Number, Ranking and Time Sequence Test Questions in English

(B) Given that $1.12.91$ is the first Sunday.
Since $1.12.91$ is a Sunday,$2.12.91$ is a Monday,and $3.12.91$ is the first Tuesday of December $91$.
To find the subsequent Tuesdays,we add $7$ days to the date of the first Tuesday:
First Tuesday: $3.12.91$
Second Tuesday: $3 + 7 = 10.12.91$
Third Tuesday: $10 + 7 = 17.12.91$
Fourth Tuesday: $17 + 7 = 24.12.91$
Therefore,the fourth Tuesday of December $91$ is $24.12.91$.

(C) Let $T$ be today. The day before yesterday is $T-2$. The day after the day before yesterday is $(T-2)+1 = T-1$ (yesterday).
Given that five days ago, yesterday was Thursday, so $T-5-1 = \text{Thursday}$. Thus, $T-6 = \text{Thursday}$, which means $T = \text{Wednesday}$.
Now, we need to find how many days ago $(x)$ Sunday was three days before the day after tomorrow.
The day after tomorrow is $T+2$. Three days before that is $(T+2)-3 = T-1$ (yesterday).
We want to know when yesterday was Sunday. Since today is Wednesday, yesterday was Tuesday.
To have yesterday be Sunday, we need to go back from Wednesday to Sunday, which is $4$ days ago.

(C) Given that $25^{th}$ August is a Thursday.
Subtracting $3$ days from Thursday gives Monday $(25 - 3 = 22)$. So,$22^{nd}$ August is a Monday.
Since there are $7$ days in a week,the Mondays in the month occur on dates: $22 - 7 = 15^{th}$,$15 - 7 = 8^{th}$,$8 - 7 = 1^{st}$,and $22 + 7 = 29^{th}$.
Thus,the Mondays in August fall on the $1^{st}, 8^{th}, 15^{th}, 22^{nd},$ and $29^{th}$.
Therefore,there are $5$ Mondays in that month.

(C) October has $31$ days in total.
If $1^{st}$ October is a Sunday,then the dates $1^{st}, 8^{th}, 15^{th}, 22^{nd},$ and $29^{th}$ October are also Sundays.
Since $29^{th}$ October is a Sunday,$30^{th}$ October is a Monday and $31^{st}$ October is a Tuesday.
Therefore,the next day,$1^{st}$ November,will be a Wednesday.

(B) Given that $3^{rd}$ December $1990$ is a Sunday.
Since December has $31$ days,the dates that are Sundays in December $1990$ are: $3, 10, 17, 24,$ and $31$.
Since $31^{st}$ December $1990$ is a Sunday,the next day,$1^{st}$ January $1991$,is a Monday.
Following this,$2^{nd}$ January $1991$ is a Tuesday,and $3^{rd}$ January $1991$ is a Wednesday.

(C) The year $1996$ is a leap year because it is divisible by $4$. Therefore,February has $29$ days.
Given that February $1$ is a Wednesday,we can find the dates of all Wednesdays in February by adding multiples of $7$: $1, 8, 15, 22, 29$.
Since February $29$ is a Wednesday,the next day,March $1$,is a Thursday.
Following this,March $2$ is a Friday,and March $3$ is a Saturday.

(B) common year consists of $365$ days.
Since $365 = 52 \times 7 + 1$,a common year has $52$ weeks and $1$ extra day.
Therefore,the last day of a common year is the same as the first day of that year.
Since the first day was Friday,the last day of the year must also be Friday.

(C) To find the day on $18^{th}$ February,$1999$,we calculate the number of odd days between the two dates.
From $18^{th}$ February,$1997$ to $18^{th}$ February,$1998$ is $1$ year. Since $1998$ is not a leap year,it contains $365$ days,which equals $52$ weeks and $1$ odd day.
From $18^{th}$ February,$1998$ to $18^{th}$ February,$1999$ is another $1$ year. Since $1999$ is not a leap year,it also contains $1$ odd day.
Total odd days = $1 + 1 = 2$ days.
Since $18^{th}$ February,$1997$ was a Tuesday,adding $2$ days to Tuesday gives: Wednesday,Thursday.
Therefore,$18^{th}$ February,$1999$ will be a Thursday.

