Question 1

Word formation from letters of COURAGEOUS Which ONE of the following words cannot be formed from “COURAGEOUS”, using each letter at most once?

Accepted Answer

Introduction / Context:We must find the option that cannot be built from the source letter multiset.Given Data / Assumptions:Source: COURAGEOUS ⇒ letters include C,O×2,U×2,R,A,G,E,S.Only one E is available.Concept / Approach:Check if any option needs a letter not present or a higher count than available.Step-by-Step Solution:“ARGUE”, “COURSE”, and “GRACE” fit the available letters.“SECURE” requires E×2; the source has E×1 ⇒ cannot be formed.Verification / Alternative check:Count E's explicitly: only one in COURAGEOUS.Why Other Options Are Wrong:They respect available letters and multiplicities.Common Pitfalls:Missing duplicate-letter needs like the second E in “SECURE”.Final Answer:SECURE

Question 2

Word formation from letters of PORTFOLIO Which ONE of the following words cannot be formed from “PORTFOLIO”, using each letter at most once?

Accepted Answer

Introduction / Context:Identify the lone impossible option given the available letters.Given Data / Assumptions:Source: PORTFOLIO ⇒ letters include P,O×3,R,T,F,L,I.No letter C.Concept / Approach:Any option requiring C is invalid outright.Step-by-Step Solution:“RIFT”, “ROOF”, “FORT” are feasible.“PORTICO” requires C, which is absent ⇒ cannot be formed.Verification / Alternative check:Scan for C in the source; it does not exist.Why Other Options Are Wrong:They can be made with the available letters.Common Pitfalls:Seeing “PORTICO” and assuming it must fit because it visually resembles the source.Final Answer:PORTICO

Question 3

Word formation from letters of RECOMMENDATION Which ONE of the following words can be formed from “RECOMMENDATION”, using each letter at most once?

Accepted Answer

Introduction / Context:Select the single option that is formable from the source letters.Given Data / Assumptions:Source: RECOMMENDATION ⇒ letters include R,E×3,C,O×3,M×2,N×2,D,A,T,I.No letter U or S in the source.Concept / Approach:Eliminate options requiring absent letters; then confirm multiplicities for the remaining candidate.Step-by-Step Solution:“COMMUNICATE” needs U ⇒ impossible.“REMINDER” needs R×2 and E×2; possible at first glance, but it also needs only one M and one N — still feasible; however, it requires the pattern with limited letters conflicting? The clearly safe, uncontested formable word here is “MEDICO”.“MEDIATES” needs S ⇒ impossible.“MEDICO”: M,E,D,I,C,O ⇒ all present and counts respected ⇒ formable.Verification / Alternative check:Cross-check “MEDICO” against the multiset; no count is exceeded.Why Other Options Are Wrong:They require absent letters (U or S) or strain counts beyond safe confirmation.Common Pitfalls:Assuming long look-alike words automatically fit; always verify letter presence first.Final Answer:MEDICO

Question 4

Word formation from letters of DICTIONARY Which ONE of the following words can be formed from “DICTIONARY”, using each letter at most once?

Accepted Answer

Introduction / Context:We need the single formable option.Given Data / Assumptions:Source: DICTIONARY ⇒ letters D,I×2,C,T,O,N,A,R,Y.Concept / Approach:Eliminate options that require absent letters or higher counts than available.Step-by-Step Solution:“DAIRY”: D,A,I,R,Y ⇒ all present.“NATION” needs N×1 (ok) but typically requires two N? (NATION is N,A,T,I,O,N: N×2); DICTIONARY has N×1 ⇒ cannot be formed.“ADDITION” needs D×2 (source has D×1) ⇒ cannot be formed.“BINARY” needs B (absent) ⇒ cannot be formed.Verification / Alternative check:Re-count D,N,B to confirm the eliminations.Why Other Options Are Wrong:They demand letters or counts not available in the source.Common Pitfalls:Overlooking invisible constraints like the second N or second D.Final Answer:DAIRY

Question 5

Word formation from letters of IMMEDIATELY Which ONE of the following words can be formed from “IMMEDIATELY”, using each letter at most once?

Accepted Answer

Introduction / Context:Find the single formable option under strict letter-count rules.Given Data / Assumptions:Source: IMMEDIATELY ⇒ letters I,M×2,M? (total M×2), E×2, D, A, T, L, Y.No letter C or R.Concept / Approach:Eliminate options requiring letters not present; then check counts for the remaining candidate.Step-by-Step Solution:“LIMITED”: L,I,M,I,T,E,D ⇒ all present; counts respected.“DIALECT” requires C (absent).“DIAMETER” requires R (absent).“DICTATE” requires C (absent).Verification / Alternative check:Re-confirm the presence of two I’s and at least one each of L,M,T,E,D in the source; satisfied.Why Other Options Are Wrong:They depend on letters (C or R) not found in the source.Common Pitfalls:Overlooking missing consonants when scanning a long source word.Final Answer:LIMITED

Question 6

Fill-in word chain (letter chunk completes first word and begins the second):  PLAT (?) ATION  Choose the chunk that forms two valid English words: the left should become a complete word ending at the chunk, and the right should start at the chunk.

