Grouping Questions: Reasoning Practice

Ans: A

Solution:The provided information states that Sameer, Gita, and Mihir are good in History and Biology. It also states that Gita and Kusum are good in Marathi and Maths. To find who is good in both Biology and Marathi, we need to look for an overlap between these two groups.

From the first statement, Gita is good in Biology.
From the second statement, Gita is good in Marathi.

Therefore, Gita is good in both Biology and Marathi.

Ans: C

Solution:We are given seven books: P, Q, R, S, T, U, and V. We need to identify which of these are old Gazetteers with blue covers. Let’s break down the information provided:

Cover Color:
* Blue covers: R, Q, T
* Red covers: P, S, U, V

Condition:
* New books: S, U
* Old books: P, Q, R, T, V

Type:
* Law reports: P, R, S
* Gazetteers: Q, T, U, V

Now, let’s find the books that meet all three criteria: “old”, “Gazetteers”, and “blue covers”.

* Old Books: P, Q, R, T, V
* Gazetteers: Q, T, U, V
* Blue Covers: R, Q, T

We need books that appear in all three lists.
* Looking at “Old Books” and “Gazetteers”, we have Q, T, and V.
* Now, from this subset (Q, T, V), we need to find those with “Blue Covers”.
* Q has a blue cover.
* T has a blue cover.
* V has a red cover.

Therefore, the books that are old Gazetteers with blue covers are Q and T.

Ans: A

Solution:Let’s break down the given information to determine each girl’s expertise:

* Anita: Expert in dancing and music.
* Geeta: Expert in music and painting.
* Seeta: Expert in music and painting.
* Anita: Expert in debate and dancing. (This tells us Anita is expert in dancing, and also debate)
* Neeta: Expert in painting and debate.
* Neeta: Expert in Painting and debate.
* Seeta: Expert in Painting and debate. (This tells us Seeta is expert in painting, and also debate)

Now let’s list each girl and their known expertise:

* Anita: Dancing, Music, Debate.
* Geeta: Music, Painting.
* Seeta: Music, Painting, Debate.
* Neeta: Painting, Debate.

Based on this, let’s check who is NOT an expert in painting:

* Anita: Not mentioned as an expert in painting.
* Geeta: Expert in painting.
* Seeta: Expert in painting.
* Neeta: Expert in painting.

Therefore, Anita is the only girl who is not listed as an expert in painting.

The final answer is $\boxed{Anita only}$.

Ans: C

Solution:To determine the number of people who are not wearing glasses and are short, we can use the provided information about the six individuals (A, B, C, D, E, and F).

First, let’s identify who is not wearing glasses. The problem states that A, C, and D are wearing glasses, which means B, E, and F are not wearing glasses.

Next, let’s identify who is short. The problem states that D and F are tall, and the rest are short. Therefore, A, B, C, and E are short.

Now, we need to find the individuals who satisfy both conditions: not wearing glasses AND short.
From the list of people not wearing glasses (B, E, F), we check which of them are also short.
– B is short.
– E is short.
– F is tall.

Therefore, the individuals who are not wearing glasses and are short are B and E.

The question asks “How many are Not wearing glass and short?”. Based on our analysis, there are two such people: B and E.

Looking at the options, Option C lists “B, E”.

Ans: B

Solution:We need to form a team of four members from four girls (G1, G2, G3, G4) and five boys (B1, B2, B3, B4, B5) adhering to three conditions:
I. The team must include at least two boys.
II. If B3 is selected, G2 cannot be selected.
III. G3 and B5 must be selected together.

Let’s evaluate each option against these conditions:

* Option A: B1, B2, G4, G3
* Condition I (Minimum two boys): Met (B1, B2).
* Condition II (If B3 is selected, G2 is not): Met (B3 is not selected).
* Condition III (G3 and B5 together): Not met (B5 is not selected).
* Therefore, Option A cannot be formed.

