Number, Ranking and Time Sequence Test Questions in English

(D) The code $4$ represents 'sit down'.
We need to count the number of times the digit $4$ appears in the given sequence:
Sequence: $1, 2, 3, 4, 2, 3, 1, 4, 4, 3, 2, 2, 1, 2, 4, 3, 1, 4, 4, 1, 2$
Identifying the occurrences of $4$:
$1, 2, 3, \mathbf{4}, 2, 3, 1, \mathbf{4}, \mathbf{4}, 3, 2, 2, 1, 2, \mathbf{4}, 3, 1, \mathbf{4}, \mathbf{4}, 1, 2$
Counting the occurrences:
1st: $4$ (at position $4$)
2nd: $4$ (at position $8$)
3rd: $4$ (at position $9$)
4th: $4$ (at position $15$)
5th: $4$ (at position $18$)
6th: $4$ (at position $19$)
Wait,let us re-examine the sequence provided: $1, 2, 3, 4, 2, 3, 1, 4, 4, 3, 2, 2, 1, 2, 4, 3, 1, 4, 4, 1, 2$.
Counting the $4$s: There is one $4$ at index $4$,two $4$s at indices $8$ and $9$,one $4$ at index $15$,and two $4$s at indices $18$ and $19$.
Total count = $1 + 2 + 1 + 2 = 6$.
However,checking the options provided $(2, 3, 4, 5)$,it appears there might be a typo in the question sequence or the expected answer. Based on the provided sequence,the digit $4$ appears $6$ times. If the sequence was meant to be shorter or if the code $44$ was intended as a single command,the result would differ. Given the standard interpretation of such logic puzzles,we count the occurrences of the digit $4$.

(D) The numbers from $1$ to $45$ that are exactly divisible by $3$ are multiples of $3$.
These numbers in ascending order are: $3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45$.
To find the ninth number from the top,we count the terms in the sequence:
$1^{st}: 3$
$2^{nd}: 6$
$3^{rd}: 9$
$4^{th}: 12$
$5^{th}: 15$
$6^{th}: 18$
$7^{th}: 21$
$8^{th}: 24$
$9^{th}: 27$
Thus,the ninth number from the top is $27$.

(D) The numbers from $5$ to $85$ that are divisible by $5$ are: $5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85$.
Arranging these numbers in descending order gives: $85, 80, 75, 70, 65, 60, 55, 50, 45, 40, 35, 30, 25, 20, 15, 10, 5$.
Counting from the bottom (the smallest number),the $1$st is $5$,the $2$nd is $10$,the $3$rd is $15$,the $4$th is $20$,the $5$th is $25$,the $6$th is $30$,the $7$th is $35$,the $8$th is $40$,the $9$th is $45$,the $10$th is $50$,and the $11$th is $55$.
Therefore,the eleventh number from the bottom is $55$.

(A) First,list all numbers from $1$ to $100$ that are exactly divisible by $4$: $4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100$.
Next,identify which of these numbers 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 find there are $7$ such numbers.

(D) number is divisible by $9$ if it is a multiple of $9$. The multiples of $9$ in the range $9$ to $54$ are $9, 18, 27, 36, 45, 54$.
By definition,any number that is divisible by $9$ must also be divisible by $3$,because $9 = 3 \times 3$.
Therefore,it is impossible for a number to be divisible by $9$ and not be divisible by $3$.
Thus,there are $0$ such numbers.

(B) First,identify all numbers from $11$ to $50$ that are divisible by $7$: $14, 21, 28, 35, 42, 49$.
Next,identify which of these numbers are also divisible by $3$: $21$ $(7 \times 3)$ and $42$ $(14 \times 3)$.
Exclude these numbers from the list.
The remaining numbers are $14, 28, 35, 49$.
Therefore,there are $4$ such numbers.

