Number, Ranking and Time Sequence Test Questions in English

(B) To find the number of $7$s that are immediately preceded by $6$ and not immediately followed by $4$,we examine the series:
$7, 4, 2, 7, 8, 4, 3, 6, 7, 5, 3, 5, 7, 8, 4, 3, 7, 6, 7, 2, 4, 0, 6, 7, 4, 3$
$1$. Identify all $7$s in the series.
$2$. Check the digit before each $7$ to see if it is $6$.
$3$. Check the digit after each $7$ to see if it is $NOT$ $4$.
- The first $7$ preceded by $6$ is at the $9$th position: $6, 7, 5$. Here,$7$ is preceded by $6$ and followed by $5$ (not $4$). This satisfies the condition.
- The second $7$ preceded by $6$ is at the $19$th position: $6, 7, 2$. Here,$7$ is preceded by $6$ and followed by $2$ (not $4$). This satisfies the condition.
- The third $7$ preceded by $6$ is at the $24$th position: $6, 7, 4$. Here,$7$ is preceded by $6$ but followed by $4$. This does not satisfy the condition.
Thus,there are $2$ such $7$s.

(A) To find the number of $18$s that are immediately preceded by $9$ but not immediately followed by $7$,we look for the pattern $9, 1, 8, X$,where $X \neq 7$.
$1$. Let's examine the sequence: $7, 1, 9, 1, 1, 7, 1, 8, 9, 1, 7, 1, 2, 1, 3, 1, 4, 5, 7, 1, 3, 9, 1, 7$.
$2$. Identify all occurrences of $18$ in the sequence: The sequence contains only one $8$ at the $8$th position,which is preceded by $1$,not $9$.
$3$. Re-evaluating the sequence for the pattern $9, 1, 8$: The sequence contains no instance where $18$ is preceded by $9$.
$4$. Since there are no instances of $9, 1, 8$ in the given sequence,the count is $0$.

(A) We need to find the occurrences of the patterns $9, 1, 8$ or $8, 1, 9$ in the given series.
The series is: $2, 1, 9, 8, 1, 9, 8, 3, 7, 1, 9, 7, 8, 1, 2, 9, 1, 9, 8, 1, 8, 2, 1, 2$.
Let's scan the series:
$1$. $2, 1, 9, 8, 1, 9, 8, 3, 7, 1, 9, 7, 8, 1, 2, 9, 1, 9, 8, 1, 8, 2, 1, 2$
Scanning for $9, 1, 8$ or $8, 1, 9$:
- At index $4, 5, 6$: $1, 9, 8$ (No,$1$ is not in the middle)
- At index $16, 17, 18$: $9, 1, 9$ (No)
- At index $17, 18, 19$: $1, 9, 8$ (No)
- Let's re-examine the sequence: $2, 1, 9, 8, 1, 9, 8, 3, 7, 1, 9, 7, 8, 1, 2, 9, 1, 9, 8, 1, 8, 2, 1, 2$
- Looking for $9, 1, 8$ or $8, 1, 9$:
- Sequence segment: $... 9, 1, 9, 8 ...$ (No)
- Sequence segment: $... 9, 1, 8 ...$ (Found at the end: $9, 1, 8$)
- Checking again: $2, 1, 9, 8, 1, 9, 8, 3, 7, 1, 9, 7, 8, 1, 2, 9, 1, 9, 8, 1, 8, 2, 1, 2$
- The pattern $9, 1, 8$ appears once at the end of the sequence $(... 9, 1, 8 ...)$.
- There are no other occurrences of $9, 1, 8$ or $8, 1, 9$ in the provided series.
- Therefore,the correct answer is $1$.

(B) To find the number of students who passed, we use the formula: $\text{Total passed} = (\text{Rank from top} + \text{Rank from bottom}) - 1$.
Given: Rank from top $= 13$, Rank from bottom $= 26$.
Number of students passed $= (13 + 26) - 1 = 39 - 1 = 38$.
Given: Number of students who failed $= 6$.
Total number of students in the class $= \text{Number of students passed} + \text{Number of students failed}$.
Total students $= 38 + 6 = 44$.

(A) To find the total number of persons in a row when the position of one person is given from both ends,we use the formula:
Total number of persons $= (\text{Position from front}) + (\text{Position from back}) - 1$
Given:
Position from front $= 12$
Position from back $= 47$
Total number of persons $= 12 + 47 - 1$
Total number of persons $= 59 - 1 = 58$
Therefore,there are $58$ persons in the row.

(D) The Managing Director entered the room $10 \, \text{minutes}$ before $12:30$, which means he arrived at $12:20$.
He arrived $20 \, \text{minutes}$ before the chairman, so the chairman arrived at $12:20 + 20 \, \text{minutes} = 12:40$.
The chairman was $30 \, \text{minutes}$ late for the interview.
Therefore, the scheduled time for the interview was $12:40 - 30 \, \text{minutes} = 12:10$.

(C) We need to find the number of $8$s that satisfy two conditions:
$1$. The digit immediately before $8$ must be $7$.
$2$. The digit immediately after $8$ must $NOT$ be $4$.
Let's analyze the sequence: $2, 3, 8, 2, 5, 7, 8, 3, 7, 8, 4, 6, 9, 8, 4, 3, 2, 7, 8, 9, 5, 7, 8, 1, 5, 2, 9$
- First $8$ (at index $3$): Preceded by $3$. (Does not satisfy condition $1$)
- Second $8$ (at index $7$): Preceded by $7$,followed by $3$. (Satisfies both conditions)
- Third $8$ (at index $10$): Preceded by $7$,followed by $4$. (Does not satisfy condition $2$)
- Fourth $8$ (at index $14$): Preceded by $9$. (Does not satisfy condition $1$)
- Fifth $8$ (at index $19$): Preceded by $7$,followed by $9$. (Satisfies both conditions)
- Sixth $8$ (at index $23$): Preceded by $7$,followed by $1$. (Satisfies both conditions)
Thus,there are $3$ such $8$s.

