SSC CGL (Tier II) Practice Test - 6 - SSC CGL MCQ

# SSC CGL (Tier II) Practice Test - 6 - SSC CGL MCQ

Test Description

## 150 Questions MCQ Test SSC CGL Tier II Mock Test Series 2024 - SSC CGL (Tier II) Practice Test - 6

SSC CGL (Tier II) Practice Test - 6 for SSC CGL 2024 is part of SSC CGL Tier II Mock Test Series 2024 preparation. The SSC CGL (Tier II) Practice Test - 6 questions and answers have been prepared according to the SSC CGL exam syllabus.The SSC CGL (Tier II) Practice Test - 6 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 - 6 below.
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SSC CGL (Tier II) Practice Test - 6 - Question 1

### What is the sum of all natural numbers between 100 and 400 which are divisible by 13?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 1
The first number is 104 which is divisible by 13.

So, the numbers are 104, 117, 130, ... 390.

Thus, the number of terms in the series is 23.

Sum of this AP

(2 × 104 + 22 × 13) = 5681

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

### What is the volume of the given solid?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 2
Let the radius of the base of the cylinder be r.

Radius of the base of the cone = Radius of the base of the cylinder = r

Let the height of the cone be h1 and height of the cylinder be h2.

r = 1 cm (Given in the figure)

h2 = 4 cm (Given in the figure)

h1 + h2 = 7 cm (Given in the figure)

Hence, h1 = 3 cm

Volume of the solid = Volume of the cone + Volume of the cylinder

Substituting the variables by their numeric values in the equation, we get

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

### In a certain test, 4 students each had a score of 80, 10 students each had a score of 70, 4 students each had a score of 75 and 2 students each had a score of 50. What was the average score of 20 students?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 3
Total score of all students = 4 × 80 + 10 × 70 + 4 × 75 + 2 × 50 = 1420

Required average = 1420/20 = 71

SSC CGL (Tier II) Practice Test - 6 - Question 4

Find the third proportional to 16 and 20.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 4
Let x be the third proportional.

This means- 16 : 20 = 20 : x

16x = 400

x = 25

SSC CGL (Tier II) Practice Test - 6 - Question 5

If two dice are thrown, what is the probability that the number appeared on second dice is greater than that on the first dice?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 5
Total number of events = 36

If first dice gives 1, the second dice can be any number from 2 to 6.

Similarly, the favorable events are (1, 2) … (1, 6), (2, 3)...(2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6).

So total number of favorable events = 15

Required Probability = 15/36 = 5/12

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

If point P(x, y) divides the line segment joining the points (2, 3) and (2, 7) internally in the ratio 1 : 3, then the coordinates of point P are

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 6
Ratio in which P(x, y) divides (2, 3) and (2, 7) is 1 : 3.

SSC CGL (Tier II) Practice Test - 6 - Question 7

In how many minimum number of years will a sum of money be more than double at 15% compound interest rate?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 7
Rate = 15%

Let the principal money be P and time be T years.

According to the question,

(1.15)T > 2

Now, substituting values of T

T = 2,

(1.15)2 = 1.3225

T = 3,

(1.15)3 = 1.5208

T = 5,

(1.15)5 = 2.011

For T = 5,

(1.15)T > 2

So, time = 5 years

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

x, y and z are positive numbers. If 3x > 9y and 2y > 4z, then which of the following is TRUE?

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

3x > 9y

3x > 32y

x > 2y

x > y

Also,

2y > 4z

2y > 22z

y > 2z

y > z

Thus, x > y > z

Hence answer option 1 is correct.

SSC CGL (Tier II) Practice Test - 6 - Question 9

Directions: Identify the missing number in the given series.

