Question 1

From the proverb “Fortune favors the brave”, what can we logically infer about risk-taking and success?

Accepted Answer

Introduction / Context: This question is based on interpreting a proverb: 'Fortune favors the brave.' We are asked which conclusion logically follows from this statement. The first conclusion says that risks are necessary for success. The second conclusion says that cowards die many times before their death. The task is to relate each conclusion to the meaning of the proverb and see which one truly follows from it. Given Data / Assumptions: Statement: 'Fortune favors the brave.' Conclusion 1: Risks are necessary for success. Conclusion 2: Cowards die many times before their death. We interpret 'fortune' as success or good outcomes in life, and 'brave' as those who take bold actions, often involving risk. Concept / Approach: Proverb-based reasoning questions require understanding the underlying idea expressed by the proverb. 'Fortune favors the brave' suggests that success tends to come to people who act courageously rather than to those who are timid or overly cautious. This has a natural connection to risk-taking: bravery often means facing risk. We must check which conclusion captures this core idea and which conclusion introduces unrelated or extra information. Step-by-Step Solution: Step 1: Understand the proverb. 'Fortune favors the brave' means that success or good luck is more likely to come to those who act with courage, initiative, and willingness to take chances, rather than to those who are passive or fearful. Step 2: Examine Conclusion 1: 'Risks are necessary for success.' Bravery usually involves taking risks or facing danger rather than avoiding it. If fortune (success) favors the brave, it implies that taking courageous steps, which often include risk, is an important part of achieving success. The proverb therefore supports the idea that some level of risk-taking is necessary to gain fortune. Step 3: While the proverb does not use the word 'risk' directly, its central message is that success comes to those who dare. So, interpreting this as 'Risks are necessary for success' is consistent with the meaning of the proverb. Hence, Conclusion 1 reasonably follows. Step 4: Examine Conclusion 2: 'Cowards die many times before their death.' This line is essentially a different proverb or idea, often associated with the notion that fearful people suffer repeatedly in their imagination. However, this specific wording is not stated nor clearly implied by 'Fortune favors the brave.' Step 5: While one might infer that cowards do not get fortune, the statement about 'dying many times before their death' goes beyond the original proverb and introduces an additional concept not contained in the given statement. Hence, Conclusion 2 does not logically follow from the statement alone. Verification / Alternative check: To see if Conclusion 2 is independent, note that it could be true or false regardless of whether 'Fortune favors the brave' is true. The original statement does not talk about how many times cowards 'die' or suffer; it only describes how fortune relates to bravery. Why Other Options Are Wrong: Option A (Neither conclusion follows) is wrong because Conclusion 1 clearly connects to the core meaning of the proverb, which encourages bravery and, by implication, risk-taking to achieve success. Option B (Only conclusion 2 follows) is wrong because Conclusion 2 is an unrelated proverb and not implied by the given statement. Option C (Either 1 or 2 follows) is wrong as both are not equally supported; only Conclusion 1 captures the meaning of the original statement. Common Pitfalls: A common error is to treat any proverb about bravery and cowardice as interchangeable, even though they express subtly different ideas. Another mistake is to accept a conclusion simply because it sounds wise or familiar, instead of checking whether it directly follows from the given statement. Final Answer: Since the proverb directly encourages bravery as a condition for attaining fortune (success), it supports the idea that taking risks is necessary to succeed. Therefore, only conclusion 1 follows.

Question 2

Given that all poets are readers and no reader is wise, what must we conclude about poets and wisdom?

Accepted Answer

Introduction / Context: This is another syllogism question involving relationships between three groups: poets, readers and wise people. The statements tell us how these sets overlap or do not overlap, and we must decide which conclusion logically follows. Such questions demand careful translation of statements like 'No reader is wise' and 'All poets are readers' into simple set relations and then reasoning from them without adding extra assumptions. Given Data / Assumptions: Statement 1: All poets are readers. Statement 2: No reader is wise. Conclusion 1: No poet is wise. Conclusion 2: All readers are poets. Concept / Approach: We can treat each group as a set: P = set of all poets R = set of all readers W = set of all wise people The statement 'All poets are readers' means P ⊆ R. The statement 'No reader is wise' means R and W are disjoint sets (R ∩ W = ∅). We then check which conclusions must always be true given these relations. Step-by-Step Solution: Step 1: Translate the statements into set language. 'All poets are readers' means every poet is a reader, so P is a subset of R (P ⊆ R). Step 2: 'No reader is wise' means there is no person who is both a reader and wise. Hence, R ∩ W = ∅ (the intersection of readers and wise people is empty). Step 3: Consider Conclusion 1: 'No poet is wise.' Since every poet is a reader (P ⊆ R), and no reader is wise (R ∩ W = ∅), it follows that no poet can be in W. In other words, poets are a subset of readers, and readers have no overlap with wise people, so poets also have no overlap with wise people. Step 4: Formally, if P ⊆ R and R ∩ W = ∅, then P ∩ W = ∅. This is exactly the content of Conclusion 1: no poet is wise. Therefore, Conclusion 1 must follow. Step 5: Consider Conclusion 2: 'All readers are poets.' This would mean R ⊆ P. But we are only told that P ⊆ R and nothing about whether every reader must be a poet. It is entirely possible that there are readers who are not poets, such as scientists, engineers or students who read but do not write poetry. Step 6: Because we can imagine many scenarios where there are readers who are not poets while still satisfying the given statements, Conclusion 2 does not logically follow. Verification / Alternative check: Example: Suppose there are 10 readers, of which 3 are poets. All 3 poets are readers, and none of the 10 readers is wise. This satisfies all given statements. Yet not all readers are poets, so Conclusion 2 is false in this valid scenario, confirming that it does not follow. Why Other Options Are Wrong: Option A (Neither conclusion follows) is wrong because Conclusion 1 is clearly and directly implied by the given statements. Option B (Only conclusion 2 follows) is wrong since Conclusion 2 is not supported and can be false even when the statements are true. Option C (Either 1 or 2 follows) is wrong because only one of them (Conclusion 1) is guaranteed; the other is not. Common Pitfalls: A frequent error is to misuse the direction of the subset relation: from 'All poets are readers' some people wrongly conclude 'All readers are poets', which is logically incorrect. Another mistake is to ignore the implication of 'No reader is wise', which completely excludes readers (and thus poets) from the wise group. Final Answer: Since poets are readers and no reader is wise, poets cannot be wise either. Therefore, only conclusion 1 follows.

Question 3

To determine Guna's exact age, which of the given pieces of information are necessary and sufficient?

Accepted Answer

Introduction / Context: This is a data sufficiency question about ages. The main question is: 'What is Guna's age?' Two separate statements give relationships among the ages of Guna, Vinay, Keshav and Arjun. We have to decide whether each statement alone is enough, whether both together are needed, or whether even together they are not sufficient. The focus is on algebraic reasoning and understanding when information is sufficient to find a unique value. Given Data / Assumptions: Main Question: What is Guna's age? Statement 1: Guna, Vinay and Keshav are all of the same age. Statement 2: The sum of ages of Vinay, Keshav and Arjun is 32, and Arjun is as old as Vinay and Keshav together. We assume all ages are in years and are positive integers, as is typical in such questions. Concept / Approach: The technique is to express each statement in algebraic form, see how many unknowns there are and whether we can solve for Guna's age uniquely. If one statement alone gives multiple possibilities, it is not sufficient. If combining the two statements allows us to find a single numerical value for Guna's age, then both are needed. Step-by-Step Solution: Step 1: Assign variables. Let G be Guna's age, V be Vinay's age, K be Keshav's age and A be Arjun's age. Step 2: Translate Statement 1: 'Guna, Vinay and Keshav are all of the same age.' This means G = V = K. Step 3: Check sufficiency of Statement 1 alone. If G = V = K, we still do not know any actual numerical value. Many values are possible (e.g., all three could be 8, or 10, or 15). So Statement 1 alone is not sufficient to find Guna's age. Step 4: Translate Statement 2: 'Sum of ages of Vinay, Keshav and Arjun is 32 and Arjun is as old as Vinay and Keshav together.' In symbols, V + K + A = 32 and A = V + K. Step 5: Use the condition A = V + K in V + K + A = 32. Substituting, we get V + K + (V + K) = 32, that is 2(V + K) = 32, so V + K = 16. Then A = V + K = 16. Step 6: Check sufficiency of Statement 2 alone. From Statement 2, we know V + K = 16, but we do not know the individual values of V and K, and we have no information relating G to them. Guna's age G could be anything; we cannot determine it. So Statement 2 alone is also not sufficient. Step 7: Now combine Statement 1 and Statement 2. From Statement 1, we have G = V = K. From Statement 2, we already found V + K = 16. Since V = K, we can write V + K = G + G = 2G = 16, which gives G = 8. Step 8: Thus, when both statements are used together, Guna's age is uniquely determined as 8 years. Verification / Alternative check: We can verify with concrete numbers: If G = V = K = 8, then V + K = 8 + 8 = 16 and A = 16. The sum V + K + A is 8 + 8 + 16 = 32, satisfying Statement 2. So both statements are consistent and produce a unique age for Guna. Why Other Options Are Wrong: Option A (Neither 1 nor 2 needed) is wrong because we do need information from the statements to determine Guna's age. Option B (Only statement 2 is needed) is wrong because Statement 2 does not involve G directly and gives no way to find his age alone. Option D (Only statement 1 is needed) is wrong because Statement 1 only tells us that G shares an age with Vinay and Keshav but not what that age is. Common Pitfalls: A typical mistake is to think that since Statement 2 contains actual numbers (like 32), it must be sufficient. But without a link to G, it is not. Another pitfall is not combining the equations correctly or forgetting that G, V and K are equal, which is crucial to solving the problem. Final Answer: We can find Guna's age only when we use the equality of ages from Statement 1 together with the numerical relations from Statement 2, giving G = 8. Hence, both statements 1 and 2 are needed.

