Explanation : 

Given sequence: 2, 7, 22, 67, 202, X, 1822

Pattern: Each number is generated by multiplying the previous number by 3 and adding 1.

Let’s verify each step:
2 → 7: 2 × 3 + 1 = 7
7 → 22: 7 × 3 + 1 = 22
22 → 67: 22 × 3 + 1 = 67
67 → 202: 67 × 3 + 1 = 202
202 → X: 202 × 3 + 1 = 607
607 → 1822: 607 × 3 + 1 = 1822

Answer: X = 607

Explanation :

Replacing the incorrect term by correct term

Let’s solve this sequence problem step by step:

1. Original sequence:
3, 2, 7, 4, 13, 10, 21, 18, 31, 28, 43, 40

2. Analysis of odd-positioned numbers (1st, 3rd, 5th…):
3, 7, 13, 21, 31, 43
Differences: 4, 6, 8, 10, 12
Pattern: Each difference increases by 2

3. Analysis of even-positioned numbers (2nd, 4th, 6th…):
2, 4, 10, 18, 28, 40
Differences: 2, 6, 8, 10, 12
Problem: First difference is 2, breaking the pattern

4. Correction needed:
Starting from 3 (first number)
For consistent pattern, next number should create difference of 4
Therefore, second number should be 0 (not 2)
Because: 3 → 0 gives difference of 3
And: 0 → 4 gives difference of 4

5. After correction:
Odd terms differences: 4, 6, 8, 10, 12
Even terms differences: 4, 6, 8, 10, 12

Answer: (a) 0 is correct.

The sequence becomes perfect when 2 is replaced with 0, making both odd and even terms follow same pattern of differences increasing by 2.

Explanation : 

Seven Book P, Q, R, S, T, U & V Arrangement

From the given information, we know that R, Q and T have blue covers and other books have red covers. We also know that only S and U are new books and the rest are old. Finally, we know that P, R and S are law reports; the rest are Gazetteers.

Based on this information, we can conclude that Q and T are old Gazetteers with blue covers. This is because they both have blue covers (as stated in the first sentence), they are both old (since only S and U are new), and they are both Gazetteers (since P, R and S are law reports).

Explanation : 

Arrangement of English Alphabet

The English alphabet is: ABCD EFGH IJKL MNOP QRST UVWX YZ

As per the question, the letters are arranged as follows:

DCBA HGFE LKJI PONM TSRQ XWVU ZY

The 13th letter in this arrangement is P.

The 4th letter to its right is T.

Explanation : 

Finding the Heaviest Person Between A, B, C, D, E

Analysis of Statement 1:
– Tells us either A or D is heaviest among A, B, E, and F
– No information about C
– Cannot determine the answer alone

Analysis of Statement 2:
– Tells us C is heavier than A
– No information about how C compares to B, E, F, or D
– Cannot determine the answer alone

Combining Both Statements:
1. From Statement 1:
– Either A or D is heaviest among A, B, E, F

2. From Statement 2:
– C is heavier than A
– C is heavier than D

3. Combined Logic:
– If C is heavier than both A and D
– And A or D was heaviest among A, B, E, F
– Then C must be heaviest among all (A, B, C, D, E)

Conclusion:
Both statements are necessary to conclude that C is the heaviest.

Explanation : 

In a group of 120 persons 80 are indian

Total Information:
– Total number of persons = 120
– Number of Indians = 80
– Number of Foreigners = 40
– Total English speakers = 70

Analysis:

1. Given that all 40 foreigners can speak English:
– Foreigners who speak English = 40
– Remaining English speakers must be Indians
– Indians who speak English = Total English speakers – Foreigners
– Indians who speak English = 70 – 40 = 30

Conclusion:
Therefore, at least 30 Indians must be able to speak English.

Explanation : 

Priya is 4 ranks below seema

Statement-1: Priya is 4 ranks below Seema and is 31st from the bottom.
In a class of 40 students, Seema’s rank is 31 + 4 = 35th.

Statement-2: Ena is 2 ranks above Seema and is 37th from the bottom.
In a class of 40 students, Seema’s rank is 37 – 2 = 35th.

Either Statement-1 alone or Statement-2 alone is sufficient to answer the question.

Explanation : 

Problem: Find (PQ) × 3 = RQQ, where P, Q, and R are distinct digits and R ≠ 0

Step 1: Express the Equation
(10P + Q) × 3 = 100R + 10Q + Q
30P + 3Q = 100R + 11Q
30P = 100R + 8Q
15P = 50R + 4Q

Step 2: Analyze the Equation
15P – 50R = 4Q
15P ≡ 50R (mod 4)

Step 3: Find the Values
Case 1: P = 8, R = 2
15 × 8 = 120 = 50 × 2 + 4Q
120 = 100 + 4Q
4Q = 20
Q = 5

Verification:
85 × 3 = 255
This confirms P = 8, R = 2, Q = 5 are distinct digits and R ≠ 0

Step 4: Calculate (P + R)/Q
(8 + 2)/5 = 10/5 = 2

Final Answer: 2

Explanation : 

3P + 4P + PP + PP = RQ2

Breaking it down:
(30 + P) + (40 + P) + (10P + P) + (10P + P) = 100R + 10Q + 2

Simplifying:
24P + 70 = 100R + 10Q + 2

Further Breaking Down:
20P + 70 + 4P = 100R + 10Q + 2

Finding Value of P:
– Unit digit is 2
– Must come from 4 × P
– Therefore, P must be 3 or 8

Case 1 (P = 3):
24P + 70 = (24 × 3) + 70
= 72 + 70
= 142

Case 2 (P = 8):
24P + 70 = (24 × 8) + 70
= 192 + 70
= 262

Arithmetic Mean:
(142 + 262) ÷ 2 = 202

Answer: 202

Explanation : 

Objective type test of 90 questions

Given Information:
Total questions = 90
Marks for correct answer = +5
Marks for incorrect answer = -2
Total marks obtained = 387

Let’s solve:
x = number of correct answers
y = number of incorrect answers

Equation 1 (Total Questions):
x + y = 90

Equation 2 (Total Marks):
5x – 2y = 387

Solving the System of Equations:
From Equation 1: x + y = 90
x = 90 – y
Substitute in Equation 2:
5(90 – y) – 2y = 387
450 – 5y – 2y = 387
450 – 7y = 387
-7y = -63
y = 9

Therefore: Incorrect answers = 9
Correct answers = 81 (90 – 9)