price of the TV = -10-10+100/100=-19%

Cost Price of the TV after two years = 20000-(19/100 X 20000)=Rs.16200

RRB JE ECE (CBT I) Mock Test- 7 - Question 21
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In the following diagram ABCD is a square inscribed in a circle of centre O. If perimeter of the square ABCD is 40 cm, then what is the area of the blue portion?(take π = 3.14)
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 21
Side of the square = Perimeter/4 = 40/4 =10 cm

Diagonal AC of square = Side x √2 =10√2 cm

AC is also the diameter of a circle.

Radius of circle = 5√2 cm

Area of circle = 3.14 x (5√2)2 = 3.14 x 50 = 157 cm2

Area of square = 100 cm2

Area of blue portion = 157-100 = 57 cm2.

RRB JE ECE (CBT I) Mock Test- 7 - Question 22
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Ramesh salary got 2 salary hikes, one of 10%, the next one of 25%. If the total increase in Ramesh salary was 24000, what is the increase in his salary from the first hike to the second Hike?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 22
Let Ramesh's initial salary be Rs 100x

His salary after 1st Hike = (110/100)*100x = Rs 110x

His salary after 2nd Hike = (125/100)*110x = Rs 137.5x

Given 137.5x - 100x = 24000

=> 37.5x = 24000

=> x = 640

increase in his salary from the first hike to the second Hike

= 137.5x - 110x

= 27.5x

= Rs 17,600

RRB JE ECE (CBT I) Mock Test- 7 - Question 23
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Which of the following is irrational?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 23

Option (1):

= Product of 3rd root of 2197 and √6889

= 13 x 83

= 1079

Option (2):

= Product of 4th root of 4096 and √1377

= 8 x 9√17

= 72√17 (irrational)

Option (3):

= Product of 3rd root of 3375 and √5929

= 15 x 77

= 1155

Option (4):

= Product of 4th root of 2401 and √1444

= 7 x 38

= 266

RRB JE ECE (CBT I) Mock Test- 7 - Question 24
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A car wiper can clean 693 sq. inch area in its one sweep. If its length is decreased by 3.5 inch, then how much area will it clean in one sweep? (Car wiper covers a semi-circle in one sweep)
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 24
Let original length a car wiper be L inch.

Area cleaned by the car wiper in 1 sweep

= (1/2) x (22/7) x (L)2

L2 = 693 x (7/11) = 63 x 7

L = 21 inch.

Area cleaned by car wiper in a sweep if its length decreased by 3.5 inch.

= (1/2) x (22/7) x (17.5) x (17.5)

= 481.25 sq. inch.

RRB JE ECE (CBT I) Mock Test- 7 - Question 25
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A bus reached Nagpur to Amravati in 2:30 hr with an average speed of 70 km/hr. If the average speed increased by 30 km/hr. How much time it will take to cover the same distance.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 25
Distance = speed *time

= (70 * 150)/60 min

= 175 km

New speed = (70 + 30) km/hr = 100 km/hr

Time = (Distance )/speed

= (175 * 60)/100

= 105 min

RRB JE ECE (CBT I) Mock Test- 7 - Question 26
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Two poles of equal height are on either side of a 100 m wide road. From a point on the road, the angles of elevation of their tops are 30° and 60°. Find the height of each pole.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 26
Let the height of poles be h m.

Tan 60 = h/x

=> √3 = h/x

Or h = √3 x

Similarly,

1/√3 = h/(100-x)

=> 100-x = √3 h = 3x

=> x = 25

Therefore, h = 25√3 m

RRB JE ECE (CBT I) Mock Test- 7 - Question 27
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The sales for the month of January was 500 units. Sales increased by 50 units every month for the next 11 months. Find the average monthly sales for the year?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 27
Total sales for the year = 500 + 550 + 600 + 650 +700 +750 + 800 +850 +900+ 950 + 1000 +1050 = 9300

Average monthly sales for the year = 9300/12 = 775

RRB JE ECE (CBT I) Mock Test- 7 - Question 28
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Select the odd one from the following options.

a = 2.5 x 2.4 - 8 x 0.5

b = 12 x 4 x 0.5 - 16

c = 21 - 30 x 0.5
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 28
a = 2.5 x 2.4 - 8 x 0.5 = 2

b = 12 x 4 x 0.5 - 16 = 8

c = 21 - 30 x 0.5 = 6

RRB JE ECE (CBT I) Mock Test- 7 - Question 29
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A grocer bought two types of rice A and B, at Rs.20 per kg and Rs.25 per kg. If no profit and no loss occurs on selling both the items together at Rs.24 per kg, then find the ratio of quantity of rice A and B.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 29
Let quantity of rice A and B are a and b respectively.

Total selling price = 20a + 25b

Total cost price = 24a + 24b

No profit, no loss occurred. So,

Total selling price = total cost price

20a + 25b = 24a + 24b

4a = b

a: b = 1: 4

RRB JE ECE (CBT I) Mock Test- 7 - Question 30
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In an election contested by two candidates, a candidate who got 54% of the total votes won by a margin of 288 votes. Total votes are.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 7 - Question 30
Let total number of votes cast = x

54% of x = 46% of x + 288

8% of x = 288

x = 288 x 100/8 = 3600. Hence, the answer option is (2).

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