3, 3, 5, ?, 15, 23, 33

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 9
02 + 3 = 3

12 + 2 = 3

22 + 1 = 5

32 + 0 = 9

42 - 1 = 15

52 - 2 = 23

62 - 3 = 33

SSC CGL (Tier II) Practice Test - 6 - Question 10

The square root of is

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 10
Expression

∴ Required square root = √225 = 15

SSC CGL (Tier II) Practice Test - 6 - Question 11

What is the value of (x + y) in the given figure, if the perimeter is 34 cm?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 11
Perimeter = Sum of all sides

34 = 3 + x + y + 3 + 3 + 3 + 3 + 3 + 4

34 = 22 + x + y

x + y = 12 cm

SSC CGL (Tier II) Practice Test - 6 - Question 12

Find the value of

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 12

SSC CGL (Tier II) Practice Test - 6 - Question 13

What is the value of cos 15° + cos 105°?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 13
cos15° + cos105°

SSC CGL (Tier II) Practice Test - 6 - Question 14

Two places A and B are 100 km apart on a highway. One car starts from A and the other from B at the same time. If the cars travel in the same direction at a constant speed, they meet in 5 hours. If the cars travel towards each other, they meet in 1 hour. What is the speed of the car running faster?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 14

When they are moving in the same direction (x > y), then (x – y) x 5 = 100

⇒x – y = 20 …..(i)

When they are moving in opposite directions, then (x + y) = 100 ….(ii)

2x = 120

⇒ x = 120/2 = 60 km/hr and y = 40 km/hr

SSC CGL (Tier II) Practice Test - 6 - Question 15

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 was the expenditure on education in the year 1986?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 15
Income in 1986 = Rs. 2,00,000

Expenditure in 1986 is 75% of the income i.e.

= Rs. 1,50,000

Therefore, expenditure on education

SSC CGL (Tier II) Practice Test - 6 - Question 16

The ratio between the length and breadth of a rectangular field is 5 ∶ 4. If the breadth is 20 metres less than the length, the perimeter of the field is:

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 16

Given:

The ratio between the length and breadth of a rectangular field is 5 ∶ 4.

The breadth is 20 metres less than the length.

Concept used:

Perimeter of a rectangle = 2(Length + Breadth)

Calculation:

Let the length and breadth of the field be 5q and 4q meter respectively.

According to the question,

5q - 4q = 20

⇒ q = 20

Length = 5 × 20 = 100m

Breadth = 4 × 20 = 80m

Now, the perimeter of the field = 2(100 + 80)

⇒ 360m

∴ The perimeter of the field is 360m.

SSC CGL (Tier II) Practice Test - 6 - Question 17

5 cot θ + 12 = 0, where π / 2 < θ < π . Find the value of 36 tan θ + 26 sin θ - 39 cos θ

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 17

Given:

5 cot θ + 12 = 0, where π / 2 < θ < π .

Concept used:

θ lies in the 2nd quadrant, So the value of tan θ, cot θ, and cos θ is negative and

the value of sin θ is only positive hear.

Calculation:

5 cot θ + 12 = 0

⇒ 5 cot θ = - 12

⇒ cot θ = - 12 / 5

Value of tanθ = - 5 / 12

In triangle ABC,

AB = 5 , BC = 12 so value of AC = 13

So sinθ = 5 / 13 and cos θ = - (12 / 13)

So, the value of 36 tan θ + 26 sin θ - 39 cos θ

⇒ 36 (- 5 / 12) + 26 (5 / 13) - 39 {- (12 / 13)}

⇒ - 15 + 10 + 36 = 31

SSC CGL (Tier II) Practice Test - 6 - Question 18

Pipes A and B can fill a tank in 12 hrs and 16 hrs respectively. Whereas pipe C can empty it in 20 hrs. Pipe A and C first opened for 5 hrs and then Pipe A closed and B opened. In how many hours the tank will be filled?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 18

Given:

Pipes A and B can fill a tank in 12 hrs and 16 hrs respectively.

The pipe C can empty it in 20 hrs.

Pipes A and C first opened for 5 hrs

then Pipe A closed and B opened.

Concept used:

Efficiency = Total work / Total time taken

Calculation:

Let the total volume of the tank is LCM of 12, 16, and 20 = 240

The efficiency of Pipe A = 240 / 12

⇒ 20 units

The efficiency of Pipe B = 240 / 16

⇒ 15 units

The efficiency of Pipe C = 240 / 20

⇒ 12 units

Pipes A and C first opened for 5 hrs

So, In 5 hrs A pipe filled = 20 × 5 units

⇒ 100 units

In 5 hrs C pipe emty = 12 × 5

⇒ 60 units

So total filled = 100 - 60

⇒ 40 units

Remaining Amount = 240 - 40

⇒ 200 units.