Question 4

Given that no purse is made of cloth and all purses are made of leather, which conclusions about leather, cloth and purses must follow?

Accepted Answer

Introduction / Context: This question again involves categorical statements (syllogisms) about three sets: purses, leather items and cloth items. We are told that no purse is cloth and that all purses are leather. We must determine which conclusions about leather, cloth and purses are logically forced by these statements. This tests your ability to reason about set inclusion and exclusion without mixing up directions or making invalid generalisations. Given Data / Assumptions: Statement 1: No purse is cloth. Statement 2: All purses are leather. Conclusion I: No leather is cloth. Conclusion II: Some leather items are cloth. Conclusion III: Some leather items are purses. Concept / Approach: Let us represent the sets: P = set of all purses L = set of all leather items C = set of all cloth items 'All purses are leather' means P ⊆ L. 'No purse is cloth' means P ∩ C = ∅. We must check each conclusion in turn and see whether it must hold in every possible arrangement of these sets that respects the given statements. Step-by-Step Solution: Step 1: From 'All purses are leather' (P ⊆ L), every purse belongs to the leather set. However, this statement does not say anything about items in L that are not purses. Step 2: From 'No purse is cloth' (P ∩ C = ∅), we know that there is no item which is both a purse and cloth. But again, this does not tell us about other leather items or other cloth items outside the purse category. Step 3: Check Conclusion I: 'No leather is cloth.' This claims that the entire leather set L and cloth set C do not overlap (L ∩ C = ∅). The given information only tells us that P, which is a subset of L, does not overlap with C. It is still possible that some other leather items (not purses) could be made of cloth in some hypothetical scenario, depending on how we interpret 'leather' and 'cloth' in the abstract. Logically, nothing stops L ∩ C from containing items that are not purses. Therefore, Conclusion I is not forced by the statements. Step 4: Check Conclusion II: 'Some leather items are cloth.' This claims that L and C do overlap (L ∩ C is non-empty). But as we just saw, the given information is compatible with both possibilities: they might overlap or they might not. So we cannot assert this as a must. Conclusion II also does not logically follow. Step 5: Check Conclusion III: 'Some leather items are purses.' Since all purses are leather, if we assume that purses exist at all, then at least some leather items are purses. Exam questions of this type usually assume that the subject set (here, purses) is non-empty, otherwise the statements would be trivial. So, if P is non-empty and P ⊆ L, then there is at least one element in both P and L, meaning some leather items are purses. Step 6: Therefore, Conclusion III follows: there exist leather items that are purses. Verification / Alternative check: Example where Conclusions I and II fail but the statements hold: Let there be leather jackets that are partly cloth-lined, counted in both L and C. Also let some purses be pure leather and not cloth. Then 'All purses are leather' and 'No purse is cloth' remain true, but some leather items are cloth, so Conclusion I is false and Conclusion II is true. This shows that nothing in the statements forbids such a situation, hence Conclusions I and II do not necessarily follow. However, as long as at least one purse exists and all purses are leather, then some leather items are purses. Conclusion III must hold under the usual exam assumption that the set of purses is non-empty. Why Other Options Are Wrong: Option A (Only conclusion I follows) is wrong because we cannot rule out the possibility that some leather items are also cloth, as long as they are not purses. Option B (Only conclusion II follows) is wrong because the existence of leather items that are cloth is not guaranteed by the statements. Option D (All the above) is wrong because Conclusions I and II are not logically forced. Common Pitfalls: A very common mistake is to jump from 'No purse is cloth' and 'All purses are leather' to 'No leather is cloth', which overgeneralises the restriction placed only on purses. Another frequent error is to ignore the usual assumption of existence (that there is at least one purse), which is needed to make sense of 'All purses are leather' in the context of typical exam questions. Final Answer: The only statement that must be true in all valid scenarios is that some leather items are purses. Therefore, only conclusion III follows.

Question 5

Given that nutritious food is delicious and good for health, and that honey is nutritious, what can we conclude about honey and delicious food?

Accepted Answer

Introduction / Context: This question tests basic logical inference using simple everyday statements. We are told that nutritious food is both delicious and good for health, and that honey is nutritious. We must judge which conclusions follow about honey's effect on health and about whether every delicious food must be nutritious. The key is to track the direction of implications and not to assume the reverse without justification. Given Data / Assumptions: Statement 1: Nutritious food is delicious and good for health. Statement 2: Honey is nutritious. Conclusion I: Honey is good for health. Conclusion II: Delicious foods are nutritious. Concept / Approach: We can interpret the main statement as: If a food is nutritious, then it is delicious and good for health. In logical form, Nutritious ⇒ Delicious and Nutritious ⇒ Good for health. Honey is stated to be nutritious, so we can apply these implications directly to honey. But we must be careful not to flip the direction, that is, not to assume Delicious ⇒ Nutritious without evidence. Step-by-Step Solution: Step 1: Let N(x) mean 'x is nutritious', D(x) mean 'x is delicious', and H(x) mean 'x is good for health'. Step 2: The statement 'Nutritious food is delicious and good for health' can be read as: For any food x, if N(x) is true, then D(x) and H(x) are both true. Symbolically, N(x) ⇒ D(x) and N(x) ⇒ H(x). Step 3: We are told 'Honey is nutritious.' That is N(honey) is true. Step 4: Applying the implication N(x) ⇒ H(x) to honey, we get H(honey) is true. In words, honey is good for health. So Conclusion I follows directly and logically. Step 5: Now consider Conclusion II: 'Delicious foods are nutritious.' Our original statement only asserts that nutritious foods are delicious, not that every delicious food must be nutritious. There may be foods that are delicious but not nutritious at all, such as sugary junk food. The implication works only one way: Nutritious ⇒ Delicious, not Delicious ⇒ Nutritious. Step 6: Since we cannot conclude that all delicious foods are nutritious from the given information, Conclusion II does not logically follow. Verification / Alternative check: Think of a concrete example: Suppose a cake is very delicious but full of unhealthy ingredients. It is not nutritious, yet it is delicious. This does not contradict the given statement, which only talks about nutritious foods being delicious, not about all delicious foods being nutritious. On the other hand, if honey is nutritious and all nutritious foods are good for health, then honey must be good for health. Any scenario that satisfies the premises will satisfy this conclusion. Why Other Options Are Wrong: Option B (Only conclusion II follows) is wrong because we have just seen that Delicious ⇒ Nutritious is not supported by the given information. Option C (Neither conclusion follows) is wrong because Conclusion I clearly and directly follows from the premises. Option D (Both conclusions follow) is wrong since Conclusion II is not logically forced by the statements. Common Pitfalls: A common mistake is to treat an implication as reversible. Students see 'Nutritious food is delicious' and incorrectly assume 'Delicious food is nutritious' as well, which is not logically valid. Another error is overlooking the special case of honey: people may focus only on the general statement and forget to apply it specifically to honey, which is explicitly said to be nutritious. Final Answer: Because honey is nutritious and all nutritious foods are good for health, we can conclude that honey is good for health. But we cannot conclude that every delicious food is nutritious. Hence, only conclusion I follows.

Question 6

In the following logical reasoning question, two categorical statements about mirrors, glass, and reflective objects are given as premises. Treat both statements as true, even if they appear to contradict real world facts, and then decide which of t…

