SSC CGL (Tier II) Practice Test - 8 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 8 - 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 - 8

SSC CGL (Tier II) Practice Test - 8 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 8 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 8 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 - 8 below.
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SSC CGL (Tier II) Practice Test - 8 - Question 1

### The interior angle of a regular polygon exceeds its exterior angle by 108°. The number of sides of the polygon is

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 1
Suppose the interior and exterior angles of a triangle are i and e. So from question,

i - e = 108° ...(i)

i + e = 180° ...(ii)

So, from equation (i) and (ii)

i = 144°

From polygon formula, if n is the number of sides of polygon, then

(n - 2)180° = 144°n

So, n =

So, number of sides of polygon = 10

SSC CGL (Tier II) Practice Test - 8 - Question 2

### Mohit is standing at some distance from a 60-meter tall building. Mohit is 1.8-metre tall. When Mohit walks towards the building, the angle of elevation from his head becomes 60° from 45°. How much distance (in meters) does Mohit cover towards the building?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 2

Height of the building from the head of Mohit = 60 - 1.8 = 58.2 m

Case I: When the angle of elevation is 45°

Distance from building (AC) = 58.2 m

Case II: When the angle of elevation is 60°

Distance from the building (BC) =

Distance covered by Mohit (AB) = AC - BC =

= (19.4 × 3) - (19.4 × √3)

= 19.4 (3 - √3)m

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

### In 4 years, at simple interest, the principal increases by 12%. Calculate the amount (in Rs) received at the end of 2 years on Rs 20,000 at the same rate if compounded annually?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 3
Given,

RT = 12

Solving which we get:

R = 3% (T = 4 years)

The rate of interest is 3% in this situation.

Condition 2:

P = Rs. 20,000

R = 3%

T = 2 years annually compounded interest

Formula used:

A = P [1 + (R/100)]t

A = 20,000(1 + 0.03)²

A = 20,000(1.03)²

A = Rs. 21,218

SSC CGL (Tier II) Practice Test - 8 - Question 4

Two cards are drawn in succession from a pack of 52 cards. The first card should be a Queen and the second should be a King. What is the probability of doing so if the first card is replaced?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 4
Probability of drawing a Queen = 4/52 = 1/13

Since the first card is replaced, the pack will again have 52 cards. So, the probability of drawing a King = 4/52 = 1/13

Both the events are independent, hence the probability of drawing both cards in succession = 1/13 x 1/13 = 1/169

SSC CGL (Tier II) Practice Test - 8 - Question 5

Directions: Study the following line graph to answer the given question:

Sales of Drinks and Food Over the Five Days (Sales in Thousands)

When the sales are modelled mathematically, the growth of sales from Monday to Tuesday is expressed in linear form for both drinks and food. The difference between their growth would be (in thousand)

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 5
Growth of the sales of food from Monday to Tuesday = 50 - 40 = 10 thousand

Growth of the sales of drinks from Monday to Tuesday = 45 - 30 = 15 thousand

Difference between the growths = 15 - 10 = 5 thousand

Hence, this is the required result.

SSC CGL (Tier II) Practice Test - 8 - Question 6

A person has left an amount of Rs. 1,20,000 to be divided between his two sons, aged 14 years and 12 years such that they get equal amounts when each attains 18 years of age. If the amount gets a simple interest of 5% per annum, the younger son's share at present is

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 6
Let the share that the elder son gets be Rs. x.

Share that younger son gets = Rs. (1,20,000 - x)

According to the question,

On solving, we get

x = Rs. 62,400

Share that younger son gets = Rs. (1,20,000 - 62,400) = Rs. 57,600

SSC CGL (Tier II) Practice Test - 8 - Question 7

A cylinder with base radius 8 cm and height 2 cm is melted to form a cone of height 6 cm. The radius of the cone will be

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 7
If radius of cylinder = r1

Volume of the cylinder = πr12h = 3.14 × 8 × 8 × 2 = 402 cm3

Let radius of the cone = r

So, from question,

Thus, r2 = 64

Hence, r = 8 cm

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

Giving two successive discounts of 40% is equal to giving one discount of ________%.

