SSC CGL (Tier II) Practice Test - 3 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 3 - SSC CGL MCQ

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## 30 Questions MCQ Test SSC CGL Tier II Mock Test Series 2024 - SSC CGL (Tier II) Practice Test - 3

SSC CGL (Tier II) Practice Test - 3 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 3 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 3 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for SSC CGL (Tier II) Practice Test - 3 below.
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SSC CGL (Tier II) Practice Test - 3 - Question 1

### If sin x = 1/2 and sin y = 2/3, then what is the value of [(6 cos2 x - 4 cos4 x)/(18 cos2 y - 27 cos4 y)]?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 1
Given expression: (6 cos2 x - 4 cos4 x)/(18 cos2 y - 27 cos4 y)

Putting the values in the given equation,

SSC CGL (Tier II) Practice Test - 3 - Question 2

### The value of the expression tan 1° tan 2° tan 3° ... tan 89° is equal to

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 2
tan 1° tan 2° tan 3° ... tan 88° tan 89°

= (tan 1° tan 89°) (tan 2° tan 88°) × (tan 3° tan 87°) ... (tan 45°)

= (tan 1° cot 1°) (tan 2° cot 2°) ... (tan 45°) = 1.1.1 ... 1 = 1

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SSC CGL (Tier II) Practice Test - 3 - Question 3

### A certain number of men can finish a piece of work in 100 days. If there were 10 men less, it would take 10 days more for the work to be finished. How many men were there?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 3
Let the original number of men be "M" and the original number of days be D.

We have, D = 100 (given)

Old: M x 100 = 100M

New: (M – 10) × (100 + 10) = 100M

Thus, 110(M – 10) = 100M

M = 110

SSC CGL (Tier II) Practice Test - 3 - Question 4

Reena and Shaloo are partners in a business. Reena invests Rs. 35,000 for 8 months and Shaloo invests Rs. 42,000 for 10 months. Out of a profit of Rs. 31,570, Reena's share will be

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 4
Ratio of the share of Reena to that of Shaloo

Therefore, Reena's share = (2/5) × 31, 570 = Rs. 12,628

SSC CGL (Tier II) Practice Test - 3 - Question 5

The length of a rectangle is 4 cm more than its breadth. If the perimeter of the rectangle is 28 cm, find the length of the rectangle.

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 5
Perimeter = 28 cm

2(x + x + 4) = 28

4x + 8 = 28

x = 5 cm

So, length of rectangle = 5 + 4 = 9 cm

SSC CGL (Tier II) Practice Test - 3 - Question 6

A cylindrical tank is filled up to 40% of its capacity. If 160 gallons of water is added to it, then 6/7 of it will be filled. The height of the tank is 4 metres. If the tank is filled with 140 gallons of water, what will be the height of water in the tank?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 6
160 gallons of water increases the height from 40%, i.e. (2/5) to (6/7)

So, (6/7 - 2/5) of the tank = 160 gallons

16/35th of the tank = 160 gallons

Total volume =

So, 140 gallons means 2/5th of the total volume.

Answer = 2/5 × 4 = 1.6 metres

SSC CGL (Tier II) Practice Test - 3 - Question 7

Directions: Study the information carefully to answer the question that follows.

Number of students that appeared in MBA entrance exam from 2008-2011 = 8,00,000

Percentage breakup of number of students that appeared for MBA entrance exam in 2008:

(Note: 0-1 year experienced means the person is having work experience greater than 0 years and up to 1 year.)

Q. If the total students who appeared for MBA entrance exam from 2008-2011 were freshers and 6,15,620 in number, then what percent of them appeared for exam in 2008?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 7
Total fresher students that appeared for MBA entrance exam from 2008-2011 = 6,15,620

Total fresher students that appeared for MBA entrance exam in 2008 = 0.32 × 0.61 × 8,00,000 = 1,56,160

Percent = 1,56,160 × = 25.37

SSC CGL (Tier II) Practice Test - 3 - Question 8

In a club, there are 12 wrestlers. When a wrestler, whose weight is 90 kg, leaves the club, he is replaced by a new wrestler. Then, the average weight of this 12-member club increases by 0.75 kg. What is the weight (in kg) of the new wrestler who joins the club?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 8
Let the original average be a and original weight of 12 members be W.