(B) To calculate the total number of days from $26^{th}$ January $1996$ to $15^{th}$ May $1996$ (inclusive),we count the days in each month:
January: $31 - 26 + 1 = 6$ days.
February: Since $1996$ is a leap year (divisible by $4$),February has $29$ days.
March: $31$ days.
April: $30$ days.
May: $15$ days.
Total number of days = $6 + 29 + 31 + 30 + 15 = 111$ days.

(C) Two months have the same calendar if the total number of days between the start of the first month and the start of the second month is exactly divisible by $7$.
$(a)$ June to October: $30 (June) + 31 (July) + 31 (Aug) + 30 (Sep) = 122$ days. $122 / 7$ leaves a remainder of $3$.
$(b)$ April to November: $30 (Apr) + 31 (May) + 30 (Jun) + 31 (Jul) + 31 (Aug) + 30 (Sep) + 31 (Oct) = 214$ days. $214 / 7$ leaves a remainder of $4$.
$(c)$ April to July: $30 (Apr) + 31 (May) + 30 (Jun) = 91$ days. Since $91 / 7 = 13$,the number of days is divisible by $7$. Thus,April and July have the same calendar.
$(d)$ October to December: $31 (Oct) + 30 (Nov) = 61$ days. $61 / 7$ leaves a remainder of $5$.

(D) Suman is $3^{rd}$ among the girls,which means there are $2$ girls ranked above her.
Since Suman is one rank below Amit,Amit must be ranked immediately above Suman.
If Amit is $5^{th}$ among the boys,there are $4$ boys ranked above him.
Total students ranked above Amit = (Number of girls ranked above Suman) + (Number of boys ranked above Amit) = $2 + 4 = 6$.
Since $6$ students are ranked above Amit,Amit's rank in the class is $6 + 1 = 7^{th}$.

(B) Let the six members be $P, Q, R, G, S, M$ sitting around a circular table facing the centre.
From condition $(i)$,$R$ is between $G$ and $P$. This implies the sequence is $G-R-P$ or $P-R-G$.
From condition $(ii)$,$M$ is between $P$ and $S$. This implies the sequence is $P-M-S$ or $S-M-P$.
Combining these,we get the arrangement $G-R-P-M-S$. Since there are $6$ members,the remaining member $Q$ must occupy the space between $S$ and $G$.
Thus,the circular order is $G-R-P-M-S-Q$ (or vice versa).
Therefore,$Q$ is positioned between $G$ and $S$.

(D) From the given information:
$1$. Mohan is older than Prabir: $Mohan > Prabir$
$2$. Suresh is younger than Prabir: $Prabir > Suresh$
$3$. Mihir is older than Suresh but younger than Prabir: $Prabir > Mihir > Suresh$
Combining these,we get the order: $Mohan > Prabir > Mihir > Suresh$
Therefore,Suresh is the youngest among the four.

(B) Based on the given information,we can determine the order of the boys in the race:
$1$. Prabir finished before Mohit but behind Mihir: $Mihir > Prabir > Mohit$.
$2$. Suresh finished before Sanchit but behind Mohit: $Mohit > Suresh > Sanchit$.
Combining these two sequences,we get the final order: $Mihir > Prabir > Mohit > Suresh > Sanchit$.
Therefore,Mihir finished first and won the race.

(A) From the given information,we can derive the following inequalities:
$1$. $M > R$
$2$. $Q < R$ and $Q < N$ (which implies $R > Q$ and $N > Q$)
$3$. $N < M$ (since $N$ is not as old as $M$)
Combining these,we get the order: $M > N > R > Q$ or $M > R > N > Q$ (depending on the relative age of $N$ and $R$).
In both scenarios,$M$ is greater than $N, R$,and $Q$.
Therefore,$M$ is the oldest.

(B) Total number of students = $40$.
Samir's rank from the top = $12^{th}$.
Alok is $8$ ranks below Samir,so Alok's rank from the top = $12 + 8 = 20^{th}$.
To find the rank from the bottom,we use the formula: $\text{Rank from bottom} = (\text{Total students} - \text{Rank from top}) + 1$.
Substituting the values: $\text{Rank from bottom} = (40 - 20) + 1 = 20 + 1 = 21^{st}$.
Therefore,Alok's rank from the bottom is $21^{st}$.

