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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  RRB JE ECE (CBT I) Mock Test- 9 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 9 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT I) Mock Test- 9

RRB JE ECE (CBT I) Mock Test- 9 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT I) Mock Test- 9 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT I) Mock Test- 9 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- 9 below.
Solutions of RRB JE ECE (CBT I) Mock Test- 9 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & RRB JE ECE (CBT I) Mock Test- 9 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- 9 | 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- 9 - Question 1
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The height of a square base right prism is 15 cm. If the total surface area of the prism is 608 cm2, then what will be the volume of the prism?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 1
Let the side of the square base be ‘a’.

Now, total surface area = Perimeter of square base × Height

+ 2 × Area of square base.

⇒ 608 = 4a × 15 + 2 × a2 ⇒ a2 + 30a –304 = 0 ⇒ a = 8

∴Volume of prism = Area of base × Height

= a2 × 15 = 8 × 8 × 15 = 960 cm3

RRB JE ECE (CBT I) Mock Test- 9 - Question 2
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If the angles of a pentagon are in the ratio 1 : 3 : 6 : 7 : 10, then the smallest angle is:

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 2
Suppose the angles of a pentagon are

x°, 3x°, 6x°, 7x°, 10x°

But x° + 3x° + 6x° + 7x° + 10x° = (2x5 -4) x 90°

or 27x° = 6 x 90°

or x = 20°

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RRB JE ECE (CBT I) Mock Test- 9 - Question 3
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What will be the value of x in the infinite series

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 3

Or x2 = 6 + x

Or x2 - x 6 = 0

Or (x-3) (x+2) = 0

∴ x = 3

RRB JE ECE (CBT I) Mock Test- 9 - Question 4
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If a (tanθ + cotθ) = 1, sinθ + cosθ = b with 0° < θ < 90°, then the relation between a and b is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 4

a (tanθ + cotθ) = 1[Given]

sinθ + cosθ = b[Given]

a = sinθ cosθ.................(i)

Also, sinθ + cosθ = b

Squaring both sides,

sin2θ + cos2θ + 2sinθ cosθ = b2

1 + 2 a = b2 ⇒ 2a = b2 – 1[From ⇒ equ. (i)]

RRB JE ECE (CBT I) Mock Test- 9 - Question 5
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The driver of an ambulance sees a school bus 40 meters ahead of him. After 20 seconds, the school bus is 60 meters behind. If the speed of the ambulance is 30 km/hr, then what is the speed of the school bus?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 5
Relative Speed,

= (Total.distance/Total.time)

=60+40/20

= 5 m/s =5×18/5 = 18 kmph

Relative Speed = (speed of ambulance - speed of school bus)

Speed of school bus = speed of ambulance - relative speed.

= 30-18 = 12 kmph.

RRB JE ECE (CBT I) Mock Test- 9 - Question 6
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At what percent above the cost price must an article be marked so that the seller gains 33% after allowing the customer a discount of 5% ?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 6
Let the marked price be ₹x

Let the Cost price be ₹y

Selling Price = x - 5% of x

x = 95x/100 = 19x/20

y + 33% of y = 19x/20

7y = 5x

x = 7/5 y = y + 2/5 y = y + 40% of y

RRB JE ECE (CBT I) Mock Test- 9 - Question 7
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If (x11+1) is divided by (x+1), then the remainder is :
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 7

p(x) = x11 +1

Divisor = x+1

∴ Remainder = p(−1)=(−1)11 +1= −1 + 1 = 0

RRB JE ECE (CBT I) Mock Test- 9 - Question 8
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Evaluate : cos80o/sin10o + cos 59o cosec 31o
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 8

RRB JE ECE (CBT I) Mock Test- 9 - Question 9
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The measurement of each angle of a polygon is 160°. The number of its sides is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 9
Let n be the sides of the polygon

or 16n = (2n - 4) x 9

or 16n = 18n - 36

2n = 36 ⇒ n = 18

∴ the number of sides of polygon are 18.

