SSC CGL (Tier II) Practice Test - 2 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 2 - 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 - 2

SSC CGL (Tier II) Practice Test - 2 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 2 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 2 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 - 2 below.
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SSC CGL (Tier II) Practice Test - 2 - Question 1

### If = (343)y – 1, then y is equal to

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 1
Let x =

On squaring both sides, x2 = 7x

⇒ x2 – 7x = 0

⇒ x(x – 7) = 0

⇒ x = 7

⇒ 7 = (73)y – 1 = 73y – 3

⇒ 3y – 3 = 1

⇒ 3y = 4

⇒ y = 4/3

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

### Find the value of

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

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

### In the given figure, PQ is the diameter of a circle with radius 5 cm. If the length of RS is 6 cm, then what is the area of the trapezium PQRS?

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

RT = 3 cm, OR = 5 cm

Area of trapezium PQRS

SSC CGL (Tier II) Practice Test - 2 - Question 4

If xy = 11 and x – y = 5, then x3y2 – x2y3 equals

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 4
x3y2 – x2y3

= x2y2 (x – y)

= (11)2 (5)

= 121 × 5

= 605

SSC CGL (Tier II) Practice Test - 2 - Question 5

Aman and Farhan can complete a work in 30 days. Aman works for 16 days and Farhan finishes remaining work in 44 days. In how many days will Farhan complete the work alone?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 5
Let Aman's 1 day's work = x and Farhan's 1 day's work = y.

Then, x + y = 1/30 and 16x + 44y = 1

Solving these two equations, we get

x = 1/60 and y = 1/60

Farhan's 1 day's work = 1/60

Hence, Farhan alone will finish the whole work in 60 days.

SSC CGL (Tier II) Practice Test - 2 - Question 6

Four strips of paneling, each 40 cm long and 4 cm wide, are arranged to form a square as shown below. What is the area of the inner square (in cm2)?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 6
Side of inner square = 44 - 8 = 36 cm

Area = 362 = 1296 cm2

SSC CGL (Tier II) Practice Test - 2 - Question 7

When 80% of a number is added to 80, the total is the same number again. Which of the following is that number?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 7
Let the number be "N".

Thus, we have

80% of N + 80 = N

0.8N + 80 = N

80 = 0.2N

N = 400

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

A 50 m high pole stands on a 250 m high building. To an observer at a height of 300 m, the building and the pole subtend equal angles. The distance of the observer from the top of the pole is

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

From the figure, we find that,

tanθ = 50/x (From triangle OMP) ... (1)

Also, tan 2θ = 300/x (From triangle OBM) ... (2)

⇒ 3(x2 - 2500) = x2

⇒ x2 = 3x2 - 7500

⇒ 2x2 = 7500

⇒ x2 = 3750 = 625 × 6

⇒ x = 25√6

Hence, distance of the observer from the top of the pole is 25.

SSC CGL (Tier II) Practice Test - 2 - Question 9

If then the value of

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 9

Dividing by 2,

SSC CGL (Tier II) Practice Test - 2 - Question 10

If an amount of \$665 is distributed among A, B and C in the ratio (2/3) : (3/4) : (4/5), how much would be the share of A?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 10
Let the shares (in dollars) of A, B and C be (2/3)X, (3/4)X and (4/5)X, respectively.

Then, we get,

(2/3)X + (3/4)X + (4/5)X = 665

⇒ X = \$300

A's share = (2/3) × 300 = \$200

SSC CGL (Tier II) Practice Test - 2 - Question 11

Three angles of a quadrilateral are in the ratio 10 : 8 : 7, and the fourth angle measures 60°. What is the difference between the largest and the smallest angles?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 11
Sum of all the angles of a quadrilateral = 360°

Or, 10x + 8x + 7x + 60° = 360°

Or, 25x = 300 or x = 12°

Difference between the largest and smallest angles = 120° - 60° = 60°

Therefore, the 4th option is the answer.

SSC CGL (Tier II) Practice Test - 2 - Question 12

In a cricket match, Rahul and Sachin respectively scored 1/9th and 2/11th parts of the total runs. What part of the total runs did they collectively score?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 12
Let the total runs scored be x.

Runs scored by Rahul = x/9

Runs scored by Sachin

Total runs scored by them

Part of total runs scored by them collectively

SSC CGL (Tier II) Practice Test - 2 - Question 13

A number, when divided successively by 4, 5 and 6, leaves remainders 2, 3 and 4, respectively. The least such number is

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 13
Let p be the number.

Suppose,

p ÷ 4 = q, remainder = 2

q ÷ 5 = r, remainder = 3

r ÷ 6 = s, remainder = 4

Then,

r = 6s + 4

q = 5r + 3 = 5(6s + 4) + 3 = 30s + 23

p = 4q + 2 = 4(30s + 23) + 2 = 120s + 94

Suppose the final quotient = 1

Then, p = 120 x 1 + 94 = 214.

