Puzzle Test Questions in English

(D) Step $1$: Determine the initial order of the books from top to bottom.
From $(ii)$ and $(iii)$,the top three books are: $1.$ Sociology,$2.$ Education,$3.$ Accountancy.
From $(iv)$ and $(v)$,the remaining books are Economics,Psychology,and Hindi. Since Economics is immediately above Psychology and Hindi is immediately below Psychology,the order is: $4.$ Economics,$5.$ Psychology,$6.$ Hindi.
The remaining book is English,which must be at position $7$.
Initial order: $1.$ Sociology,$2.$ Education,$3.$ Accountancy,$4.$ Economics,$5.$ Psychology,$6.$ Hindi,$7.$ English.
Step $2$: Apply the interchanges.
- Sociology and English swap: English is at $1$,Sociology is at $7$.
- Accountancy and Hindi swap: Hindi is at $3$,Accountancy is at $6$.
- Education and Psychology swap: Psychology is at $2$,Education is at $5$.
New order: $1.$ English,$2.$ Psychology,$3.$ Hindi,$4.$ Economics,$5.$ Education,$6.$ Accountancy,$7.$ Sociology.
Step $3$: Identify the book between Psychology and Sociology.
In the new order,Psychology is at position $2$ and Sociology is at position $7$. The books between them are Hindi,Economics,Education,and Accountancy.
Since none of these are in the options,the correct answer is None of these.

(D) $1$. The topmost row $(1^{st})$ contains jam bottles.
$2$. Cakes are at the bottom ($6^{th}$ row).
$3$. Biscuits are above chocolates but below chips: (Chips > Biscuits > Chocolates).
$4$. Peppermints are below chocolates: (Chocolates > Peppermints).
$5$. Combining these,the sequence from top to bottom is: $1^{st}$: Jam bottles,$2^{nd}$: Chips,$3^{rd}$: Biscuits,$4^{th}$: Chocolates,$5^{th}$: Peppermints,$6^{th}$: Cakes.
$6$. Therefore,the bottles of peppermints are in the $5^{th}$ row from the top.

(B) Given conditions:
$1$. Every novel $(n)$ has a drama $(d)$ next to it.
$2$. Every story-book $(s)$ has a comic $(c)$ next to it.
$3$. No story-book $(s)$ is next to a novel $(n)$.
$4$. $A$ novel $(n)$ is at the top.
Based on these conditions,if we place a novel at the top,the next book must be a drama $(d)$ to satisfy condition $1$. Since a story-book cannot be next to a novel,the next book after the drama must be a story-book $(s)$,followed by a comic $(c)$ to satisfy condition $2$. This creates the repeating sequence $n, d, s, c$. Thus,the order is $ndsc$.

(C) Let the positions be $1$ to $7$ from top to bottom.
From $(iii)$,Geography is $4^{th}$ from bottom,so it is at position $4$. English is $5^{th}$ from top,so it is at position $5$.
From $(ii)$,History and Civics are adjacent $(H, C)$.
From $(iv)$,there are two books between Civics and Economics. If Civics is at $2$,Economics is at $5$ (occupied by English),which is impossible. If Civics is at $3$,Economics is at $6$. If Civics is at $6$,Economics is at $3$ (occupied by Geography),impossible.
Thus,Civics must be at $2$ and History at $1$. Then Economics is at $5$ (impossible as English is there). Let's re-evaluate: If Civics is at $6$,then Economics is at $3$ (occupied). If Civics is at $1$,Economics is at $4$ (occupied).
Given the constraints,we need to identify the position of the Science book. Option $(c)$ states there is one book between English $(5)$ and Science. If Science is at $7$,there is one book $(6)$ between them. This allows us to fix the positions and calculate the gap between Civics and Science.

(NONE) Let the positions be $1$ to $7$ from top to bottom.
From $(iii)$,Geography is $4^{th}$ from bottom,so it is at position $7 - 4 + 1 = 4^{th}$ from top.
English is $5^{th}$ from top.
From $(ii)$,History is immediately above Civics.
From $(iv)$,there are two books between Civics and Economics.
Arranging these: If History is $1^{st}$ and Civics is $2^{nd}$,then Economics must be $5^{th}$ (since two books are between $2$ and $5$). But English is $5^{th}$.
If History is $2^{nd}$ and Civics is $3^{rd}$,then Economics must be $6^{th}$.
Thus,the order from top to bottom is: $1$: Science,$2$: History,$3$: Civics,$4$: Geography,$5$: English,$6$: Economics,$7$: (Remaining book).
The three books above English are History,Civics,and Geography.
Since the arrangement is already determined by the given statements $(i)$ to $(iv)$,no additional information is required.