(B) To find the number of $3$s that are immediately preceded by $6$ but not immediately followed by $7$,we examine the sequence:
$2, 3, 7, 4, 3, 5, 6, 3, 7, 4, 6, 3, 8, 9, 6, 5, 1, 8, 3, 7, 2, 4, 2, 8, 6, 3, 9$
$1$. Look for the pattern $6, 3, X$ where $X \neq 7$.
$2$. In the sequence,we identify the occurrences of $6, 3$:
- $6, 3, 7$ (This does not count because it is followed by $7$)
- $6, 3, 8$ (This counts because it is followed by $8$)
- $6, 3, 9$ (This counts because it is followed by $9$)
Thus,there are $2$ such $3$s.

(B) First,find the numbers between $11$ and $50$ that are divisible by $7$.
The multiples of $7$ in this range are: $14, 21, 28, 35, 42, 49$.
There are $6$ such numbers.
Next,identify which of these are also divisible by $3$ (i.e.,multiples of $21$).
The numbers divisible by both $7$ and $3$ are $21$ and $42$.
To find the numbers divisible by $7$ but not by $3$,subtract the common multiples from the set of multiples of $7$.
Required numbers = ${14, 28, 35, 49}$.
Therefore,the count of such numbers is $4$.

(A) To find the numbers from $1$ to $100$ that are divisible by $4$ and contain the digit $4$,we list the multiples of $4$ up to $100$: $4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100$.
Now,we identify those that contain the digit $4$:
$4$ (contains $4$)
$24$ (contains $4$)
$40$ (contains $4$)
$44$ (contains $4$)
$48$ (contains $4$)
$64$ (contains $4$)
$84$ (contains $4$)
Counting these,we get $7$ such numbers.
Therefore,the required number of values is $7$.

(C) To find the rank from the last, we use the formula: $\text{Rank from last} = (\text{Total number of students} - \text{Rank from first}) + 1$.
Given, $\text{Total students} = 31$ and $\text{Rank from first} = 13$.
Substituting the values: $\text{Rank from last} = 31 - 13 + 1 = 18 + 1 = 19$.
Therefore, Abhishek's rank from the last is $19^{th}$.

(D) Initially, Savitha is $10$th from the left and Anita is $9$th from the right.
After interchanging, Savitha's new position is $15$th from the left.
Since Savitha now occupies Anita's original position, we know that the $15$th position from the left is the same as the $9$th position from the right.
To find the total number of girls in the row, we use the formula: $\text{Total} = (\text{Position from left} + \text{Position from right}) - 1$.
Substituting the values: $\text{Total} = (15 + 9) - 1 = 24 - 1 = 23$.
Therefore, there are $23$ girls in the row.

(B) To calculate the total number of days including both the start and end dates:
January: From $26^{th}$ to $31^{st} = 6$ days.
February: Since $1988$ is a leap year $(1988 \div 4 = 497)$,February has $29$ days.
March: $31$ days.
April: $30$ days.
May: $15$ days.
Total days $= 6 + 29 + 31 + 30 + 15 = 111$ days.

(B) According to Ram,Lakshman's birthday is on one of the days among $20^{th}$ and $21^{st}$ November.
According to Anil,Lakshman's birthday is on one of the days among $21^{st}$,$22^{nd}$,and $23^{rd}$ November.
The day common to both statements is $21^{st}$ November.
Therefore,Lakshman's birthday is on $21^{st}$ November.

(D) We need to find the number of $5$s that satisfy two conditions:
$1$. It is not immediately preceded by $3$ (i.e.,the pattern is not $3, 5$).
$2$. It is immediately followed by $7$ (i.e.,the pattern is $5, 7$).
Let's examine the series: $1, 5, 7, 3, 5, 7, 4, 7, 3, 7, 2, 5, 6, 5, 8, 5, 7, 4, 5, 6, 5, 5, 7, 1, 5, 7, 7, 5, 5$
- The first $5$ is followed by $7$ and preceded by $1$. (Matches)
- The second $5$ is followed by $7$ but preceded by $3$. (Does not match)
- The third $5$ is followed by $8$. (Does not match)
- The fourth $5$ is followed by $7$ and preceded by $8$. (Matches)
- The fifth $5$ is followed by $6$. (Does not match)
- The sixth $5$ is followed by $5$. (Does not match)
- The seventh $5$ is followed by $7$ and preceded by $5$. (Matches)
- The eighth $5$ is followed by $7$ and preceded by $1$. (Matches)
- The ninth $5$ is followed by $5$. (Does not match)
- The tenth $5$ is followed by nothing. (Does not match)
Counting the matches,we find there are $4$ such $5$s.

(D) To find the instances where an even number is followed by two odd numbers,we look for the pattern: (Even,Odd,Odd).
Given series: $1, 8, 5, 7, 2, 9, 8, 4, 3, 6, 2, 7, 5, 1, 8, 9, 4, 3, 6, 5, 9$.
Checking the sequence:
$1$. $8, 5, 7$ (Even,Odd,Odd) - Yes.
$2$. $2, 9, 8$ (No,$8$ is even).
$3$. $6, 2, 7$ (No,$2$ is even).
$4$. $2, 7, 5$ (Even,Odd,Odd) - Yes.
$5$. $8, 9, 4$ (No,$4$ is even).
$6$. $6, 5, 9$ (Even,Odd,Odd) - Yes.
There are $3$ such instances.