Accepted Answer

Introduction / Context: This logical reasoning problem tests your understanding of categorical statements and how valid conclusions are derived from them. We are given two universal negative statements about mirrors, glass, and reflective objects, and we must examine whether either of the two proposed universal positive conclusions logically follows. The key skill here is to convert the verbal statements into set relationships and then check if the conclusions are forced by those relationships or if alternative diagrams are still possible. Given Data / Assumptions: Statement I: No mirrors are glass. In set terms, the set of mirrors and the set of glass objects do not overlap. Statement II: No reflective objects are mirrors. The set of reflective objects and the set of mirrors do not overlap. Conclusion I: All glass objects are reflective. Conclusion II: All reflective objects are glass. All statements are to be treated as logically true premises, regardless of real world knowledge. Concept / Approach: The correct approach is to interpret each statement as a relationship between sets and then test whether the proposed conclusions are logically necessary. A conclusion is valid only if it holds in every possible diagram that satisfies all the given statements. If we can draw even one logically consistent diagram where a conclusion fails, that conclusion does not follow. Universal statements like “all A are B” or “no A are B” must be handled carefully, because they require strong evidence from the premises. Step-by-Step Solution: Step 1: Represent Statement I as a Venn diagram idea. The set Mirrors and the set Glass do not intersect at all. That is, Mirrors ∩ Glass = ∅. Step 2: Represent Statement II similarly. The set Reflective and the set Mirrors do not intersect. So Reflective ∩ Mirrors = ∅. Step 3: Observe that both statements talk about how Mirrors relate to Glass and Reflective, but there is no direct relation given between Glass and Reflective. Step 4: Test Conclusion I: “All glass objects are reflective.” The premises allow at least two possibilities: (a) Glass is completely inside Reflective, or (b) Glass is partly or completely outside Reflective. Since both arrangements are compatible with the premises, Conclusion I is not forced and therefore does not logically follow. Step 5: Test Conclusion II: “All reflective objects are glass.” Similarly, the premises do not restrict Reflective objects to being inside the Glass set. Reflective objects simply are not mirrors. They could be glass, plastic, metal, or any other material, as long as they are not mirrors. Therefore, it is possible to have reflective objects that are not glass, so Conclusion II also does not follow. Verification / Alternative check: A quick way to verify is to construct a concrete example. Consider some mirrors (M), some glass objects (G), and some reflective objects (R) such that: Mirrors are made of a special material that is not glass or counted as reflective in this simplified model. Glass objects include drinking glasses and glass blocks, none of which are mirrors. Reflective objects include polished metal plates, which are not mirrors and not made of glass. All given statements are satisfied, but many glass objects are not reflective and many reflective objects are not glass. This shows that neither conclusion is logically compelled by the premises. Why Other Options Are Wrong: Option a: “Only conclusion I follows” is incorrect because we have seen that Glass does not have to be a subset of Reflective. Glass could be non-reflective in some possible diagrams. Option b: “Only conclusion II follows” is incorrect because Reflective objects are only restricted from being Mirrors. They are not required to be Glass, so they can lie outside the Glass set. Option c: “Both conclusions I and II follow” is incorrect because neither conclusion is forced, let alone both. Option e: “Cannot be determined from the information given” is weaker than the correct logical description. It is not that the information is missing; it is that the conclusions simply do not follow logically from what is provided. Common Pitfalls: A frequent mistake is to rely on real world knowledge, such as believing that mirrors are made of glass and are reflective, and then trying to derive the conclusions from that intuition. In standard reasoning questions, you must ignore real world facts and treat the premises as complete and absolute. Another mistake is assuming that if a term appears in two premises, there must be a direct relation between the other two sets, which is not necessarily true. Always test conclusions by constructing possible diagrams and checking whether the conclusion is forced in every case. Final Answer: Neither conclusion I nor conclusion II follows from the given statements.

Question 7

In this logical deduction question, you are given a universal statement and a specific case. Treat the following statement as true and decide which of the given conclusions logically follows from it. Statement: All boys are tall. Rajiv is a boy. Con…

Accepted Answer

Introduction / Context: This problem is a classic example of a syllogism in logical reasoning. You are given a general rule about a group, in this case all boys, and then a specific individual who belongs to that group. Your task is to decide whether each conclusion about that individual follows logically from the general rule. This type of question checks whether you can correctly apply universal statements to particular instances without being confused by contradictory options. Given Data / Assumptions: Statement: All boys are tall. Additional fact: Rajiv is a boy. Conclusion I: Rajiv is tall. Conclusion II: Rajiv is not tall. All statements are to be treated as logically true premises, regardless of real life heights. Concept / Approach: The key concept is that a universal affirmative statement of the form “All A are B” means every member of set A is guaranteed to be in set B. When we are told that a specific individual belongs to set A, we can directly conclude that the individual also belongs to set B. Any conclusion that directly contradicts this must be rejected. This is a direct application of basic deductive reasoning, where we move from the general to the particular. Step-by-Step Solution: Step 1: Interpret the given universal statement. “All boys are tall” means that for every person x, if x is a boy, then x is tall. No exception is allowed by this statement. Step 2: Note the specific information about Rajiv. We are explicitly told that Rajiv is a boy. Therefore Rajiv belongs to the set of boys. Step 3: Apply the universal rule to Rajiv. Since Rajiv is in the set of boys and all members of that set are tall, it must follow that Rajiv is tall. Step 4: Evaluate Conclusion I. It states exactly what we deduced: Rajiv is tall. Therefore Conclusion I logically follows from the given statement. Step 5: Evaluate Conclusion II. It says Rajiv is not tall. This directly contradicts the result that Rajiv must be tall. A valid conclusion can never directly contradict a logically necessary result of the premises, so Conclusion II does not follow. Verification / Alternative check: You can verify this with a simple substitution example. Imagine a group where every boy is at least 180 cm tall. If we now say Rajiv is a boy in that group, it is impossible for Rajiv to be shorter than 180 cm. Any statement that says “Rajiv is not tall” is inconsistent with the given rule. This confirms that only the first conclusion aligns with the premises and the second is impossible if the statements are true. Why Other Options Are Wrong: Option b: “Only conclusion II follows” is wrong because it asserts the exact opposite of what the premises demand. Option c: “Both conclusion I and conclusion II follow” is impossible because two directly contradictory conclusions cannot both be correct at the same time. Option d: “Neither conclusion I nor conclusion II follows” is incorrect because we have seen that Conclusion I is a direct and compulsory result of the given statement. Option e: “The relationship cannot be determined” is also incorrect. The relationship is fully determined by the universal statement and the classification of Rajiv as a boy. Common Pitfalls: Learners sometimes think that because they see both “Rajiv is tall” and “Rajiv is not tall”, the answer must be “either or neither”. However, syllogism questions are not about what could be true in some world, but what must be true given the premises. Since the given statement leaves no room for any boy to be not tall, the conclusion that Rajiv is tall is unavoidable. Never treat mutually exclusive conclusions as both correct when the premises clearly support only one of them. Final Answer: Only conclusion I follows from the given statement.

Question 8

In this logical reasoning problem, two related statements about luck, hard work, and success are given. Treat both statements as true and then decide which of the following conclusions logically follow. Statement I: If we are lucky, then things auto…

Accepted Answer

Introduction / Context: This question examines your ability to connect conditional statements and derive meaningful conclusions. The ideas of luck, hard work, and success are tied together in two premises. From these, you must decide which conclusions necessarily follow. While the language is everyday, the underlying structure is logical: one statement describes what happens when we are lucky, and the other describes when luck appears. Taken together, they give a chain from hard work to success. Given Data / Assumptions: Statement I: If we are lucky, then things automatically fall into place. This suggests that luck leads directly to favorable outcomes or success. Statement II: Luck follows only when we work hard. This means hard work is a necessary condition for luck to appear. Conclusion I: Hard work is the key to success. Conclusion II: If you do not work hard, luck will not follow you. All statements are to be treated as logically true premises, ignoring any real world exceptions. Concept / Approach: Both statements can be expressed as conditional relationships. Statement I links Luck to Success: Luck implies Success. Statement II links Hard Work to Luck: only when you work hard does luck appear, which translates to Luck implies Hard Work or equivalently, no Hard Work implies no Luck. By combining these, we can trace a path from Hard Work to Luck and then to Success, and we can also derive the contrapositive about what happens when there is no Hard Work. The conclusions must be checked against these derived relationships. Step-by-Step Solution: Step 1: Rewrite Statement I in logical form: Luck leads to things falling into place, which we can interpret as success or favorable outcomes. So Luck implies Success. Step 2: Rewrite Statement II in logical form: “Luck follows only when we work hard” means that having Luck requires that Hard Work has been done. In logic, this is Luck implies Hard Work. The contrapositive is also true: if there is no Hard Work, then Luck does not follow. Step 3: Combine the two statements. From Statement II we have Luck implies Hard Work, and from Statement I we have Luck implies Success. Hard Work is therefore necessary for Luck, and Luck is a pathway to Success. In everyday language, this strongly supports the idea that Hard Work is a central factor or key to success. Step 4: Check Conclusion I. It says, “Hard work is the key to success.” Since success in the premises comes only when Luck appears, and Luck appears only when there is Hard Work, Hard Work becomes a necessary precondition in the chain that leads to success. This conclusion captures the overall message of the two statements and therefore follows. Step 5: Check Conclusion II. We already derived the contrapositive of Statement II: if you do not work hard, Luck does not follow. This is exactly what Conclusion II states. Hence, Conclusion II also follows directly. Verification / Alternative check: To verify, imagine two scenarios. In the first, someone works very hard; then according to the statements, Luck may follow them, and once they are lucky, things fall into place, meaning they achieve success. In the second scenario, a person does not work hard at all. Then, according to Statement II, Luck cannot follow, so the first link in the chain to success is broken. This illustrates both that Hard Work is the key factor enabling success and that without Hard Work, Luck does not appear. Thus both conclusions are consistent with and supported by the given premises. Why Other Options Are Wrong: Option a: “Only conclusion II follows” is incorrect because it ignores the overall chain created by the two statements, which clearly elevates Hard Work as the central condition for success, justifying Conclusion I as well. Option c: “Neither conclusion I nor conclusion II follows” is wrong because we have clearly derived both from the original premises. Option d: “Only conclusion I follows” is incomplete because the contrapositive of Statement II exactly matches Conclusion II. Option e: Suggesting that the conclusions are unrelated is incorrect; they are directly connected to the logical structure of the statements. Common Pitfalls: A common mistake is to treat luck and hard work as independent and to assume that success can happen without any one of them, which is not allowed once you accept the premises as absolutely true. Another error is failing to use contrapositives: many learners overlook that “Luck follows only when we work hard” also implies that lack of hard work stops luck from appearing. Always convert “only when” statements into a clear conditional and then use its contrapositive to find additional valid conclusions. Final Answer: Both conclusion I and conclusion II logically follow from the given statements.