(B) Let the books from top to bottom be $B_1, B_2, ..., B_{10}$.
$1$. English is between History and Mathematics: $(H, E, M)$.
$2$. History is between Mathematics and English: $(H, E, M, H, E)$.
$3$. Hindi is between English and Mathematics: $(H, E, M, H, E, Hi, M)$.
$4$. Mathematics is between two Hindi books: $(H, E, M, H, E, Hi, M, Hi)$.
$5$. Two Hindi books are between Mathematics and History: $(H, E, M, H, E, Hi, M, Hi, Hi, H)$.
Counting the sequence: $1:H, 2:E, 3:M, 4:H, 5:E, 6:Hi, 7:M, 8:Hi, 9:Hi, 10:H$.
The $6^{th}$ book from the top is Hindi.

(D) Let the positions be $1$ to $7$ from left to right.
$1$. From $(4)$,Fiat is at one of the ends. Since it is third to the left of Ambassador,Fiat must be at position $1$ and Ambassador at position $4$.
$2$. From $(2)$,Fargo is fourth to the right of Fiat $(1+4=5)$,so Fargo is at position $5$.
$3$. From $(1)$,Cardilac is to the immediate right of Fargo,so Cardilac is at position $6$.
$4$. From $(3)$,Maruti is between Ambassador $(4)$ and Bedford. Since position $5$ is occupied by Fargo,Bedford must be at position $2$ and Maruti at position $3$.
$5$. The remaining car,Mercedes,must be at position $7$.
The final arrangement is:

Position	Car
$1$	Fiat
$2$	Bedford
$3$	Maruti
$4$	Ambassador
$5$	Fargo
$6$	Cardilac
$7$	Mercedes

Mercedes is at position $7$. Maruti is at position $3$. The position of Mercedes relative to Maruti is $7-3=4$ places to the right. Thus,Mercedes is fourth to the right of Maruti.

(B) Let the positions be $1$ to $7$ from left to right.
From $(4)$,Fiat is at one end and third to the left of Ambassador. So,Fiat is at position $1$ and Ambassador is at position $4$.
From $(2)$,Fargo is fourth to the right of Fiat $(1+4=5)$,so Fargo is at position $5$.
From $(1)$,Cardilac is to the immediate right of Fargo,so Cardilac is at position $6$.
From $(3)$,Maruti is between Ambassador $(4)$ and Bedford. Since position $5$ is occupied by Fargo,Bedford must be at position $2$ and Maruti at position $3$.
The remaining car,Mercedes,must be at position $7$.
The final arrangement is:

Position	Car
$1$	Fiat
$2$	Bedford
$3$	Maruti
$4$	Ambassador
$5$	Fargo
$6$	Cardilac
$7$	Mercedes

Clearly,Cardilac (position $6$) is to the immediate left of Mercedes (position $7$).

(C) Let the positions be $1$ to $7$ from left to right.
From $(4)$,Fiat is at one end and is third to the left of Ambassador. Thus,Fiat is at position $1$ and Ambassador is at position $4$.
From $(2)$,Fargo is fourth to the right of Fiat. Since Fiat is at $1$,Fargo is at position $1 + 4 = 5$.
From $(1)$,Cardilac is to the immediate right of Fargo. Thus,Cardilac is at position $6$.
From $(3)$,Maruti is between Ambassador and Bedford. Since Ambassador is at $4$,the only available spots for Maruti and Bedford are $2$ and $3$. Thus,Bedford is at $2$ and Maruti is at $3$.
The remaining car,Mercedes,must be at position $7$.
The final arrangement is:

$1$. Fiat	$2$. Bedford
$3$. Maruti	$4$. Ambassador
$5$. Fargo	$6$. Cardilac
$7$. Mercedes	-

Cardilac is at position $6$. The cars on its immediate sides are Fargo (at $5$) and Mercedes (at $7$).