(C) Let the number be $x$.
According to the first condition,the number is greater than $3$ and less than $8$,which can be written as $3 < x < 8$.
According to the second condition,the number is greater than $6$ and less than $10$,which can be written as $6 < x < 10$.
To find the common range,we combine both inequalities: $6 < x < 8$.
The integer satisfying this condition is $7$.

(C) Given the pattern: $16 \times 85 = 8651$.
Here,the digits of the result are formed by taking the tens digit of the first number $(1)$,the units digit of the second number $(5)$,the units digit of the first number $(6)$,and the tens digit of the second number $(8)$.
Wait,let's re-examine the pattern: $16 \times 85 = 8651$.
Digits of $16$ are $1, 6$. Digits of $85$ are $8, 5$.
The result $8651$ is formed by: (tens of $85$)(units of $16$)(units of $85$)(tens of $16$).
Applying this to $73 \times 42$:
Tens of $42$ is $4$,units of $73$ is $3$,units of $42$ is $2$,tens of $73$ is $7$.
Thus,the result is $4327$.

(B) Total number of boys $= 16$.
When Pramod shifted $2$ places to the left,his new position from the left end is $7^{th}$.
This means his original position from the left end was $7 + 2 = 9^{th}$.
To find his earlier position from the right end,we use the formula: $\text{Position from right} = \text{Total} - \text{Position from left} + 1$.
Substituting the values: $\text{Position from right} = 16 - 9 + 1 = 8^{th}$.

(D) The buses leave every $30 \,minutes$.
Since the next bus is at $9:35 \,am$,the previous bus must have left at $9:35 \,am - 30 \,minutes = 9:05 \,am$.
The clerk informed the passenger that the bus had left $10 \,minutes$ ago.
Therefore,the time of information is $9:05 \,am + 10 \,minutes = 9:15 \,am$.

(D) Anita's position after shifting is $12^{th}$ from the left end.
Since she shifted $4$ places to the right, her original position was $12 - 4 = 8^{th}$ from the left end.
The total number of girls in the row is $21$.
The formula to find the position from the right end is: $\text{Position from right} = (\text{Total number of girls} - \text{Position from left}) + 1$.
Substituting the values: $\text{Position from right} = (21 - 8) + 1 = 13 + 1 = 14^{th}$.

(C) The condition is to find the number $5$ such that it is followed by $3$ (pattern: $5, 3$) but not preceded by $7$ (pattern: $7, 5, 3$ is excluded).
Let's analyze the sequence: $8, 9, 5, 3, 2, 5, 3, 8, 5, 5, 6, 8, 7, 3, 3, 5, 7, 7, 5, 3, 6, 5, 3, 3, 5, 7, 3, 8$
$1$. The first $5$ is followed by $3$ and preceded by $9$. (Valid)
$2$. The second $5$ is followed by $3$ and preceded by $2$. (Valid)
$3$. The third and fourth $5$'s are not followed by $3$.
$4$. The fifth $5$ is followed by $3$ and preceded by $6$. (Valid)
$5$. The sixth $5$ is followed by $3$ and preceded by $3$. (Valid)
Wait,let's re-examine the sequence carefully:
Sequence: $8, 9, (5, 3), 2, (5, 3), 8, 5, 5, 6, 8, 7, 3, 3, 5, 7, 7, (5, 3), 6, (5, 3), 3, 5, 7, 3, 8$
Checking the conditions for each $5$ followed by $3$:
- $9, 5, 3$: Preceded by $9$ (Valid)
- $2, 5, 3$: Preceded by $2$ (Valid)
- $7, 5, 3$: Preceded by $7$ (Invalid)
- $6, 5, 3$: Preceded by $6$ (Valid)
There are $3$ such numbers.

(C) We need to find even numbers that are preceded by an even number and followed by an odd number. The pattern is: (Even) (Target Even) (Odd).
Let's analyze the sequence: $8, 6, 7, 6, 8, 9, 3, 2, 7, 5, 3, 4, 2, 2, 3, 5, 5, 2, 2, 8, 1, 1, 9$.
$1$. $8, 6, 7$: $6$ is preceded by $8$ (even) and followed by $7$ (odd). (Count: $1$)
$2$. $6, 8, 9$: $8$ is preceded by $6$ (even) and followed by $9$ (odd). (Count: $2$)
$3$. $4, 2, 3$: $2$ is preceded by $4$ (even) and followed by $3$ (odd). (Count: $3$)
$4$. $2, 2, 3$: $2$ is preceded by $2$ (even) and followed by $3$ (odd). (Count: $4$)
$5$. $2, 8, 1$: $8$ is preceded by $2$ (even) and followed by $1$ (odd). (Count: $5$)
There are $5$ such numbers. Therefore,the correct option is $C$.

(C) To find the pairs of successive numbers with a difference of $2$,we examine each adjacent pair $(a, b)$ such that $|a - b| = 2$:
$1$. $(6, 4) \rightarrow |6 - 4| = 2$
$2$. $(4, 2) \rightarrow |4 - 2| = 2$
$3$. $(5, 3) \rightarrow |5 - 3| = 2$
$4$. $(8, 6) \rightarrow |8 - 6| = 2$
$5$. $(3, 1) \rightarrow |3 - 1| = 2$
$6$. $(4, 2) \rightarrow |4 - 2| = 2$
Checking the series: $6, 4, 1, 2, 2, 8, 7, 4, 2, 1, 5, 3, 8, 6, 2, 1, 7, 1, 4, 1, 3, 2, 8, 6$
Pairs: $(6,4), (4,2), (5,3), (8,6), (3,1), (4,2)$.
Thus,there are $6$ such pairs.