Now pipe A will be closed and Pipe B will be opened

So, In 1 hr Both Pipe B and C filled 15 - 12 = 3 units

So a total of 200 units will be filled in (200 / 3) = 66.66 hrs

The total tank will be filled in (66.66 + 5) = 71.66 hrs

71.66 hours the tank will be filled

SSC CGL (Tier II) Practice Test - 6 - Question 19

Chand started a business by investing Rs. 8560, after 12 months Sooraj invested Rs. 6420. If the total profit earned by Sooraj and Chand at the end of two years is Rs. 1815, then find the difference in the profit earned by Sooraj and Chand.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 19

Given:

The amount invested by Chand = Rs. 8560

The time of investment for Chand = 24 months

The amount invested by Sooraj = Rs. 6420

The time of investment for Sooraj = 12 months

The total profit earned = Rs. 1815

Concept:

Profit ∝ Amount of investment

Profit ∝ Time of investment

Therefore,

The ratio of profit earned by two people is equal,

To the ratio of the product of the Amount of investment and the Time of investment.

Calculation:

Let S and Chand be the profit earned by Sooraj and Chand

According to the question and as per the above concept,

The profit earned by Sooraj = 3/11 × 1815 = Rs. 495

The profit earned by Chand = 8/11 × 1815 = Rs. 1320

The required difference = Rs. 1320 - Rs. 495 = Rs. 825

Therefore, 'Rs. 825' is the required answer.

SSC CGL (Tier II) Practice Test - 6 - Question 20

If , where x + y + z ≠ 0, then find the value of x.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 20

Given:

Calculation:

According to the question,

Add 3 on both sides of the equation.

Since x + y + z ≠ 0,

Therefore,

SSC CGL (Tier II) Practice Test - 6 - Question 21

Study the given graph and answer the question that follows.

Q. Which of the following statement is correct?

I. The total number of computers sold by dealer A on Monday, Tuesday and Friday is 240.

II. The total number of computers sold by dealer B on Wednesday, Thursday and Saturday is 215.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 21

Calculation:

The total number of computers sold by dealer A on Monday, Tuesday, and Friday = 75 + 80 + 78 = 233

Thus, statement I is incorrect.

The total number of computers sold by dealer B on Wednesday, Thursday, and Saturday = 60 + 80 + 75 = 215

Thus, statement II is correct.

∴ Only statement II is correct.

SSC CGL (Tier II) Practice Test - 6 - Question 22

Which of the following will have the lowest discount percentage?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 22

Given :

Marked price : Cost price = 7 ∶ 3 and Cost price : Selling price = 3 ∶ 5

Marked price : Cost price = 7 ∶ 2 and Cost price : Selling price = 1 ∶ 2

Marked price : Cost price = 3 ∶ 1 and Cost price : Selling price = 5 ∶ 6

Marked price : Cost price = 8 ∶ 3 and Cost price : Selling price = 2 ∶ 3

Formula used:

Discount % =

Where, S.P = Selling Price

M.P = Marked Price, C.P = Cost Price

Calculation:

From the option,

Step 1: Option 1

Marked price : Cost price = 7 ∶ 3 -----(1)

Cost price : Selling price = 3 ∶ 5 ------(2)

Marked Price : Cost Price : Selling Price = 7 : 3 : 5

Let the value of Marked Price , Cost Price and Selling Price be 7x , 3x and 5x respectively

Discount % =

= 200/7

= 28.57 %

Step 2: Option 2

Marked price : Cost price = 7 ∶ 2 -----(3)

Cost price : Selling price = 1 ∶ 2------(4)

multiply(4) by 2 to make the Cost Price equal

Marked Price : Cost Price : Selling Price = 7 : 2 : 4

Let the value of Marked Price , Cost Price and Selling Price be 7x , 2x and 4x respectively

Discount % =

= 300 / 7

= 42.85 %(approx.)