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 8
Single equivalent discount =

= (80 – 16)% = 64%

SSC CGL (Tier II) Practice Test - 8 - Question 9

A beaker contains acid and water in the ratio 1 : x, respectively. When 300 ml of the mixture and 50 ml of water are mixed, the ratio of acid to water becomes 2 : 5. What is the value of x?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 9
In 300 ml of mixture

Therefore,

14x + 2 = 30

14x = 28

x = 2

SSC CGL (Tier II) Practice Test - 8 - Question 10

Directions: Study the following line graph to answer the given question:

Sales of Drinks and Food Over the Five Days (Sales in Thousands)

Q. What is the ratio of the differences of the highest sales and the lowest sales of drinks and the food (approx.)?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 10
Highest sales of drinks = 65,000

Lowest sales of drinks = 15,000

Difference = 65,000 - 15,000

50,000

Highest sales of food = 50,000

Lowest sales of food = 30,000

Difference = 50,000 - 30,000 = 20,000

Ratio = 50,000/20,000

= 5 : 2

Thus, the ratio is 5 : 2

Hence, this is the required result.

SSC CGL (Tier II) Practice Test - 8 - Question 11

A man buys two chairs for a total cost of Rs. 900. By selling one for 4/5 of its cost and the other for 5/4 of its cost, he makes a profit of Rs. 90 on the whole transaction. The cost of the lower priced chair is

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 11
Let the cost price of 1 chair be Rs. x.

CP of other chair = Rs. (900 - x)

4/5x + 5/4 (900 - x) = 900 + 90

4/5x + 1125 - 5x/4 = 990

9x/20 = 135

x = 135 × 20/9 = Rs. 300

CP of the lower priced chair is Rs. 300.

SSC CGL (Tier II) Practice Test - 8 - Question 12

What is the value of [(sin 59° cos 31° + cos 59° sin 31°) ÷ (cos 20° cos 25° - sin 20° sin 25°)]?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 12
We know that sin A cos B + cos A sin B = sin (A + B)

sin 59° cos 31° + cos 59° sin 31° = sin (59 + 31) = sin 90°

cos A cos B - sin A sin B = cos (A + B)

cos 20° cos 25° - sin 20° sin 25° = cos (20 + 25) = cos 45°

sin 90° ÷ cos 45° = 1 ÷ 1/√2 = √2

SSC CGL (Tier II) Practice Test - 8 - Question 13

In the given right-angled triangle ABC, AB = 12 cm and AC = 15 cm. A square is inscribed in the triangle. One of the vertices of the square coincides with a vertex of the triangle. What is the maximum possible area (in cm2) of the square?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 13

AB = 12 cm, AC = 15 cm

Triangle ADE and ABC are similar (By AA)

Let side of the square or DB or BF = x cm.

So, AD = (12 - x) cm, FC = (9 - x)

After putting the values above, we get side of the square = 36/7 cm

Hence, area of the square = (36/7)2

SSC CGL (Tier II) Practice Test - 8 - Question 14

Directions: Identify the missing number in the series provided.

25, 65, 145, 265, 425, 625, ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 14
(4 + 12) × 5 = 25

(4 + 32) × 5 = 65

(4 + 52) × 5 = 145

(4 + 72) × 5 = 265

(4 + 92) × 5 = 425

(4 + 112) × 5 = 625

(4 + 132) × 5 = 865

SSC CGL (Tier II) Practice Test - 8 - Question 15

If and then which of the following can be the value of X + y?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 15
X = 1/(1 + (1/(1 + X)))

Or X = 1/((1 + X + 1)/(1 + X))

Or X = (1 + X)/(2 + X)

Or X(2 + X) = 1 + X Or X2 + X - 1 = 0

Or X = (-1 - 51/2)/2 or (-1 + 51/2)/2

y = 2/(2 + (1/(1 + y)))

Or y = 2/((2y + 2 + 1)/(1 + y))

Or y = (2y + 2)/(2y + 3)

Or y(2y + 3) = 2y + 2

Or 2y2 + y - 2 = 0

y = (-1 - 171/2)/4 or (-1 + 171/2)/4

Now, taking X = (51/2 - 1)/2 and y = (171/2 - 1)/4, the value of X + y is obtained as the value that appears in answer option (2).