(D) To find the number of $8$'s that are divisible by both the preceding and succeeding numbers,we examine each $8$ in the series:
$1$. The first $8$ (at index $8$): Preceded by $2$,succeeded by $4$. $8$ is divisible by $2$ $(8/2 = 4)$ and $8$ is divisible by $4$ $(8/4 = 2)$. This is a valid $8$.
$2$. The second $8$ (at index $10$): Preceded by $4$,succeeded by $4$. $8$ is divisible by $4$ $(8/4 = 2)$. This is a valid $8$.
$3$. The third $8$ (at index $14$): Preceded by $2$,succeeded by $2$. $8$ is divisible by $2$ $(8/2 = 4)$. This is a valid $8$.
$4$. The fourth $8$ (at index $18$): Preceded by $9$,succeeded by $4$. $8$ is not divisible by $9$. Not valid.
$5$. The fifth $8$ (at index $22$): Preceded by $4$,succeeded by $3$. $8$ is not divisible by $3$. Not valid.
$6$. The sixth $8$ (at index $25$): Preceded by $2$,succeeded by $4$. $8$ is divisible by $2$ $(8/2 = 4)$ and $8$ is divisible by $4$ $(8/4 = 2)$. This is a valid $8$.
$7$. The seventh $8$ (at index $29$): Preceded by $1$,succeeded by $3$. $8$ is not divisible by $3$. Not valid.
Thus,there are $4$ such $8$'s.

(B) The given sequence is: $1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 4, 6, 8, 9, 7, 5, 3, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1$.
Total number of terms $n = 27$.
The middle term is the $\frac{n+1}{2} = \frac{27+1}{2} = 14^{th}$ term.
Counting from the left,the $14^{th}$ term is $9$.
The sequence is: $1, 2, 3, 4, 5, 6, 7, 8, 9, 2, 4, 6, 8, (9), 7, 5, 3, 1, 9, 8, 7, 6, 5, 4, 3, 2, 1$.
The third number to the left of $9$ is found by moving three positions back: $1^{st}$ is $8$,$2^{nd}$ is $6$,$3^{rd}$ is $4$.
Therefore,the required number is $4$.

(B) We need to find the number of $3$'s that satisfy two conditions:
$1$. It is not preceded by $6$ (i.e.,the digit before $3$ is not $6$).
$2$. It is not followed by $9$ (i.e.,the digit after $3$ is not $9$).
Let's analyze the sequence: $9, 3, 6, 6, 3, 9, 5, 9, 3, 7, 8, 9, 1, 3, 9, 6, 3, 9$
- The first $3$ is preceded by $9$ and followed by $6$. (Condition satisfied: $9, 3, 6$)
- The second $3$ is preceded by $6$ and followed by $9$. (Condition not satisfied)
- The third $3$ is preceded by $9$ and followed by $7$. (Condition satisfied: $9, 3, 7$)
- The fourth $3$ is preceded by $1$ and followed by $9$. (Condition not satisfied)
- The fifth $3$ is preceded by $6$ and followed by $9$. (Condition not satisfied)
Thus,there are $2$ such $3$'s that satisfy the given conditions.

(A) We need to find the number of $7$'s that satisfy two conditions:
$1.$ The digit immediately preceding $7$ is $NOT$ $5$.
$2.$ The digit immediately following $7$ is either $2$ or $3$.
Let's examine the sequence: $5, 7, 2, 6, 5, 7, 3, 8, 3, 7, 3, 2, 5, 7, 2, 7, 3, 4, 8, 2, 6, 7, 8$
- The first $7$ is preceded by $5$. (Does not satisfy condition $1$)
- The second $7$ is preceded by $5$. (Does not satisfy condition $1$)
- The third $7$ is preceded by $3$ and followed by $3$. (Satisfies both conditions)
- The fourth $7$ is preceded by $5$. (Does not satisfy condition $1$)
- The fifth $7$ is preceded by $2$ and followed by $3$. (Satisfies both conditions)
- The sixth $7$ is preceded by $6$ and followed by $8$. (Does not satisfy condition $2$)
Thus,there are $2$ such $7$'s.

(C) We need to find the number of $6$'s that satisfy two conditions:
$1$. Preceded by $7$ (i.e.,the pattern is $7, 6$).
$2$. Not immediately followed by $9$ (i.e.,the pattern is $NOT$ $7, 6, 9$).
Let's examine the series: $6, 7, 9, 5, 6, 9, 7, 6, 8, 7, 6, 7, 8, 6, 9, 4, 6, 7, 7, 6, 9, 5, 7, 6, 3$.
Identify all occurrences of $7, 6$:
- $7, 6, 8$ (Preceded by $7$,followed by $8$ - Valid)
- $7, 6, 7$ (Preceded by $7$,followed by $7$ - Valid)
- $7, 6, 9$ (Preceded by $7$,followed by $9$ - Invalid)
- $7, 6, 3$ (Preceded by $7$,followed by $3$ - Valid)
Counting the valid occurrences:
$1$. $7, 6, 8$ (First instance)
$2$. $7, 6, 7$ (Second instance)
$3$. $7, 6, 3$ (Third instance)
There are $3$ such $6$'s.

(B) We need to find the number of $7$'s that satisfy two conditions:
$1$. Preceded by $8$ (i.e.,the pattern $8, 7$).
$2$. Not followed by $3$ (i.e.,the pattern $8, 7, x$ where $x \neq 3$).
Let's analyze the series: $8, 9, 8, 7, 6, 2, 2, 6, 3, 2, 6, 9, 7, 3, 2, 8, 7, 2, 7, 7, 8, 7, 3, 7, 7, 9, 4$
Identify all $7$'s preceded by $8$:
- The first $7$ is preceded by $8$ $(8, 7, 6)$. Here,$7$ is followed by $6$ (not $3$). This satisfies the condition.
- The second $7$ is preceded by $8$ $(8, 7, 2)$. Here,$7$ is followed by $2$ (not $3$). This satisfies the condition.
- The third $7$ is preceded by $8$ $(8, 7, 3)$. Here,$7$ is followed by $3$. This does not satisfy the condition.
Thus,there are $2$ such $7$'s.