Step 3: Option 3

Marked price : Cost price = 3 ∶ 1 -----(5)

Cost price : Selling price = 5 ∶ 6 ------(6 )

multiply (5) × 5 to make the Cost Price equal

Marked Price : Cost Price : Selling Price = 15 : 5 : 6

Let the value of Marked Price , Cost Price and Selling Price be 15x , 5x and 6x respectively

Discount % =

= 900/15

= 60%

Step 4: Option 4

Marked price : Cost price = 8 ∶ 3 -----(7)

Cost price : Selling price = 2 ∶ 3------(8)

multiply (8) × 3 and (7) by 2 to make the Cost Price equal

Marked Price : Cost Price : Selling Price = 16 : 6 : 9

Let the value of Marked Price , Cost Price and Selling Price be 16x , 6x and 9x respectively

Discount % =

= 700 / 16

= 43.75%

Hence, Option A has the lowest discount.

Answer is "Marked price : Cost price = 7 ∶ 3 and Cost price : Selling price = 3 ∶ 5" .

SSC CGL (Tier II) Practice Test - 6 - Question 23

If x and y are the LCM and HCF of the numbers, 30, 45, 90 and 135 respectively, then what is the value of x - 7y?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 23

Given:

x and y are the LCM and HCF of the numbers, 30, 45, 90 and 135 respectively

Concept used:

HCF - Highest common factor

LCM - Least common multiple

Calculation:

Let's find factors of given numbers,

30 = 2 × 3 × 5

45 = 3 × 3 × 5

90 = 2 × 3 × 3 × 5

135 = 3 × 3 × 3 × 5

So, 3 and 5 are highest common factors of all the given numbers.

HCF (30, 45, 90 and 135) = 15

Y = 15

And,

30 = 2 × 3 × 5

45 = 3 × 3 × 5

90 = 2 × 3 × 3 × 5

135 = 3 × 3 × 3 × 5

So, the least common factor of all the given numbers = 3 × 3 × 3 × 2 × 5

Hence, LCM (30, 45, 90 and 135) = 270

X = 270

Value of x - 7y,

x - 7y = 270 - 7 (15)

= 270 - 105

= 165

SSC CGL (Tier II) Practice Test - 6 - Question 24

If then find the value of .

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 24

Given:

Calculations:

(let k is the constant)

x = 4k, y = 5k, z = 3k

Now,

= 4

SSC CGL (Tier II) Practice Test - 6 - Question 25

Study the following table to answer the questions that are given below.

Expenditures of a company per annum over the given year.

Q. The ratio between the total expenditure on taxes for all the years and the total expenditure on fuel and transport for all years respectively is approximately.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 25

Calculation:

Total expenditure on taxes for all years = 76 + 105 + 84 + 85 + 89 = 439

Total expenditure on fuel and transport for all years = 96 + 156 + 143 + 124 + 134 = 653

Ratio between the total expenditure on taxes for all the years and the total expenditure on fuel and transport for all years = 439 : 653

= 10.2(approx.) × 43 : 15.2(approx.) × 43

= 10.2 : 15.2

= 10 : 15 (approx.)

SSC CGL (Tier II) Practice Test - 6 - Question 26

If the number 678967# is exactly divisible by 72, the minimum value of # is:

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 26

Given:

The number 678967# is exactly divisible by 72

Concept used:

If the number is divisible by 72 that means it has to be divisible by 8 and 9 as these are the two factors of 72.

Divisibility rules of 8 : when the number made by last three digits of a number is divisible by 8 then the number is also divisible by 8. Apart from this if the last 3 or more digits of a number are zero then the number is divisible by 8.

Divisibility rules of 9 : when the sum of all the digits of a number is divisible by 9 then the number is also divisible by 9.

Calculations:

The number is 678967#, so if we apply the divisibility rules of 8, the last 3 digits have to be divisible by 8.

Now, 67# has to be divisible by 8, so if we put 2 in place # it becomes 672 which is divisible by 8.

Again, applying divisibility rules of 9, by adding all the numbers = 6 + 7 + 8 + 9 + 6 + 7 + # = 43 + #

So, if we take 2 again in place of # the sum of the numbers become = 45 which is divisible by 9.