* Option B: G3, B2, G2, B5
* Condition I (Minimum two boys): Met (B2, B5).
* Condition II (If B3 is selected, G2 is not): Met (B3 is not selected).
* Condition III (G3 and B5 together): Met (G3 and B5 are both selected).
* Therefore, Option B can be formed.

* Option C: B4, B1, B5, G2
* Condition I (Minimum two boys): Met (B4, B1, B5).
* Condition II (If B3 is selected, G2 is not): Met (B3 is not selected).
* Condition III (G3 and B5 together): Not met (G3 is not selected, but B5 is).
* Therefore, Option C cannot be formed.

* Option D: G3, G1, B2, B4
* Condition I (Minimum two boys): Met (B2, B4).
* Condition II (If B3 is selected, G2 is not): Met (B3 is not selected, G2 is not selected).
* Condition III (G3 and B5 together): Not met (B5 is not selected).
* Therefore, Option D cannot be formed.

Based on the analysis, only Option B satisfies all the given conditions.

Ans: D

Solution:The problem states that eleven people work in three departments: HR, IT, and Finance, with each department having between three and five members. We are given several clues about who works with whom and in which department.

1. K is in HR. (Clue 1)
2. C and D are in the same department, which is not IT. (Clue 2)
3. HR has at most three people. (Clue 3)
4. G is with D. Since C and D are together and not in IT, and G is with D, then C, D, and G are in the same department. This department cannot be HR (only 3 people max) or IT. Therefore, C, D, and G are in Finance.
5. B is not with G or K. Since G is in Finance and K is in HR, B must be in IT.
6. F is not with G or B. Since G is in Finance and B is in IT, F must be in HR.
7. H is with F. Since F is in HR, H is also in HR.
8. HR department members are K, F, and H. This confirms HR has exactly 3 members.
9. A is not in IT. A cannot be in HR (already full). Therefore, A is in Finance.
10. E is with J. E and J cannot be in Finance (as that would make it 5 people already: C, D, G, A, and potentially E, J). They also cannot be in HR (already full). Therefore, E and J are in IT.
11. E is not with I. Since E is in IT, I cannot be in IT. I also cannot be in HR. Thus, I is in Finance.

Based on this, the IT department has B, E, and J.

Ans: B

Solution:The problem states there are 7 family members, with 4 males and 3 females, identified as L, M, N, O, P, Q, and R. We are given several clues about their relationships and professions.

* M is a businessman and P’s father, confirming M is male.
* Q is a doctor and P’s grandfather, confirming Q is male.
* N is a housewife and O’s daughter-in-law, confirming N is female.
* R is an advocate and P’s uncle, confirming R is male.

From these deductions, we have identified M, Q, and R as males. N is identified as female. The genders of L and O are not explicitly stated. P’s gender is also not explicitly stated, but since M is P’s father and Q is P’s grandfather, P is their child/grandchild.

The question asks for a possible group of males. We know M, Q, and R are males. Let’s examine the options:

* Option A: ORPM. This includes O, R, P, and M. We know R and M are male. If O is male and P is male, this group would consist of 4 males.
* Option B: QMRP. This includes Q, M, R, and P. We know Q, M, and R are male. If P is also male, this group would consist of 4 males.
* Option C: LQNR. This includes L, Q, N, and R. We know Q and R are male, and N is female. L’s gender is unknown. This group includes a female (N), so it cannot be a group of all males.

Given that M, Q, and R are confirmed males, and the family has 4 males in total, P must be the fourth male for a group of four males to exist. This would make “QMRP” a valid group of males. Additionally, if O were male and P were male, then “ORPM” would also be a valid group. However, the provided solution specifically selects “QMRP”. This implies P must be male, and O’s gender is not necessarily male for this option to be correct.

The solution concludes that “QMRP” is a group of males, which aligns with our identified males (M, Q, R) and the deduction that P must be male to complete the group of 4 males in the family.

Ans: D

Solution:The problem requires us to form a team of five members from a group of five boys and four girls, subject to specific conditions. We are told that Rachel is already selected as a team member.