(B) We need to find the number of $1$'s that are followed by $2$,provided that the $2$ is not followed by $3$. The condition is $(1, 2, \text{not } 3)$.
Let's analyze the sequence: $1, 2, 1, 3, 4, 5, 1, 2, 3, 5, 2, 1, 2, 6, 1, 4, 5, 1, 1, 2, 4, 1, 2, 3, 2, 1, 7, 5, 2, 1, 2, 5$.
$1$. The first $1$ is followed by $2$,and $2$ is followed by $1$ (not $3$). This is a valid pair: $(1, 2, 1)$.
$2$. The next $1$ is followed by $3$ (invalid).
$3$. The next $1$ is followed by $2$,and $2$ is followed by $3$ (invalid).
$4$. The next $1$ is followed by $2$,and $2$ is followed by $6$ (not $3$). This is a valid pair: $(1, 2, 6)$.
$5$. The next $1$ is followed by $4$ (invalid).
$6$. The next $1$ is followed by $1$ (invalid).
$7$. The next $1$ is followed by $2$,and $2$ is followed by $4$ (not $3$). This is a valid pair: $(1, 2, 4)$.
$8$. The next $1$ is followed by $2$,and $2$ is followed by $3$ (invalid).
$9$. The next $1$ is followed by $7$ (invalid).
$10$. The next $1$ is followed by $2$,and $2$ is followed by $5$ (not $3$). This is a valid pair: $(1, 2, 5)$.
Counting the valid occurrences,we have $4$ such $1$'s.

(D) We need to find the pattern $X, 6, 7$ where $X \neq 8$.
Let's analyze the series: $8, 7, 6, 7, 8, 6, 7, 5, 6, 7, 9, 7, 6, 1, 6, 7, 7, 6, 8, 8, 6, 9, 7, 6, 8, 7$.
$1$. $8, 7, 6, 7$: Here $6$ is preceded by $7$,not $8$. This fits the criteria $(X=7)$.
$2$. $8, 6, 7$: Here $6$ is preceded by $8$. This does not fit.
$3$. $5, 6, 7$: Here $6$ is preceded by $5$,not $8$. This fits the criteria $(X=5)$.
$4$. $1, 6, 7$: Here $6$ is preceded by $1$,not $8$. This fits the criteria $(X=1)$.
$5$. $8, 6, 9, 7$: No $6, 7$ sequence.
$6$. $6, 7, 7$: Here $6$ is preceded by $7$,not $8$. This fits the criteria $(X=7)$.
$7$. $8, 6, 9, 7, 6, 8, 7$: Here $6$ is preceded by $7$,not $8$. This fits the criteria $(X=7)$.
Counting the valid occurrences,we find $4$ such instances.

(C) We need to find the pattern $(X, 2, 1)$ where $X \neq 4$.
Given sequence: $4, 2, 1, 2, 1, 4, 2, 1, 1, 2, 4, 4, 4, 1, 2, 2, 1, 2, 1, 4, 4, 2, 1, 4, 2, 1, 2, 1, 2, 4, 1, 4, 2, 1, 2, 4, 1, 4, 6$.
Let's identify all occurrences of $2$ followed by $1$:
$1$. $(4, 2, 1)$ - Preceded by $4$ (Exclude)
$2$. $(2, 2, 1)$ - Preceded by $2$ (Include)
$3$. $(1, 2, 1)$ - Preceded by $1$ (Include)
$4$. $(4, 2, 1)$ - Preceded by $4$ (Exclude)
$5$. $(4, 2, 1)$ - Preceded by $4$ (Exclude)
$6$. $(2, 2, 1)$ - Preceded by $2$ (Include)
$7$. $(2, 2, 1)$ - Preceded by $2$ (Include)
$8$. $(4, 2, 1)$ - Preceded by $4$ (Exclude)
Counting the valid occurrences: $2, 3, 6, 7$ correspond to the pattern $(X, 2, 1)$ where $X \neq 4$.
There are $4$ such occurrences.

(C) We need to find the occurrences of the pattern $9, 7, 6$ in the given sequence.
Given sequence: $7, 8, 9, 7, 6, 5, 3, 4, 2, 8, 9, 7, 2, 4, 5, 9, 2, 9, 7, 6, 4, 7$.
Let's scan the sequence for the pattern $9, 7, 6$:
$1$. The first occurrence is at the beginning: $7, 8, \mathbf{9, 7, 6}, 5, ...$
$2$. Scanning further: $..., 8, 9, 7, 2, 4, 5, 9, 2, \mathbf{9, 7, 6}, 4, 7$.
There are $2$ such occurrences in the sequence.
Therefore,the correct option is $C$.

(D) First,count the frequency of each number in the series:
$2$ appears $3$ times $(2, 2, 2)$
$3$ appears $1$ time $(3)$
$4$ appears $3$ times $(4, 4, 4)$
$5$ appears $2$ times $(5, 5)$
$6$ appears $2$ times $(6, 6)$
$7$ appears $5$ times $(7, 7, 7, 7, 7)$
$8$ appears $2$ times $(8, 8)$
$9$ appears $4$ times $(9, 9, 9, 9)$
Comparing the frequencies,the numbers $5, 6,$ and $8$ each appear $2$ times. Therefore,they have equal frequency.