(C) According to the first condition,the number is greater than $3$ but less than $8.$ The set of such integers is ${4, 5, 6, 7}.$
According to the second condition,the number is greater than $6$ but less than $10.$ The set of such integers is ${7, 8, 9}.$
The required number must satisfy both conditions simultaneously.
The intersection of the two sets ${4, 5, 6, 7} \cap {7, 8, 9}$ is ${7}.$
Therefore,the number is $7.$

(B) To find the total number of students in the class,we use the formula:
Total students = (Rank from the top) + (Rank from the bottom) - $1$
Given:
Rank from the top = $9$
Rank from the bottom = $38$
Total students = $9 + 38 - 1$
Total students = $47 - 1 = 46$
Therefore,there are $46$ students in the class.

(D) Let Monika's initial position from the left end be $x$.
When she is shifted by $4$ places to the right, her new position from the left end becomes $x + 4$.
It is given that her new position is $12^{th}$ from the left end.
So, $x + 4 = 12$, which gives $x = 8$.
Thus, Monika's initial position was $8^{th}$ from the left end.
To find her position from the right end, we use the formula: $\text{Position from right} = (\text{Total number of girls} - \text{Position from left}) + 1$.
$\text{Position from right} = (21 - 8) + 1 = 13 + 1 = 14$.
Therefore, her earlier position from the right end was $14^{th}$.

(C) Initially,Deepak is $7th$ from the left and Madhu is $12th$ from the right.
After interchanging positions,Deepak becomes $22nd$ from the left.
Since Deepak is now at the position where Madhu was earlier (which is $12th$ from the right),we know the position of the same boy from both ends.
Total number of boys = (Position from left + Position from right) - $1$.
Total number of boys = $(22 + 12) - 1 = 34 - 1 = 33$ boys.

(B) Let the position of the tree be $P$.
Since the tree is $5^{th}$ from the left end,there are $4$ trees to its left.
Since the tree is $5^{th}$ from the right end,there are $4$ trees to its right.
Total number of trees in the row = (Number of trees to the left) + (The tree itself) + (Number of trees to the right).
Total number of trees = $4 + 1 + 4 = 9$ trees.
Alternatively,using the formula: Total = (Position from left) + (Position from right) - $1$.
Total = $5 + 5 - 1 = 9$ trees.

(C) Total number of persons in the queue = $50$.
Amrita's position from the front = $10^{th}$.
Mukul's position from behind = $25^{th}$.
Number of persons between Amrita and Mukul = $50 - (10 + 25) = 50 - 35 = 15$.
Mamta is exactly in the middle of these $15$ persons.
Position of Mamta from Amrita = $(15 + 1) / 2 = 8^{th}$ position after Amrita.
Since Amrita is at the $10^{th}$ position from the front,Mamta's position from the front = $10 + 8 = 18^{th}$.

(A) To find the total number of students in the class, we use the formula: $\text{Total} = (\text{Rank from top} + \text{Rank from bottom}) - 1$.
Given, Raman's rank from the top is $16$ and from the bottom is $49$.
Therefore, $\text{Total} = (16 + 49) - 1$.
$\text{Total} = 65 - 1 = 64$.
Thus, there are $64$ students in the class.

(A) To find the total number of students in the class,we use the formula:
Total students = (Rank from the top) + (Rank from the bottom) - $1$
Given:
Rank from the top = $7$
Rank from the bottom = $28$
Total students = $7 + 28 - 1$
Total students = $35 - 1 = 34$
Therefore,there are $34$ students in the class.

(B) Atul's position from the right is $12$th, which means there are $11$ boys to his right.
Atul's position from the left is $4$th, which means there are $3$ boys to his left.
Total number of boys in the line = $(\text{Boys to the right} + \text{Atul} + \text{Boys to the left}) = (11 + 1 + 3) = 15$ boys.
To have $28$ boys in the line, the number of boys to be added = $28 - 15 = 13$.

(D) First,calculate the number of students who passed the examination.
Number of students who passed = (Rank from top + Rank from bottom - $1$)
Number of students who passed = $(16 + 29 - 1) = 44$.
Now,add the number of students who did not participate and those who failed to find the total number of students in the class.
Total students = (Passed students) + (Did not participate) + (Failed students)
Total students = $44 + 6 + 5 = 55$.