By taking other possible values for X and y, values of X + y are obtained that exist in none of the other answer options.

Hence, answer option B is correct.

SSC CGL (Tier II) Practice Test - 8 - Question 16

The average of 5 numbers is 4.25. The average of two of them is 4.1, while the average of the other two is 3.62. What is the average of the remaining number?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 16

Given:

The average of 5 numbers is 4.25

The average of two of them is 4.1

The average of the other two is 3.62

Formula used:

Total Sum = Average × Number of terms

Calculation:

Total Sum of 5 numbers = 4.25 × 5

= 21.25

Sum of first two numbers = 4.1 × 2

= 8.2

Sum of other two numbers = 3.62 × 2

= 7.24

Remaining number = Total Sum of 5 numbers - Sum of first two numbers - Sum of other two numbers

= 21.25 - 8.2 - 7.24

= 5.81

SSC CGL (Tier II) Practice Test - 8 - Question 17

If the sales tax is reduced from , then what difference does it make to a person who purchases an article with marked price of Rs. 7200.

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 17

Given:

the sales tax reduced from and

Marked Price = Rs. 7200

Formula used:

Discount = Discount % × Marked Price

Calculation:Discount 1 =

Discount 2 =

Difference in discounts =

= 11 × 12

= 132

SSC CGL (Tier II) Practice Test - 8 - Question 18

From the following equations x2 + y2 = 41 and x4 - y4 = 369 the value of x and y will be respectively

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 18

Calculation:

x2 + y2 = 41

x4 - y4 = 369

(x2 + y2) (x2 - y2) = 369

41 (x2 - y2) = 369

(x2 - y2) = 9

(x + y) (x - y) = 9

Now, check from options,

x = +5 or -5 and y = +4 or -4

Hence, option 4 is correct.

SSC CGL (Tier II) Practice Test - 8 - Question 19

If the compound interest on a sum for 2 years at per annum is ₹340, then the simple interest on the same sum at the same rate for the same period of time is

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 19

Calculation:

Compound Interest = 340

Time = 2 years

Rate of interest = (25/2)% = 1/8

Let principal be 64

For First year, Simple and Compound Interest are same, here it is 8.

Second year, Compound interest = 8 + 1 = 9

Total Compound Interest = 8 + 9 = 17

Let compound interest be 17x

17x = 340

1x = 20

64x = 1280

For Simple interest, effective rate = 2 × 25/2% = 25%

Simple Interest = 25% of 1280 = 320

Hence, option 1 is correct.

SSC CGL (Tier II) Practice Test - 8 - Question 20

A runs thrice as fast as B and B runs 4 times as fast as C. the distance covered by C in 96 minutes will be covered by A in:

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 20

Given:

Speed of A = 3 × Speed of B ... (1)

Speed of B = 4 × Speed of C ... (2)

Concept used:

Speed = Distance / time

For constant distance, clearly speed is inversely proportional to time.

Calculation:

Using (1) and (2),

Speed of A = 3 × 4 × Speed of C

Speed of A = 12 × Speed of C.

Speed of A / Speed of C = 12 : 1.

Ratio of time taken will be inverse to the ratio of speeds of A and C as the distance is constant.

Time taken by A / Time taken by C = 1 : 12.

Time taken by A to cover a distance which is covered by C in 96 minutes

Time taken by A = 96 / 12

Time taken by A = 8 minutes.

∴ The time taken by A will be 8 minutes.

Note - In the official answer key, the answer was given as "12'' but in the calculation it was found to be "8". So, the option has been changed.