Thus, two rules satisfies the condition and the number is 2.

SSC CGL (Tier II) Practice Test - 6 - Question 27

In ΔPQR, ∠PQR = 90°, PQ = 7 cm, QR = 24 cm. Calculate the radius of the incircle.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 27

Given∶

ΔPQR is a right angle triangle, ∠PQR = 90°, PQ = 7 cm, QR = 24 cm.

Formula Used∶

In radius of the circle = Area of triangle /Semi perimeter of triangle.

Pythagoras theorem

Semi- perimeter of a triangle = (a + b +c)/2, where a, b and c are the sides of the triangle.

Area of triangle = 1/2 × Base × Height

Calculation∶

In ΔPQR, ∠PQR = 90°, PQ = 7cm and QR = 24 cm

So, PR2 = PQ2 + QR2 [Pythagoras theorem]

⇒ PR2 = 72 + 242

⇒ PR2 = 49 + 576 = 625

⇒ PR = 25 cm

Semi perimeter of ΔPQR,

S = (PQ + QR + PR)/2

⇒ S = (7 + 24 + 25) /2

⇒ S = 28 cm

Now, Area of ΔPQR = 1/2 × PQ × QR

Area = 1/2 × 7 × 24 = 84 cm2

⇒ In radius, r = Area/semi perimeter

⇒ r = 84/28 = 3 cm

∴ The radius of the incircle is 3 cm.

SSC CGL (Tier II) Practice Test - 6 - Question 28

In diagram in ΔPQR, T is a point on the side QR and PT is perpendicular to QR. S is a point on PT in such a way that PS : ST = 3 : 1. If ∠QPT = 30° and tan ∠PRQ = 4 × tan ∠SQT, then find ∠PRQ.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 28

Given:

PT is perpendicular to QR.

PS : ST = 3 : 1.

∠QPT = 30° and tan ∠PRQ = 4 × tan ∠SQT

Formula Used:

If θ is an angle of a right angled triangle, then tan θ = Perpendicular/Base

Calculation:

∠ PQT = 180° - (∠PTQ + ∠QPT) = 180° - (90° + 30°) = 60°

Now, tan ∠PRQ = PT/TR ----(1)

and tan ∠SQT = ST/QT ----(2)

Now multiplying (1) with (2) we get,

tan ∠PRQ/tan ∠SQT = PT/TR × QT/ST

⇒ 4 × ST/PT = QT/TR

⇒ 4 × 1/4 = QT/TR

⇒ QT = TR

It means PT bisects QR in a ratio 1 : 1

So, ∠PRQ = 60° (∵dQT/TR = PQ/PR) and thus ΔPQR is an equilateral triangle.

∴ Value of ∠PRQ is 60°

SSC CGL (Tier II) Practice Test - 6 - Question 29

Find the value of 1/(4 - √15) – 1/(√15 - √14) + 1/(√14 - √13) -1/√(13 - √12) + 1/√(12 - √11) – 1/(√11 - √10) + 1/(√10 – 3)

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 29

Formula used∶

Rationalization of surds, (a + b) (a – b) = a2 – b2

Calculation∶

1/(4 - √15) – 1/(√15 - √14) + 1/(√14 - √13) -1/√(13 - √12) + 1/√(12 - √11) – 1/(√11 - √10) + 1/(√10 – 3)

⇒ [1/(4 - √15) × {(4 +√15)/ (4 +√15)}] = [(4 +√15)/{42 – (√15)2}] = (4 +√15)/(16 – 15) = (4 +√15)/1 = (4 + √15)

⇒ [{1/(√15 - √14)} × {(√15 + √14)/ (√15 + √14)}] = [{(√15 + √14)}/{(√15)2 – (√14)2}] = [√15 + √14)/(15 – 14) = (√15 + √14)/1 = √15 + √14

⇒ [{1/(√14 - √13) × {(√14 + √13)/ (√14 + √13)}] = [{(√14 + √13)}/{(√14)2 – (√13)2}] = [(√14 + √13)/(14 – 13)] = [(√14 + √13)/1] = √14 + √13