Let’s analyze the conditions and their implications when Rachel is in the team:

1. Ross and Janice have to be together. This means either both Ross and Janice are in the team, or neither of them is.
2. Phoebe will not be in the same team as Rachel. Since Rachel is in the team, Phoebe cannot be.
3. Mark and Monika will not be in the same team.
4. Chandler and Richard have to be together. This means either both Chandler and Richard are in the team, or neither of them is.
5. Rachel cannot be with Joey. Since Rachel is in the team, Joey cannot be.

Given that Rachel is on the team, we can immediately exclude Phoebe (from condition 2) and Joey (from condition 5).

Now, we need to select four more members to complete the team of five.

Let’s consider condition 1: Ross and Janice have to be together.
Let’s consider condition 4: Chandler and Richard have to be together.

Since Rachel is on the team, and Phoebe and Joey are excluded, we have the remaining potential members as:
Boys: Ross, Chandler, Mark, Richard
Girls: Monika, Janice

We need to select four more members from these.

Consider the paired conditions:
* Ross and Janice must be together.
* Chandler and Richard must be together.

If we select the pair (Ross, Janice) and the pair (Chandler, Richard), this adds four members to the team. Since Rachel is already in the team, this makes a team of five. Let’s check if this combination satisfies all other conditions:

* Team: {Rachel, Ross, Janice, Chandler, Richard}
* Condition 1 (Ross and Janice together): Satisfied.
* Condition 2 (Phoebe not with Rachel): Phoebe is not in the team, so satisfied.
* Condition 3 (Mark and Monika not together): Mark is not in the team, and Monika is not in the team, so satisfied.
* Condition 4 (Chandler and Richard together): Satisfied.
* Condition 5 (Rachel not with Joey): Joey is not in the team, so satisfied.

This combination of {Rachel, Ross, Janice, Chandler, Richard} is a valid team. The other members besides Rachel are Janice, Ross, Chandler, and Richard.

Let’s look at the options provided:
Option A: Monika, Janice, Ross, Mark (Team: Rachel, Monika, Janice, Ross, Mark).
* Condition 3: Mark and Monika are together, which violates the condition.
Option B: Monika, Janice, Chandler, Richard (Team: Rachel, Monika, Janice, Chandler, Richard).
* Condition 1: Ross and Janice are not together (Ross is not selected). This does not explicitly violate condition 1 as it implies they are either both in or both out. However, if Janice is in, Ross should be in. This option doesn’t include Ross with Janice. Also, this would mean we select Monika from the remaining girls and then need to select 3 more boys from Ross, Chandler, Richard, Mark. If we select Chandler and Richard together, we still need one more boy.
Option C: Phoebe, Janice, Ross, Mark (Team: Rachel, Phoebe, Janice, Ross, Mark).
* Condition 2: Phoebe is with Rachel, which violates the condition.
Option D: Janice, Ross, Chandler and Richard (Team: Rachel, Janice, Ross, Chandler, Richard).
* This matches our derived valid team.
Option E: nan

The solution states that the other team members are “Janice, Ross, Chandler, and Richard”. This leads to Option 4. The statement “There is no need for 3rd statement” seems to imply that condition 3 (Mark and Monika will not be in the same team) is irrelevant to finding the correct answer given the other constraints and the selection of Rachel. Our analysis confirms this. The derived team satisfies all conditions.

Ans: A

Solution:The problem describes the order in which six people (P, Q, R, S, T, and U) read a magazine. We need to determine who read it first and last.

Let’s break down the clues:

1. “The one who read it last had taken it from R.” This implies that R read the magazine *before* the last person. So, R is not the last person to read.

2. “S was neither the first nor the last to read it.” This means S read the magazine in positions 2, 3, 4, or 5.

3. “S read it sometime before Q, who read it sometime after U.” This gives us relative orderings:
* S < Q (S read before Q)
* U < Q (U read before Q)
Combining these, we know Q is not among the first two readers, and S is not the last reader. Also, U is not the last reader.