(A) The position of $P$ from the left is $14^{th}$.
The position of $Q$ from the right is $7^{th}$.
There are $4$ boys between $P$ and $Q$.
Total number of boys = (Position of $P$ from left) + (Number of boys between $P$ and $Q$) + (Position of $Q$ from right).
Total number of boys = $14 + 4 + 7 = 25$.

(C) The total number of students in the class is $46$.
Aruna's rank from the beginning is $12$.
To find the rank from the last,we use the formula: $\text{Rank from last} = (\text{Total students} - \text{Rank from beginning}) + 1$.
Substituting the values: $\text{Rank from last} = (46 - 12) + 1 = 34 + 1 = 35$.
Therefore,Aruna's rank from the last is $35$.

(A) The formula to find the rank from the bottom when the rank from the top and the total number of students are given is: $\text{Rank from bottom} = (\text{Total students} - \text{Rank from top}) + 1$.
For Manoj:
Rank from top = $7$
Total students = $31$
Rank from bottom = $(31 - 7) + 1 = 24 + 1 = 25^{th}$.
For Sachin:
Rank from top = $11$
Total students = $31$
Rank from bottom = $(31 - 11) + 1 = 20 + 1 = 21^{st}$.
Thus,their respective ranks from the bottom are $25^{th}$ and $21^{st}$.

(C) Total number of students in the class $= 39$.
Sumit's rank from the last $= 17^{th}$.
Ravi is $7$ ranks ahead of Sumit, so Ravi's rank from the last $= 17 + 7 = 24^{th}$.
To find Ravi's rank from the start, we use the formula: $\text{Rank from start} = (\text{Total students} - \text{Rank from last}) + 1$.
$\text{Ravi's rank from start} = (39 - 24) + 1 = 15 + 1 = 16^{th}$.

(C) Let the number of boys be $x$. Then, the number of girls $= 2x$.
Total students $= x + 2x = 60$, which implies $3x = 60$, so $x = 20$.
Thus, the number of boys is $20$ and the number of girls is $40$.
Kamal is ranked $17^{th}$ from the top, meaning there are $16$ students ahead of him.
Since there are $9$ girls ahead of him, the number of boys ahead of him is $16 - 9 = 7$.
Total boys $= 20$. Boys behind Kamal $= (\text{Total boys}) - (\text{Boys ahead of him}) - (\text{Kamal himself, if he is a boy})$.
Assuming Kamal is a boy: Boys behind him $= 20 - 7 - 1 = 12$.

(B) Total number of boys in the row = $10$.
After shifting $2$ places to the left, Rohit's new position is $7th$ from the left end.
To find his position from the right end, we use the formula: $\text{Position from right} = (\text{Total} - \text{Position from left}) + 1$.
$\text{Position from right} = (10 - 7) + 1 = 3 + 1 = 4th$ from the right.
Since he was shifted $2$ places to the left to reach this position, his earlier position was $2$ places to the right of the current position.
$\text{Earlier position from right} = 4 - 2 = 2nd$ from the right end.

(A) Total number of persons in the queue $= 48$.
Vijay's position from the front $= 14^{th}$.
Jack's position from the end $= 17^{th}$.
Number of persons between Vijay and Jack $= 48 - (14 + 17) = 48 - 31 = 17$.
Since Mary is exactly in between Vijay and Jack,she occupies the middle position among these $17$ persons.
The number of persons between Vijay and Mary is given by $\frac{17 - 1}{2} = \frac{16}{2} = 8$.

(B) Initially,Rita is $9^{th}$ from the right and Monika is $10^{th}$ from the left.
After interchanging,Rita is $17^{th}$ from the right and Monika is $18^{th}$ from the left.
Since Rita takes Monika's original position,her new position from the right $(17^{th})$ corresponds to Monika's original position from the left $(10^{th})$.
Total number of girls = (Position from left + Position from right - $1$) = $(10 + 17 - 1) = 26$.
Alternatively,using Monika's new position: Total = $(18 + 9 - 1) = 26$.

