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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  RRB JE ECE (CBT I) Mock Test- 2 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 2 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT I) Mock Test- 2

RRB JE ECE (CBT I) Mock Test- 2 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT I) Mock Test- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT I) Mock Test- 2 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- 2 below.
Solutions of RRB JE ECE (CBT I) Mock Test- 2 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & RRB JE ECE (CBT I) Mock Test- 2 solutions in Hindi for Electronics and Communication Engineering (ECE) 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- 2 | 100 questions in 90 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
RRB JE ECE (CBT I) Mock Test- 2 - Question 1
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A 6 m-long vertical pole casts a shadow of 4 m. At the same time, a tower casts a shadow 25 m long on the ground. The height of the tower is :

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 1
Let the angle of elevation of the top of the tower from the endpoint of the shadow be θ

RRB JE ECE (CBT I) Mock Test- 2 - Question 2
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Two circles of radius 37 cm and 20 cm intersect each other at P, and Q. O and O’ are the centres of the circles. If the length of PQ is 24 cm, then the distance between their centre is:

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

OP= 37 cm and O'P = 20 cm

PQ = 24 cm

PM = MQ = 12 cm

O'M = 16 cm

OO' = OM + 0'M

OO' = 35 + 16 = 51 cm

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RRB JE ECE (CBT I) Mock Test- 2 - Question 3
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Concentrations of three solutions A, B and C, are 20%, 30% and 40%, respectively. They are mixed in the ratio 3: 5: x resulting in a solution of 30% concentration. Find x

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 3
3 × 0.2 + 5 × 0.3 + x × 0.4 = 0.3 (3+5+x)

⇒ 2.1 + 0.4x = 0.3(8+x)

⇒ 2.1 + 0.4x = 2.4 + 0.3x

or, x = 3

RRB JE ECE (CBT I) Mock Test- 2 - Question 4
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In which of the following is p > q?

I. (0.9)p > (0.9)q

II. (1.8)p < (1.8)q

III. (8.5)p > (8.5)q

IV. (1/2)p < (12)q
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 4
Use the following results.

If (N)p > (N)q when N > 0 and positive integral

= p > q

If (N)p > (N)q when N > 0 and fractional

= p < q

If (N)P < (N)q when N > 0 and positive integral

= P < q

< If (N)p < (N)q when N > 0 and fractional

= P > q

Using these results, we get option C is the correct answer.

RRB JE ECE (CBT I) Mock Test- 2 - Question 5
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There are some rabbits and some pigeons in a zoo. If their heads and feet are counted, the number of heads is 90, and their feet are counted, then their number of feet is 224. Tell me, how many rabbits are there in this zoo?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 5

Ratio = 22 : 68

That is, the number of rabbits is 22.

RRB JE ECE (CBT I) Mock Test- 2 - Question 6
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A shopkeeper marks the price of an article at Rs. 320. Find the cost price if, after allowing a discount of 10%, he still gains 20% on the cost price.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 6
Market price = 320 Rs.

Discount= 10%

Selling price = 320 x 90/100 = 288 Rs.

Profit is given = 20%

So , C.P. = [100/ (100 + Gain%)] x S.P

CP = (100/120 )x 288 = 240 Rs.

RRB JE ECE (CBT I) Mock Test- 2 - Question 7
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With the increase in the price of a pressure cooker by 20%, its sale decreases by 50%. How much of a shopkeeper will earn less or more income from the past?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 7

= - 30 - 10

= - 40%

Percentage decrease in income = 40%

RRB JE ECE (CBT I) Mock Test- 2 - Question 8
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If '-' stands for division, '+' for multiplication, ' ÷ ' for subtraction and 'x' for addition, which one of the following equations is correct?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 8
18 − 3 + 2 × 8 ÷ 6 = 14

After replacing signs,

18 ÷ 3 × 2 + 8 − 6

⇒ 18/3 × 2 + 8 − 6

⇒ 12 + 8 − 6

⇒ 14

RRB JE ECE (CBT I) Mock Test- 2 - Question 9
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From a container, 5 litres of wine was drawn out and was replaced by soda. Again 5 litres of the mixture was drawn out & replaced by the soda. Thus the quantity of wine and soda in the container after these two operations is 9:16. The quantity of the mixture is
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 9

k = 12.5 litres

RRB JE ECE (CBT I) Mock Test- 2 - Question 10
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A number consists of two digits. If the digits interchange places and the new number is added to the original number, then the resulting number will be divisible by:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 10
Let the ten's digit be x, and the unit's digit be y.