Question 9

In the following comparison puzzle, assume all the inequality statements are true and then determine which of the given conclusions must be true. Statements: P > Q ≤ R > Z; Y > X > P ≥ U; Q ≥ S < T Conclusions: I. P > S II. Z < T

Accepted Answer

Introduction / Context: This question is a typical inequality or comparison problem often asked in logical reasoning sections. Several variables are connected with greater than, less than, and less than or equal to symbols. Your task is to determine which conclusions must be true in every possible arrangement that satisfies the given statements. This tests your ability to combine chains of inequalities and to distinguish between what is definitely true and what is only possibly true. Given Data / Assumptions: Statements: P > Q Q ≤ R R > Z Y > X X > P P ≥ U Q ≥ S S < T Conclusion I: P > S. Conclusion II: Z < T. All inequalities are consistent and refer to the same scale of comparison, such as height, marks, or any ordered quantity. Concept / Approach: The main technique is to build chains of inequalities using transitivity. If A > B and B ≥ C, then A > C. But if two elements do not share a chain, their relationship may remain undetermined. To check a conclusion, you must see whether it holds in all valid numerical assignments that satisfy the given statements. If you can construct even a single valid counterexample where a conclusion fails, that conclusion is not definitely true. Step-by-Step Solution: Step 1: Use P and Q. From P > Q and Q ≥ S, we can combine them: since Q is at least as large as S, P is definitely greater than Q and so also greater than S. Step 2: Formally, we have P > Q and Q ≥ S, therefore P > S. This directly supports Conclusion I as a definite truth. Step 3: Now check the parts involving Z and T. From the first chain, R > Z, so Z is less than R. From the last chain, S < T and Q ≥ S, so T is greater than S and S is at most Q. However, we are not given any link between Z and S, or Z and T, through a continuous chain of inequalities. Step 4: Try to see if Z < T must always hold. To disprove a universal conclusion, it is enough to find one possible arrangement where all the given statements are satisfied, but Z < T is not true. Step 5: Create a counterexample. Take numbers P = 9, Q = 7, R = 10, Z = 8, Y = 15, X = 11, U = 5, S = 5, T = 6. Then: P > Q is 9 > 7 (true) Q ≤ R is 7 ≤ 10 (true) R > Z is 10 > 8 (true) Y > X is 15 > 11 (true) X > P is 11 > 9 (true) P ≥ U is 9 ≥ 5 (true) Q ≥ S is 7 ≥ 5 (true) S < T is 5 < 6 (true) Here Z = 8 and T = 6, so Z > T, which directly contradicts Conclusion II. Verification / Alternative check: The counterexample confirms that while all the given statements hold, Z can be greater than T. Therefore, Z < T is not guaranteed by the premises. On the other hand, the relationship P > Q ≥ S always leads to P > S, no matter what valid numbers you choose. This means Conclusion I is robust across all scenarios, but Conclusion II is not. Checking multiple sample assignments is a practical way to gain confidence in your reasoning. Why Other Options Are Wrong: Option a: “None of the conclusions is true” is incorrect because Conclusion I has been shown to be true in every valid arrangement. Option c: “Only conclusion II is true” is wrong because we have explicitly built a situation where Z > T, so Conclusion II fails. Option d: “Either conclusion I or conclusion II is true” is incorrect because both can be checked separately, and we find that I is always true while II is not necessarily true. Option e: “Both conclusions I and II are true” is incorrect since Conclusion II does not hold in all possible cases. Common Pitfalls: Many candidates mistakenly assume that if two variables appear in the same group of inequalities, there must be a fixed relationship between them. However, unless there is a continuous chain connecting two variables, their order may remain undetermined. Another frequent error is not using counterexamples; if a relationship is not obvious, trying real numbers that satisfy the conditions can quickly show whether a conclusion is always true or only sometimes true. Final Answer: Only conclusion I is true.

Question 10

In the following symbolic inequality problem, assume all statements are true and decide which of the given conclusions must follow. Statements: Y ≤ K < D = S; D B II. Y < Z

Accepted Answer

Introduction / Context: This question belongs to the family of symbolic inequality reasoning problems. You are given several chained comparisons using symbols such as less than, greater than, and equal to. From these, you must determine which conclusions are definitely true in every possible arrangement of the variables. This tests your ability to track relative positions and to distinguish between what must be true and what may or may not be true. Given Data / Assumptions: Y ≤ K < D = S D B Conclusion II: Y < Z All variables represent comparable quantities on the same scale, and the inequality relations are consistent. Concept / Approach: To solve such problems, we form chains by linking inequalities that share common variables. This allows us to infer new relationships. However, it is important to check whether a conclusion is forced by the chain or if there is freedom to assign values such that the conclusion may fail. If we can construct even one valid example where the conclusion does not hold, then that conclusion is not definitely true. When dealing with symbols like ≤ and ≥, remember that equality is also possible, not just strict inequality. Step-by-Step Solution: Step 1: Start from Y ≤ K < D. We know D = S, but for the conclusions we mainly need D and Z. From the third statement we have D < Z, so Z is greater than D and everything that is less than or equal to D. Step 2: Link Y to Z. Since Y ≤ K and K < D, we can write Y ≤ K < D < Z. This means that Y is less than D, which is itself less than Z. Therefore Y must be less than Z. That directly supports Conclusion II: Y < Z is definitely true. Step 3: Now analyze Conclusion I, which claims A > B. From the data we know A ≥ D and D < B. This implies that B is greater than D, while A is at least as large as D. There are two possible cases: Case 1: A = D. Then we have D D. In this case A could be greater than B or still less than B, depending on the actual values, because we only know that B is above D, not how far above it. Since we can easily choose numbers where A is not greater than B, Conclusion I is not guaranteed. Step 4: Construct a concrete counterexample for Conclusion I. Take D = 5, so K = 4, Y = 3, S = 5, B = 7, O = 9, Z = 8, and A = 5. Then: Y ≤ K < D = S gives 3 ≤ 4 < 5 = 5, which is true. D B is false. However, Y = 3 and Z = 8, hence Y < Z is true. Verification / Alternative check: The counterexample satisfies all the given statements but makes Conclusion I false. Therefore, Conclusion I cannot be considered definitely true. In contrast, the chain Y ≤ K < D < Z always leads to Y < Z no matter which valid values you choose. There is no way to make Y greater than or equal to Z while keeping all the inequalities valid. This comparison confirms that only Conclusion II is logically guaranteed by the premises. Why Other Options Are Wrong: Option a: “None of the conclusions is true” is incorrect because Conclusion II always holds. Option b: “Only conclusion I is true” is clearly wrong because we have constructed an example where A is not greater than B. Option d: “Either conclusion I or conclusion II is true” is incorrect because there is no such either or situation; Conclusion II is always true, while Conclusion I is not. Option e: “Both conclusions I and II are true” fails because Conclusion I can be false in some valid arrangements. Common Pitfalls: A typical mistake is to assume that A ≥ D and D B. This would only be justified if A were strictly greater than D and D were very close to B in a way that forced A to exceed B, which is not given. Another pitfall is ignoring the possibility of equality in symbols like ≥ and ≤. Always consider the equality case when these symbols appear; it often provides the counterexample needed to disprove a conclusion. Final Answer: Only conclusion II is true.

Question 11

In the following inequality based reasoning question, assume all statements are true and then determine which of the given conclusions must follow. Statements: A > C; G > E; G ≤ C; R ≤ I; K ≤ I Conclusions: I. A > G II. C > E

Accepted Answer

Introduction / Context: This problem is another example of symbolic inequality reasoning. Several variables are compared with greater than and less than or equal to symbols. Your goal is to combine these relationships logically and decide which conclusions must be true in every possible arrangement that satisfies the given statements. This evaluates your comfort with chaining inequalities and using transitivity to infer new comparisons. Given Data / Assumptions: A > C G > E G ≤ C R ≤ I K ≤ I Conclusion I: A > G Conclusion II: C > E All variables represent quantities on a common ordered scale. Concept / Approach: The strategy is to focus on the parts of the data that are relevant to the conclusions. Here, R, I, and K are not involved in the conclusions, so they can be ignored. We then connect the remaining inequalities using transitivity. If we find that a conclusion holds under all possible numeric assignments consistent with the statements, then it is definitely true. If there exists any valid assignment where the conclusion fails, it is not definite. Step-by-Step Solution: Step 1: Look at Conclusion I, A > G. We know A > C and G ≤ C. From G ≤ C, G is at most C, and from A > C, A is strictly greater than C. Step 2: Combine these. Since A is greater than C and C is greater than or equal to G, A must also be greater than G. More formally: A > C and C ≥ G implies A > G. So Conclusion I is logically forced by the given inequalities. Step 3: Next, check Conclusion II, C > E. We know G > E and G ≤ C. From G > E, E is less than G. From G ≤ C, C is at least as large as G. Step 4: Apply transitivity again. If C is greater than or equal to G, and G is greater than E, then C must be greater than E. Expressed as a chain, C ≥ G > E implies C > E. This directly supports Conclusion II. Step 5: Confirm there is no counterexample. To see that both conclusions always hold, consider arbitrary numbers satisfying the original inequalities. For example, let C = 10, A = 12, G = 9, E = 5. Then: A > C becomes 12 > 10, true. G > E becomes 9 > 5, true. G ≤ C becomes 9 ≤ 10, true. Now check the conclusions: A > G is 12 > 9 (true) and C > E is 10 > 5 (true). Trying other consistent choices gives the same pattern: both conclusions remain true. Verification / Alternative check: Another way to verify is to reason symbolically. For Conclusion I: From A > C and C ≥ G, substitute C with any value at least G. A is greater than that value, so A must be greater than G. For Conclusion II: From G > E and G ≤ C, G is sandwiched between E and C. Because G is greater than E and C is no smaller than G, C must be greater than E. Since these chains rely only on transitivity and the definitions of the inequality symbols, there is no room to assign values in a way that breaks the conclusions while keeping the premises true. Why Other Options Are Wrong: Option b: “Only conclusion I is true” is incorrect because it ignores the strong chain C ≥ G > E that guarantees C > E. Option c: “Only conclusion II is true” is wrong since A > C ≥ G forces A > G. Option d: “Either conclusion I or conclusion II is true” suggests that sometimes only one might hold, but we have seen that both must hold together. Option e: “None of the conclusions is true” is clearly false because both conclusions are directly supported by the premises. Common Pitfalls: Learners sometimes fail to spot that when a variable is bounded on both sides, as G is between E and C, it can create strong relationships between the outer variables. Another common mistake is to think that a non-strict inequality like ≤ or ≥ is too weak to support strict conclusions. In fact, when combined correctly with strict inequalities, they often do yield strict results, as in C ≥ G and G > E giving C > E. Always trace the direction of inequalities carefully and remember that transitivity is your main tool in such questions. Final Answer: Both conclusions I and II are true.