(D) Initially, Shilpa is $8^{th}$ from the left and Reena is $17^{th}$ from the right.
After interchanging positions, Shilpa becomes $14^{th}$ from the left.
Since Shilpa now occupies the position previously held by Reena, her new position is $14^{th}$ from the left and $17^{th}$ from the right.
The total number of girls in the row is given by the formula: $(\text{Position from left} + \text{Position from right} - 1)$.
Total girls = $(14 + 17 - 1) = 30$.

(C) Let the initial positions be: Kashish is $5^{th}$ from the left and Mona is $6^{th}$ from the right.
After interchanging,Kashish becomes $13^{th}$ from the left.
Since Kashish is now at the position where Mona was initially,the total number of children in the queue is given by: $(\text{Kashish's new position from left} + \text{Mona's original position from right} - 1) = (13 + 6 - 1) = 18$.
Now,we need to find Mona's new position from the right. Mona is now at the position where Kashish was initially ($5^{th}$ from the left).
Using the formula: $\text{Position from right} = (\text{Total children} - \text{Position from left} + 1)$.
Mona's new position from the right = $(18 - 5 + 1) = 14^{th}$.

(B) Let the total number of boys be $N$.
Initially,Kapil is $8^{th}$ from the right and Nikunj is $12^{th}$ from the left.
After interchanging positions,Nikunj is at the position previously held by Kapil,which is $21^{st}$ from the left.
Since this position is $8^{th}$ from the right,the total number of boys $N = (\text{Position from left} + \text{Position from right} - 1) = (21 + 8 - 1) = 28$.
Now,Kapil is at the position previously held by Nikunj,which is $12^{th}$ from the left.
Kapil's position from the right = $(N - \text{Position from left} + 1) = (28 - 12 + 1) = 17^{th}$.

(C) To find the minimum number of persons, we consider the arrangement where the positions overlap as much as possible.
Given: $3$ persons ahead of $C$, $8$ persons between $B$ and $C$, $5$ persons between $A$ and $B$, and $21$ persons behind $A$.
Case $I$: Order $C, B, A$. Total $= 3 (\text{ahead of } C) + 1 (C) + 8 (\text{between } C, B) + 1 (B) + 5 (\text{between } B, A) + 1 (A) + 21 (\text{behind } A) = 40$.
Case $II$: Order $C, A, B$. Since there are $8$ persons between $B$ and $C$ and $5$ between $A$ and $B$, the number of persons between $A$ and $C$ is $8 - 5 - 1 = 2$. Total $= 3 (\text{ahead of } C) + 1 (C) + 2 (\text{between } C, A) + 1 (A) + 21 (\text{behind } A) = 28$.
Comparing the two cases, the minimum number of persons is $28$.

(B) According to Satish,the brother's birthday is on one of the days between $15^{th}$ and $18^{th}$ February,which are $16^{th}$ and $17^{th}$ February.
According to Kajal,the brother's birthday is on one of the days between $16^{th}$ and $19^{th}$ February,which are $17^{th}$ and $18^{th}$ February.
To find the common date,we look for the day present in both sets: {$16^{th}, 17^{th}$} and {$17^{th}, 18^{th}$}.
The common day is $17^{th}$ February.
Therefore,the brother's birthday is on $17^{th}$ February.

(D) The buses leave every $30$ minutes.
The next bus is scheduled for $9:35 \,a.m.$
Therefore,the previous bus must have departed at $9:35 \,a.m. - 30 \,minutes = 9:05 \,a.m.$
The clerk stated that the bus left $10$ minutes ago.
Thus,the time when the information was given is $9:05 \,a.m. + 10 \,minutes = 9:15 \,a.m.$

(A) Given that the $7^{th}$ day of the month is $3$ days earlier than Friday.
Counting $3$ days back from Friday: Thursday ($1^{st}$ day back),Wednesday ($2^{nd}$ day back),Tuesday ($3^{rd}$ day back).
So,the $7^{th}$ day is a Tuesday.
Since the $7^{th}$ day is a Tuesday,the $14^{th}$ day $(7+7)$ is also a Tuesday.
Counting forward from the $14^{th}$ day to the $19^{th}$ day: $15^{th}$ is Wednesday,$16^{th}$ is Thursday,$17^{th}$ is Friday,$18^{th}$ is Saturday,and $19^{th}$ is Sunday.
Therefore,the $19^{th}$ day of the month is Sunday.