Then, W = 12a --- (i) (Total = Average × Number of people)

Let weight of new person be x.

Then, after person leaves and new one joins,

W - 90 + x = 12a + 0.75 × 12 --- (ii)

Putting value of 12a from (i), we get:

W - 90 + x = W + 0.75 × 12

-90 + x = 0.75 × 12

x = 90 + 9

x = 99 kg 108

SSC CGL (Tier II) Practice Test - 3 - Question 9

Directions: Study the information given below and answer the question that follows.

Expenditure is 70%, 75%, 80% and 90% of the income in the years 1982, 1986, 1990 and 1994, respectively.

Q. What is the percentage increase in the income from 1982 to 1990?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 9
Required percentage increase =

4/5 × 100 = 80%

SSC CGL (Tier II) Practice Test - 3 - Question 10

A jogger covered a certain distance at some speed. Had he moved 3 km/hr faster, he would have taken 20 minutes less. If he had moved 1 km/hr slower, he would have taken 10 minutes more. What is the distance (in km) that he jogged?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 10
Let distance = x km

Speed = y kmph

Time taken when moving at normal speed - time taken when moving 3 kmph faster = 20 minutes

Now,

Time taken when moving 1 kmph slower - time taken when moving at normal speed = 10 minutes

(ii) ÷ (i) gives:

9y - 9 = 6y + 18

3y = 27

y = 9 km/hr

SSC CGL (Tier II) Practice Test - 3 - Question 11

What would be the cost of building a fence around a circular field with area equal to 32378.5 sq. metres, given that the price per metre for building the fence was Rs. 154?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 11
Area of field = 32378.5 sq. metres

Let radius of field be r metres.

πr2 = 32378.5

(22/7) × r2 = 32378.5

r2 = 32378.5 x 7/22 = 10302.25

r = 101.5 m

Circumference = 2 × (22/7 )× 101.5 = 638 m

Cost of fencing 1 m = Rs. 154

Cost of fencing 638 m = Rs. 154 638 = Rs. 98,252

SSC CGL (Tier II) Practice Test - 3 - Question 12

If xyz = 1, yzx = 125, and zyx = 243 (where x, y and z are natural numbers), then what is the value of 9x + 10y - 18z?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 12
xyz = 153, yzx = 531, zyx = 351

So, x = 1, y = 5 and z = 3

Thus, 9x + 10y - 18z = 9 x 1 + 10 x 5 - 18 x 3 = 9 + 50 - 54 = 5

Hence, answer option (d) is correct.

SSC CGL (Tier II) Practice Test - 3 - Question 13

In the given figure, PQRS is a square with side 2 cm. There are 5 equal circles with centres P, Q, R, S and O. Find the area (in sq. cm) of the unshaded part of square PQRS.

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 13

Option (d) is correct.

SSC CGL (Tier II) Practice Test - 3 - Question 14

Consider the triangle ABC as shown, where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of triangle ADC to that of triangle BDC?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 14

Let AD = x cm and AC = y cm

∠BCD = ∠BAC

In ΔABC and ΔBDC,

∠ABC = ∠DBC

∠BAC = ∠BCD

⇒ ∠BDC = ∠BCA

ΔABC ~ ΔCBD

x = 7

Also, BC/BD = AC/CD or 12/9 = y/6

y = 8

Perimeter of ΔADC = AC + AD + DC = (8 + 7 + 6) cm = 21 cm

Perimeter of ΔBDC = BD + BC + DC = (9 + 12 + 6) cm = 27 cm

Ratio of perimeter of ΔADC to perimeter of ΔBDC = 7/9

SSC CGL (Tier II) Practice Test - 3 - Question 15

A number, when divided by 296, gives 75 as the remainder. If the same number is divided by 37 then the remainder will be

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 15
If the first divisor is a multiple of second divisor, then

The remainder obtained by diving the given number by second divisor = remainder obtained by dividing the first remainder by the second divisor

Here, 296 ÷ 37 = 8

∴ Required remainder = remainder obtained by dividing 75 by 37 = 1

SSC CGL (Tier II) Practice Test - 3 - Question 16

What is the compound interest on RS. 30,00,000 for years at 12% per annum compounded half - yearly?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 16

Principle = Rs. 30,00,000, rate = 12%, time = 1.5 years

Formula used:

CI = principle[(1 + rate / 100)time - 1]

Calculations:

When compounded half yearly, rate becomes half, time becomes double.