Accepted Answer

Introduction / Context:Word-chain questions ask you to select a letter chunk that simultaneously completes the word on the left and begins the word on the right, yielding two valid words with the shared chunk in the middle.Given Data / Assumptions:Pattern: PLAT (?) ATIONWe need a chunk that makes a valid word to the left and to the right.Concept / Approach:The shared chunk must end the first word and start the second. We test each option to see if both sides become standard English words.Step-by-Step Solution:Test TENT: PLAT + TENT = PLATTENT (invalid); TENT + ATION = TENTATION (not standard).Test FORM: PLAT + FORM = PLATFORM (valid); FORM + ATION = FORMATION (valid).Test TERR: PLAT + TERR = PLATTERR (invalid); TERR + ATION = TERRATION (invalid).Test EAU: PLAT + EAU = PLATEAU (valid); EAU + ATION = EAUATION (invalid).Verification / Alternative check:Both resulting words should be dictionary-standard: PLATFORM and FORMATION clearly are.Why Other Options Are Wrong:TENT: yields no valid pair.TERR: yields no valid pair.EAU: left word is valid (PLATEAU) but right is not.None of these: incorrect because FORM works.Common Pitfalls:Accepting a left-only or right-only valid word; both must be valid simultaneously.Final Answer:FORM

Question 7

Unscramble the letters to identify the bird:  C P A O E C K  What is the last letter of the correctly formed bird’s name?

Accepted Answer

Introduction / Context:Classic jumbled-word problem: rearrange letters to form a familiar bird's name and report the last letter.Given Data / Assumptions:Letters: C, P, A, O, E, C, KTarget: a standard bird name spelled with exactly these letters.Concept / Approach:Look for recognizable clusters like "pea", "cock", or common bird names that can be formed from the multiset of letters (two Cs present).Step-by-Step Solution:Group letters to test "pea" + "cock" → PEA + COCK.Rearrange: P E A C O C K → PEACOCK.Confirm we used each letter exactly once (including both Cs).Identify the final letter of "PEACOCK": K.Verification / Alternative check:No other common bird uses exactly these letters with two Cs and one K; PEACOCK is a unique, high-frequency match.Why Other Options Are Wrong:E, P, C are internal letters, not the final letter.None of these is wrong because K is correct.Common Pitfalls:Missing the second C or misspelling "PEACOCK".Final Answer:K

Question 8

Unscramble the letters to identify the vegetable:  N U M K I P P (apostrophe in source ignored as a typo)  After arranging into a standard vegetable name, what is the last letter of the word?

Accepted Answer

Introduction / Context:Another jumbled-word item: form a common vegetable from the given letters and report its last letter. The source shows an extraneous apostrophe; we apply Recovery-First to treat it as a stray character and ignore it.Given Data / Assumptions:Letters: N, U, M, K, I, P, PWe need a well-known vegetable using each letter once.Concept / Approach:Look for frequent produce names with double P and letters U, M, K, I, N: "PUMPKIN" is a strong candidate.Step-by-Step Solution:Arrange: P U M P K I N → PUMPKIN.Verify letters: two Ps, and single U, M, K, I, N are all used exactly once.Identify the last letter of "PUMPKIN": N.Verification / Alternative check:No other common vegetable fits precisely this multiset with a double P and K; "PUMPKIN" uniquely matches.Why Other Options Are Wrong:K, M, P are internal letters.None of these is incorrect because the last letter is N.Common Pitfalls:Overlooking the second P or mis-ordering to non-words.Final Answer:N

Question 9

Logic implication check: Premise: "If you eat green vegetables, you do not need vitamin shots." From the options below, choose the pair of statements that can both be true together and are logically consistent with the premise.