(D) Let us represent the heights based on the given information:
$1$. Geeta is taller than Seeta: $Geeta > Seeta$.
$2$. Geeta is not shorter than Radha: $Geeta \geq Radha$.
$3$. Radha and Rani are of the same height: $Radha = Rani$.
$4$. Geeta is shorter than Paru: $Paru > Geeta$.
Combining these,we get the sequence: $Paru > Geeta \geq Radha = Rani > Seeta$.
Therefore,Paru is the tallest among all the girls.

(C) From the given information:
$1$. Mahesh is taller than Karan but not Yash: $Yash > Mahesh > Karan$.
$2$. Hrithik is taller than Yash but not Abhishek: $Abhishek > Hrithik > Yash$.
Combining these two sequences,we get the overall height order:
$Abhishek > Hrithik > Yash > Mahesh > Karan$.
Since they are standing in increasing order of their heights,the shortest person will be first in line.
The order from shortest to tallest is: $Karan < Mahesh < Yash < Hrithik < Abhishek$.
Therefore,Karan will be first in line.

(D) Akshay's position from the left = $16^{th}$.
Avinash is $11^{th}$ to the right of Akshay,so Avinash's position from the left = $16 + 11 = 27^{th}$.
Avinash is $3^{rd}$ to the right of Vijay,which means Vijay is $3^{rd}$ to the left of Avinash.
Vijay's position from the left = $27 - 3 = 24^{th}$.
We are given that Vijay is $18^{th}$ from the right end.
Total number of boys = (Vijay's position from left) + (Vijay's position from right) - $1$.
Total number of boys = $24 + 18 - 1 = 41$.

(A) Given: $M$ is older than only $P$. This implies that $M$ is the second youngest,and $P$ is the youngest. The order is $M > P$.
We are also given that $T$ is older than $R$.
Combining these,we have $T > R > M > P$.
Therefore,$T$ is the oldest among them.

(C) $1$. Ritesh's position from the left end = $12^{th}$.
$2$. Sudhir is $4^{th}$ to the right of Ritesh. Therefore,Sudhir's position from the left end = $12 + 4 = 16^{th}$.
$3$. Sudhir's position from the right end is given as $22^{nd}$.
$4$. The total number of children in the row is calculated using the formula: $\text{Total} = (\text{Position from left} + \text{Position from right}) - 1$.
$5$. $\text{Total} = (16 + 22) - 1 = 38 - 1 = 37$.

(A) Given conditions:
$1$. $A > B$ and $A < C$,so $C > A > B$.
$2$. $B$ is taller than only $E$,which means $B > E$ and $B$ is at the fourth position from the top.
$3$. Combining these: $C > A > B > E$.
$4$. Since $C$ is not the tallest,$D$ must be the tallest.
$5$. The final order is $D > C > A > B > E$.
Therefore,$A$ is in the middle.

(D) Total number of children in the row = $40$.
Madhu's position from the left = $18^{th}$.
Sandhu's position from the right = $11^{th}$.
To find the number of children between them, we use the formula: $\text{Number of children between} = \text{Total} - (\text{Position from left} + \text{Position from right})$.
$\text{Number of children between} = 40 - (18 + 11) = 40 - 29 = 11$.
Wait, the formula for non-overlapping cases is $\text{Total} - (\text{Left} + \text{Right})$.
$40 - (18 + 11) = 40 - 29 = 11$.
Therefore, there are $11$ children between them.

(D) Jaya's rank among the girls is $4$,which means there are $3$ girls ahead of her.
Jaya's rank in the total class is $18$,meaning there are $17$ students ahead of her in total.
Since there are $3$ girls ahead of her,the number of boys ahead of her is $17 - 3 = 14$.
Therefore,Jaya's rank among the boys is $14 + 1 = 15^{th}$.

(A) $1$. From the statement '$S$ is older than $R$ but not as old as $T$',we get the order: $T > S > R$.
$2$. From the statement '$Q$ is older than only $P$',it implies that $Q$ is older than $P$ but younger than everyone else. This gives the order: $Q > P$.
$3$. Combining these,we get the final sequence: $T > S > R > Q > P$.
$4$. Therefore,$P$ is the youngest among them.

(C) To find the total number of students in the class,we use the formula:
Total students = (Position from top) + (Position from bottom) - $1$
Given:
Position from top = $16$
Position from bottom = $12$
Total students = $16 + 12 - 1 = 28 - 1 = 27$
Therefore,there are $27$ students in the class.