Question 12

In this inequality reasoning question, assume all the statements are true and then decide which of the given conclusions must follow. Statements: Z ≤ K < D = S; D < A < O; G ≥ D < R Conclusions: I. G > A II. Z < R

Accepted Answer

Introduction / Context: This problem continues the theme of symbolic inequality reasoning. You are given several chains of inequalities relating different variables, and you are asked to determine which conclusions must always be true. The challenge is to merge the information from each chain, track the relative ordering of variables, and then test whether the proposed conclusions are logically forced or only sometimes true. Given Data / Assumptions: Z ≤ K < D = S D < A < O G ≥ D and D < R Conclusion I: G > A Conclusion II: Z < R All statements describe the relative order of the variables on a common scale. Concept / Approach: We rely on transitivity to connect variables that appear in overlapping inequalities. When you see a sequence like Z ≤ K < D and then D < R, you can form a longer chain Z ≤ K < D < R and deduce relationships between Z and R. For each conclusion, ask whether there is enough information to fix the order of the two variables involved. If you can construct a valid assignment of numbers that violates a conclusion while keeping all the statements true, that conclusion is not definite. Step-by-Step Solution: Step 1: Start with the first chain: Z ≤ K < D = S. This tells us that K is less than D and Z is less than or equal to K. Hence Z is also less than D. Step 2: From the third chain we know D < R. Combining this with the relationships from Step 1 gives Z ≤ K < D < R. This implies Z is less than R, since everything up to D is below R and Z is no greater than K, which is below D. Step 3: Therefore, Conclusion II (Z < R) is supported by a continuous chain: Z ≤ K < D < R, so Z is strictly less than R. Even if Z = K, Z is still less than D and thus less than R. Step 4: Now examine Conclusion I, G > A. From the data we have G ≥ D and D < A. So A is above D, while G is at least as large as D. There are several possibilities for G: Case 1: G = D. Then since D < A, we have G = D < A, which means G is not greater than A. Case 2: G > D. In this case G could be either greater than A or still less than A, depending on the actual numbers, because there is no direct comparison between G and A beyond the fact that both are above D. Because we can choose values where G ≤ A while all statements remain true, Conclusion I is not guaranteed. Step 5: Construct a counterexample for Conclusion I. Let D = 5, S = 5, K = 4, Z = 3, R = 8, A = 7, O = 9, and G = 5. Then: Z ≤ K < D = S gives 3 ≤ 4 < 5 = 5 (true). D < A < O gives 5 < 7 < 9 (true). G ≥ D < R gives 5 ≥ 5 < 8 (true). Here G = 5 and A = 7, so G > A is false, while Z = 3 and R = 8, so Z < R is true. Verification / Alternative check: By testing multiple sets of values, you will find that Z is always less than R because the chain Z ≤ K < D < R cannot be violated without breaking the given statements. In contrast, the relationship between G and A is underdetermined. Sometimes G can be greater than A, but other times it can be equal to or less than A, depending on how much larger than D we choose them to be. This confirms that only Conclusion II is logically forced by the premises. Why Other Options Are Wrong: Option a: “None of the conclusions is true” is incorrect because Conclusion II is always true. Option b: “Only conclusion I is true” is wrong because we have produced an example where G is not greater than A. Option d: “Either conclusion I or conclusion II is true” is incorrect because there is no such either or situation; Conclusion II is always true and Conclusion I is not. Option e: “Both conclusions I and II are true” fails because Conclusion I can be false in a valid arrangement. Common Pitfalls: A common error is assuming that G ≥ D and D < A automatically makes G > A. This is not valid because G could be exactly equal to D, which might be significantly less than A. Another pitfall is forgetting to propagate the inequalities properly, for example, not extending Z ≤ K < D with the further inequality D < R. Always make sure to connect chains fully before judging the truth of a conclusion. Final Answer: Only conclusion II is true.

Question 13

In the following set of symbolic inequalities, assume all statements are true and then decide which of the given conclusions must follow. Statements: C ≤ R ≤ N = M ≥ F; Q ≥ M < O; D ≥ L; C ≥ D ≤ S ≥ Z Conclusions: I. R ≤ F II. C < Q

Accepted Answer

Introduction / Context: This is a more involved symbolic inequality question that includes several variables and mixed relations using ≤, ≥, and = symbols. You are asked to identify which of the proposed conclusions must be true in all possible arrangements that satisfy the given statements. The challenge comes from properly chaining only the relevant inequalities and recognizing when the available information is not strong enough to fix the relative order of two variables. Given Data / Assumptions: C ≤ R ≤ N = M ≥ F Q ≥ M and M < O D ≥ L C ≥ D and D ≤ S and S ≥ Z Conclusion I: R ≤ F Conclusion II: C < Q All inequalities refer to the same underlying ordered quantity. Concept / Approach: When working with many variables, it is helpful to focus on the ones that appear in the conclusions. For Conclusion I, we care about R and F; for Conclusion II, we care about C and Q. We use transitivity to form chains involving these variables. A conclusion is definite only if every allowed assignment of numerical values keeps it true. If even one assignment satisfies all statements but breaks a conclusion, that conclusion is not definitely true. Step-by-Step Solution: Step 1: Analyze the chain involving R and F. From C ≤ R ≤ N = M ≥ F, we know R is less than or equal to N, and N equals M, which is greater than or equal to F. Step 2: Write this explicitly: R ≤ N and N = M and M ≥ F. This simplifies to R ≤ M and M ≥ F. There is no direct relation that compares R and F. They are both related to M but in different ways: R is at most M, F is at most M, yet either one could be smaller than the other. Step 3: Construct a counterexample for Conclusion I (R ≤ F). Take M = N = 10, R = 9, C = 8, F = 5. Then C ≤ R ≤ N = M ≥ F becomes 8 ≤ 9 ≤ 10 = 10 ≥ 5, which is true. In this example R = 9 and F = 5, so R is greater than F, and thus R ≤ F is false. This shows that Conclusion I is not always true. Step 4: Now look at Conclusion II involving C and Q. From C ≤ R ≤ N = M we know C ≤ M. From C ≥ D and D ≤ S and S ≥ Z, we know C is at least as large as D, but this does not directly affect Q. From Q ≥ M we know Q is at least as large as M, and M is at least as large as C. Step 5: Combine what we have: C ≤ M and Q ≥ M. This gives C ≤ M ≤ Q, so C ≤ Q. However, the conclusion claims C < Q, which is a strict inequality. Because equality is allowed, it is possible for C and Q to be equal. Step 6: Construct a counterexample for Conclusion II. Let M = N = Q = C = 10, R = 10, F = 8, O = 12, D = 9, L = 7, S = 11, Z = 6. Check: C ≤ R ≤ N = M ≥ F becomes 10 ≤ 10 ≤ 10 = 10 ≥ 8, true. Q ≥ M < O becomes 10 ≥ 10 < 12, true. D ≥ L is 9 ≥ 7, true. C ≥ D ≤ S ≥ Z becomes 10 ≥ 9 ≤ 11 ≥ 6, true. In this arrangement, C = Q = 10, so C < Q is false. This confirms that Conclusion II is not definite. Verification / Alternative check: The counterexamples prove that neither conclusion is forced by the premises. For Conclusion I, you can always let F be much smaller than both M and R while still keeping M ≥ F and R ≤ M valid. For Conclusion II, setting C and Q equal to M automatically satisfies all the inequalities but violates the strict inequality C < Q. Therefore, the safest and correct judgment is that none of the proposed conclusions must always hold. Why Other Options Are Wrong: Option b: “Only conclusion I is true” is wrong because we found a configuration where R is greater than F. Option c: “Only conclusion II is true” is incorrect due to the equal value example where C equals Q. Option d: “Either conclusion I or conclusion II is true” is not valid since we demonstrated cases where both conclusions fail simultaneously. Option e: “Both conclusions I and II are true” is clearly false; neither holds in all valid assignments. Common Pitfalls: A common mistake is to assume that if C ≤ M and Q ≥ M, then C must be strictly less than Q, ignoring the possibility of equality. Another pitfall is overextending chains and inferring relationships that are not supported, such as assuming R is less than or equal to F just because both are tied to M. Always test your intuition with sample values to verify whether a conclusion is unavoidable or if alternative arrangements exist that break it. Final Answer: None of the conclusions is true.