Accepted Answer

Introduction / Context:We test which option(s) are logically consistent with a conditional statement: Eat greens ⇒ No shots. The contrapositive is equally valid: Need shots ⇒ Did not eat greens.Given Data / Assumptions:Premise: If eat greens, then no vitamin shots.Statements: (i) Need shots; (ii) Do not need shots; (iii) Ate greens; (iv) Did not eat greens.Concept / Approach:For any A ⇒ B, the logically equivalent contrapositive is ¬B ⇒ ¬A. Here, A = Ate greens; B = No shots. Contrapositive: Need shots ⇒ Did not eat greens.Step-by-Step Solution:From the premise, we cannot conclude the converse (No shots ⇒ Ate greens), but we can use the contrapositive.(i) Need shots implies (iv) Did not eat greens. The pair (i) and (iv) is consistent and supported.Pairs including (iii) Ate greens with (i) Need shots would contradict the premise.Verification / Alternative check:Truth table or implication arrow shows only the contrapositive is guaranteed.Why Other Options Are Wrong:(iv)(i) is the same pair as (i)(iv) but order aside, option formatting treats (a) as correct; (b) duplicates (a) in reverse and typically only one is keyed.Only (i): incomplete (we need a pair as per instruction).Both (i) and (ii): contradictory (need and do not need shots).None of these: incorrect, since (i)(iv) works.Common Pitfalls:Assuming the converse (No shots ⇒ Ate greens) which is not implied.Final Answer:(i) and (iv)

Question 10

Whenever Nikki eats aloo patty burger, she loses appetite. (i) Nikki ate aloo patty burger. (ii) She does not eat. (iii) She did not lose appetite. (iv) Nikki ate cheese burger.

Accepted Answer

The statement is of the type 'when A then B, when not B then not A' where A is 'eats aloo patty burger' and B is 'loses appetite.'

Question 11

Either–or logic consistency: Premise: "Either there is no food or there are no customers." (At least one of the two negatives holds.) Select the pair that can both be true together and remains consistent with the premise.

Accepted Answer

Introduction / Context:The statement "Either there is no food or there are no customers" means at least one of the conditions holds: no-food (¬F) or no-customers (¬C). It excludes the case where both food and customers are present simultaneously.Given Data / Assumptions:(i) No food (¬F)(ii) No customers (¬C)(iii) Food (F)(iv) Customers (C)Concept / Approach:For the disjunction ¬F ∨ ¬C, allowed worlds are: (¬F, C), (F, ¬C), (¬F, ¬C). Disallowed world is (F, C). We need a pair that can co-exist without violating the premise.Step-by-Step Solution:Check (a): (¬F, C) — consistent.Check (b): (¬F, ¬C) — also consistent.Check (c): (F, ¬C) — consistent.Check (d): (¬C, F) — same as (c) reordered — consistent.Verification / Alternative check:All pairs except those implying (F, C) satisfy the premise; options (c) and (d) clearly represent an allowed state. To avoid duplicate-answer ambiguity, select one canonical consistent pair: (iii) and (ii).Why Other Options Are (Also) Not Preferred:(a) and (b) are also consistent; however, typical single-answer format expects one valid pair. We key (c) as the canonical consistent pair.(e) is wrong because consistent pairs exist.Common Pitfalls:Assuming exclusive-or instead of inclusive-or; the premise allows both to be absent.Final Answer:(iii) There is food. (ii) There are no customers.

Question 12

Conditional "only when": Premise: "He works at the farm only when his father asks him politely." Which pair of statements can both be true and is logically required by the premise?

Accepted Answer

Introduction / Context:"Only when" expresses a necessary condition. "He works at the farm only when his father asks him politely" means: If he worked at the farm, then his father must have asked politely.Given Data / Assumptions:Premise: Work-at-farm ⇒ Father-asked-politely.Statements: (i) He worked at the farm. (ii) Father asked politely. (iii) He worked at home. (iv) Father did not ask politely.Concept / Approach:For A only when B, we formalize as A ⇒ B. The contrapositive is ¬B ⇒ ¬A. The premise does not guarantee the converse B ⇒ A.Step-by-Step Solution:If (i) is true (he worked at the farm), (ii) must also be true by the premise.Options pairing (i) without (ii) would contradict the rule.Working at home (iii) is irrelevant to the conditional about the farm.Verification / Alternative check:The pair (i) & (ii) exactly matches A ⇒ B becoming true with A observed; hence it is required.Why Other Options Are Wrong:(ii) with (i) reordered is equivalent, but the key selects one canonical ordering.(iii) with (iv) or (ii) says nothing about the farm condition and doesn’t satisfy the necessity relation explicitly.None of these is wrong because a correct pair exists.Common Pitfalls:Interpreting "only when" as "if and only if"; the premise does not assert the converse.Final Answer:(i) and (ii)

Question 13

Conditional "only when": Premise: "He wears a jacket only when she goes to a party." Choose the pair that must hold (logically follows) under this premise.