SSC CGL (Tier II) Practice Test - 2 - Question 14

Two trains, 100 m and 120 m long, travelling in opposite directions cross each other in 4 seconds. If the speed of one of the trains 90 km/hr, what is the speed of the other?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 14
Speed = (100 + 120) ÷ 4 = 55 m/s = 198 km/hr

When the two trains are crossing each other (in opposite directions), the effective speed would be the sum of the individual speeds of the trains.

Speed of the other train = 198 - 90 = 108 km/hr

SSC CGL (Tier II) Practice Test - 2 - Question 15

If P = 229 × 321 × 58, Q = 227 × 321 × 58, R = 226 × 322 × 58 and S = 225 × 322 × 59, then which of the following is TRUE?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 15
Let A = 225 x 321 x 58

So, P = 24 x A = 16A

Q = 22 x A = 4A

R = 2 x 3 x A = 6A

S = 3 x 5 x A = 15A

Clearly, as A > 0, P > S > R > Q.

Hence, answer option (1) is correct.

SSC CGL (Tier II) Practice Test - 2 - Question 16

Direction: Read the given statements and conclusions carefully. Assuming that the information given in the statements is true, even if it appears to be at variance with commonly known facts, decide which of the given conclusions logically follow(s) from the statements.

Statements:

1) All cars are equipment.

2) Some bikes are cars.

3) Some cars are begs.

4) All begs are books.

Conclusions:

1) Some bikes are begs.

2) Some books are bikes.

3) All begs are equipment.

4) Some books are equipment.

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 16

Statements:

1) All cars are equipment.

2) Some bikes are cars.

3) Some cars are begs.

4) All begs are books.

The least possible diagram for the given statements is as follows

Conclusions:

1) Some bikes are begs → False (It is possible but not definite)

2) Some books are bikes → False (It is possible but not definite)

3) All begs are equipment → False (It is possible but not definite)

4) Some books are equipment → True (because some cars are begs and all begs are books so some books are also equipment)

Hence, “Only conclusions 4 follows”.

SSC CGL (Tier II) Practice Test - 2 - Question 17

How many four letter meaningful words can be made from the 3rd, 5th, 6th and the last letter of the word EVENTFUL?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 17
The 3rd, 5th, 6th and the last letters of the word EVENTFUL are E, T, F, and L.

The four letter meaningful words that can be made from the letters E, T, F, and L are FELT and LEFT.

There are two four letter meaningful words possible using the given letters.

Therefore, two is the correct answer.

SSC CGL (Tier II) Practice Test - 2 - Question 18

Select the number from among the given options that can replace the question mark (?) in the following matrix.

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 18

The logic is:

(Column 1 - Column 2) = (Column 3 - Column 4)

Row 1: 14 - 12 = 2 and 10 - 8 = 2

Row 2: 10 - 8 = 2 and 6 - 4 = 2

Row 3: 8 - 6 = 2 and 4 - 2 = 2

Similarly,

Row 4:

Let the missing number be 'x'

So, x - 14 = 2

⇒ x = 2 + 14

⇒ x = 16

and 12 - 10 = 2

Hence, ‘16’ is the correct answer.

SSC CGL (Tier II) Practice Test - 2 - Question 19

A clock is set right at 5 am. The clock loses 16 minutes in 24 hours. What will be the true time when clock indicates 10 pm on 4th day?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 19

Number of hours from 5 AM of a particular day to 10 PM on 4th day = (19 + 24 + 24 + 22) = 89 hours

Clock loses 16 minutes in 24 hours

So, in 1 hour the clock loses 16/24 minutes

So, in 89 hours, the clock loses (16/24) × 89 minutes = 59.34 minutes which is nearly equal to 1 hour.

So, the original time on the 4th day is 11 PM when the clock shows 10 PM.

Hence, ’11 PM’ is the correct answer.

SSC CGL (Tier II) Practice Test - 2 - Question 20

In a row of people, Raghav is standing at the 14th position from the extreme left end and Naresh is at the 16th position from the extreme right end. If they interchange their respective positions, Raghav now becomes 23rd from the extreme left end.

Q. What is the position of Naresh from the extreme right end after the interchange?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 20

Given:

1) Raghav is standing at the 14th position from the extreme left end and Naresh is at the 16th position from the extreme right end.

2) After interchange their respective positions, Raghav now becomes 23rd from the extreme left end.

⇒ The total number of persons = Raghav position from left + Raghav position from right - 1

⇒ The total number of persons = 23 + 16 - 1 = 38

Naresh position from the right:

⇒ The total number of persons = Naresh position from left + Naresh position from right - 1

⇒ 38 = 14 + Naresh position from right - 1

⇒ Naresh position from right = 38 - 14 + 1 = 39 - 14 = 25

Hence, Naresh position from right is “25”.