Question 14

In this syllogism style reasoning question, treat the following statements as true and decide which of the given conclusions must follow. Statements: No chair is a table. All chairs are benches. Some chairs are boxes. Conclusions: I. No table is a b…

Accepted Answer

Introduction / Context: This is a classic syllogism question involving three sets: chairs, tables, benches, and boxes. You are given categorical statements about how these sets overlap or do not overlap, and you must decide which conclusions follow beyond doubt. The challenge is to visualize or mentally model the relationships and distinguish certain conclusions from those that are merely possible or not supported at all. Given Data / Assumptions: Statement 1: No chair is a table. So the sets chairs and tables do not intersect. Statement 2: All chairs are benches. The set of chairs is completely inside the set of benches. Statement 3: Some chairs are boxes. At least one object is both a chair and a box. Conclusion I: No table is a box. Conclusion II: Some benches are boxes. All statements are to be treated as logically true, irrespective of real furniture. Concept / Approach: Syllogism questions are best handled either by drawing simple Venn diagrams or by thinking in terms of subsets and intersections. A conclusion must hold in every possible diagram that satisfies the statements. If you can construct even one diagram in which a conclusion fails while all statements remain true, that conclusion does not follow. Also, when we know that all elements of one set are contained in another, any overlap that the smaller set has with a third set will be inherited by the larger set. Step-by-Step Solution: Step 1: From Statement 2, all chairs are benches. So if something is a chair, it is definitely a bench. Step 2: From Statement 3, some chairs are boxes. That means there is at least one object that is both a chair and a box. Step 3: Combine Step 1 and Step 2: the object that is both a chair and a box must also be a bench, because every chair is a bench. Therefore, at least one bench is also a box. Step 4: This chain directly supports Conclusion II: some benches are boxes is definitely true. Step 5: Now examine Conclusion I: no table is a box. From the premises, we only know that no chair is a table, and some chairs are boxes. This allows us to say that those particular boxes that are chairs are not tables, but it tells us nothing about other boxes that might not be chairs. It is entirely possible that some non-chair boxes are tables. Step 6: Construct a counterexample for Conclusion I. Imagine one object that is a chair, a bench, and a box, but not a table (to satisfy the given statements). Also imagine a different object that is both a table and a box. This arrangement satisfies all the statements: No chair is a table: the chair object is not a table. All chairs are benches: the chair object is also a bench. Some chairs are boxes: the chair object is also a box. However, there is also a table that is a box, which makes Conclusion I false. Verification / Alternative check: The reasoning confirms that Conclusion II must always hold, because any overlap between chairs and boxes automatically becomes an overlap between benches and boxes due to the subset relationship. On the other hand, the relationship between tables and boxes is not constrained by the premises except that the chairs that are boxes cannot be tables. There may still be other boxes that are tables, so we cannot assert “no table is a box” as a universal truth. Why Other Options Are Wrong: Option a: “Only conclusion I follows” is incorrect because Conclusion I does not follow at all, while Conclusion II does. Option c: “Either conclusion I or conclusion II follows” is wrong because there is no such either or condition; Conclusion II always follows and Conclusion I does not. Option d: “Neither conclusion I nor conclusion II follows” fails because we have shown that Conclusion II definitely follows. Option e: “Both conclusion I and conclusion II follow” is false because Conclusion I is not logically forced. Common Pitfalls: One typical mistake is to confuse “some chairs are boxes” with “all boxes are chairs” or “all boxes are benches”, which are not given. Another common error is overgeneralizing from the fact that some boxes are chairs (and therefore not tables) to claim that no box can be a table. Always remember that the word “some” in logic means “at least one” and does not by itself restrict the rest of the set. Drawing quick sketches of circles often prevents these overgeneralizations. Final Answer: Only conclusion II follows.

Question 15

In this Venn diagram and possibility based reasoning question, treat the statements as true and decide which conclusion logically follows. Statements: Some rings are bracelets. All bracelets are earrings. All earrings are necklaces. Conclusions: I. …

Accepted Answer

Introduction / Context: This question tests your understanding of syllogisms with a focus on possibility based conclusions. You are given three nested set relationships among rings, bracelets, earrings, and necklaces. The conclusions talk about what is possible, not what is always true. Your task is to see which of the stated possibilities are compatible with the given statements and which are directly contradicted by them. Given Data / Assumptions: Statement 1: Some rings are bracelets. Statement 2: All bracelets are earrings. Statement 3: All earrings are necklaces. Conclusion I: All bracelets that are earrings can never be rings is a possibility. Conclusion II: There is a possibility that all rings are necklaces. All statements and conclusions are to be interpreted in the standard logical reasoning sense. Concept / Approach: Syllogism with possibilities requires checking compatibility. If a conclusion is a possibility, then we must see whether there exists at least one Venn diagram that satisfies all the given statements and also satisfies the content of that conclusion. If the conclusion contradicts any statement, even in the best arranged diagram, then it cannot be considered a valid possibility. The phrase “some rings are bracelets” forces an overlap between rings and bracelets, which must be respected in every diagram. Step-by-Step Solution: Step 1: Interpret the statements: Some rings are bracelets: there is at least one object that is both a ring and a bracelet. All bracelets are earrings: every bracelet lies completely within the earring set. All earrings are necklaces: the earring set lies completely within the necklace set. Step 2: From these, we can deduce that all bracelets are earrings and all earrings are necklaces, so all bracelets are necklaces. Furthermore, since some rings are bracelets, those rings are also earrings and necklaces. Thus some rings are earrings and some rings are necklaces. Step 3: Check Conclusion I carefully. It essentially states that there is a possible arrangement in which no bracelet that is an earring is a ring. However, from Statement 1 we know that some rings are bracelets. Because all bracelets are earrings, those rings that are bracelets are automatically also earrings. That means in every valid arrangement, there is at least one bracelet which is also a ring. Step 4: Conclusion I therefore contradicts the given fact that some rings are bracelets. You can never arrange the sets to satisfy both “some rings are bracelets” and “no bracelet is a ring” at the same time. Hence, Conclusion I is not a valid possibility. Step 5: Now examine Conclusion II: There is a possibility that all rings are necklaces. From the statements, we already know that some rings are necklaces (those that are bracelets), but we are not told anything about rings that are not bracelets. They could lie inside or outside the necklace set without violating any statement, as long as some rings overlap bracelets. Step 6: To see that Conclusion II is a valid possibility, construct a diagram where the entire set of rings lies inside the necklace set, and some portion of the rings intersects the bracelet set. Because bracelets are inside earrings and earrings are inside necklaces, this setup satisfies all the given statements and also makes every ring a necklace. Therefore, Conclusion II is indeed a possible situation. Verification / Alternative check: You can verify this by imagining concrete objects. Let there be three rings: R1, R2, and R3. Make R1 and R2 also bracelets. Since all bracelets are earrings and all earrings are necklaces, R1 and R2 are necklaces. Now place R3 somewhere inside the necklace set but outside the bracelet and earring sets. This arrangement preserves the fact that some rings are bracelets and ensures that all rings are necklaces. There is no conflict with any given statement, which confirms that Conclusion II is a valid possibility. Why Other Options Are Wrong: Option a: “Only conclusion I follows” is incorrect, since Conclusion I cannot be true in any arrangement that respects “some rings are bracelets”. Option c: “Either conclusion I or conclusion II follows” is wrong because only Conclusion II is compatible with the data. Option d: “Neither conclusion I nor conclusion II follows” is incorrect because we have just seen that Conclusion II is possible. Option e: “Both conclusion I and conclusion II follow” is impossible because Conclusion I contradicts the basic premise that some rings are bracelets. Common Pitfalls: A frequent mistake is misreading “some rings are bracelets” as “some bracelets are rings” and then casually assuming that it might be turned off in another arrangement. In logical reasoning, once “some rings are bracelets” is given as a premise, it must hold in every valid model of the situation. Another pitfall is assuming that a statement about possibility can override a hard given fact, which it cannot. Always start from the premises and only then test possible arrangements that could satisfy the conclusions. Final Answer: Only conclusion II follows.

Question 16

In this syllogism question with possibility based conclusions, treat the following statements as true and then decide which conclusion logically follows. Statements: Some pages are papers. No page is a book. All books are pencils. Conclusions: I. Al…