⇒ [{1/(√13 -√12) × {(√13 + √12)/ (√13 + √12)}] = [{(√13 +√12)}/{(√13)2 – (√12)2}] = [(√13 + √12)/(13 – 12)] = [(√13 + √12)/1] = √13 + √12

⇒ [{1/(√12 -√11) × {(√12 + √11)/ (√12 + √11)}] = [{(√12 +√11)}/{(√12)2 – (√11)2}] = [(√12 + √11)/(12 – 11)] = [(√12 + √11)/1] = √12 + √11

⇒ [{1/(√11 -√10) × {(√11 + √10)/ (√11 + √10)}] = [{(√11 +√10)}/{(√11)2 – (√10)2}] = [(√11 + √10)/(11 – 10)] = [(√11 + √10)/1] = √11 + √10

⇒ [{1/(√10 -3) × {(√10 + 3)/ (√10 + 3)}] = [{(√10 +3)}/{(√10)2 – (3)2}] = [(√10 + 3)/(10 – 9)] = [(√10 + 3)/1] = √10 + 3

Now, put all the values in their required places in the given equation, we get

⇒ (4 + √15) – (√15 + √14) +(√14 + √13) – (√13 + √12) + (√12 + √11) – (√11 + √10) + (√10 + 3)

⇒ 4 + √15 – √15 - √14 + √14 + √13 – √13 - √12 + √12 + √11 – √11 - √10 + √10 + 3 = 7

∴ The required value is 7.

SSC CGL (Tier II) Practice Test - 6 - Question 30

In the following questions, identify the reasoning between the two terms to the left of the mark (∶∶) and select the correct option for each question by applying the same reasoning to both the terms to the right of the mark (∶∶) in each question.

CIGARETTE ∶ GICERAETT ∶∶ DIRECTION ∶ _______

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 30

In certain coding language,

CIGARETTE is written as GICERAETT as follows,

Similarly,

Direction can be written as follows,

Hence, ‘RIDTCENOI’ is the correct answer.

SSC CGL (Tier II) Practice Test - 6 - Question 31

Directions: In the following question, there is a certain relationship between two numbers given on one side of : :. One number is given on the other side of : :, while the other number is to be found from the given alternatives. Find the missing number, which has the same relationship with this number as that between the numbers of the other given pair.

8 : 27 : : 125 : ___

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 31
The relation between the numbers is

23 : (next prime)3 :: 53 : (Next prime)3

8 : 27 :: 125 : 343

Therefore, option 2 is the correct answer.

SSC CGL (Tier II) Practice Test - 6 - Question 32

Four words have been given, out of which three are alike in some manner and one is different. Select the word that is different.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 32
Out of the given four words, plotter is the odd one out as it is an output device whereas the other three, i.e. keyboard, webcam and microphone are input devices of the computer. Hence, option 1 is the answer. Do you have a better or a different solution?
SSC CGL (Tier II) Practice Test - 6 - Question 33

Directions: A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 33

Option (a) is correct.

SSC CGL (Tier II) Practice Test - 6 - Question 34

Directions: Select the term which will replace the question mark (?) in the given alphanumeric series.

G4T, ?, M20P, P43N, S90L

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 34
Pattern 1 (First letter of each term): G + 3 = J, J + 3 = M, M + 3 = P and P + 3 = S

Pattern 2 (Last letter of each term): T - 2 = R, R - 2 = P, P - 2 = N and N - 2 = L

Pattern 3 (Middle number of each term): 4 x 2 + 1 = 9, 9 x 2 + 2 = 20, 20 x 2 + 3 = 43 and 43 x 2 + 4 = 90

Therefore, the required arrangement is "J9R".

SSC CGL (Tier II) Practice Test - 6 - Question 35

If it is possible to make a meaningful word with the third, fifth, eighth and tenth letters of the word DISTRIBUTE, which of the following will be the third letter of that word?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 35
The third, fifth, eighth and tenth letters of the word DISTRIBUTE are S, R, U and E, respectively. The word formed is 'SURE' and its third letter is R.
SSC CGL (Tier II) Practice Test - 6 - Question 36

If HOTEL is coded as 300, the code for BAR would be _____.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 36
The numbers are assigned according to the position of letters in the English alphabet.