Then, number = 10x + y.

Number obtained by interchanging the digits = 10y + x.

∴ (10x + y) + (10y + x) = 11(x + y), which is divisible by 11.

RRB JE ECE (CBT I) Mock Test- 2 - Question 11
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The sum of the 28th, 29th and 30th terms of an Arithmetic Progression, which has 57 terms, is 111. Find the arithmetic mean of the progression.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 11
Since the progression has an odd number of terms, the average of the three middlemost terms will be its arithmetic mean.

Hence, the required mean 111/3 = 37

RRB JE ECE (CBT I) Mock Test- 2 - Question 12
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S =

Find S.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 12

RRB JE ECE (CBT I) Mock Test- 2 - Question 13
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Mahesh grows potatoes in his backyard, which is in the shape of a square. Each potato takes 1 cm2 in his backyard. This year, he has been able to grow 137 more potatoes than last year. The shape of the backyard remained a square. How many potatoes did Mahesh produce this year?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 13
Let the area of backyard be x2 this year and previous year was y2

x2 − y2 = 137

(x+y) (x-y) = 137 x 1

x + y = 137 ---- (i)

x - y = 1 ---- (ii)

On solving eq. (i) & (ii), we get

x = 69, y = 68

No of potatoes produced this year = 692 = 4761

RRB JE ECE (CBT I) Mock Test- 2 - Question 14
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The dimensions of a luggage box are 80 cm, 60 cm and 40 cm. How many sq. cm of cloth is required to cover the box?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 14
Surface area of a cuboid= 2(lb+bh+lh),

where l = length,

b = breadth,

h = height of the cuboid

Let the length, breadth and height of the luggage be l , b, h respectively

Here, l = 80 cm, b = 60 cm, h = 40 cm

Therefore, area= 2(80 x 60+60 x 40+40 x 80) cm2

= 2(4800+2400+3200) cm2

= 2 x 10400 cm2

= 20800 cm2

20800sq. cm of cloth is required to cover the box

RRB JE ECE (CBT I) Mock Test- 2 - Question 15
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In how many years will 800 rupees 10% annual compound interest rate be 926.10 if interest is biannually combined?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 15
Supposed t is the half-yearly year

t = 3

Hence Required year = 3/2 = 1.5 year.

RRB JE ECE (CBT I) Mock Test- 2 - Question 16
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A shopkeeper earns a profit of 12% on selling a book at a 10% discount on the printed price. The ratio of the cost price and the printed price of the book is
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 16
Let the printed price of the book be Rs x.

and the cost price of the book will be Rs. y.

According to the question,

90% of x = 112% of y

Hence, required ratio = 45 : 56

RRB JE ECE (CBT I) Mock Test- 2 - Question 17
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What is the mode of the following list of numbers:

52, 54, 55, 56, 55, 54, 53, 55, 53, 51, and 57 ?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 17
The "mode" is the value that occurs most often. If no number in the list is repeated, then there is no mode for the list.

So 55 is the mode as it comes 3 times.

RRB JE ECE (CBT I) Mock Test- 2 - Question 18
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In an election, there were only two candidates. One of the candidates secured 40% of the votes and was defeated by the other candidate by 298 votes. The total number of votes polled is?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 18

Let the total number of votes = 100

20 units = 298

1 unit = 298/20

100 units = 298/20 x 100 = 1490

RRB JE ECE (CBT I) Mock Test- 2 - Question 19
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If sin x = 4/5, then sec2 x−1 = ?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 19

RRB JE ECE (CBT I) Mock Test- 2 - Question 20
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Read the following graph carefully and answer the question given below.

Q. For which year, the percent increase of foreign exchange reserves over the previous year is the highest?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 20

RRB JE ECE (CBT I) Mock Test- 2 - Question 21
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Read the following graph carefully and answer the question given below.

Q. The ratio of the number of years in which the foreign exchange reserves are above the average reserves to those in which the reserves are below the average reserves is?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 21
Average foreign exchange reserves over the given period = 3480 million US $.

The country had reserves above 3480 million US $ during the years 1992-93, 1996-97 and 1997-98, i.e., for 3 years and below 3480 million US $ during the years 1991-92, 1993-94, 1994-95, 1995-56 and 1998-99, i.e., for 5 years.

Hence, required ratio = 3 : 5.

RRB JE ECE (CBT I) Mock Test- 2 - Question 22
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Read the following graph carefully and answer the question given below.