Accepted Answer

Introduction / Context: This is a syllogism problem involving four sets: pages, papers, books, and pencils. You are given three categorical statements and asked to decide which of the two possibility based conclusions follows. The main challenge is to understand what is rigidly fixed by the statements and what is still flexible, allowing different valid Venn diagrams. A conclusion that talks about possibility is valid if there exists at least one arrangement satisfying all the statements in which that conclusion also holds. Given Data / Assumptions: Some pages are papers. No page is a book. All books are pencils. Conclusion I: All books can never be papers. Conclusion II: It is possible that all pencils are papers. All statements are assumed to be logically true, regardless of real documents. Concept / Approach: To handle such questions, draw or imagine Venn diagrams representing the sets and their relationships. “Some pages are papers” enforces an overlap between pages and papers. “No page is a book” forces the page set to be completely outside the book set. “All books are pencils” makes the book set a subset of the pencil set. A possibility conclusion is valid if you can draw at least one diagram where it holds without breaking any given statement. Step-by-Step Solution: Step 1: Interpret the statements: Some pages are papers: at least one element lies in both the page and paper sets. No page is a book: the page set and book set do not overlap. All books are pencils: book set is a subset of the pencil set. Step 2: Notice that nothing in the statements explicitly links books to papers or pencils to papers. The only restriction is that pages and books are disjoint, and some pages lie inside papers. Step 3: Examine Conclusion I: “All books can never be papers.” For this to be a definite conclusion, it would have to be impossible for any arrangement to make all books papers. But the only restriction we have is that pages are not books. Books might still be papers, as long as the books lie in a region of the paper set where there are no pages. Step 4: Construct an arrangement to test this. Place the paper set as a large region. Inside it, draw a small region for pages on one side, and another disjoint region for books on another side. Ensure that the page region does not overlap the book region to satisfy “no page is a book”. Because all books lie in papers in this diagram, all books are papers without breaking any conditions. This shows that it is not true that “all books can never be papers”. So Conclusion I does not follow. Step 5: Now check Conclusion II: “It is possible that all pencils are papers.” From the statements, we know that all books are pencils, but we do not know anything about pencils that are not books. There is nothing preventing the pencil set from being placed completely inside the paper set, as long as the page and book constraints are respected. Step 6: Build a diagram where all pencils are papers. Let the paper set be the largest set. Place the pencil set entirely inside the paper set. Inside the pencil set, place the book set. On a separate part of the paper set, place the pages such that they overlap with papers but do not touch books. This satisfies: Some pages are papers: the page region overlaps with papers. No page is a book: the page region does not overlap with the book region. All books are pencils: the book region lies inside the pencil region. All pencils are papers: the pencil region lies inside the paper region. Thus Conclusion II is a valid possibility. Verification / Alternative check: The key is to see that the statements say nothing that forbids pencils from being entirely inside the paper set. Nor do they forbid books from being papers, as long as they are not pages. Therefore, the most flexible set is the pencil set, which you can place wholly inside the paper set without conflict. This confirms that Conclusion II, which talks about a possible arrangement, is logically correct, while Conclusion I makes a strong impossibility claim that the data do not support. Why Other Options Are Wrong: Option a: “Only conclusion I follows” is incorrect because Conclusion I has been shown not to be forced and, in fact, is contradicted by a valid arrangement. Option c: “Either conclusion I or conclusion II follows” is wrong because only Conclusion II passes the possibility test. Option d: “Neither conclusion I nor conclusion II follows” is incorrect since there clearly exists an arrangement where all pencils are papers. Option e: “Both conclusion I and conclusion II follow” is impossible because Conclusion I is not supported by the premises. Common Pitfalls: Learners often confuse “no page is a book” with “no book is a paper”, which is not stated. Another frequent error is assuming that a lack of direct connection between two sets means they can never overlap, when actually it means they may or may not overlap. For possibility questions, always ask “can I draw at least one picture where this holds and all statements are still true” rather than “is it always true”. Final Answer: Only conclusion II follows.

Question 17

In this advanced syllogism question with four statements, assume all statements are true and decide which of the given conclusions logically follow. Statements: All papers are books. All pencils are pens. No paper is a pencil. No stationary item is …

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Introduction / Context: This question extends standard syllogism reasoning by adding more sets and multiple universal negative statements. You must judge two conclusions that refer to possibility and combined membership across sets. The objective is to strictly follow the given relationships and see whether any diagram can satisfy both the statements and the conclusions. A conclusion fails if it violates even one of the premise relationships in every possible arrangement. Given Data / Assumptions: All papers are books. All pencils are pens. No paper is a pencil. No stationary item is a book. Conclusion I: Some papers are stationary items is a possibility. Conclusion II: Some pens are books as well as papers. All sets (papers, books, pencils, pens, stationary items) are to be interpreted as sets in a Venn diagram sense. Concept / Approach: The main tools are subset and disjoint set reasoning. Statements like “All papers are books” mean the papers set lies completely inside the books set. “No stationary item is a book” means stationary items and books are disjoint. To test a possibility conclusion, you must look for at least one Venn diagram that satisfies all premises and also satisfies the conclusion. If such a diagram cannot be drawn because the conclusion always conflicts with the premises, then that conclusion does not follow. Step-by-Step Solution: Step 1: Translate the first two statements: All papers are books: Papers ⊂ Books. All pencils are pens: Pencils ⊂ Pens. Step 2: Translate the next two statements: No paper is a pencil: Papers and Pencils are disjoint. No stationary item is a book: Stationary and Books are disjoint. Step 3: Evaluate Conclusion I: “Some papers are stationary items is a possibility.” If some papers were stationary items, that would mean the intersection of Papers and Stationary is non empty. But all papers are books, and stationary items are disjoint from books. If an element lies in both Papers and Stationary, it would lie in Books (because all papers are books) and also in Stationary, which contradicts “No stationary item is a book.” Step 4: Because any overlap between Papers and Stationary would violate the rule that Stationary and Books are disjoint, there is no diagram where some papers are stationary items. Therefore, Conclusion I is not a valid possibility. Step 5: Now examine Conclusion II: “Some pens are books as well as papers.” For an object to be a paper and a book, it must be in the Papers subset, which lies inside Books. The conclusion also says that this object is a pen. Step 6: However, we must also respect “No paper is a pencil” and “All pencils are pens.” Nothing directly forbids some pens from being books or papers, but we must check for conflicts. If a paper is also a pen, that paper is not a pencil (by the third statement). There is no direct contradiction here, so at first glance it may seem possible. Step 7: Look more carefully. The given statements do not guarantee that any pen overlaps with books or papers at all. Pens might form a completely separate set from books. Because a conclusion in standard syllogism must follow in every valid arrangement, we must test if there is a valid diagram where no pen is a book or paper. Step 8: Construct such a diagram: place Papers inside Books, and Stationary outside Books. Place Pencils inside Pens, and then place the entire Pens set completely outside the Books set. This satisfies: All papers are books: Papers ⊂ Books. All pencils are pens: Pencils ⊂ Pens. No paper is a pencil: Papers and Pencils are disjoint. No stationary item is a book: Stationary is disjoint from Books. In this arrangement, there is no pen that is also a book or a paper. Therefore, it is not necessary that some pens are books and papers. Verification / Alternative check: The key is to distinguish “can be” from “must be”. The premises allow pens to be totally separate from books, so we cannot insist that some pens are books and papers. That makes Conclusion II not logically necessary. Meanwhile, any attempt to make some papers stationary items immediately creates an element lying in both Books and Stationary, which contradicts the given “No stationary item is a book.” This shows that Conclusion I is also impossible. Hence neither conclusion follows from the premises. Why Other Options Are Wrong: Option a: “Only conclusion I follows” is false because Conclusion I contradicts the disjointness of Stationary and Books. Option b: “Only conclusion II follows” is wrong because we have shown a valid model where no pen is a book, so the existence of a pen that is also a paper is not required. Option c: “Either conclusion I or conclusion II follows” is incorrect since both can fail simultaneously in the same diagram. Option e: “Both conclusion I and conclusion II follow” is clearly false for the same reasons. Common Pitfalls: Students often confuse logical necessity with mere plausibility. Just because it seems natural that pens, books, and papers might overlap in everyday life does not mean the premises force that overlap. Another mistake is overlooking how one universal negative (no stationary item is a book) interacts with a universal positive (all papers are books), which together completely forbid any paper from being stationary. Always check for hidden contradictions that arise when you try to enforce a possibility conclusion. Final Answer: Neither conclusion I nor conclusion II follows.

Question 18

Six persons Akhilesh, Bhavesh, Cindy, Divakar, Erika, and Farheen have different heights. Using the given statements, determine who is the tallest person and decide which data statements are sufficient to answer the question. Question: Who is the ta…