4. “There were as many readers between P and T as there were between R and P.” This is a key clue about the spacing between these three individuals. Let’s denote the number of readers between two people X and Y as `Dist(X, Y)`. So, `Dist(P, T) = Dist(R, P)`.
This condition implies that P is exactly in the middle of R and T, or R is exactly in the middle of P and T, or T is exactly in the middle of R and P. For example, if R reads at position 1, P at 3, then T must read at 5, because `Dist(R, P) = 1` (only one person between them) and `Dist(P, T) = 1` (only one person between them). Or, if P reads at 1, R at 3, then T must read at 5. Or, if T reads at 1, P at 3, then R must read at 5.

Let’s combine the information to find the sequence.
From clue 1, R is not last. From clue 2, S is not first or last. From clue 3, U is not last.

Consider clue 4: `Dist(P, T) = Dist(R, P)`. This means P is equidistant from R and T. Given there are 6 positions, this can happen in these scenarios for R, P, T or T, P, R:
* R at 1, P at 2, T at 3 (Dist=0)
* R at 1, P at 3, T at 5 (Dist=1)
* R at 1, P at 4, T at 7 (not possible as only 6 people)
* R at 2, P at 3, T at 4 (Dist=0)
* R at 2, P at 4, T at 6 (Dist=1)
* R at 3, P at 4, T at 5 (Dist=0)
* And so on.

Let’s re-examine the solution’s statement: “As per the above statement, we can say that number of readers between P and T is equal to the number of readers between R and P. here Between T and P only one reader which is U. similarly, between P and R only one reader which is S.”
This implies the sequence R – S – P – U – T or T – U – P – S – R. In both these cases, `Dist(R, P) = 1` and `Dist(P, T) = 1`.

Let’s test the sequence R – S – P – U – T, positions 1 to 5.
* R at 1, S at 2, P at 3, U at 4, T at 5.
* If this is the sequence of 5, we need to fit Q in.

Let’s consider the solution’s conclusion that T read first and Q read last. This means the order starts with T and ends with Q: T – ? – ? – ? – ? – Q.

If T is first and Q is last, let’s check the clues:
* Clue 1: Last person (Q) took it from R. This means R read before Q. So R cannot be last. This is consistent.
* Clue 2: S is not first or last. Consistent if S is in positions 2, 3, 4, or 5.
* Clue 3: S before Q, U before Q. Consistent if S and U are before the last position (Q).
* Clue 4: `Dist(P, T) = Dist(R, P)`.
If T is at 1 and Q is at 6: T – ? – ? – ? – ? – Q.
Let’s assume the sequence implied by the solution: T – U – P – S – R – Q.
* T (1st)
* U (2nd)
* P (3rd)
* S (4th)
* R (5th)
* Q (6th – last)

Check clues with T-U-P-S-R-Q:
1. Last (Q) took it from R. Yes, R is before Q.
2. S is neither first nor last. Yes, S is 4th.
3. S before Q (4th before 6th), U before Q (2nd before 6th). Yes.
4. `Dist(P, T) = Dist(R, P)`?
* P is at 3, T is at 1. `Dist(P, T) = 1` (U is between them).
* P is at 3, R is at 5. `Dist(R, P) = 1` (S is between them).
This condition is satisfied.

Therefore, the order is T, U, P, S, R, Q.
The first person to read was T.
The last person to read was Q.

The final answer is $\boxed{T and Q}$.

Ans: A

Solution:
The problem requires us to find a set of three additional team members, given that Abhinav and Chirag are already selected, and adhere to several selection criteria. The team must have a total of five members, chosen from five boys (Abhinav, Hitesh, Chirag, Deva, Jawahar) and four girls (Preeti, Shanti, Rekha, Kirtana).

The given selection criteria are:
(i) Abhinav and Shanti must be together.
(ii) Deva and Kirtana cannot be together.
(iii) Preeti and Rekha cannot be together.
(iv) Chirag and Jawahar must be together.
(v) Rekha and Hitesh cannot be together.