Rate is 12%, when compounded half yearly rate becomes (12 ÷ 2)% = 6% and time is 1.5 years, when compound did half yearly time is (1.5 × 2)years = 3 years

= 24 × 23877

= Rs. 573048

∴ The answer is Rs. 573048

SSC CGL (Tier II) Practice Test - 3 - Question 17

The circumference of the base of a right circular cylinder is 176 cm and its height is 12 cm. Find the total surface area (in cm2) of the cylinder. (Use π = 22/7)

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 17

Circumference of the base (C) = 176 cm

Height (h) = 12 cm

Formula used:

Circumference of the base = 2 × π × r

Total surface area (TSA) of cylinder = 2 × π × r × (r + h)

Calculation:

According to the question,

Circumference of the base = 176

176 = 2 × π × r

⇒ 176 = 2 × 22 / 7 × r

⇒ 176 × 7 / 2 × 22 = r

⇒ 8 × 7 / 2 = r

⇒ r = 56 / 2

⇒ r = 28

Now T.S.A.:

2 × π × 28 × (28 + 12)

⇒ 2 × π × 28 × 40

⇒ 2 × π × 1120

⇒ 2240 × π = 7040 cm2.

∴ The Total Surface Area of the cylinder will be 7040 cm2.

SSC CGL (Tier II) Practice Test - 3 - Question 18

The difference between the compounded interest and the simple interest earned on a certain sum of money in two years at 9% interest per annum is ₹97.2. Find the sum invested.

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 18

Given:

Compound Interest (CI) - Simple Interest (SI) = Rs. 97.2

Principle (P) =?

Rate of interest (R) = 9% per annum

Time period (T) = 2 years

Concept Used:

(CI - SI) = P(R)T/(100)T.

Calculations:

According to the question,

⇒ (97.2) = P(9)2/(100)2.

⇒ 81P = 972000

⇒ P = 972000/81 = Rs. 12000

∴ The sum invested is Rs. 12000.

SSC CGL (Tier II) Practice Test - 3 - Question 19

The following pie diagram shows information about tyres manufacturing data from various companies in India during 2017. If the total tyres produced in India during 2017 was 1,80,000 units, how many units of tyres were produced by company ‘D’?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 19
The total tyres produced in India during 2017 was = 1,80,000 units,

Tyres produced by company ‘D’ in 2017 = 15% of 1,80,000 = 1,80,000 × (15/100) = 27,000

SSC CGL (Tier II) Practice Test - 3 - Question 20

Rakesh secured 92% in a test and Kiran secured 96% in the same test. If the test is conducted out of 375 marks, what is the sum of the marks secured by both Rakesh and Kiran?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 20

Rakesh got 92 % and

Kiran got 96 % out of 375 marks

Concept :

X % = X/100

Calculations :

Marks obtained by Rakesh

⇒ 92% of 375

⇒ (92/100)× 375

⇒ 345

Marks obtained by Kiran

⇒ 96 % of 375

⇒ (96/100) × 375

⇒ 360

Now, the sum of marks obtained by Rakesh and Kiran

⇒ (345 + 360)

⇒ 705

∴ The required sum is 705.

SSC CGL (Tier II) Practice Test - 3 - Question 21

The ratio of the length and breadth of the rectangle is 24 : 7. If the diagonal length of the rectangle is 50 cm, then find the area of the rectangle. (in cm2)

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 21

Given:

The ratio of the length and breadth of the rectangle is 24 : 7.

The diagonal length of the rectangle is 50 cm

Formula used:

(1) The diagonal length = d

d2 = l2 + b2

(2) The area of the rectangle = l × b

Where,

l = length

Calculation:

The length of the rectangle = 24k

The breadth of the rectangle = 7k

The diagonal length of the rectangle = 50 cm

Therefore,

⇒ (24k)2 + (7k)2 = 502

⇒ 576k2 + 49k2 = 2500

⇒ 625k2 = 2500

⇒ k2 = 4

⇒ k = ± 2

Neglect the negative value of k.