(D) The number of days from $17^{th}$ December $1982$ to $17^{th}$ December $1983$ is $365$ days. Since $365 = 52 \times 7 + 1$,there is $1$ odd day. Thus,$17^{th}$ December $1983$ is Saturday $+ 1 = \text{Sunday}$.
From $17^{th}$ December $1983$ to $17^{th}$ December $1984$,the period includes $1984$ which is a leap year ($366$ days). Since $366 = 52 \times 7 + 2$,there are $2$ odd days. Thus,$17^{th}$ December $1984$ is Sunday $+ 2 = \text{Tuesday}$.
Now,we need to find the day for $22^{nd}$ December $1984$. The difference between $17^{th}$ December and $22^{nd}$ December is $22 - 17 = 5$ days.
Adding $5$ days to Tuesday: $\text{Tuesday} + 5 = \text{Sunday}$.

(B) According to Kailash,Deepak's birthday is between $20^{th}$ May and $28^{th}$ May,which means the possible dates are $21^{st}, 22^{nd}, 23^{rd}, 24^{th}, 25^{th}, 26^{th},$ and $27^{th}$ May.
According to Geeta,Deepak's birthday is between $12^{th}$ May and $22^{nd}$ May,which means the possible dates are $13^{th}, 14^{th}, 15^{th}, 16^{th}, 17^{th}, 18^{th}, 19^{th}, 20^{th},$ and $21^{st}$ May.
The common date in both sets is $21^{st}$ May.
Therefore,Deepak's birthday falls on $21^{st}$ May.

(D) According to Sangeeta,the father's birthday is between $8^{th}$ and $13^{th}$ December,which means the possible dates are $9^{th}, 10^{th}, 11^{th}, 12^{th}$ December.
According to Natasha,the father's birthday is between $9^{th}$ and $14^{th}$ December,which means the possible dates are $10^{th}, 11^{th}, 12^{th}, 13^{th}$ December.
The common dates in both sets are $10^{th}, 11^{th}, 12^{th}$ December.
Since there is no unique date identified,the data is inadequate to determine the exact birthday.

(C) According to Amit,the distance $d$ is $10 < d < 15$.
According to Sunita,the distance $d$ is $12 < d < 14$.
To satisfy both conditions,the distance must be in the intersection of these two intervals.
The intersection of $(10, 15)$ and $(12, 14)$ is $(12, 14)$.
Among the given options,only $13 \ km$ lies within the range $(12, 14)$.

(B) Ashish leaves his house at $6:40 \text{ a.m.}$ (since $20 \text{ minutes}$ to $7$ is $6:40$).
He reaches Kunal's house in $25 \text{ minutes}$,which is $6:40 + 25 \text{ minutes} = 7:05 \text{ a.m.}$
They finish their breakfast in another $15 \text{ minutes}$,so they leave Kunal's house at $7:05 + 15 \text{ minutes} = 7:20 \text{ a.m.}$

(B) Ajay reached the bus stop at $8:40 \,a.m.$
Since it takes $10 \,minutes$ to reach the stop,he left home at $8:40 \,a.m. - 10 \,minutes = 8:30 \,a.m.$
He left home $15 \,minutes$ earlier than his usual time.
Therefore,his usual time to leave home is $8:30 \,a.m. + 15 \,minutes = 8:45 \,a.m.$

(B) Anuj reached the meeting place $15 \,minutes$ before $08:30 \,hours$,which is $08:15 \,hours$.
Anuj arrived $30 \,minutes$ (half an hour) earlier than the man who was $40 \,minutes$ late.
This means the man who was late arrived at $08:15 \,hours + 30 \,minutes = 08:45 \,hours$.
Since this man was $40 \,minutes$ late,the scheduled time of the meeting was $08:45 \,hours - 40 \,minutes = 08:05 \,hours$.

