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RRB JE ECE (CBT I) Mock Test- 8 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT I) Mock Test- 8

RRB JE ECE (CBT I) Mock Test- 8 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT I) Mock Test- 8 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT I) Mock Test- 8 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- 8 below.
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RRB JE ECE (CBT I) Mock Test- 8 - Question 1

Ashok remembers his birthday is after 21st December. While his mother remembers his birthday is before 23rd December. On which date of December is his birthday?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 1
As Ashok remembers, his birthday is after the 21st and his mother remembers his birthday is before 23rd, therefore, his birthday is on 22nd December.
RRB JE ECE (CBT I) Mock Test- 8 - Question 2

If in a frequency distribution, the mean and median are 21 and 22 respectively, then its mode is approximately-

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 2
Mode + 2 Mean = 3 Median

Mode = 3 × 22 - 2 × 21 = 66 - 42 = 24.

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RRB JE ECE (CBT I) Mock Test- 8 - Question 3

O is the center of the circle. AC and BD are two chords of the circle intersecting each other at P. If ∠APB=300, ∠APB=300, and ∠AOB=450∠AOB=450 then ∠DOC∠DOC is equal to ?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 3

RRB JE ECE (CBT I) Mock Test- 8 - Question 4

A does 60% of the work in 45 days. He then calls B, and they together finish the remaining work in 18 days. How long B alone would take to do the whole work?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 4

Time of A to do complete work

= 45/60 x 100 = 75

Time of A+B to do complete work

= 18/40 x 100 = 45

Let time of B to do complete work alone is T

= 1/45 1/75 = 1/T

T = 112.5 days

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

Abhinash gets 32% marks out of 300 in computer Science. How much marks out of 200 he should get in java language paper so that out of total marks his percentage becomes 46% -

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 5
Abhinash get marks in computer science = 300×32/100=96

Total marks in both subjects = 500×46/100=230

In java = 230−96=134

Required percentage = =134/200×100=67%

RRB JE ECE (CBT I) Mock Test- 8 - Question 6

There are two taps P and Q that can fill a tank in 12 and 24 minutes respectively. Both of them are opened together and after 4 minutes Q is closed. How long does it take for P alone to fill the tank?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 6

Part of a tank filled in 1 Minute by tank P = 1/12

Part of a tank filled in 1 Minute by tank Q = 1/24

Part of tank Filled by P and Q in 4 minutes = 4 x (1/12 + 1/24) = 4 x 3/24 = 1/2

Remaining part = 1 - 1/2 = ½

Equating Time and part filled, = 1/12 x 2/1 = 1/x

X = 6 minutes

RRB JE ECE (CBT I) Mock Test- 8 - Question 7

Find the compound interest on a sum of Rs 425 compounded half yearly for 2 years at the rate of 4%.

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 7

The compound interest for the given condition can be given by,

Therefore, the compound interest is Rs 35.03.

RRB JE ECE (CBT I) Mock Test- 8 - Question 8

Average age of 6 boys is 14 years. The average age of 11 girls is 12 years. What is the average age (in years) of all boys and girls?

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

Average of 6 boys = Sum of the age of 6 boys/6

⇒ 14 = sum of ages of 6 boys/6

⇒ Sum of the age of 6 boys = 84

Now,

Average of 11 girls = sum of age of 11 girls/11

⇒ 12 = sum of age of 11 girls/11

⇒ Sum of the age of 11 girls = 132

Now,

Sum of age of all girls and boys together = 84 + 132 = 216

∴ Average = 216/17 = 12.7

RRB JE ECE (CBT I) Mock Test- 8 - Question 9

The value of tan 40. tan 430 .tan 470 .tan 860 is-

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 9

tan40. tan430. tan470 tan860

=. tan40 tan430. cot40 cot430

= 1

∴ (tan(900 - 9) = cotθ)

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

Profit obtained from selling an article for Rs 310 equals the loss incurred on selling that article for Rs 230. What will be the loss percentage when the selling price is Rs 180?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 10
As the percentage of profit is equal to percentage of loss,

Profit/loss = (310 - 230) / 2

⇒ 80 / 2 = 40

Cost price = (230 + 40) or (310 - 40) = 270

Loss percentage when selling price is 180 = {(270 - 180) / 270} x 100

⇒ (90/270) x 100 = 100/3 = 33 1/3

Hence, option C is the correct answer.