(B) To find the number of $6$s that are preceded by $9$ but not followed by $4$,we examine the sequence: $5, 6, 4, 3, 2, 9, 6, 3, 1, 6, 4, 9, 6, 4, 2, 1, 5, 9, 6, 7, 2, 1, 4, 7, 4, 9, 6, 4, 2$.
$1$. Identify all occurrences of $6$ in the sequence:
- The first $6$ is preceded by $5$ and followed by $4$ $(5, 6, 4)$.
- The second $6$ is preceded by $9$ and followed by $3$ $(9, 6, 3)$. This satisfies the condition (preceded by $9$,not followed by $4$).
- The third $6$ is preceded by $1$ and followed by $4$ $(1, 6, 4)$.
- The fourth $6$ is preceded by $9$ and followed by $4$ $(9, 6, 4)$. This does not satisfy the condition (it is followed by $4$).
- The fifth $6$ is preceded by $9$ and followed by $7$ $(9, 6, 7)$. This satisfies the condition (preceded by $9$,not followed by $4$).
- The sixth $6$ is preceded by $9$ and followed by $4$ $(9, 6, 4)$. This does not satisfy the condition (it is followed by $4$).
Thus,there are $2$ such $6$s that meet the criteria. The correct option is $B$.

(A) We need to find the pattern where $7$ is in the middle and $1$ and $3$ are on either side. This means we are looking for the sequences $(1, 7, 3)$ or $(3, 7, 1)$.
Let's analyze the given series: $2, 9, 7, 3, 1, 7, 3, 7, 7, 1, 3, 3, 1, 7, 3, 8, 5, 7, 1, 3, 7, 7, 1, 7, 3, 9, 0, 6$.
$1$. The first occurrence is $(1, 7, 3)$ at index $4-6$: $2, 9, 7, 3, \mathbf{1, 7, 3}, 7, 7, 1, 3, 3, 1, 7, 3, 8, 5, 7, 1, 3, 7, 7, 1, 7, 3, 9, 0, 6$.
$2$. The second occurrence is $(1, 7, 3)$ at index $12-14$: $2, 9, 7, 3, 1, 7, 3, 7, 7, 1, 3, 3, \mathbf{1, 7, 3}, 8, 5, 7, 1, 3, 7, 7, 1, 7, 3, 9, 0, 6$.
$3$. The third occurrence is $(1, 7, 3)$ at index $18-20$: $2, 9, 7, 3, 1, 7, 3, 7, 7, 1, 3, 3, 1, 7, 3, 8, 5, 7, \mathbf{1, 3}, 7, 7, 1, 7, 3, 9, 0, 6$ (Wait,this is $1, 3$,not $1, 7, 3$).
Let's re-examine carefully:
Series: $2, 9, 7, 3, \mathbf{1, 7, 3}, 7, 7, 1, 3, 3, \mathbf{1, 7, 3}, 8, 5, 7, 1, 3, 7, 7, \mathbf{1, 7, 3}, 9, 0, 6$.
There are exactly $3$ such occurrences.

(C) To find pairs of alternate numbers with a difference of $2$,we examine the sequence by skipping one number at a time.
The sequence is: $6, 4, 1, 2, 2, 8, 7, 4, 2, 1, 6, 3, 8, 6, 2, 1, 7, 1, 4, 1, 3, 2, 8, 6$.
Pairs of alternate numbers $(a_n, a_{n+2})$ are:
$(6, 1), (4, 2), (1, 2), (2, 8), (2, 7), (8, 4), (7, 2), (4, 1), (2, 6), (1, 3), (6, 8), (3, 6), (8, 2), (6, 1), (2, 7), (1, 1), (7, 4), (1, 1), (4, 3), (1, 2), (3, 8), (2, 6)$.
Now,we calculate the absolute difference $|a_n - a_{n+2}|$ for each pair:
$|6-1|=5, |4-2|=2, |1-2|=1, |2-8|=6, |2-7|=5, |8-4|=4, |7-2|=5, |4-1|=3, |2-6|=4, |1-3|=2, |6-8|=2, |3-6|=3, |8-2|=6, |6-1|=5, |2-7|=5, |1-1|=0, |7-4|=3, |1-1|=0, |4-3|=1, |1-2|=1, |3-8|=5, |2-6|=4$.
The pairs with a difference of $2$ are $(4, 2), (1, 3),$ and $(6, 8)$.
Thus,there are $3$ such pairs.

(D) We need to find even numbers that satisfy two conditions:
$1$. Preceded by an even number.
$2$. Followed by an odd number.
Let's analyze the sequence: $8, 6, 7, 6, 8, 9, 3, 2, 7, 5, 3, 4, 2, 2, 3, 5, 5, 2, 2, 8, 1, 1, 9$.
- $6$ (preceded by $8$,followed by $7$): Even,Even,Odd. (Matches)
- $8$ (preceded by $6$,followed by $9$): Even,Even,Odd. (Matches)
- $2$ (preceded by $4$,followed by $3$): Even,Even,Odd. (Matches)
- $2$ (preceded by $2$,followed by $8$): No.
- $8$ (preceded by $2$,followed by $1$): Even,Even,Odd. (Matches)
Checking the sequence again:
$(8, 6, 7)$ -> $6$ is preceded by $8$ (even) and followed by $7$ (odd).
$(6, 8, 9)$ -> $8$ is preceded by $6$ (even) and followed by $9$ (odd).
$(4, 2, 3)$ -> $2$ is preceded by $4$ (even) and followed by $3$ (odd).
$(2, 8, 1)$ -> $8$ is preceded by $2$ (even) and followed by $1$ (odd).
Total count is $4$. Since $4$ is not in the options,the correct answer is $D$.