SSC CGL (Tier II) Practice Test - 8 - Question 21

A completed the school project tin 45 days. How many days will B take to complete the same work, if he is 40% more efficient than A?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 21

Given:

Time taken by A to complete the school project = 45 days

B is 40% more efficient than A

Concept used:

If X is y% more efficient than Y, and time taken by X to complete the work is x units, time taken by Y to complete the work

Calculation:

As B is 40% more efficient than A and efficiency is inversely proportional to time taken, time taken by B to complete the same work

∴ Time taken by B to complete the work is days.

SSC CGL (Tier II) Practice Test - 8 - Question 22

How many bullets can be made out of a lead cylinder 45 m high and the radius 18 cm, each bullet being 4.5 cm in diameter?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 22

Given:

Height of lead cylinder = 45 m = 4500 cm.

The radius of the cylinder = 18 cm.

Diameter of the bullet = 4.5 cm.

Concept used:

The volume of sphere = (4 / 3) πr3

where r = radius of the sphere

Volume of cylinder = πr'2h

where

r = radius of the cylinder

h = height of the cylinder

Calculation:

Radius of sphere = diameter / 2 = 4.5 / 2 = 2.25 cm.

Let the number of bullets be N.

N × Volume of sphere = Volume of cylinder

N × (4 / 3) πr3 = πr2h

N × 4r3 = 3r'2h

N × 4 × (2.25)3 = 3 × (18)2 × 4500

N = 96000.

Note - In the official answer key, the answer was given as "900'' but in the calculation, it was found to be "96000". So, the option has been changed.

SSC CGL (Tier II) Practice Test - 8 - Question 23

By selling a car for Rs. 2,56,000, Raj gains 28%. If the profit is reduced to 16% then the selling price will be:

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 23

Given:

Profit% for selling price Rs. 256000 = 28%

Concept used:

Profit% = [(SP - CP) / CP] × 100

where

SP = Selling price

CP = Cost price

Calculation:

For 28% profit, SP = Rs. 256000

Profit% = [(SP - CP) / CP] × 100

28 × CP = 25600000 - 100 × CP

128 × CP = 25600000

CP = 25600000 / 128

CP = 200000.

For 16% profit,

Profit% = [(SP - CP) / CP] × 100

32000 = SP - 200000

SP = 232000.

∴ The selling price will be Rs. 2,32,000.

SSC CGL (Tier II) Practice Test - 8 - Question 24

The radius of a right circular cylinder is 21 cm and its curved surface area is 5280 cm2. What is its volume?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 24

Given:

Radius of cylinder (r) = 21 cm.

Curved Surface Area (CSA) of cylinder = 5280 cm2.

Formula used:

Curved Surface Area (CSA) of cylinder = 2πrh

where

h = height of cylinder.

Volume of cylinder = π × r2 × h

Calculation:

CSA of cylinder

CSA = 2πrh

5280 = 2 × (22 / 7) × 21 × h

5280 = 2 × 22 × 3 × h

h = 5280 / 132

h = 40 cm.

Volume of cylinder

= π × r2 × h

= (22 / 7) × 21 × 21 × 40

= 22 × 21 × 3 × 40

= 55440 cm3.

∴ The volume of the cylinder is 55440 cm3.

SSC CGL (Tier II) Practice Test - 8 - Question 25

Perimeter of a triangular field is 540 meter. The ratio of its arms is 5 : 12 : 13. The area of this field is

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 25

Given:

Perimeter = 540 m

Ratio of sides = 5 : 12 : 13

Calculation:

Let the lengths be 5x, 12x and 13x

Perimeter of triangle = 5x + 12x + 13x = 30x

30x = 540

x = 18

This is a right angled triangle as the sides are 5x, 12x and 13x satisfy pythagoras theorem.

Let base be 5x and height be 12x.

Area of right angled triangle = 1/2 × base × height

= 1/2 × 5x × 12x = 30x2

Put the value of x, we get

Area = 30x2 = 30 × (18)2 = 9720 m2

Hence, option 3 is correct.

SSC CGL (Tier II) Practice Test - 8 - Question 26

A man observes a train passing over a bridge 1 km long. The length of the train is half that of bridge. If train clears the bridge in 2 minutes, find the speed of the train.