(B) The next bell is scheduled for $7:45 \, \text{a.m.}$
Since the bells ring at regular intervals of $45 \, \text{minutes}$, the previous bell rang at $7:45 \, \text{a.m.} - 45 \, \text{minutes} = 7:00 \, \text{a.m.}$
The priest stated that the last bell was rung $5 \, \text{minutes}$ ago.
Therefore, the time when the priest gave this information is $7:00 \, \text{a.m.} + 5 \, \text{minutes} = 7:05 \, \text{a.m.}$

(D) The interval between two consecutive trains is $2 \,hours \,30 \,minutes$ $(2.5 \,hours)$.
The next train is scheduled to depart at $18:00 \,hrs$.
Therefore,the previous train must have departed at $18:00 - 2:30 = 15:30 \,hrs$.
The announcement states that the train left $40 \,minutes$ ago.
Thus,the time of the announcement is $15:30 + 40 \,minutes = 16:10 \,hrs$.

(C) The desk officer disposed of the application on Friday.
Since the desk officer received it on the same day the senior clerk put it up,the senior clerk must have put it up on Friday.
The senior clerk was on leave the day before,which was Thursday.
The inward clerk forwarded the application to the senior clerk on the day the senior clerk was on leave,which was Thursday.
Therefore,the inward clerk must have received the application on the day before that,which is Wednesday.

(B) To determine when the computers are used most,we identify the time interval where the maximum number of employees are present simultaneously.
Group $1$ ($5$ people): $8:00 \,A.M. - 2:00 \,P.M.$
Group $2$ ($10$ people): $10:00 \,A.M. - 4:00 \,P.M.$
Group $3$ ($5$ people): $12 \,noon - 6:00 \,P.M.$
Between $12 \,noon$ and $2:00 \,P.M.$,all three groups are present in the office ($5 + 10 + 5 = 20$ people). Therefore,the demand for the $3$ computers is highest during this interval.

(C) The monkey climbs $30$ feet in the first part of the hour and slips $20$ feet,resulting in a net gain of $10$ feet per hour.
To reach $120$ feet,we must consider the final jump. In the last hour,the monkey will climb $30$ feet and reach the target without slipping back.
Therefore,the monkey needs to reach $120 - 30 = 90$ feet first.
At a rate of $10$ feet per hour,it takes $9$ hours to reach $90$ feet.
Starting at $8:00 \, a.m.$,$9$ hours later is $5:00 \, p.m.$
In the next hour (from $5:00 \, p.m.$ to $6:00 \, p.m.$),the monkey climbs $30$ feet,reaching $90 + 30 = 120$ feet.
Thus,the monkey touches the flag at $6:00 \, p.m.$

(A) Let us analyze the availability of each brother:
$1.$ Kamal: Tuesday,Thursday,Sunday ($12$ $p.m.$ to $4$ $p.m.$)
$2.$ Navin: Monday,Thursday,Friday,Sunday ($10$ $a.m.$ to $2$ $p.m.$)
$3.$ Rajiv: Monday,Wednesday,Thursday ($9$ $a.m.$ to $12$ $p.m.$) and Friday,Saturday,Sunday ($2$ $p.m.$ to $4$ $p.m.$)
Now,let us check the common days:
- On Thursday: Kamal is available ($12$ $p.m.$ to $4$ $p.m.$),Navin is available ($10$ $a.m.$ to $2$ $p.m.$),and Rajiv is available ($9$ $a.m.$ to $12$ $p.m.$). There is no common time slot where all three overlap.
- On Sunday: Kamal is available ($12$ $p.m.$ to $4$ $p.m.$),Navin is available ($10$ $a.m.$ to $2$ $p.m.$),and Rajiv is available ($2$ $p.m.$ to $4$ $p.m.$). Again,there is no common time slot where all three overlap.
Therefore,there is no day when all three brothers are available at the same time.