Now,

The length of the rectangle = 24 × 2 = 48 cm

The breadth of the rectangle = 7 × 2 = 14 cm

Therefore,

The area of the recatangle = 48 × 14 = 672 cm2

∴ The required answer is 672 cm2.

SSC CGL (Tier II) Practice Test - 3 - Question 22

Direction: A line graph is given below, study the graph carefully and answer the following questions.

Based on the given graph, the difference in the earnings was the highest between which of the two consecutive days:

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 22

Earning on Sunday = 100

Earning on Monday = 500

Earning on Tuesday = 300

Earning on Wednesday = 600

Earning on Thursday = 650

Earning on Friday = 300

Earning on Saturday = 350

∴ The difference in earnings was highest between Sunday and Monday.

SSC CGL (Tier II) Practice Test - 3 - Question 23

The total number of white balls and pink balls is five times the number of blue balls and the total number of pink balls and blue balls are twice the number of white balls. If there are 33 pink balls, then find the total number of balls in the bag.

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 23

Given:

The total number of white balls and pink balls is five times the number of blue balls.

The total number of pink balls and blue balls is twice the number of white balls.

There are 33 pink balls.

Calculation:

Let W, P and B are the numbers of white balls, pink balls and blue balls.

According to the question,

The number of pink balls, P = 33

⇒ W + P = 5B

⇒ W + 33 = 5B ----(1)

⇒ P + B = 2W

⇒ 33 + B = 2W

⇒ B = 2W - 33 ----(2)

Now, substitute the equation (2) in equation (1).

⇒ W + 33 = 5(2W - 33)

⇒ W + 33 = 10W - 165

⇒ 9W = 198

⇒ W = 22

Substitute the above value in equation (1).

⇒ 22 + 33 = 5B

⇒ 5B = 55

⇒ B = 11

Therefore,

The number of blue balls = 11

The total number of balls in the bag = 33 + 22 + 11 = 66

∴ The required answer is 66.

SSC CGL (Tier II) Practice Test - 3 - Question 24

A batsman scored 84 runs in his 12th innings, thereby improving his average score per innings by 4 runs. What is the average score per innings of the batsman after the 12th innings?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 24

Given:

The batsman scored 84 runs in his 12th innings, thereby improving his average score per innings by 4 runs.

Concept used:

Total = Average × Number of entities

Calculation:

Let the average of the batman in the first 11 innings be N.

Total run scored by the batsman in the first 11 innings = 11 × N = 11N

Total run scored by the batsman in all 12 innings = 11N + 84

New average score after 12th innings = N + 4

According to the question,

(11N + 84)/12 = (N + 4)

⇒ (11N + 84) = (N + 4) × 12

⇒ 11N + 84 = 12N + 48

⇒ 12N - 11N = 84 - 48

⇒ N = 36

Now, N + 4 = 36 + 4

⇒ 40

∴ 40 is the average score per innings of the batsman after the 12th innings.

SSC CGL (Tier II) Practice Test - 3 - Question 25

The table shows the production of different types of cars (in thousands).

If the data related to the production of cars of type E is represented by a pie chart, then the central angle of the sector representing the data of production of cars in 2013 will be:

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 25

Total production of type E cars over the year = 20 + 42 + 40 + 35 + 43 = 180

Production of type E cars in 2013 = 42 thousands

180 thousands represents = 360°

∴ 42 thousands will represent = [360°/180] × 42 = 84°

SSC CGL (Tier II) Practice Test - 3 - Question 26

If x (x + y + z) = 30, y (x + y + z) = 64, z (x + y + z) = 50, then find the value of 2(x + y + z), where x, y, z > 0.

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 26

Given:

x(x + y + z) = 30 -----(1)

y(x + y + z) = 64 -----(2)

z(x + y + z) = 50 -----(3)

Where x, y, z > 0.

Calculation:

By adding the equations (1), (2), and (3) we get

[x (x + y + z)] + [y (x + y + z)] + [z (x + y + z)] = 144

Taking (x + y + z) common,

⇒ (x + y + z) (x + y + z) = 144

⇒ (x + y + z)2 = 144

⇒ (x + y + z) = 12

Multiplying 2 on both sides,

⇒ 2 (x + y + z) = 24

∴ The value of 2(x + y + z) is 24.