RRB JE ECE (CBT I) Mock Test- 9 - Question 10
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Two towers of the same height stand on opposite sides of a road 100 m wide. At a point on the road between the towers, the elevations of the towers are 30° and 45°. Find the height of the towers and the position of the point from the farthest tower.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 10

Let AB, CD are towers

E is the point between road BC.

Then in △DCE

CD = EC = X = AB

IN △ABE

BE= √3x then

Position of the farthest tower from point = 63.4 mtr.

RRB JE ECE (CBT I) Mock Test- 9 - Question 11
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If (a/b)x−1 = (b/a)x−3 ,then the value of x is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 11

(a/b)x−1 = (b/a)x−3 = (a/b)−(x−3) = (a/b)3−x

∴ x - 1 = 3 - x or 2x = 4

∴ x = 2

RRB JE ECE (CBT I) Mock Test- 9 - Question 12
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If the perimeter of an equilateral triangle is 36.9 metre and the length of an altitude is 8 metre, then the area of the triangle is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 12
Perimeter of the equilateral triangle of side x = 36.9 m

∴ side of equilateral triangle x = 36.9/3 = 12.3 m

height = 8

hence area = 1/2×12.3×8 = 49.2 m2

RRB JE ECE (CBT I) Mock Test- 9 - Question 13
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In what ratio Martina should mix two varieties of juice worth Rs. 52 per liter and Rs. 74 per liter, so that by selling the mixture at Rs.71.28 per liter she can earn a profit of 8%?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 13

RRB JE ECE (CBT I) Mock Test- 9 - Question 14
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40% of the students in a school are girls. If the number of boys is 1680, then what is the total number of students?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 14

% of boys = 100 – 40 = 60%

60% of x = 1680

x = (1680 × 100)/60

x = 2800

RRB JE ECE (CBT I) Mock Test- 9 - Question 15
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Prem invested a certain sum of money in a simple interest. It amounts to Rs 300 at the end of 3 years and to Rs 400 at the end of 8 year. What was the rate of interest in which he invested his sum?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 15

Let principal be P and rate of interest be r%

At the end of another 5 years means 3 + 5 = 8 Years.

On subtracting Eq. (i) from Eq. (ii)

∴ p x r = 2,000

240 x r = 2000

r = 8.33%

RRB JE ECE (CBT I) Mock Test- 9 - Question 16
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10 men can complete a work in 8 days. 20 women can complete the same work in 6 days. In how many days 16 men and 18 women can complete the same work, working together ?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 16

10m → 8 days ⇒ 20m → 4 days ⇒ 1m → 80 days

20w → 6 days ⇒ 20w → 6 days ⇒ 1w → 120 days

∴ 2m = 3w

Thus, 16m + 18w = 24w + 18w = 42w

20w x 6 = 42w x

(Where x is the number of days taken by 42 women)

∴ x = 20/7 days ⇒ 2(6/7) days

RRB JE ECE (CBT I) Mock Test- 9 - Question 17
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461 + 462 + 463 + 464 + 465 is divisible by:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 17

461 [1+4+42+43+44] = 461 [1+4+ 16 + 64 +256] = 461 x 341

Since 341 is divisible by 11, then the given expression is also divisible by 11.

RRB JE ECE (CBT I) Mock Test- 9 - Question 18
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A cane contains a mixture of two liquids 'A' and 'B' in the ratio 7 : 5. When 9 liters of mixture are drawn off and the cane is filled with 'B', the ratio of 'A' and 'B' becomes 7 : 9 liters. Liter of Liquid 'A' contained by the cane initially was?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 18

According to the question

4 units = 9 litres

1 unit = 9/4 litres

12 units = 94×12 = 27 litres

Initially solution was = 27+9 = 36litres

Quantity of liquid 'A' initially = ‘712×36 = 21 litres

RRB JE ECE (CBT I) Mock Test- 9 - Question 19
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

What is the average number of students appearing for aptitude tests from all the cities together?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 19

(30 + 22.5 + 37.5 + 17.5 + 32.5)/5 = 28

RRB JE ECE (CBT I) Mock Test- 9 - Question 20
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

What is the ratio of the number of students appearing for the aptitude test for city B to that of city A?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 20

22.5 : 30

3 : 4

RRB JE ECE (CBT I) Mock Test- 9 - Question 21
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).