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 26

Given:

Length of bridge (l) = 1 km

Length of train (l') = 1/2 km = 0.5 km

Time taken by train to cross the bridge (t) = 2 minutes = (2 / 60) hour = (1 / 30) hour.

Concept used:

Speed of train = (length of bridge + length of train) / time

Calculation:

Speed of train

=

= 1.5 × 30

= 45 km/hr.

∴ Speed of train is 45 km/hr.

SSC CGL (Tier II) Practice Test - 8 - Question 27

A company has four departments Human Resource, Marketing, Sales and Information Technology.

The expenses of all departments are shown in the table given below (in thousands).

Q. What is the total expenditure of Human Resource, Marketing and Information Technology departments of company?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 27

Given:

Table shows the expenses of different departments of a company.

Concept used:

Calculation:

Total expenditure of Human Resource, Marketing and Information Technology is (in thousands)

= Expenditure of Human Resource + Expenditure of Marketing + Expenditure of Information Technology

= [(20 + 15 + 10 + 32) + (45 + 50 + 40 + 35) + (80 + 70 + 55 + 43)]

= [77 + 170 + 248]

= 495.

∴ Total expenditure of Human Resource, Marketing and Information Technology is 495000.

SSC CGL (Tier II) Practice Test - 8 - Question 28

The following pie-chart and table represent the percentage-wise distribution of number Laptops, say brand A and brand B sold by six shops during six months July – December in the year 2020.

Total number of Laptops sold : 2400

Q. If 25% of the laptops sold by brand B during the month September were sold at a discount, then how many Laptops of brand B were sold without a discount during the month September?

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 28

Given:

Total number of Laptops sold : 2400

Calculation:

Total number of Laptops sold = 100%

100% = 2400

1% = 24

Number of laptops sold in September is 25%,

= 25 × 24

= 600

Ratio of Brand A and Brand B laptop in September is 3 : 2

3 units + 2 units = 600

5 units = 600

1 unit = 120

Brand B = 120 × 2

= 240

25% of the laptops sold by brand B during the month September were sold at a discount,

Then, 75% of laptops were sold without discount,

= 75% × 240

= 180 laptops

SSC CGL (Tier II) Practice Test - 8 - Question 29

AB is a line segment of length 2a, with M as mid-point. Semicircles are drawn on one side with AM, MB and AB as diameter as shown in the above figure. A circle with centre O and radius r is drawn such that this circle touches all the three semicircles. The value of r is

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 29

Given:

AB is a line segment of length 2a.

Concept:

Pythagoras theorem: In an right - angled triangle

Hypotenuse2 = Base2 + Perpendicular2

Fomula used:

(a - b)2 = a2 - 2ab + b2

(a + b)2 = a2 + 2ab + b2

Calculation:

We know that, two circles touch each other externally if

Distance between their centres = Sum of their radii

⇒ OL = r + a/2 -----(1)

⇒ OM = CM - OC

⇒ OM = a - r -----(2)

⇒ LM = a/2 -----(3)

In △OML, using Pythagoras theorem

⇒ (OL)2 = (OM)2 + (LM)2

Using the formula disucssed above

∴ The value or r is a/3.

SSC CGL (Tier II) Practice Test - 8 - Question 30

Select the correct option that indicates the arrangement of the given words in the order in which they appear in an English dictionary.

1. Impersonal

2. Impertinent

3. Impersonate

4. Impermanent

5. Imperturbable

6. Impertinence

7. Imperfection

Detailed Solution for SSC CGL (Tier II) Practice Test - 8 - Question 30
The arrangement of the given words in the order in which they appear in an English dictionary is as shown below

7. Imperfection

4. Impermanent

1. Impersonal

3. Impersonate

6. Impertinence

2. Impertinent

5. Imperturbable

Hence, the correct answer is "7, 4, 1, 3, 6, 2, 5".

## SSC CGL Tier II Mock Test Series 2024

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## SSC CGL Tier II Mock Test Series 2024

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