(D) Based on the given information:
$1$. $Geeta > Shilpa$
$2$. $Deepa > Geeta$
Combining these,we get: $Deepa > Geeta > Shilpa$.
$3$. $Reepa > Gayatri$
$4$. $Fatima$ is the seniormost person.
Since we have two separate chains ($Deepa > Geeta > Shilpa$ and $Reepa > Gayatri$) and no information linking the two groups,we cannot determine whether $Shilpa$ or $Gayatri$ is the most junior. Therefore,the data is inadequate.

(B) Let the four friends be $S$ (Sachin),$M$ (Meena),$B$ (Bharti),and $P$ (Parveen).
$1$. Sachin is sitting to the immediate left of Meena: The arrangement is $(S, M)$ in clockwise order.
$2$. Sachin is not next to Bharti: This means Bharti cannot be to the left of Sachin.
$3$. Parveen is sitting to the right of Bharti: This implies the sequence $(B, P)$ in clockwise order.
$4$. Combining these,if we place $M$ at the top,$S$ is to her immediate left (clockwise). Since $B$ cannot be next to $S$,$B$ must be to the right of $M$. Then $P$ must be to the right of $B$.
$5$. The circular arrangement in clockwise order is: $M \rightarrow S \rightarrow P \rightarrow B$.
$6$. Checking the conditions: $S$ is to the immediate left of $M$ (correct). $S$ is not next to $B$ (correct,$P$ is between them). $P$ is to the right of $B$ (correct,in a circle $P$ follows $B$).
$7$. Therefore,the person sitting to the immediate right of Meena is Bharti.

(A) Total number of students in the row = $40$.
Soman's position from the left end = $30^{th}$.
Kailash is $6^{th}$ to the left of Soman.
Therefore, Kailash's position from the left end = $30 - 6 = 24^{th}$.
To find Kailash's position from the right end, we use the formula: $\text{Position from right} = (\text{Total students} - \text{Position from left}) + 1$.
Position from right = $(40 - 24) + 1 = 16 + 1 = 17^{th}$.

(B) $1$. Since $P$ is to the immediate left of $N$,if we place $N$ at the bottom,$P$ is to its left.
$2$. $L$ is between $N$ and $M$. Since $P$ is already to the left of $N$,$L$ must be to the right of $N$.
$3$. Therefore,the circular arrangement in clockwise order is $N, P, M, L$.
$4$. In this arrangement,$M$ is to the immediate right of $P$ and to the immediate left of $L$.

(D) Given conditions:
$1$. $R > M$ and $R > T$
$2$. $P > N$
Combining these,we have two groups: $(R, M, T)$ and $(P, N)$.
There is no information provided to establish a relationship between the two groups (e.g.,whether $R > P$ or $P > R$).
Since the relative positions of all five individuals cannot be determined,the third highest marks cannot be identified.
Therefore,the data is inadequate.

(A) Alisha's rank from the top is $15^{th}$.
Manav is $4$ ranks above Alisha, so Manav's rank from the top is $15 - 4 = 11^{th}$.
The total number of students in the class is $20$.
The formula to find the rank from the bottom is: $\text{Rank from bottom} = (\text{Total students} - \text{Rank from top}) + 1$.
Substituting the values: $\text{Rank from bottom} = (20 - 11) + 1 = 9 + 1 = 10^{th}$.
Thus, Manav's rank from the bottom is $10^{th}$.

(D) Given conditions:
$1$. $D > A$ and $D > E$
$2$. $C > B$
Combining these,we have two separate groups: $(D, A, E)$ and $(C, B)$.
Since there is no information comparing the weights of the members of the first group with the members of the second group (e.g.,we do not know if $D > C$ or $C > D$),it is impossible to determine who is the heaviest among all five.
Therefore,the data provided is inadequate to identify the heaviest person.

(D) Given: $F$ is heavier than only $J$. This means $J$ is the lightest, and $F$ is the second lightest.
Order: $J < F < \dots < \dots < \dots$
$B$ is heavier than $F$ and $W$, but not as heavy as $K$. This implies $K > B > W > F$.
Combining these, the order of weight from heaviest to lightest is: $K > B > W > F > J$.
Therefore, the third heaviest person is $W$.