RRB JE ECE (CBT I) Mock Test- 8 - Question 11

Diameter of the cylindrical vessel is 14 cm. having some water. When a sphere of iron is totally submerged in this water level of water increases by 913 cm. What is the radius of the sphere?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 11
Let R be the radius of the ball.

Then,

The volume of water displaced = Volume of iron ball

⇒ R3 = 73

⇒ R = 7.

RRB JE ECE (CBT I) Mock Test- 8 - Question 12

The given figure, AB = 16 cm, AC = 12 cm, BC = 14 cm, then the length of median AD will be -

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 12

(16)2 + (12)2 = 2(AD2 + CD2)

256 + 144 = 2AD2 + 98

AD = √151

RRB JE ECE (CBT I) Mock Test- 8 - Question 13

Directions: Each question is followed by 2 statements. Go through these questions and answer the questions as per the instructions : choose your answer as: (A) If the question can be answered by using only statement (I), but not by using statement (II) alone. (B) If the question can be answered by using only statement (II), but not by using statement (I) alone. (C) If the question can be answered by using both the statements together, but cannot be answered by using either statement alone. (D) If the question cannot be answered by using both the statements together.

What is the number?

A. The product of two digits of a number is 27.

B. The sum of the two digits is 12. The ratio of the ten's digit to the unit digit is 3:1.

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 13

Let the ten's and unit's digit be x and y respectively.

From A, xy = 27

From B, x + y = 12 and

Now,

3y +y = 12

4y = 12 ⇒ = 3

x = 3y = 9 Number = 93

Thus, B alone gives the answer and A alone does not give the answer

RRB JE ECE (CBT I) Mock Test- 8 - Question 14

Two places R and S are 800 km apart from each other. Two persons start from R towards S at an interval of 2 hours. Whereas A leaves R for S before B. The speeds of A and B are 40 km/h and 60 km/h respectively. B overtakes A at M, which is on the way from R to S. What is the distance from R, where B overtakes A?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 14
Distance between R and M = 4 * 60 = 240
RRB JE ECE (CBT I) Mock Test- 8 - Question 15

The area of a square is 72.25 cm2 . Find its perimeter (in cm).

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 15
Area of square = side side =72.25

Side = 8.5

Perimeter = 4 side = 4 8.5 = 34cm

RRB JE ECE (CBT I) Mock Test- 8 - Question 16

Arrange the fractions 3/4, 5/12, 13/16, 16/29, 3/8 in their ascending order of magnitude.

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 16

RRB JE ECE (CBT I) Mock Test- 8 - Question 17

The concentration of acid in two solutions A and B is 19 percent and 9 percent respectively.In what ratio solution A and B should be mixed to get a solution having 17 percent acid?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 17
Using Alligation method-

The required ratio = 8:2 = 4:1

RRB JE ECE (CBT I) Mock Test- 8 - Question 18

Find the greatest number which on dividing 1580 and 3800 leaves remainders 8 and 1 respectively

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 18
HCF of (1580-8) and (3800-1) will be the greatest number

So HCF of 1572 and 3799

So, 131 Is the number.

RRB JE ECE (CBT I) Mock Test- 8 - Question 19

What must be added to each term of the ratio 2 : 5 so that it may equal to 5 : 6 ?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 19

Suppose x must be added.

Then, 2+x/5+x=5/6

12 + 6x = 25 + 5x

x = 13

RRB JE ECE (CBT I) Mock Test- 8 - Question 20

The angle of elevation of the top of a tower at a distance of 25 m from its foot is 600. The approximate height of the tower is -

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 20
We know that

tanθ = p/b

Therefore

tan60 = √3 = p/25 → p = 25√3

= 25 x 1.73

= 43.3 m

RRB JE ECE (CBT I) Mock Test- 8 - Question 21

How many whole numbers are there between 52 and 356 which are exactly divisible by 6?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 21
First whole number which is divisible by 6 between 52 and 356 is 54

Last whole number which is divisible by 6 between 52 and 356 is 354

So, by using AP formula:-

Last term = first term + (number of terms-1) × Common difference

354 = 54 + (n-1) × 6

300 = (n-1) × 6

50= (n-1)

So, n =51

Therefore,

There are 51 whole numbers between 52 and 356 which are exactly divisible by 6.

RRB JE ECE (CBT I) Mock Test- 8 - Question 22

Directions: In the following problems, there is one question and some statements given below the question. You have to decide whether the data given in the statements is sufficient to answer the question. Read all the statements carefully and find out which of the statements is/are sufficient to answer the given question. Choose the correct alternative for each question.