(A) Let the positions be $1$ to $7$ from left to right.
From $(4)$,Fiat is at one of the ends and is third to the left of Ambassador. Thus,Fiat is at position $1$ and Ambassador is at position $4$.
From $(2)$,Fargo is fourth to the right of Fiat $(1+4=5)$,so Fargo is at position $5$.
From $(1)$,Cardilac is to the immediate right of Fargo,so Cardilac is at position $6$.
From $(3)$,Maruti is between Ambassador $(4)$ and Bedford. Since position $2$ and $3$ are available,and Maruti must be between Ambassador and Bedford,the sequence must be Bedford $(2)$,Maruti $(3)$,Ambassador $(4)$.
Remaining car Mercedes must be at position $7$.
Final arrangement:

$1$. Fiat

$2$. Bedford

$3$. Maruti

$4$. Ambassador

$5$. Fargo

$6$. Cardilac

$7$. Mercedes

Comparing with options,Maruti $(3)$ is to the immediate left of Ambassador $(4)$. Thus,option $A$ is correct.

(C) Let the positions be from $1$ to $7$ (left to right).
From $(4)$,Fiat is at one of the ends and is third to the left of Ambassador. Thus,Fiat is at position $1$ and Ambassador is at position $4$.
From $(2)$,Fargo is fourth to the right of Fiat $(1+4=5)$,so Fargo is at position $5$.
From $(1)$,Cardilac is to the immediate right of Fargo,so Cardilac is at position $6$.
From $(3)$,Maruti is between Ambassador $(4)$ and Bedford. Since position $5$ is occupied by Fargo,Bedford must be at position $2$ and Maruti at position $3$.
The remaining car,Mercedes,must be at position $7$.
The final arrangement is:

$1$. Fiat

$2$. Bedford

$3$. Maruti

$4$. Ambassador

$5$. Fargo

$6$. Cardilac

$7$. Mercedes

The cars to the right of the Ambassador are Fargo,Cardilac,and Mercedes.

(A) Let us denote the five friends by the first letter of their names: $S$ (Sachin),$K$ (Kunal),$M$ (Mohit),$A$ (Anuj),and $R$ (Rohan).
From the given information:
$1$. Sachin is shorter than Kunal but taller than Rohan: $R < S < K$.
$2$. Mohit is the tallest: $M$ is at the top.
$3$. Anuj is shorter than Kunal and taller than Sachin: $S < A < K$.
Combining these inequalities,we get the order:
$R < S < A < K < M$.
Comparing the heights,Rohan is the shortest.

(D) Let us denote the five friends by the first letter of their names: $S$ (Sachin),$K$ (Kunal),$M$ (Mohit),$A$ (Anuj),and $R$ (Rohan).
From the given information:
$1$. Sachin is shorter than Kunal but taller than Rohan: $R < S < K$.
$2$. Mohit is the tallest: $M$ is at the top.
$3$. Anuj is shorter than Kunal and taller than Sachin: $S < A < K$.
Combining these inequalities,we get the order of heights as:
$R < S < A < K < M$.
Arranging them from shortest to tallest: Rohan,Sachin,Anuj,Kunal,Mohit.
Therefore,Anuj is in the middle.

(B) Let us denote the five friends by the first letter of their names: $S$ (Sachin),$K$ (Kunal),$M$ (Mohit),$A$ (Anuj),and $R$ (Rohan).
From the given conditions:
$1$. Sachin is shorter than Kunal but taller than Rohan: $R < S < K$.
$2$. Mohit is the tallest: $M$ is at the end.
$3$. Anuj is shorter than Kunal but taller than Sachin: $S < A < K$.
Combining these,we get the order of heights as: $R < S < A < K < M$.
In the order of increasing heights (shortest to tallest),the sequence is:
1st: Rohan
2nd: Sachin
3rd: Anuj
4th: Kunal
5th: Mohit
Therefore,the second person in the order of increasing heights is Sachin.

(D) Let us denote the five friends by the first letter of their names: $S$ (Sachin),$K$ (Kunal),$M$ (Mohit),$A$ (Anuj),and $R$ (Rohan).
From the given information:
$1$. $Sachin$ is shorter than $Kunal$ but taller than $Rohan$: $R < S < K$.
$2$. $Mohit$ is the tallest: $M$ is at the top.
$3$. $Anuj$ is shorter than $Kunal$ and taller than $Sachin$: $S < A < K$.
Combining these inequalities,we get the order of heights as:
$R < S < A < K < M$.
Thus,the order from shortest to tallest is $Rohan < Sachin < Anuj < Kunal < Mohit$.
The second tallest person is $Kunal$.