The number of students appearing for the aptitude test from city D is approximately what percentage of the number of students appearing for the aptitude test from city C?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 21

17.5×100/37.5 = 47%

RRB JE ECE (CBT I) Mock Test- 9 - Question 22
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DIRECTIONS: DIRECTIONS: Study the following graph carefully and answer the questions that follow:

No. of students appearing for aptitude test from various cities (in thousands).
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 22

55 : 80

= 11 : 16

RRB JE ECE (CBT I) Mock Test- 9 - Question 23
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The batting average of a batsman in 57 innings is 58 runs. He was out for a duck in 7 innings. His batting average for remaining innings is :
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 23
Total runs of 57 innings = 57 x 58

∴ Average of remaining (57-7) or 50 innings

= 57×58/50 = 66.12

RRB JE ECE (CBT I) Mock Test- 9 - Question 24
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If sin(x+y)/sin(x−y) = a+b/a−b , then the value of tan x/tan y is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 24

RRB JE ECE (CBT I) Mock Test- 9 - Question 25
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A and B can do a piece of work in 15 days. B and C can do the same work in 12 days and C and A in 10 days. How many days will A take to complete the work alone?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 25

Thus, A completes 2.5 unit work in a day.

Thus, total time required by A = 60/25 = 24 days.

RRB JE ECE (CBT I) Mock Test- 9 - Question 26
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If 2 men and 3 women can do a piece of work in 8 days and 3 men and 2 women in 7 days. In how many days can the work be done by 5 men and 4 women working together?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 26

|2 x 8 - 3 x 7| men’s work = |3x 8 - 2 x 7| women’s work.

∴ 5 men’s work = 10 women’s work

∴ 1 men’s work = 2 women’s work

2 men’s work = 4 women’s work

∴ Days taken by 5 men + 4 women = 14 women are 7x8/14 = 4 days

RRB JE ECE (CBT I) Mock Test- 9 - Question 27
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If a and b are two distinct positive numbers, then which of the following is true?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 27

Let α = 16 & b = 9

Option (α) : LHS =

Hence, option (α) is wrong.

Option (α) : LHS =

Hence, option (b) is wrong.

RRB JE ECE (CBT I) Mock Test- 9 - Question 28
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5 year ago, the father’s age was 3 times his son’s age. 5 years hence the ratio between the ages of the father and the son becomes 11: 5. Find the father’s present age.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 28

Father Son

Old ratio 3 : 1

New ratio 11 : 5

Let, 5 years age father’s age was 3x years, where

x=(5+5)×(11−5)/3×5−1×11 =10×6/4 =15

∴ Father’s age (5 years ago) = 3 × 15 = 45 years

∴ His present age = 45 + 5 = 50 years

RRB JE ECE (CBT I) Mock Test- 9 - Question 29
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P and Q starting simultaneously from two different places proceed towards each other at a speed of 20 km/hour and 30 km/hour respectively. By the time they meet each other, Q has covered 36 km more than P. The distance (in k.m.) between the two places is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 29
Let the distance = x km

Let them meet when P travels y km.

then, Q travels in the same time x – y km

from question,

(x – y) – y = 36

x – 2y = 36 .......(i)

Also, 20 × y = 30 (x–y) [∵Time is same]

∴y/x−y−=3/2

2y = 3x – 3y

5y = 3x ........(ii)

Putting the value of y from (ii) to (i),

x−6x/5=36⇒−x/5=36⇒x=180 km

RRB JE ECE (CBT I) Mock Test- 9 - Question 30
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A man walks from A to B at the rate of 5 km per hour and returns back at the rate of 3 kmph. What is the average speed for the whole journey ?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 9 - Question 30
We know that

Average speed = 2xy / x+y

Here, x= 5 km per hour

y = 3 km per hour

therefore, Average speed = 2*5*3 / 5+3 = 30/8 = 15 kmph.

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