We are given that Abhinav and Chirag are selected. Let’s analyze how this impacts the other criteria and then evaluate each option.

Since Abhinav is selected, from criterion (i), Shanti must also be selected.
Since Chirag is selected, from criterion (iv), Jawahar must also be selected.

So, our team currently consists of Abhinav, Chirag, Shanti, and Jawahar. We need to select one more member to complete the team of five.

Let’s re-examine the criteria with the current selected members:
Abhinav, Chirag, Shanti, Jawahar are selected.

Now let’s look at the options and see which combination of three additional members, when combined with Abhinav and Chirag, forms a valid team of five.

Option A: Hitesh, Jawahar, Shanti
If we select these three, the full team would be Abhinav, Chirag, Hitesh, Jawahar, Shanti.
Let’s check the criteria:
(i) Abhinav and Shanti are together (Satisfied).
(ii) Deva and Kirtana are not in the team, so this is not applicable.
(iii) Preeti and Rekha are not in the team, so this is not applicable.
(iv) Chirag and Jawahar are together (Satisfied).
(v) Rekha is not in the team. Hitesh is in the team, so this is satisfied.
This option seems plausible as it doesn’t violate any explicit rules. However, the option lists three *additional* members, and our current deduction requires only *one* additional member. This option is presenting a full set of 3 additional members. Let’s assume the question implies that the option members, when *added* to Abhinav and Chirag, should form a valid team of 5.

Let’s re-interpret the options. Each option provides three *other* members. So, the total team would be Abhinav, Chirag, and the three members from the option.

Analyzing the given solution’s approach:
The solution states: “Abhinav and Chirag are selected. Let us check each option”. This implies the options are the *remaining* members to complete the team of 5.

Let’s apply this interpretation:
Team = {Abhinav, Chirag} + {Members from option}

Option A: Hitesh, Jawahar, Shanti
Team = {Abhinav, Chirag, Hitesh, Jawahar, Shanti}
Check criteria:
(i) Abhinav and Shanti: Both are in the team. (Satisfied)
(ii) Deva and Kirtana: Neither is in the team. (Satisfied)
(iii) Preeti and Rekha: Neither is in the team. (Satisfied)
(iv) Chirag and Jawahar: Both are in the team. (Satisfied)
(v) Rekha and Hitesh: Rekha is not in the team, Hitesh is. (Satisfied)
This option is valid.

Option B: Preeti, Rekha, Deva
Team = {Abhinav, Chirag, Preeti, Rekha, Deva}
Check criteria:
(iii) Preeti and Rekha cannot be placed together. Both are in this team. (Violated)
This option is invalid.

Option C: Preeti, Kirtana, Jawahar
Team = {Abhinav, Chirag, Preeti, Kirtana, Jawahar}
Check criteria:
(i) Abhinav and Shanti: Abhinav is selected, but Shanti is not. This violates the rule that Abhinav and Shanti have to be put together. (Violated)
This option is invalid.

Option D: Shanti, Kirtana, Preeti
Team = {Abhinav, Chirag, Shanti, Kirtana, Preeti}
Check criteria:
(i) Abhinav and Shanti: Both are in the team. (Satisfied)
(iv) Chirag and Jawahar: Chirag is selected, but Jawahar is not. This violates the rule that Chirag and Jawahar are to be kept together. (Violated)
This option is invalid.

Revisiting the provided solution’s reasoning for Option C and D:
The solution’s reasoning for Option C: “If they are selected, then the team members will be Abhinav, Chirag, Preeti, Kirtana, Jawahar. But we are given that Abhinav and Shanti must be together. This is not possible.” This matches our reasoning.

The solution’s reasoning for Option D: “If they are selected, then the team members will be Abhinav, Chirag, Shanti, Kirtana, Preeti. But we are given that Chirag and Jawahar must be together. This is not possible.” This also matches our reasoning.

The solution states that Option A is the correct answer and its reasoning is “Hitesh, Jawahar, Shanti -> These people can be selected together.” This aligns with our check.