(A) Let us denote the friends by the first letter of their names: $S$ $(Sachin)$,$K$ $(Kunal)$,$M$ $(Mohit)$,$A$ $(Anuj)$,and $R$ $(Rohan)$.
From the given information:
$1$. $Sachin$ is shorter than $Kunal$ but taller than $Rohan$: $R < S < K$.
$2$. $Mohit$ is the tallest: $M$ is at the top.
$3$. $Anuj$ is shorter than $Kunal$ and taller than $Sachin$: $S < A < K$.
Combining these,we get the order: $R < S < A < K < M$.
Comparing the heights,$Kunal$ $(K)$ is taller than $Anuj$ $(A)$ and shorter than $Mohit$ $(M)$.

(A) Given the conditions:
$1$. $X > A$
$2$. $Y = B$
$3$. $Z < C$
$4$. $A > Y$
Combining these inequalities:
Since $X > A$ and $A > Y$,we have $X > A > Y$.
Since $Y = B$,we have $X > A > Y = B$.
From the given information,we know $Z < C$. While the exact position of $Z$ and $C$ relative to $X, A, Y, B$ is not explicitly defined,$X$ remains at the top of the hierarchy established by the chain $X > A > Y = B$.
Therefore,$X$ is the most knowledgeable person among the group.

(C) Given information:
$1$. Child $A$ is less intelligent than child $B$ $(A < B)$.
$2$. Child $C$ is less intelligent than child $D$ $(C < D)$.
$3$. Child $B$ is less intelligent than child $C$ $(B < C)$.
$4$. Child $A$ is more intelligent than child $E$ $(E < A)$.
Combining these inequalities,we get the sequence:
$E < A < B < C < D$.
From the sequence,it is clear that child $D$ is the most intelligent.

(A) Let us represent the heights as inequalities based on the given information:
$1$. Vineet is taller than Manick: $Vineet > Manick$
$2$. Vineet is not as tall as Ravi: $Ravi > Vineet$
$3$. Jacob is taller than Dilip: $Jacob > Dilip$
$4$. Jacob is shorter than Manick: $Manick > Jacob$
Combining these inequalities,we get:
$Ravi > Vineet > Manick > Jacob > Dilip$
Therefore,Ravi is the tallest among the five boys.

(A) Based on the given conditions:
$(i)$ $P > Q$
$(ii)$ $P > R$
$(iii)$ $Q > S > T$
Combining these inequalities,we get the relationship: $P > Q > S > T$ and $P > R$.
Regardless of the relative height of $R$ compared to $Q$ or $S$,$P$ remains taller than all other individuals.
Therefore,$P$ is the tallest.

(A) Let the rank be represented such that a higher value means a better rank.
From the given information:
$1$. Rohit was ranked lower than Sanjay: $Rohit < Sanjay$.
$2$. Vikas was ranked higher than Dinesh: $Dinesh < Vikas$.
$3$. Kamal's rank was between Rohit and Vikas: $Rohit < Kamal < Vikas$ or $Vikas < Kamal < Rohit$.
Combining these,we get the sequence: $Dinesh < Vikas < Kamal < Rohit < Sanjay$.
Thus,Sanjay holds the highest rank.

(C) We can analyze the information as follows:
$1$. Raj got more marks than Mukesh but not as many as Priya: $Priya > Raj > Mukesh$.
$2$. Priya got more marks than Gaurav and Kavita: $Priya > Gaurav$ and $Priya > Kavita$.
$3$. Gaurav got less marks than Mukesh but his marks are not the lowest: $Mukesh > Gaurav > Kavita$.
Combining these,the descending order of marks is: $Priya > Raj > Mukesh > Gaurav > Kavita$.
Clearly,Raj is second in the descending order of marks.

(D) Based on the given information,we can represent the weights as follows:
$1$. Ashish is heavier than Govind: $Ashish > Govind$
$2$. Mohit is lighter than Jack: $Mohit < Jack$
$3$. Pawan is heavier than Jack but lighter than Govind: $Jack < Pawan < Govind$
Combining these inequalities,we get the sequence:
$Mohit < Jack < Pawan < Govind < Ashish$
Therefore,Ashish is the heaviest among them.

(D) Let us represent the size relationship as follows:
$1$. $Pune > Jhansi$
$2$. $Sitapur > Chittor$
$3$. $Jhansi > Raigarh$
$4$. $Raigarh > Sitapur$
Combining these inequalities,we get:
$Pune > Jhansi > Raigarh > Sitapur > Chittor$
Therefore,$Chittor$ is the smallest.

(B) Let us represent the height relationships as follows:
$1$. $Anand < Rohan < Seema$
$2$. $Pushpa < Krishna < Anand$
$3$. $Krishna < Dhiraj < Seema$
Combining these inequalities,we get the overall sequence:
$Pushpa < Krishna < Dhiraj < Anand < Rohan < Seema$
Thus,$Seema$ is the tallest among them.