(D) To find the odd numbers that are immediately followed by another odd number,we examine the sequence pair by pair:
$5, 1$ (Odd,Odd) - Yes
$1, 4$ (Odd,Even) - No
$4, 7$ (Even,Odd) - No
$7, 3$ (Odd,Odd) - Yes
$3, 9$ (Odd,Odd) - Yes
$9, 8$ (Odd,Even) - No
$8, 5$ (Even,Odd) - No
$5, 7$ (Odd,Odd) - Yes
$7, 2$ (Odd,Even) - No
$2, 6$ (Even,Even) - No
$6, 3$ (Even,Odd) - No
$3, 1$ (Odd,Odd) - Yes
$1, 5$ (Odd,Odd) - Yes
$5, 8$ (Odd,Even) - No
$8, 6$ (Even,Even) - No
$6, 3$ (Even,Odd) - No
$3, 8$ (Odd,Even) - No
$8, 5$ (Even,Odd) - No
$5, 2$ (Odd,Even) - No
$2, 2$ (Even,Even) - No
$2, 4$ (Even,Even) - No
$4, 3$ (Even,Odd) - No
$3, 4$ (Odd,Even) - No
$4, 9$ (Even,Odd) - No
$9, 6$ (Odd,Even) - No
The pairs are: $(5, 1), (7, 3), (3, 9), (5, 7), (3, 1), (1, 5)$.
Counting these,we find $6$ such occurrences. Therefore,the correct option is $D$.

(C) We need to find an even number $(E)$ such that the pattern is (Odd,Even,Even) i.e.,$(O, E, E)$.
Let's analyze the sequence: $5, 1, 4, 7, 3, 9, 8, 5, 7, 2, 6, 3, 1, 5, 8, 6, 3, 8, 5, 2, 2, 4, 3, 4, 9, 6$.
$1$. $(7, 2, 6)$: Here $2$ is an even number,preceded by $7$ (odd) and followed by $6$ (even). This is a valid pair.
$2$. $(5, 8, 6)$: Here $8$ is an even number,preceded by $5$ (odd) and followed by $6$ (even). This is a valid pair.
$3$. $(5, 2, 2)$: Here $2$ is an even number,preceded by $5$ (odd) and followed by $2$ (even). This is a valid pair.
$4$. $(3, 4, 9)$: Here $4$ is an even number,but it is followed by $9$ (odd). This is not a valid pair.
Thus,there are $3$ such even numbers.

(D) We need to find odd numbers that satisfy the pattern: (Even number) - (Odd number) - (Even number).
Let us examine the sequence: $5, 1, 4, 7, 3, 9, 8, 5, 7, 2, 6, 3, 1, 5, 8, 6, 3, 8, 5, 2, 2, 4, 3, 4, 9, 6$.
$1$. In $4, 7, 3$,the number $7$ is preceded by $4$ (even) and followed by $3$ (odd). (Does not fit)
$2$. In $8, 5, 7$,the number $5$ is preceded by $8$ (even) and followed by $7$ (odd). (Does not fit)
$3$. In $2, 6, 3$,the number $3$ is preceded by $6$ (even) and followed by $1$ (odd). (Does not fit)
$4$. In $6, 3, 1$,the number $3$ is preceded by $6$ (even) and followed by $1$ (odd). (Does not fit)
$5$. In $8, 6, 3$,the number $3$ is preceded by $6$ (even) and followed by $8$ (even). (Fits: $6, 3, 8$)
$6$. In $6, 3, 8$,the number $3$ is preceded by $6$ (even) and followed by $8$ (even). (Fits: $6, 3, 8$)
$7$. In $8, 5, 2$,the number $5$ is preceded by $8$ (even) and followed by $2$ (even). (Fits: $8, 5, 2$)
$8$. In $4, 3, 4$,the number $3$ is preceded by $4$ (even) and followed by $4$ (even). (Fits: $4, 3, 4$)
There are $4$ such odd numbers: $3$ (in $6, 3, 8$),$3$ (in $6, 3, 8$),$5$ (in $8, 5, 2$),and $3$ (in $4, 3, 4$).

(B) We need to find numbers that satisfy the pattern: (Odd number divisible by $3$ or $5$) $\rightarrow$ (Odd number) $\rightarrow$ (Even number).
Let's analyze the series: $12, 19, 21, 3, 25, 18, 35, 20, 22, 21, 45, 46, 47, 48, 9, 50, 52, 54, 55, 56$.
$1$. Check $21$: It is odd and divisible by $3$. The next number is $3$ (odd),and the next is $25$ (odd). (Does not fit)
$2$. Check $3$: It is odd and divisible by $3$. The next number is $25$ (odd),and the next is $18$ (even). (Fits: $3, 25, 18$)
$3$. Check $45$: It is odd and divisible by $3$ or $5$. The next number is $46$ (even). (Does not fit)
$4$. Check $9$: It is odd and divisible by $3$. The next number is $50$ (even). (Does not fit)
Only one such number $(3)$ satisfies the condition. Therefore,the correct option is $B$.

(B) We need to find even numbers $E$ such that $E$ is divisible by the preceding number $P$ $(E \pmod P = 0)$ and $E$ is $NOT$ divisible by the following number $F$ $(E \pmod F \neq 0)$.
Let's analyze the sequence: $3, 8, 4, 1, 5, 7, 2, 8, 3, 4, 8, 9, 3, 9, 4, 2, 1, 5, 8, 2$
$1$. $8$: Preceded by $3$ (not divisible).
$2$. $4$: Preceded by $8$ (not divisible).
$3$. $2$: Preceded by $7$ (not divisible).
$4$. $8$: Preceded by $2$ ($8/2 = 4$,divisible). Followed by $3$ $(8/3
eq \text{integer})$. Condition met.
$5$. $4$: Preceded by $3$ (not divisible).
$6$. $8$: Preceded by $4$ ($8/4 = 2$,divisible). Followed by $9$ $(8/9
eq \text{integer})$. Condition met.
$7$. $4$: Preceded by $9$ (not divisible).
$8$. $2$: Preceded by $4$ (not divisible).
$9$. $8$: Preceded by $5$ (not divisible).
$10$. $2$: Preceded by $8$ (not divisible).
The numbers satisfying the condition are $8$ (at index $8$) and $8$ (at index $11$).
Thus,there are $2$ such numbers.