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Introduction / Context: This problem is a data sufficiency question involving relative heights of six people: Akhilesh, Bhavesh, Cindy, Divakar, Erika, and Farheen. You must decide which statements give enough information to uniquely identify the tallest person. Importantly, the task is not to find the exact order in every case but to judge whether the information guarantees a single tallest person. This type of question evaluates your skill in combining rank information from different clues. Given Data / Assumptions: All six persons have different heights. Statement I: Exactly two persons are taller than Bhavesh. Exactly one person is shorter than Erika. Akhilesh is taller than Cindy but is not the tallest. Statement II: Divakar is taller than Akhilesh but shorter than Farheen. Akhilesh is taller than Cindy but shorter than Bhavesh. Question: Who is the tallest person? Concept / Approach: In data sufficiency, we consider each statement alone and then both together. A statement is sufficient if it leads to a unique answer in every possible arrangement satisfying that statement. When combining statements, we merge all the given constraints and see whether the added information removes any ambiguity that existed when using each statement individually. Ranking style clues often translate into fixed positions like “third tallest” or “second shortest”, which help pin down individuals in the order. Step-by-Step Solution: Step 1: Analyze Statement I alone. If only two persons are taller than Bhavesh among six people, Bhavesh must be the third tallest. If only one person is shorter than Erika, Erika must be the second shortest, that is, fifth in height order. Step 2: From Statement I, we also know Akhilesh is taller than Cindy but not the tallest. However, Divakar and Farheen are not linked to these fixed positions. Therefore, with Statement I alone, we know: Bhavesh is third tallest. Erika is fifth tallest. Akhilesh is taller than Cindy and not tallest. There are still multiple possibilities for who is tallest (it could be Divakar or Farheen, or possibly someone else depending on arrangements), so Statement I alone is not sufficient. Step 3: Consider Statement II alone. From it we have: Farheen is taller than Divakar, and Divakar is taller than Akhilesh. Akhilesh is taller than Cindy but shorter than Bhavesh. This gives a chain: Farheen > Divakar > Akhilesh > Cindy, and Bhavesh is above Akhilesh but below some unknown. Erika is not mentioned at all. There is still uncertainty about who is tallest because it could be Farheen or possibly another person depending on Erika's position. Hence Statement II alone is also not sufficient. Step 4: Now combine Statements I and II. From Statement I we know Bhavesh is third tallest and Erika is fifth tallest. From Statement II we have the chain Farheen > Divakar > Akhilesh > Cindy and Bhavesh > Akhilesh. Step 5: Use the fixed ranks to place everyone. Since Bhavesh is third tallest, there must be exactly two people taller than him and three shorter. Erika, being fifth tallest, is the second shortest, meaning only one person is shorter than her. Let us number positions from 1 (tallest) to 6 (shortest): Bhavesh: position 3 (from Statement I). Erika: position 5 (from Statement I). Step 6: Since Erika is fifth, only one person can be below her, so Cindy must be sixth (shortest). From Statement II, Akhilesh is taller than Cindy and shorter than Bhavesh. With Bhavesh fixed at 3, Akhilesh cannot be 1 or 2 (that would make him taller than Bhavesh) and cannot be 5 or 6 (because he is taller than Cindy and Erika is already at 5). Thus Akhilesh must be at position 4. Step 7: So far we have: Position 3: Bhavesh Position 4: Akhilesh Position 5: Erika Position 6: Cindy The remaining positions 1 and 2 must be occupied by Divakar and Farheen. Step 8: From Statement II, Farheen is taller than Divakar. Therefore Farheen must be in position 1 and Divakar in position 2. Thus, the tallest person is uniquely determined to be Farheen when both statements are used together. Verification / Alternative check: You can try to rearrange the six people while keeping all statements true and the rank positions fixed. You will see that the only way to satisfy all conditions is with Farheen as the tallest. Any attempt to place someone else at the top will violate at least one part of the combined clues. However, each statement taken individually leaves room for multiple tallest candidates, confirming that only the combination is sufficient. Why Other Options Are Wrong: Option a: Statement I alone is not sufficient, as it leaves the top two positions ambiguous. Option b: Statement II alone is not sufficient, as Erika's height is unknown and the tallest could be Farheen or someone else. Option d: Saying that both together are not sufficient is incorrect because together they uniquely identify Farheen as tallest. Option e: Either statement alone being sufficient is clearly false based on the analysis above. Common Pitfalls: A common mistake is to misinterpret “only two persons are taller than Bhavesh” or “only one person is shorter than Erika” and not convert them into exact rank positions. Another frequent error is failing to merge the two statements properly, for example by not noticing that who is third tallest in one statement must still align with the chain of comparisons in the other. Always write down approximate rankings and adjust step by step until the order is consistent. Final Answer: The data in both statements I and II together are sufficient to answer the question, but the data in either statement alone are not sufficient.

Question 19

In the following data sufficiency question on family relations, use the given statements to decide whether the data are sufficient to answer the question. Question: H is the mother of T. How is T related to W? Statement I: W is the only daughter of …

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Introduction / Context: This is a data sufficiency question involving family relationships. We know that H is the mother of T, and we are given two additional statements about W. The goal is not simply to find the exact relationship but to determine which statements provide enough information to answer the question, “How is T related to W?” This evaluates your ability to use relational clues and gender information in logical reasoning. Given Data / Assumptions: Base fact: H is the mother of T. Statement I: W is the only daughter of H. Statement II: W is the sister of T. All relationships are standard: mother, daughter, sister, brother, and so on. The answer required is T's relation to W (for example, brother, sister, sibling). Concept / Approach: For family relation questions, we track parent child links and gender information. A statement is sufficient if it allows a single unambiguous relationship to be inferred. We must consider each statement separately first, and then together, to decide whether the data are sufficient. Note that the given base fact, “H is the mother of T,” already tells us that T is a child of H and that H is female. Additional statements may reveal T's gender and sibling relationships. Step-by-Step Solution: Step 1: Use Statement I with the base fact. H is the mother of T, so T is a child of H. Statement I says W is the only daughter of H. Since W is the only daughter, there cannot be any other daughter of H. Step 2: T is already known to be a child of H. If W is the only daughter of H, then T cannot be a daughter, otherwise W would not be the only daughter. That means T must be a son of H. Step 3: From the base fact and Statement I together, H is the mother of both W (daughter) and T (son). This makes W and T siblings, with W as the sister and T as the brother. The question asks, “How is T related to W?” The answer is: T is W's brother. Step 4: Therefore, Statement I alone allows us to uniquely determine T's relation to W. So Statement I is sufficient. Step 5: Now use Statement II alone with the base fact. We know H is the mother of T, and Statement II says W is the sister of T. Knowing that W is T's sister tells us that W is female and that W and T share at least one parent. Step 6: However, Statement II alone does not tell us whether H is also W's mother or only T's mother. The question is still, “How is T related to W?” Since we do not know T's gender from Statement II alone (T could be male or female), T could be W's brother or W's sister. Thus T's exact relation to W is ambiguous with Statement II alone. Verification / Alternative check: To confirm, construct two possible scenarios consistent with Statement II: Scenario 1: T is male and H is mother of both T and W. Then W is T's sister and T is W's brother. Scenario 2: T is female and H is mother of both T and W. Then W is T's sister and T is also W's sister. Since both scenarios satisfy the base statement and Statement II but give different answers to how T is related to W, Statement II alone is not sufficient. Why Other Options Are Wrong: Option b: Claiming that Statement II alone is sufficient is wrong because T's gender remains unknown and two different relationships are possible. Option c: Saying that both statements together are required is unnecessary, because Statement I already gives a complete answer by itself. Option d: Suggesting that even both statements together are not sufficient is incorrect, as with Statement I we already pinpoint T's relationship as brother. Option e: Claiming that either statement alone is sufficient is false because Statement II alone is not sufficient. Common Pitfalls: A frequent mistake is to assume that when one person is a sister, the other must be a brother, which is not logically necessary. Another common error is to ignore the phrase “only daughter” in Statement I, which is crucial for deducing that T must be male. Always carefully track both gender and parent child links when dealing with such questions, and check whether more than one relationship remains possible. Final Answer: The data in statement I alone are sufficient to answer the question, but the data in statement II alone are not sufficient.

Question 20

In the following syllogism question, treat the statements as true and decide which of the given conclusions logically follow. Statements: No books are notebooks. All diaries are books. Conclusions: I. No notebooks are diaries. II. All diaries are no…

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Introduction / Context: This is a straightforward syllogism problem dealing with three sets: books, notebooks, and diaries. You must decide which of the conclusions follow from the given statements when they are treated as absolutely true. The key is to interpret the set relations accurately and see what they imply for possible overlaps between notebooks and diaries. Given Data / Assumptions: Statement 1: No books are notebooks. The sets books and notebooks are disjoint. Statement 2: All diaries are books. The diaries set is completely inside the books set. Conclusion I: No notebooks are diaries. Conclusion II: All diaries are notebooks. All statements are to be taken as logically precise. Concept / Approach: We use Venn diagram logic and subset rules. If one set is a subset of another, and that larger set is disjoint from a third set, then the subset must also be disjoint from that third set. Conversely, if a subset is proposed to lie inside a set that is known to be disjoint from its super set, the proposal contradicts the premises. We must test both conclusions against these rules. Step-by-Step Solution: Step 1: From “No books are notebooks,” we know that Books ∩ Notebooks = empty. There is no element that is both a book and a notebook. Step 2: From “All diaries are books,” we know that Diaries is a subset of Books. Every diary is definitely a book. Step 3: Consider Conclusion I: “No notebooks are diaries.” If all diaries are books, and no books are notebooks, then no diary can be a notebook, because that would make a diary both a book and a notebook. Formally, since Diaries ⊂ Books and Books ∩ Notebooks = empty, it follows that Diaries ∩ Notebooks = empty. So Conclusion I is logically correct and must follow. Step 4: Now evaluate Conclusion II: “All diaries are notebooks.” This claims that the set Diaries is a subset of Notebooks. However, we already know that Diaries are a subset of Books, and Books and Notebooks do not intersect at all. If any diary were a notebook, that diary would be forced to lie in both Books and Notebooks, which is impossible. Step 5: Because Conclusion II directly contradicts the given relationship between Books and Notebooks, it cannot follow. No diagram can satisfy both premises and the claim that all diaries are notebooks. Therefore, Conclusion II is false. Verification / Alternative check: Draw a Venn diagram with three sets: Books, Notebooks, and Diaries. Place the Diaries circle entirely inside the Books circle. Now, because Books and Notebooks are disjoint, the Notebooks circle must lie outside the Books circle with no overlap. As a result, it is visually clear that no part of the Diaries circle can touch the Notebooks circle. This means no diary is a notebook, confirming Conclusion I. At the same time, there is no way to redraw the diagram so that Diaries lies inside Notebooks while still staying inside Books and not overlapping with Notebooks. This reinforces that Conclusion II is impossible. Why Other Options Are Wrong: Option b: “Only conclusion II follows” is incorrect because Conclusion II conflicts with the premises. Option c: “Both conclusions I and II follow” is impossible, since the two conclusions contradict each other in light of the given data. Option d: “Neither conclusion I nor conclusion II follows” is wrong because Conclusion I is clearly forced by the subset and disjointness relationships. Option e: “Either conclusion I or conclusion II follows” wrongly suggests that exactly one of them holds in different scenarios, but in fact only Conclusion I is compatible with the premises. Common Pitfalls: A common error is to assume that if two sets do not overlap, a third set linked to one of them might still overlap with the other, even when the logical structure prevents it. Another mistake is not fully visualizing the subset relationships. Always remember that if a set is completely inside another set, and that larger set has no overlap with a third set, then the smaller set also cannot overlap with that third set. Final Answer: Only conclusion I follows.

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