(B) Based on the given information, we can arrange them in terms of height:
$1$. From $(A)$, we have: $\text{Kunal} < \text{Gopal} < \text{Ashok}$.
$2$. From $(B)$, we have: $\text{Navin} < \text{Kunal}$.
$3$. From $(C)$ and $(D)$, we have: $\text{Navin} < \text{Jayesh} < \text{Ashok}$.
Combining these, the overall sequence is: $\text{Navin} < \text{Kunal} < \text{Gopal} < \text{Jayesh} < \text{Ashok}$.
Therefore, Ashok is the tallest among them.

(C) To determine the relative heights of all individuals,we analyze the given statements:
$1$. From $(A)$,we have: $Ashok > Gopal > Kunal$.
$2$. From $(B)$,we have: $Kunal > Navin$.
Combining these,we get: $Ashok > Gopal > Kunal > Navin$.
$3$. From $(D)$,we have: $Ashok > Jayesh$.
$4$. Statement $(C)$ says $Jayesh > Navin$. Since we already know $Kunal > Navin$ and $Ashok > Gopal > Kunal > Navin$,the position of $Jayesh$ relative to $Navin$ is already constrained by the other relationships or is redundant for establishing the overall order if we only need to compare specific individuals. However,in the context of ordering all five,$Jayesh$ is taller than $Navin$ but shorter than $Ashok$. Since $Ashok > Gopal > Kunal > Navin$ is already established,statement $(C)$ does not provide a unique constraint that isn't already implied or unnecessary for the sequence. Thus,statement $(C)$ is not necessary.

(B) Let the age of $A$ be $x$.
Since $B$ is twice as old as $A$,the age of $B$ is $2x$.
Since $B$ is twice younger than $F$ (meaning $F$ is twice as old as $B$),the age of $F$ is $2 \times 2x = 4x$.
Since $C$ is half the age of $A$,the age of $C$ is $x/2$.
Since $C$ is twice the age of $D$ (meaning $D$ is half the age of $C$),the age of $D$ is $(x/2) / 2 = x/4$.
Comparing the ages: $F = 4x$,$B = 2x$,$A = x$,$C = 0.5x$,$D = 0.25x$.
The descending order of ages is $F > B > A > C > D$.
Therefore,$F$ is the oldest and $D$ is the youngest.

(D) Let the ages of Sudhanshu,Kokila,Praveen,and Nitin be $S, K, P,$ and $N$ respectively.
According to the problem,$S - K = P - S$,which implies $2S = P + K$ or $S = (P + K) / 2$. This means $S$ is the average age of $P$ and $K$,so $K < S < P$.
We are also given that $N = K$.
Substituting $N$ for $K$,we get the age order: $N = K < S < P$.
From this sequence,we can conclude:
$1$. Kokila is younger than Praveen ($K < P$ is true).
$2$. Nitin is younger than Praveen ($N < P$ is true).
$3$. Sudhanshu is older than Nitin ($S > N$ is true).
$4$. Praveen is the oldest,so the statement 'Praveen is not the oldest' is false.
Therefore,option $(d)$ is the wrong statement.

(D) Given conditions:
$1$. $A > B$
$2$. $C > E$ and $D > E$
$3$. $A > E > C$ or $C > E > A$
$4$. $C > B$
Combining the conditions:
Since $E$ lies between $A$ and $C$,we have two possibilities for the relative ages of $A, E,$ and $C$: $(A > E > C)$ or $(C > E > A)$.
Case $1$: If $A > E > C$,then since $C > B$,the sequence is $A > E > C > B$. In this case,$E > B$ is true.
Case $2$: If $C > E > A$,then since $A > B$,the sequence is $C > E > A > B$. In this case,$E > B$ is also true.
In both possible scenarios,$E$ is elder to $B$. Therefore,the statement '$E$ is elder to $B$' is necessarily true.

(C) Based on the given statements,we can represent the richness as follows:
$1$. $Jaya < Hitesh$
$2$. $Pritam < Mohan$
$3$. $Lalit = Jaya$
$4$. $Hitesh < Amit$
Combining these,we get the sequence: $Lalit = Jaya < Hitesh < Amit$.
Since $Lalit = Jaya$ and $Jaya < Hitesh$,it follows that $Lalit < Hitesh$.
Therefore,Lalit is poorer than Hitesh.