(D) Total number of children = $35$.
$M$ is $15^{th}$ from the right end.
There are $10$ children between $M$ and $R$.
Case $1$: If $R$ is to the left of $M$,then $R$'s position from the right is $15 + 10 + 1 = 26^{th}$.
Position from the left = $(\text{Total} - \text{Position from right}) + 1 = (35 - 26) + 1 = 10^{th}$.
Case $2$: If $R$ is to the right of $M$,then $R$'s position from the right is $15 - 10 - 1 = 4^{th}$.
Position from the left = $(35 - 4) + 1 = 32^{nd}$.
Since there are two possible positions for $R$ ($10^{th}$ or $32^{nd}$),the data is inadequate to determine a unique position.

(D) Given conditions:
$1$. $T > P$ and $T > B$
$2$. $T < A$ and $T < Q$
$3$. $P$ is not the shortest. Since $T > P > B$,$B$ must be the shortest.
Combining the inequalities: $A, Q > T > P > B$.
Since both $A$ and $Q$ are taller than $T$,and there is no information provided to compare the heights of $A$ and $Q$,we cannot determine who is the tallest.
Therefore,the data is inadequate.

(B) Initially,Kashish is $5^{th}$ from the left and Mona is $6^{th}$ from the right.
After interchanging places,Kashish is at the position previously held by Mona,which is $13^{th}$ from the left.
This means the position that was $6^{th}$ from the right is now $13^{th}$ from the left.
Total number of children in the queue = (Position from left + Position from right - $1$) = $13 + 6 - 1 = 18$.
Now,Mona is at the position previously held by Kashish,which is $5^{th}$ from the left.
To find Mona's new position from the right: Position from right = (Total children - Position from left + $1$) = $18 - 5 + 1 = 14^{th}$.

(B) $1$. Let the order of readers be $1, 2, 3, 4, 5$.
$2$. There are two readers between $B$ and $A$. This means the possible positions for $(B, A)$ are $(1, 4)$ or $(2, 5)$ or $(4, 1)$ or $(5, 2)$.
$3$. The one who reads first gives it to $C$,so $C$ is at position $2$.
$4$. The one who reads last took it from $A$,so $A$ is at position $4$.
$5$. If $A$ is at $4$,then $B$ must be at $1$ (since there are two readers between them).
$6$. $E$ is not first or last,so $E$ must be at position $3$.
$7$. The remaining person $D$ must be at position $5$.
$8$. The order is $B(1) o C(2) o E(3) o A(4) o D(5)$.
$9$. Since $B$ is the first reader and he passes it to $C$,$B$ passed the newspaper to $C$.

(C) Given that Sunita is $22^{nd}$ from the front.
Neela is $3$ places ahead of Sunita,so Neela's position is $22 - 3 = 19^{th}$ from the front.
Kamal is $11^{th}$ from the front.
The number of girls between Kamal and Neela is calculated as: $(\text{Position of Neela} - \text{Position of Kamal} - 1)$.
Number of girls $= 19 - 11 - 1 = 7$ girls.

(B) Initially, Rita is $9^{th}$ from the right and Monika is $10^{th}$ from the left.
After interchanging, Rita takes Monika's original position ($10^{th}$ from the left) and Monika takes Rita's original position ($9^{th}$ from the right).
We are given that after interchanging, Monika is $18^{th}$ from the left.
Since Monika is now at the position originally held by Rita (which was $9^{th}$ from the right), we know that Monika's new position is $18^{th}$ from the left and $9^{th}$ from the right.
Total number of girls $= (\text{Position from left} + \text{Position from right}) - 1$.
Total $= (18 + 9) - 1 = 27 - 1 = 26$.

(A) Let the positions of the flags from left to right be $1, 2, 3, 4, 5, 6$.
$1$. The flag of America is to the left of the Indian flag and to the right of the French flag: $France - America - India$.
$2$. The flag of Australia is to the right of the Indian flag but to the left of the Japanese flag: $India - Australia - Japan$.
$3$. The Japanese flag is to the left of the Chinese flag: $Japan - China$.
Combining these,the sequence is: $France, America, India, Australia, Japan, China$.
The two flags in the centre are the $3^{rd}$ and $4^{th}$ positions,which are India and Australia.