A = 1, B = 2, C = 3 …… Z = 26,

HOTEL = 8 + 15 + 20 + 5 + 12 = 60 ⇒ 60 × 5 = 300,

BAR = 2 + 1 + 18 = 21 ⇒ 21 × 3 = 63

SSC CGL (Tier II) Practice Test - 6 - Question 37

Select the correct combination of mathematical signs that can sequentially replace the * signs and make the equation correct.

135 * 15 * 3 * 2 * 13 * 16

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 37
Using symbols of option 1, placing mathematical signs at the place of * and using BODMAS rule, we get

135 ÷ 15 × 3 + 2 - 13 = 16

9 × 3 + 2 - 13 = 16

27 + 2 - 13 = 16

29 - 13 = 16

16 = 16

L.H.S = R.H.S

Hence, option 1 is correct.

SSC CGL (Tier II) Practice Test - 6 - Question 38

Directions: In the question below, some statements are given, followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follow(s) from the given statements disregarding commonly known facts.

Statements:

Some apples are cherries.

Only a few cherries are strawberries.

Conclusions:

1. Some strawberries are apples.

2. All apples can be strawberries.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 38
According to the statements,

Or

Only conclusion (b) follows.

SSC CGL (Tier II) Practice Test - 6 - Question 39

Directions: Three statements, followed by two conclusions numbered I and II, are given. You have to take the given statements to be true, even if they seem to be at variance with the commonly known facts. Read the conclusions and then decide which of them logically follow(s) from the given statements, disregarding the commonly known facts.

Statements:

All desks are benches.

No bench is chair.

Conclusions:

II. No chair is desk.

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 39

Both conclusions follow.

SSC CGL (Tier II) Practice Test - 6 - Question 40

Directions: Study the following information carefully and answer the given question.

9 friends, A, B, C, D, E, F, G, H, and I, are standing at some distance from each other. A is 6 m to the west of B, who is 8 m to the south of D. E is 4 m to the east of D. E is 12 m to the north of C, who is 8 m to the west of F. H is 2 m to the west of G, who is 4 m to the north of F. I is 8 m to the north of H.

Q. What is the direction of C as compared to G?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 40
C is in the southwest direction from G.

SSC CGL (Tier II) Practice Test - 6 - Question 41

Directions: Read the following information carefully and answer the question that follows.

Joe Gallo, Mickey Cohen, Frank Nitti, Danny Greene, Curtis Warren, Manya Surve and William Shapiro appeared for a test. Only one candidate got marks that lie between the marks obtained by Danny Greene and Frank Nitti. Four candidates obtained more marks than Curtis Warren. Mickey Cohen and Manya Surve obtained less marks than Danny Greene. Manya Surve did not obtain the least marks. Joe Gallo obtained more marks than Danny Greene, but less marks than William Shapiro. Curtis Warren obtained less marks than Joe Gallo. Frank Nitti obtained more marks than Danny Greene.

Q. Who obtained the maximum marks?

Detailed Solution for SSC CGL (Tier II) Practice Test - 6 - Question 41
Only one candidate got marks that lie between the marks obtained by Danny Greene and Frank Nitti. Frank Nitti obtained more marks than Danny Greene. Frank Nitti > __ > Danny Greene -- (i) Joe Gallo obtained more marks than Danny Greene, but less marks than William Shapiro. William Shapiro > Frank Nitti > Joe Gallo > Danny Greene -- (Case-1) William Shapiro > Joe Gallo > Frank Nitti > __ > Danny Greene --(Case-2) Mickey Cohen and Manya Surve obtained less marks than Danny Greene. Manya Surve did not obtain the least marks. Danny Greene > Manya Surve > Mickey Cohen Four candidates obtained more marks than Curtis Warren. Curtis Warren obtained less marks than Joe Gallo. Hence, Case-2 stands void and we get final arrangement as given below : William Shapiro > Frank Nitti > Joe Gallo > Danny Greene > Curtis Warren > Manya Surve > Mickey Cohen Therefore, William Shapiro obtained the maximum marks.