(B) Let the five players be $N, M, K, O, L$.
In terms of runs scored,the ascending order is $1^{st} < 2^{nd} < 3^{rd} < 4^{th} < 5^{th}$.
Given: $N$ is $1^{st}$ and $O$ is $4^{th}$.
$L$ has the first rank in one ranking and fifth in another. Since $L$ cannot be $1^{st}$ in runs (as $N$ is $1^{st}$),$L$ must be $5^{th}$ in runs.
So,the order is $N < < < O < L$.
We are given $K > M$. The remaining positions are $2^{nd}$ and $3^{rd}$.
Thus,the ascending order of runs is $N < M < K < O < L$.
Therefore,$L$ has scored the highest runs in the series.

(C) $1$. Let the five players be $N, M, K, O, L$.
$2$. For runs (ascending order): $N$ is $1^{st}$ and $O$ is $4^{th}$. Since $L$ has the $1^{st}$ rank in one ranking and $5^{th}$ in another,and $N$ is $1^{st}$ in runs,$L$ must be $5^{th}$ in runs. Given $K > M$,the order is $N < M < K < O < L$.
$3$. For wickets (descending order): The problem states $K$ replaces $O$ and $O$ replaces $L$ from the previous order,and $M$ remains unchanged. The original positions were $1:N, 2:M, 3:K, 4:O, 5:L$.
$4$. Applying the replacements: $K$ takes $O$'s place $(4^{th})$,$O$ takes $L$'s place $(5^{th})$,and $M$ stays at $2^{nd}$. This leaves $N$ at $1^{st}$ and $L$ at $3^{rd}$.
$5$. The descending order of wickets is $N > M > L > K > O$.
$6$. Therefore,$O$ has taken the lowest number of wickets.

(B) Let the order of reading be $1, 2, 3, 4, 5.$
$1$. There are two readers between $B$ and $A.$ This means the positions are $(1, 4)$ or $(2, 5).$
$2$. The one who reads last $(5)$ took it from $A.$ Thus,$A$ cannot be $5.$ If $A$ is $1,$ then $B$ is $4.$ If $A$ is $2,$ then $B$ is $5.$
$3$. If $A$ is $2$ and $B$ is $5,$ then $A$ is the one who gives it to the last person $(5)$. This fits the condition.
$4$. $E$ is not first or last,so $E$ must be $3$ or $4.$
$5$. The first person gives it to $C,$ so $C$ is $2.$ But $A$ is $2,$ which is a contradiction.
$6$. Let's try $A$ at $1$ and $B$ at $4.$ The last person $(5)$ took it from $A$ $(1)$,which is impossible.
$7$. Let's re-evaluate: $B$ and $A$ have two people between them. Possible pairs: $(B, 1, 2, A, 5)$ or $(A, 1, 2, B, 5).$
$8$. The last person $(5)$ took it from $A.$ So $A$ must be $4.$ If $A$ is $4,$ then $B$ is $1.$
$9$. The order is $B(1), ., ., A(4), \text{Last}(5).$
$10$. $E$ is not $1$ or $5,$ so $E$ is $2$ or $3.$ $C$ is the second reader (since $1$ gives to $C$). So $C$ is $2.$
$11$. The order is $B(1), C(2), E(3), A(4), D(5).$
$12$. $B$ passed the newspaper to $C.$

(D) Let the order of readers be $1, 2, 3, 4, 5.$
$1$. The one who reads first gives it to $C$,so $C$ is the $2^{nd}$ reader.
$2$. There are two readers between $B$ and $A$. The possible positions for $(B, A)$ are $(1, 4)$ or $(2, 5)$. Since $C$ is $2^{nd}$,$B$ cannot be $2^{nd}$. Thus,$B$ must be $1^{st}$ and $A$ must be $4^{th}$.
$3$. The one who reads last $(5^{th})$ took it from $A$ $(4^{th})$,which fits our sequence.
$4$. $E$ is not $1^{st}$ or $5^{th}$,so $E$ must be $3^{rd}$.
$5$. The remaining person $D$ must be $5^{th}$.
The order is: $1: B, 2: C, 3: E, 4: A, 5: D.$
Therefore,$D$ read the newspaper last.