SSC CGL (Tier II) Practice Test - 3 - Question 27

A train 225 m long is running at a speed of 145 km/hr. What is the time (in seconds) in which it will pass a man who starts from the engine running at the speed of 17 km/hr in the direction opposite to that of the train?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 27

Given:

Length of the train = 225 m

Speed of the train = 145 km/h

Speed of the man = 17 km/h

Concept used:

Time = Distance/speed

Relative speed = When two objects move opposite to each other then their relative speed is equal to the sum of their individual speeds.

When a train of x m crosses a man or a pole then it will cross its own length.

km/h × 5/18 = m /sec

Calculation:

Relative speed of the train in km/h = 145 + 17

⇒ 162 km/h

Relative speed of the train in m/sec = 162 × 5/18

⇒ 45 m/sec

According to the concept, the train will cover its own length

So, time = 225/45

⇒ 5 sec

Time taken by the train to cover the man = 5 sec

∴ In 5 sec it will pass the man.

SSC CGL (Tier II) Practice Test - 3 - Question 28

If (x + 1) ∶ (x + 5) ∶∶ (x + 17) ∶ (x + 53) then what is the mean proportional between (x + 5) and (9x – 1) where x > 0?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 28

Given:

(x + 1) ∶ (x + 5) ∶∶ (x + 17) ∶ (x + 53)

Concept used:

Mean proportion of two numbers x and y = √(x × y)

Calculation:

Here,

(x + 1)/(x + 5) = (x + 17)/(x + 53)

⇒ x2 + 53x + x + 53 = x2 + 17x + 5x + 85

⇒ x2 + 53x + x - x2 - 17x - 5x = 85 - 53

⇒ 32x = 32

⇒ x = 32/32

⇒ x = 1

Now,

(x + 5) and (9x – 1) = (1 + 5), (9 × 1 - 1)

⇒ 6, 8

Now,

mean proportion = √(6 × 8)

⇒ √48

⇒ 4√3

∴ Required mean proportion is 4√3.

SSC CGL (Tier II) Practice Test - 3 - Question 29

The value of 36 ÷ (8 × 3) - [3 ÷ {4 × {3 × 4 ÷ (5 - 9) + 6}}] lies between

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 29

Given:

36 ÷ (8 × 3) - [3 ÷ {4 × {3 × 4 ÷ (5 - 9) + 6}}]

Concept used:

Calculation:

36 ÷ (8 × 3) - [3 ÷ {4 × {3 × 4 ÷ (5 - 9) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {3 × 4 ÷ (- 4) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {3 × (- 4/4) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {3 × (- 1) + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × {- 3 + 6}}]

⇒ 36 ÷ 24 - [3 ÷ {4 × 3}]

⇒ 36 ÷ 24 - [3 ÷ 12]

⇒ 36 ÷ 24 - (1/4)

⇒ 3/2 - 1/4

⇒ (6 - 1)/4

⇒ 5/4

⇒ 1.25

So, this is lies in between 1 and 1.3

∴ Required answer is Option B.

SSC CGL (Tier II) Practice Test - 3 - Question 30

How much water should be added to 90 ml of a 38% sugar solution so that it becomes a 17.1% sugar solution?

Detailed Solution for SSC CGL (Tier II) Practice Test - 3 - Question 30

Given:

Total quantity of the initial mixture = 90 ml

Sugar percentage = 38%

Sugar percentage of the new mixture = 17.1%

Calculation:

In the mixture of 90 ml amount of sugar = 90 × 38%

⇒ 34.2 ml

Now, % of the sugar in the new mixture becomes 17.1

According to the question,

In the new mixture 34.2 ml of sugar = 17.1% of the mixture [∵ We added water so amount of sugar will unchanged]

So, 17.1% = 34.2

⇒ 1% = 2

⇒ 100% = 200

So, total quantity of new mixture = 200 ml

Water added = 200 - 90

⇒ 110 ml

∴ Required amount of water to be added is 110 ml.

## SSC CGL Tier II Mock Test Series 2024

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## SSC CGL Tier II Mock Test Series 2024

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