SSC CGL (Tier II) Practice Test - 2 - Question 21

In the following question, select the odd number pair from the given alternatives.

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 21

The logic is:

1) 7851 – 180 → (7 + 8) × (5 + 1) × 2 = 180

2) 3712 – 60 → (3 + 7) × (1 + 2) × 2 = 60

3) 2134 – 40 → (2 + 1) × (3 + 4) × 2 ≠ 40

4) 6932 – 150 → (6 + 9) × (3 + 2) × 2 = 150

Hence, ‘2134 – 40’ is the odd one out.

SSC CGL (Tier II) Practice Test - 2 - Question 22

How many triangles are present in the picture below?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 22

The number of triangles in the figure is shown below:

Hence, ‘27’ is the correct answer.

SSC CGL (Tier II) Practice Test - 2 - Question 23

If in a certain code language, SCIENCE is coded as GFRVMFU, how PHYSICS written in that code language?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 23

The logic follows here is :

Middle term changes to the reverse alphabet.

Similarly,

Hence, the correct answer is "UFMHCKR".

SSC CGL (Tier II) Practice Test - 2 - Question 24

Select the correct combination of mathematical signs to replace the * signs and to balance the given equation.

25 * 7 * 56 * 8 * 88 * 14

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 24

The logic followed here is:

By using BODMAS and going by the options

Given equation :- 25 * 7 * 56 * 8 * 88 * 14

Interchanging * with signs given in the options

Option. 1 +, -, =, ×, -

25 * 7 * 56 * 8 * 88 * 14

25 + 7 - 56 = 8 × 88 - 14

25 + 7- 56 = 704 -14

32 - 56 = 704 -14

-24 = 690

LHS ≠ RHS

Option. 2 -, +, ×, =, ÷

25 * 7 * 56 * 8 * 88 * 14

25 - 7 + 56 × 8 = 88 ÷ 14

25 - 7 + 56 × 8 = 6.285

25 - 7 + 448 = 6.285

473 - 7 = 6.285

466 = 6.285

LHS ≠ RHS

Option. 3 -, ×, ×, =, +

25 * 7 * 56 * 8 * 88 * 14

25 - 7 × 56 × 8 = 88 + 14

25 - 3136 = 88 + 14

25 - 3136 = 102

3111 = 102

LHS ≠ RHS

Option. 4 +, ×, ÷, =, -

25 * 7 * 56 * 8 * 88 * 14

25 + 7 × 56 ÷ 8 = 88 - 14

25 + 7 × 7 = 88 - 14

25 + 49 = 88 - 14

74 = 88 - 14

74 = 74

LHS = RHS

Therefore, option "+, ×, ÷, =, -" satisfies the above equation.

Hence, the correct answer is "+, ×, ÷, =, -"

SSC CGL (Tier II) Practice Test - 2 - Question 25

In each of the given number-clusters, the number on the right side of '=' (the equal to sign) is calculated by performing certain mathematical operations on the three numbers on the left of '=' (the equal to sign). All three number-clusters follow the same pattern. Select the number from among the given options that can replace the question mark (?) in the third number-cluster.

4, 3, 2 = 21

5, 6, 3 = 52

5, 8, 4 = ?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 25

Given:

4, 3, 2 = 21

5, 6, 3 = 52

5, 8, 4 = ?

The logic followed here is:

(1st number)2 + (2nd number)2 – (3rd number)2 = RHS

4, 3, 2 = 21

= 42 + 32 - 22

= 16 + 9 – 4

= 25 - 4

= 21 = RHS

And,

5, 6, 3 = 52

= 52 + 62 - 32

= 25 + 36 – 9

= 61 - 9

= 52 = RHS

Similarly,

5, 8, 4 = ?

= 52 + 82 - 42

= 25 + 64 – 16

= 89 - 16

= 73 = RHS

SSC CGL (Tier II) Practice Test - 2 - Question 26

Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R)

Assertion (A): Selecting the right man for the right job at the right pay is the right approach.

Reason (R): A wrongly selected man is a liability.

In the above context, which one of the following is correct?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 26
Assertion (A): Selecting the right man for the right job at the right pay is the right approach.

It is true that selecting the right man for the right job at the right pay is the correct approach. This will ensure smooth and efficient functioning

Reason (R): A wrongly selected man is a liability.

It is true because a wrongly selected man is a liability. A wrong selection can put a company into a risk of inefficiency.

But Reason is not sufficient to support the assertion.

Hence, Both 'A' and 'R' are true but 'R' is not correct explanation of 'A' is the correct answer.