(D) Let the order of reading be $1, 2, 3, 4, 5$.
$1$. There were two readers between $B$ and $A$. This means the possible positions for $(B, A)$ are $(1, 4)$ or $(2, 5)$.
$2$. The one who reads last $(5^{th})$ took it from $A$. This implies $A$ must be the $4^{th}$ reader,so $B$ is the $1^{st}$ reader.
$3$. The one who reads first $(B)$ gives it to $C$. So,$C$ is the $2^{nd}$ reader.
$4$. $E$ was not the first or last,so $E$ must be the $3^{rd}$ reader.
$5$. The remaining person $D$ must be the $5^{th}$ reader.
$6$. The sequence is: $1^{st}: B, 2^{nd}: C, 3^{rd}: E, 4^{th}: A, 5^{th}: D$.
Therefore,$D$ read the newspaper last.

(B) $1$. From the given information,we have the sequence $U, P, T, W$ as a block.
$2$. $R$ is the last one to reach,so the sequence ends with $R$.
$3$. The remaining executives are $S$ and $Q$.
$4$. We are told $U$ does not immediately follow $S$. If we place $S$ at the beginning,the sequence becomes $S, Q, U, P, T, W, R$.
$5$. In this arrangement,$U$ follows $Q$,not $S$,which satisfies all conditions.
$6$. Therefore,$S$ is the first to reach the office.

(B) Given the sequence conditions:
$1$. $U$ reaches immediately before $P$,so we have the block $(U, P)$.
$2$. $T$ follows immediately after $P$,and $W$ follows $T$,so we have the block $(U, P, T, W)$.
$3$. $R$ is the last one to reach,so $R$ is at position $7$.
$4$. $U$ does not immediately follow $S$. The remaining people are $S$ and $Q$.
$5$. If we place the block $(U, P, T, W)$ starting from position $3$,we get: $S, Q, U, P, T, W, R$.
$6$. Checking the condition: $U$ does not immediately follow $S$ (here $U$ follows $Q$),which is satisfied.
$7$. The sequence is $1: S, 2: Q, 3: U, 4: P, 5: T, 6: W, 7: R$.
$8$. Therefore,$P$ is the fourth person to reach the office.

(A) $1$. Prashant Arora has three children: Sangeeta,Vimal,and Ashish.
$2$. Ashish is married to Monika (daughter of Mr. and Mrs. Roy).
$3$. The Roys have four children: Monika (eldest),Roshan,Vandana,and Veena (youngest).
$4$. The Roys married their youngest daughter,Veena,to the eldest son of Mr. and Mrs. Sharma.
$5$. This couple (Veena and the eldest son of Sharma) has two children: Amit and Shashi.
$6$. Since Rashmi is the daughter of Amit,and Amit is the son of the eldest Sharma son,Rashmi carries the surname of her father,which is Sharma.

(A) $1$. Identify the relationship: Ashish is married to Monika.
$2$. Monika is the daughter of Mr. and Mrs. Roy.
$3$. Sameer is the son of Ashish and Monika.
$4$. Since Sameer is the son of Monika,and Monika is the daughter of Mr. Roy,Sameer is the grandson of Mr. Roy (Monika's father).

(C) Prashant Arora is the father of Ashish.
Ashish is married to Monika.
Sameer and Ajay are the sons of Ashish and Monika.
Since Sameer is the son of Ashish,who carries the surname Arora,the surname of Sameer is Arora.

(B) Ashish is married to Monika.
Monika is the daughter of Mr. and Mrs. Roy.
Since Ashish is the husband of Monika,the mother of Monika (Mrs. Roy) is the mother-in-law of Ashish.

(B) $1$. $P$ and $S$ are unmarried ladies and do not participate in any game.
$2$. $Q$ is the brother of $R$ and is neither a chess player nor a hockey player. Since there are only three games (football,chess,hockey) and $Q$ does not play chess or hockey,$Q$ must be the football player.
$3$. $T$ is the husband in a married couple. Since $Q$ is the football player,$T$ must be either the chess player or the hockey player.
$4$. The problem states that none of the ladies plays chess or football. Since $R$ is the only remaining person who could be the wife of $T$,and $R$ must play a game,$R$ must be the hockey player.
$5$. Therefore,$T$ is the chess player.
$6$. Summary: $P$ (unmarried lady,no game),$S$ (unmarried lady,no game),$Q$ (football player),$R$ (hockey player,wife of $T$),$T$ (chess player,husband of $R$).
$7$. Thus,$Q$ is the football player.

(C) $1$. $P$ and $S$ are unmarried ladies and do not play any games.
$2$. $Q$ is the brother of $R$ and is not a chess or hockey player. Since there are only three games (football,chess,hockey),$Q$ must be the football player.
$3$. $T$ is the husband in a married couple. Since $P$ and $S$ are unmarried,$T$ must be married to $R$.
$4$. Since $Q$ is the football player,and $T$ is the husband,$T$ must be the chess player (as the only remaining game for the male/husband is chess).
$5$. Therefore,$R$ must be the hockey player.
$6$. Conclusion: $R$ is the hockey player.