However, there’s a subtle point missed by the provided solution which might explain its own internal contradictions for Options C and D if not carefully read. The solution is presented as a sequence of checks, and its ultimate conclusion “Hence, ‘option 1’ is the correct answer” implies that only option A is valid.

Let’s reconfirm our deductions based on the initial conditions given Abhinav and Chirag are selected.
1. Abhinav selected => Shanti must be selected (Rule i).
2. Chirag selected => Jawahar must be selected (Rule iv).

So, the team *must* contain {Abhinav, Chirag, Shanti, Jawahar}. We need to select one more person from the remaining pool: {Hitesh, Deva, Preeti, Rekha, Kirtana}.

Let’s see which single person can be added to {Abhinav, Chirag, Shanti, Jawahar} without violating any rules.

Potential 5th members and team composition:
* Add Hitesh: Team = {Abhinav, Chirag, Shanti, Jawahar, Hitesh}.
* (i) Abhinav, Shanti together: Yes.
* (ii) Deva, Kirtana together: Neither present. Yes.
* (iii) Preeti, Rekha together: Neither present. Yes.
* (iv) Chirag, Jawahar together: Yes.
* (v) Rekha, Hitesh together: Rekha not present. Yes.
This is a valid team.

* Add Deva: Team = {Abhinav, Chirag, Shanti, Jawahar, Deva}.
* (ii) Deva, Kirtana together: Kirtana not present. Yes.
* (v) Rekha, Hitesh together: Neither present. Yes.
All other rules are fine. This is a valid team.

* Add Preeti: Team = {Abhinav, Chirag, Shanti, Jawahar, Preeti}.
* (iii) Preeti, Rekha together: Rekha not present. Yes.
All other rules are fine. This is a valid team.

* Add Rekha: Team = {Abhinav, Chirag, Shanti, Jawahar, Rekha}.
* (iii) Preeti, Rekha together: Preeti not present. Yes.
* (v) Rekha, Hitesh together: Hitesh not present. Yes.
All other rules are fine. This is a valid team.

* Add Kirtana: Team = {Abhinav, Chirag, Shanti, Jawahar, Kirtana}.
* (ii) Deva, Kirtana together: Deva not present. Yes.
All other rules are fine. This is a valid team.

This suggests multiple possibilities for the 5th member if we are just looking for one. However, the options provided are sets of three members. This means the options are the *remaining* members to complete the team of five.

So, the team is {Abhinav, Chirag} + {Option Members}.

Let’s re-evaluate based on the options being the other three members.

Option A: Hitesh, Jawahar, Shanti
Team = {Abhinav, Chirag, Hitesh, Jawahar, Shanti}
– Abhinav & Shanti: Together (Rule i) – YES
– Chirag & Jawahar: Together (Rule iv) – YES
– Deva & Kirtana: Not present – YES
– Preeti & Rekha: Not present – YES
– Rekha & Hitesh: Rekha not present – YES
Option A is valid.

Option B: Preeti, Rekha, Deva
Team = {Abhinav, Chirag, Preeti, Rekha, Deva}
– Preeti & Rekha: Cannot be together (Rule iii) – VIOLATED.
Option B is invalid.

Option C: Preeti, Kirtana, Jawahar
Team = {Abhinav, Chirag, Preeti, Kirtana, Jawahar}
– Abhinav & Shanti: Abhinav is present, Shanti is not. (Rule i) – VIOLATED.
Option C is invalid.

Option D: Shanti, Kirtana, Preeti
Team = {Abhinav, Chirag, Shanti, Kirtana, Preeti}
– Chirag & Jawahar: Chirag is present, Jawahar is not. (Rule iv) – VIOLATED.
Option D is invalid.

The solution correctly identifies that only Option A is valid. The initial part of the solution’s reasoning seems to imply checking the options as the *additional* members. The discrepancies it points out for Options C and D are correct.

The final answer is $\boxed{A}$.