(D) To determine the number of days where only one brother is available,we analyze the schedule day by day:
$1$. Monday: Navin $(10 \,a.m. - 2 \,p.m.)$ and Rajiv $(9 \,a.m. - 12 \,noon)$. Both are available between $10 \,a.m. - 12 \,noon$. Only one is available at other times.
$2$. Tuesday: Only Kamal $(12 \,noon - 4 \,p.m.)$.
$3$. Wednesday: Only Rajiv $(9 \,a.m. - 12 \,noon)$.
$4$. Thursday: Kamal $(12 \,noon - 4 \,p.m.)$,Navin $(10 \,a.m. - 2 \,p.m.)$,and Rajiv $(9 \,a.m. - 12 \,noon)$. Overlap exists.
$5$. Friday: Navin $(10 \,a.m. - 2 \,p.m.)$ and Rajiv $(2 \,p.m. - 4 \,p.m.)$. No overlap.
$6$. Saturday: Only Rajiv $(2 \,p.m. - 4 \,p.m.)$.
$7$. Sunday: Kamal $(12 \,noon - 4 \,p.m.)$,Navin $(10 \,a.m. - 2 \,p.m.)$,and Rajiv $(2 \,p.m. - 4 \,p.m.)$. Overlap exists.
Analyzing the availability,we find that on all $7$ days of the week,there are specific time slots where only one brother is present at home.

(D) To find when the youngest brother (Navin) and the eldest brother (Rajiv) are available at the same time,we compare their schedules:
Navin's availability: Monday,Thursday,Friday,Sunday $(10 \,a.m. - 2 \,p.m.)$
Rajiv's availability: Monday,Wednesday,Thursday $(9 \,a.m. - 12 \,noon)$ and Friday,Saturday,Sunday $(2 \,p.m. - 4 \,p.m.)$
Comparing the two:
$1$. On Monday: Navin is available $(10 \,a.m. - 2 \,p.m.)$ and Rajiv is available $(9 \,a.m. - 12 \,noon)$. They overlap between $10 \,a.m.$ and $12 \,noon$.
$2$. On Thursday: Navin is available $(10 \,a.m. - 2 \,p.m.)$ and Rajiv is available $(9 \,a.m. - 12 \,noon)$. They overlap between $10 \,a.m.$ and $12 \,noon$.
$3$. On Friday and Sunday: Navin is available $(10 \,a.m. - 2 \,p.m.)$ but Rajiv is only available from $2 \,p.m. - 4 \,p.m.$,so there is no overlap.
Therefore,they are available at the same time on both Monday and Thursday.

(C) If the day before yesterday was Thursday,then yesterday was Friday and today is Saturday.
Since today is Saturday,tomorrow will be Sunday.

(C) If the day before yesterday was Saturday,then yesterday was Sunday,and today is Monday.
Since today is Monday,tomorrow will be Tuesday.
Therefore,the day after tomorrow will be Wednesday.

(B) Mohini went to the movies $9$ days ago,and she only goes on a Thursday.
Therefore,$9$ days ago was a Thursday.
To find today's day,we calculate $9 \pmod 7 = 2$.
This means $9$ days is equivalent to $1$ week and $2$ days.
Counting $2$ days forward from Thursday (Friday,Saturday),we find that today is Saturday.

(C) Given that the $3^{rd}$ day of the month is Monday.
Since the days repeat every $7$ days,the $3^{rd} + 7 = 10^{th}$ day is also a Monday.
Similarly,the $10^{th} + 7 = 17^{th}$ day is also a Monday.
If the $17^{th}$ is a Monday,then the $18^{th}$ is Tuesday,the $19^{th}$ is Wednesday,the $20^{th}$ is Thursday,and the $21^{st}$ is Friday.
The fifth day from the $21^{st}$ is calculated as $21 + 5 = 26^{th}$ day of the month.
Since the $21^{st}$ is a Friday,we count $5$ days forward: $22^{nd}$ (Saturday),$23^{rd}$ (Sunday),$24^{th}$ (Monday),$25^{th}$ (Tuesday),and $26^{th}$ (Wednesday).
Therefore,the fifth day from the $21^{st}$ is Wednesday.