(D) Based on the provided information:
$1$. $Q$ is neither a hockey player nor a chess player,so $Q$ must be a football player.
$2$. $T$ is a husband,which implies $T$ is male and not a lady.
$3$. Since $T$ is a husband and the roles are assigned,$T$ must be the chess player.
$4$. $R$ is identified as a hockey player and $T$'s wife.
$5$. Therefore,$T$ is the chess player.

(C) $1$. $P$ and $S$ are unmarried ladies and do not play any games.
$2$. $Q$ is the brother of $R$ and is not a chess or hockey player. Since there are three games (football,chess,hockey) and $Q$ plays one,$Q$ must be the football player.
$3$. Since $Q$ is a brother,$Q$ is male.
$4$. $T$ is a husband,so $T$ is male.
$5$. We have $P, S$ (unmarried ladies) and $R$ (remaining person). Since $T$ is a husband,his wife must be one of the ladies. $P$ and $S$ are unmarried,so $R$ must be the wife of $T$.
$6$. Since $R$ is the wife of $T$,$R$ is a lady. The problem states no lady plays chess or football. Therefore,$R$ must be the hockey player.
$7$. $T$ must be the chess player.
Conclusion: $R$ is the wife of $T$.

(D) $1$. $P$ and $S$ are unmarried ladies who do not play any games.
$2$. $Q$ is the brother of $R$ and is not a chess or hockey player. Since there is one football player,one chess player,and one hockey player,$Q$ must be the football player. Being a brother,$Q$ is male.
$3$. $T$ is the husband in a married couple,so $T$ is male. Since $T$ is not the football player $(Q)$,he must be either the chess or hockey player.
$4$. None of the ladies play chess or football. Since $R$ is the only remaining person who could be a lady (as $P$ and $S$ are already identified as ladies),$R$ must be the hockey player (because ladies cannot play chess or football).
$5$. Since $T$ is the husband of $R$,$R$ is a lady. Thus,the three ladies are $P, S,$ and $R$.

(B) Given conditions:
$A + B > C + D$ ... $(1)$
$A + C = B + D$ ... $(2)$
$A = \frac{1}{2}(B + D)$ ... $(3)$
From equation $(2)$,we have $C = (B + D) - A$. Substituting the value of $A$ from equation $(3)$:
$C = (B + D) - \frac{1}{2}(B + D) = \frac{1}{2}(B + D)$.
Thus,$A = C$.
Now,substitute $A = C$ into equation $(1)$:
$A + B > A + D$,which simplifies to $B > D$.
From equation $(3)$,since $A = \frac{1}{2}(B + D)$ and $B > D$,it follows that $A$ is the average of $B$ and $D$. Since $B$ is greater than $D$,$B$ must be greater than the average $A$ (i.e.,$B > A$).
Since $A = C$ and $B > A$,it follows that $B > C$ as well.
Therefore,$B$ has the highest income.

(A) Given:
$A + B > C + D \dots(1)$
$A + C = B + D \dots(2)$
$A = \frac{1}{2}(B + D) \dots(3)$
Substituting the value of $A$ from $(3)$ into $(2)$:
$\frac{1}{2}(B + D) + C = B + D$
$C = (B + D) - \frac{1}{2}(B + D) = \frac{1}{2}(B + D)$
Thus,$A = C$.
Now,substituting $A = C$ into $(1)$:
$A + B > A + D$
$B > D$
From $(3)$,since $A = \frac{1}{2}(B + D)$,it implies $B > A$ and $D < A$ (because $B > D$).
Therefore,the relationship is $B > A = C > D$.
Thus,the statement '$A$ earns more than $B$' is false,because $B > A$.

(D) Given:
$A + B > C + D \dots(1)$
$A + C = B + D \dots(2)$
$A = \frac{1}{2}(B + D) \implies 2A = B + D \dots(3)$
From $(2)$ and $(3)$,$A + C = 2A \implies A = C$.
Given $A = 80,000$,so $C = 80,000$.
Also,the difference between $B$ and $D$ is equal to $A$'s income:
$B - D = A = 80,000 \dots(4)$
Adding $(3)$ and $(4)$:
$(B + D) + (B - D) = 160,000 + 80,000$
$2B = 240,000$
$B = 120,000$.
Thus,$B$'s income is $Rs\, 1,20,000$.