Question:

What will be the total weight of 10 poles, each of the same weight ?

Statements:

I. One-fourth of the weight of each pole is 5 kg.

II. The total weight of three poles is 20 kilograms more than the total weight of two poles.

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 22
From I, we conclude that weight of each pole = (4 x 5) kg = 20 kg.

So, total weight of 10 poles = (20 x 10) kg = 200 kg.

From II, we conclude that:

Weight of each pole = (weight of 3 poles) - (weight of 2 poles) = 20 kg.

So, total weight of 10 poles = (20 x 10) kg = 200 kg.

RRB JE ECE (CBT I) Mock Test- 8 - Question 23

B starts some business by investing Rs 90000. After 4 months, D joins the business by investing Rs 80000. At the end of the year, in what ratio will they share the profit?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 23

RRB JE ECE (CBT I) Mock Test- 8 - Question 24

Calculate the total numbers of prime factors in the expression (9)11 × (5)7 × (3)2 × (17)2

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 24

(9)11 × (5)7 × (7)5 × (3)2 × (17)2

⇒ (3²)11 × (5)7 × (7)5 × (3)2 × (17)2

⇒ (3)(22 + 2) × (5)7 × (7)5 × (17)2 = (3)24 × (5)7 × (7)5 × (17)2

∴ Total number of prime factors = sum of the powers of the expression = 24 + 7 + 5 + 2 = 38

RRB JE ECE (CBT I) Mock Test- 8 - Question 25

What is the net discount (in %) for successive discounts of 30% and 50%?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 25

Formula used:

Successive discount = X + Y - (XY)/100

Where,

X = First discount

Y = Second discount

Calculation:

Successive discount = X + Y - (XY)/100

X = 30%, Y = 50%

⇒ 30 + 50 - 1500/100

⇒ 65%

∴ The single equivalent discount is 65%.

RRB JE ECE (CBT I) Mock Test- 8 - Question 26

In a mixture of 45 litres, the ratio of liquid A and liquid B is 7 : 2. If 11 litres of liquid B is added to the mixture, then what will be the ratio of liquid A and liquid B in the new mixture?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 26
Let Liquid A and Liquid B be 7x and 2x respectively.

ATQ,

7x+2x = 45

9x = 45

So, x = 5

And, Liquid A is 35 and Liquid B is 10 litres respectively.

Liquid B becomes 21 litres.

Therefore,

Liquid A/Liquid B = 35/21 = 5/3

RRB JE ECE (CBT I) Mock Test- 8 - Question 27

- 5/3 = -1/6, then the value of x is:

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 27

RRB JE ECE (CBT I) Mock Test- 8 - Question 28

A shopkeeper offers a Rs. 19% discount on a mobile phone with a marked price of Rs. 7400. If he still earns a profit of Rs. 500, what is the cost price of the phone?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 28
Selling Price = Marked Price × (100-Discount) %

Therefore,

SP = 7400 × (100-19) %

SP=7400×81/100

SP = 5994

So,

Cost Price = Selling Price - Profit

CP = 5994-600

CP = 5494

RRB JE ECE (CBT I) Mock Test- 8 - Question 29

The following equation is incorrect. Which two signs should be interchanged to correct the equation?

12 + 8 - 25 ÷ 10 x 18 = 14

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 29

Given expression: 12 + 8 - 25 ÷ 10 × 18 = 14

1) ÷ and - → 12 + 8 ÷ 25 - 10 × 18 = - 167.68 ≠ 14

2) + and ÷ → 12 ÷ 8 - 25 + 10 × 18 = 156.5 ≠ 14

3) - and + → 12 - 8 + 25 ÷ 10 × 18 = 49 ≠ 14

4) × and - → 12 + 8 × 25 ÷ 10 - 18 = 14

Hence, “ × and - “ should be interchanged to correct the equation.

RRB JE ECE (CBT I) Mock Test- 8 - Question 30

Two pipes can fill a tank in 1hr. and 72 min respectively. Both the pipes are open together but due to some dust in the pipe, the efficiency pipes are decreased to 5/6 and 9/10 of their actual efficiency respectively. After some time the dust in both pipes get clear together and both pipes run with their full efficiency and then the tank can be filled in 31 min. Find out how much time it takes to clear dust.

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 8 - Question 30

L.C.M. --- 360

A = 60 --- 6

B = 72 --- 5

An = 6 x 5/6 = 5, Bn = 5 x 9/10 = 4.5

(6 + 5) x 31 + (5 + 435) x x = 360, x = 2

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