(D) Nitin's sequence (counting down from $32$): $32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1$.
Sumit's sequence (counting odd numbers upwards from $1$): $1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31$.
Let $n$ be the step number (starting from $1$).
Nitin's $n$-th number is $N_n = 32 - (n - 1) = 33 - n$.
Sumit's $n$-th number is $S_n = 2n - 1$.
Setting $N_n = S_n$,we get $33 - n = 2n - 1$.
$34 = 3n$,which gives $n = 34/3 = 11.33$.
Since $n$ must be an integer,they will never call out the same number at the same time.

(B) The original sequence is $5, 9, 8, 1, 3, 2, 7, 4, 3, 8$.
By interchanging the digits in pairs (first with second,third with fourth,etc.),the new sequence becomes $9, 5, 1, 8, 2, 3, 4, 7, 8, 3$.
Counting from the left,the sequence is:
1st: $9$
2nd: $5$
3rd: $1$
4th: $8$
5th: $2$
6th: $3$
7th: $4$
Therefore,the seventh digit counting from the left is $4$.

(C) The original sequence is $8, 9, 0, 3, 2, 1, 4, 6, 7, 5$.
There are $10$ digits in total.
The interchange pattern is: $1^{st} leftrightarrow 6^{th}$,$2^{nd} leftrightarrow 7^{th}$,$3^{rd} leftrightarrow 8^{th}$,$4^{th} leftrightarrow 9^{th}$,$5^{th} leftrightarrow 10^{th}$.
Applying these interchanges:
$1^{st}$ $(8)$ swaps with $6^{th}$ $(1)$ $ ightarrow$ $1, 9, 0, 3, 2, 8, 4, 6, 7, 5$
$2^{nd}$ $(9)$ swaps with $7^{th}$ $(4)$ $ ightarrow$ $1, 4, 0, 3, 2, 8, 9, 6, 7, 5$
$3^{rd}$ $(0)$ swaps with $8^{th}$ $(6)$ $ ightarrow$ $1, 4, 6, 3, 2, 8, 9, 0, 7, 5$
$4^{th}$ $(3)$ swaps with $9^{th}$ $(7)$ $ ightarrow$ $1, 4, 6, 7, 2, 8, 9, 0, 3, 5$
$5^{th}$ $(2)$ swaps with $10^{th}$ $(5)$ $ ightarrow$ $1, 4, 6, 7, 5, 8, 9, 0, 3, 2$
The new sequence is $1, 4, 6, 7, 5, 8, 9, 0, 3, 2$.
The seventh number from the right end is the $4^{th}$ number from the left end,which is $7$.

(C) Given that $P = 4$.
The difference between $P$ and $T$ is $5$,which means $|P - T| = 5$.
Since $P = 4$,we have $|4 - T| = 5$. This gives two possibilities: $4 - T = 5$ (which implies $T = -1$,not possible as integers are $1$ to $9$) or $T - 4 = 5$,which implies $T = 9$.
Now,the difference between $N$ and $T$ is $3$,which means $|N - T| = 3$.
Substituting $T = 9$,we get $|N - 9| = 3$.
This gives two possibilities: $N - 9 = 3$ (which implies $N = 12$,not possible as integers are $1$ to $9$) or $9 - N = 3$,which implies $N = 6$.
Therefore,the integer assigned to $N$ is $6$.

(C) Let $C$ denote a car and $S$ denote a scooter.
The sequence follows the pattern: $C, S, C, S, S, C, S, S, S, C, S, S, S, S, C, S, S, S, S, S, C, S, S, S, S, S, S, C, S, S, S, S, S, S, S, C$.
There are $36$ vehicles in total. The second half consists of the $19^{th}$ to $36^{th}$ positions.
Let's list the positions of the cars $(C)$:
$1^{st}$ car is at position $1$.
$2^{nd}$ car is at position $1+1+1 = 3$.
$3^{rd}$ car is at position $3+2+1 = 6$.
$4^{th}$ car is at position $6+3+1 = 10$.
$5^{th}$ car is at position $10+4+1 = 15$.
$6^{th}$ car is at position $15+5+1 = 21$.
$7^{th}$ car is at position $21+6+1 = 28$.
$8^{th}$ car is at position $28+7+1 = 36$.
Total vehicles = $36$. The second half is from position $19$ to $36$.
In the second half,the vehicles are:
Position $19$ $(S)$,$20$ $(S)$,$21$ $(C)$,$22$ $(S)$,$23$ $(S)$,$24$ $(S)$,$25$ $(S)$,$26$ $(S)$,$27$ $(S)$,$28$ $(C)$,$29$ $(S)$,$30$ $(S)$,$31$ $(S)$,$32$ $(S)$,$33$ $(S)$,$34$ $(S)$,$35$ $(S)$,$36$ $(C)$.
Counting the scooters $(S)$ in positions $19$ to $36$: there are $15$ scooters.

(D) The given sequence can be analyzed by grouping the terms as follows:
$4 / 4, 5 / 4, 5, 3 / 4, 5, 3, 1 / 4, 5, 3, 1, 2 / 4, 5 / 4, 5, 3 / 4, 5, 3, ...$
Observing the pattern,the sequence consists of repeating blocks that grow in length: $(4), (4, 5), (4, 5, 3), (4, 5, 3, 1), (4, 5, 3, 1, 2)$.
After the fifth block $(4, 5, 3, 1, 2)$,the pattern restarts from the beginning of the sequence structure. The next block after $(4, 5, 3)$ is $(4, 5, 3, 1)$.
Therefore,the next number in the sequence is $1$,which stands for 'Run'.