SSC CGL (Tier II) Practice Test - 2 - Question 27

Seven people – G, H, J, K, L, M and N are sitting in the same row at an equal distance facing towards the south, but not necessarily in the same order. K is sitting fourth to the left of N. G is sitting immediate left of K. Two people are sitting between M and G. The number of people sitting between H and J is the same as the number of people sitting between J and L. H is not sitting at any of the extreme ends. Who among the following group of people is sitting to the right of M?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 27

Seven people – G, H, J, K, L, M and N

1) K is sitting fourth to the left of N.

2) G is sitting immediate left of K.

3) Two people are sitting between M and G.

4) The number of people sitting between H and J is the same as the number of people sitting between J and L.

5) H is not sitting at any of the extreme ends.

Condition (4) is not satisfying in case I, so it gets eliminated and we have the final arrangement as follows:

Hence, "J, L, N" is the correct answer.

SSC CGL (Tier II) Practice Test - 2 - Question 28

Direction: In following question two conclusions have been given followed by 4 sets of possible Statements. You have to take the given Conclusions to be true even if they seem to be at variance with the commonly known facts and then decide for the given Conclusions logically follows from the which of the given statements disregarding commonly known facts.

Conclusions:

I. No bag is book is a possibility.

II. Some book can never be copy.

III. All book can be pen.

Statements:

I. Only a few book are pen. Some bag are book. No pen is bag. Some pen is copy.

II. No bag are book. No pen is bag. Only pen is copy. Only a few book are pen.

III. Some bag are not book. No pen is bag. Only pen is copy. Some book are pen.

IV. Only a few book are pen. Some bag are book. No pen is bag. Only pen is copy

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 28

From statement I,

Only a few book are pen. Some bag are book. No pen is bag. Some pen is copy

Minimum possible diagram,

I. No bag is book is a possibility → False (because Some bag are book)

II. Some book can never be copy → False

III. All book can be pen → False (because Only a few book are pen which means some book are pen and some book are not pen)

Therefore, Statement I will not be the result of the given conclusions.

From Statement II:

No bag are book. No pen is bag. Only pen is copy. Only a few book are pen.

I. No bag is book is a possibility → False (it is already true)

II. Some book can never be copy → True (because Only pen is copy)

III. All book can be pen → False (because Only a few book are pen which means some book are pen and some book are not pen)

Therefore, Statement II will not be the result of the given conclusions.

From Statement III:

Some bag are not book. No pen is bag. Only pen is copy. Some book are pen.

I. No bag is book is a possibility → True

II. Some book can never be copy → True (because Only pen is copy)

III. All book can be pen → True

Therefore, Statement III will be the result of the given conclusions.

From Statement IV:

Only a few book are pen. Some bag are book. No pen is bag. Only pen is copy.

I. No bag is book is a possibility → False

II. Some book can never be copy → True (because Only pen is copy)

III. All book can be pen → False (because Only a few book are pen which means some book are pen and some book are not pen)

Therefore, Statement IV will not be the result of the given conclusions.

Hence, only statement III follows.

SSC CGL (Tier II) Practice Test - 2 - Question 29

If “PURPLE” is coded as “517515” then how will “ORANGE” be coded as?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 29

Here, the pattern being followed is, the absolute difference between digits of positional values of letters has been taken.

P → 16 → I1 - 6I = 5

U → 21 → I2 - 1I = 1

R → 18 → I1 - 8I = 7

P → 16 → I1 - 6 I = 5

L → 12 → I1 - 2I = 1

E → 05 → I0 - 5I = 5

Similarly,

O → 15 → I1 - 5I = 4

R → 18 → I1 - 8I = 7

A → 01 → I0 - 1I = 1

N → 14 → I1 - 4 I = 3

G → 07 → I0 - 7I = 7

E → 05 → I0 - 5I = 5

Hence, 471375 is the code for ORANGE.

SSC CGL (Tier II) Practice Test - 2 - Question 30

At a retail shop, coins worth ₹2,160 were collected in a day, including ₹1, 2, 5, and 10 coins. If the number of ₹2 coins was 10 more than the number of ₹10 coins, 20 more than the number of ₹5 coins, but 20 less than the number of ₹1 coins, then what was the number of ₹10 coins?

Detailed Solution for SSC CGL (Tier II) Practice Test - 2 - Question 30
Let the number of ₹10 coins = x,

The number of ₹5 coins = y

The number of ₹1 coins = z

Then, the number of ₹2 coins = x + 10 = y + 20 = z - 20

So, x + 10 = y + 20

⇒ y = x - 10

and x + 10 = z - 20

⇒ z = x + 30

Then,

⇒ (x + 30) × 1 + (x + 10) × 2 + (x - 10) × 5 + x × 10 = 2160

⇒ x + 30 + 2x + 20 + 5x - 50 + 10x = 2160

⇒ 18x = 2160

⇒ x = 120

So, the number of ₹10 coins = x = 120 coins

Hence, the correct answer is Option A.

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