The foreign exchange reserves in 1997-98 were how many times that in 1994-95?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 22
Required ratio = 5040/3360 = 1.5
RRB JE ECE (CBT I) Mock Test- 2 - Question 23
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D is 4 years younger than B, and C is 4 years older than A. A and B are twins. If the average age of all the four boys is 11 years, then the sum of the ages of the youngest and the oldest boy is
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 23
Let the age of A = B = y years

Age of D = y-4 years

Age of C = y+4 years

Then,

y-4+y+y+y+4 = 4 x11

y = 11

Sum of ages of the youngest and oldest boy = 7+15 = 22 years

RRB JE ECE (CBT I) Mock Test- 2 - Question 24
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Find the LCM of the following fractions: 2/3, 8/9, 16/27, 32/81
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 24
LCM = (LCM of 2, 8, 16, 32) / (HCF of 3, 9, 27, 81) = 32/3
RRB JE ECE (CBT I) Mock Test- 2 - Question 25
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Present ages of Ajit and Amar are in the ratio of 8: 7, and after four years, the ratio of their ages will be 9: 8. Find the present age of Amar?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 25
Let the present age of Ajit and Amar be 8x and 7x

according to the question,

64x + 32 = 63x + 36

⇒ x = 4

So,

Present age of Ajit = 8 x 4 = 32 years

Present age of Amer = 7 x 4 = 28 years

RRB JE ECE (CBT I) Mock Test- 2 - Question 26
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Read the following information carefully and answer the question given below-

In the CBSE Board exams last year, 53% passed in Biology, 61% percent passed in English, 60% in Social Studies, 24% in Biology and English, 35% in English and Social Studies, 27% in Biology and Social Studies and 5% in none.

If the number of students in the class is 200, how many passed in only one subject?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 26

Let people who passed all three be x. Then: 53 + 61 + 60 - 24 - 35 - 27 + x = 95

x = 7

Percentage of student passed only in 1 subject = 9 + 9 + 5 = 23%

Now 23% of 200 = 46 students

RRB JE ECE (CBT I) Mock Test- 2 - Question 27
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Read the following graph carefully and answer the question given below.

The number of mobile sim cards in 4 states are given in multiple bar diagrams.

In Assam, the ratio of Aircel sim card and Airtel Sim card sold is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 27
In Assam, Aircel sim cards sold = 2 lakhs

In Assam, Airtel sim cards sold = 5 lakhs

The Ratio of Aircel and Airtel sim cards sold = 2/5 = 2:5

RRB JE ECE (CBT I) Mock Test- 2 - Question 28
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Read the following graph carefully and answer the question given below.

The number of mobile sim cards in 4 states are given in multiple bar diagrams.

In Which state are there the largest number of owners of Airtel sim card?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 28

Largest No. of owners of Airtel sim cards= 13 lakhs

(Tamil Nadu)

RRB JE ECE (CBT I) Mock Test- 2 - Question 29
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Marked price of articles ‘P’ and ‘Q’ are Rs. 50 and Rs. 35 respectively. With every article of ‘P’, 2 pencils worth Rs. 3 each are given free and with every article of ‘Q’, 3 pens worth Rs. 5 each are given free. If a customer buys 3 articles of ‘P’ and 2 articles of ‘Q’, what is the approximate discount percentage?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 29
With 3 articles of P, 6 pencils are given free

Total price = 50 x 3 + 6 x 3 = Rs. 168

With 2 articles of Q, 6 pens are given free

Total price = 35 x 2 + 6 x 5 = Rs. 100

Total Price of all product = Rs 268

Discount given = 18+30 = Rs 48

Discount Percentage = 48/268×100 ≈ 18%

RRB JE ECE (CBT I) Mock Test- 2 - Question 30
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The cash difference between the selling price of an article at a profit of 2% and 12% is Rs. 3. The ratio of two selling prices is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 2 - Question 30
Let the cost price (C.P.) of the article be Rs. x.

If profit is 2%:

Selling price (S.P.) of the article = x + (2% of x) = 1.02x (∵ S.P = C.P + Profit)

If profit is 12%:

Selling price (S.P.) of the article = x + (12% of x) = 1.12x

Cash difference between the above S.P’s = Rs.3

⇒ 1.12x – 1.02x = 3

⇒ 0.1x = 3

⇒ x = 30

The ratio of two selling prices = 1.02x : 1.12x = 51 : 56

∴ The ratio of two selling prices = 51 : 56

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