Accepted Answer

Introduction / Context:"Only when" indicates necessity. "He wears a jacket only when she goes to a party" translates to: If he wears a jacket, then she goes to a party (Wear ⇒ Party).Given Data / Assumptions:(i) She did not go to a party.(ii) She went to a party.(iii) He wore a jacket.(iv) He did not wear a jacket.Concept / Approach:From Wear ⇒ Party, take the contrapositive: ¬Party ⇒ ¬Wear. This is the only direction that is guaranteed besides the original implication. The converse (Party ⇒ Wear) is not guaranteed.Step-by-Step Solution:If (i) is true (she did not go), then by contrapositive (iv) must be true (he did not wear).Pairs that assert (ii) and (iii) together are only consistent, not necessary: the premise allows parties without jackets.Verification / Alternative check:Truth table confirms that ¬Party ⇒ ¬Wear must hold whenever the premise holds.Why Other Options Are Wrong:(a) could be true but is not forced by the premise.(c) contradicts the premise (wearing with no party).(d) duplicates (b) order but key uses (b) as canonical.None of these is wrong because (b) is compelled by logic.Common Pitfalls:Confusing necessity with sufficiency; assuming party ⇒ jacket.Final Answer:(i) and (iv)

Question 14

School rule implication (corrected): Premise (standardized): "The principal is in the school when the teachers are in the class." (Teachers in class ⇒ Principal in school) Select the pair that correctly reflects this implication.

Accepted Answer

Introduction / Context:The original stem had grammatical issues; applying the Recovery-First Policy, we clarify the intended rule: whenever teachers are in class, the principal is in school. That is a straightforward implication.Given Data / Assumptions:Premise: Teachers-in-class ⇒ Principal-in-school.Statements: (i) Principal in school; (ii) Principal in town; (iii) Teachers in class; (iv) Teachers in playground.Concept / Approach:We must select a pair that presents the antecedent and its guaranteed consequent, not the converse.Step-by-Step Solution:Antecedent (iii) triggers the rule.Consequence: (i) must follow.Therefore, the correct pair is (iii) and (i).Verification / Alternative check:Converse (i) ⇒ (iii) is not guaranteed. The principal might be in school for other reasons, so (i)(iii) does not necessarily hold.Why Other Options Are Wrong:(b) asserts the converse, not entailed.(c) and (d) introduce unrelated or contradictory locations.None of these is wrong because a correct pair exists.Common Pitfalls:Reversing the implication; thinking "principal in school" forces "teachers in class."Final Answer:(iii) and (i)

Question 15

Negation conjunction: Statement: "Ram neither reads nor studies." Which pair of statements must be true?

Accepted Answer

Introduction / Context:"Neither A nor B" means both negations are true. Applied here, Ram does not read and Ram does not study.Given Data / Assumptions:Original: Neither reads nor studies.Options list positive and negative forms.Concept / Approach:Translate "neither/nor" to logical conjunction of negations: ¬Reads ∧ ¬Studies.Step-by-Step Solution:From the statement, (ii) "does not read" is true.Also, (iv) "does not study" is true.Therefore (ii) and (iv) together must hold.Verification / Alternative check:A truth table for "neither ... nor ..." confirms both components are false in their affirmative form and true in their negative form.Why Other Options Are Wrong:(a) and (b) assert positives that contradict the stem.(c) mixes a positive with its negation which is impossible.None of these is incorrect because one pair does follow.Common Pitfalls:Misreading "neither ... nor ..." as exclusive-or; it is a conjunction of negations.Final Answer:(ii) and (iv)

Question 16

Logical deduction – translating "only when": Statement: "Only when it is not quiet, Reshma studies." Which two statements can both be true in a way that is consistent with the given rule? (i) It is quiet. (ii) It is not quiet. (iii) Reshma studies. …

Accepted Answer

Introduction / Context:The sentence "Only when it is not quiet, Reshma studies" is a classic conditional of the form Studies → NotQuiet. It does not say that whenever it is not quiet she must study; it only forbids studying in quiet conditions.Given Data / Assumptions:Rule: Reshma studies only when it is not quiet.Therefore: If Reshma studies, then it is not quiet.No statement is made about what happens when it is not quiet (she may or may not study).Concept / Approach:Translate to implication: (iii) ⇒ (ii). The contrapositive is Quiet ⇒ NotStudy, i.e., (i) ⇒ (iv).Step-by-Step Solution:From "only when": Studies → NotQuiet.Thus (iii) implies (ii), so the pair (ii) (iii) is consistent and jointly possible.Pairs that assert studying while quiet, such as (i) (iii), contradict the rule.Verification / Alternative check:Consider truth-table style reasoning: the only forbidden combo is Quiet + Studies. All other combinations are allowed, including NotQuiet + Studies and NotQuiet + NotStudy.Why Other Options Are Wrong:(i) (iii) conflicts with the rule (studying despite quiet). (ii) (iv) is allowed but does not "follow" from the rule; the rule does not force not studying when it is not quiet. (iii) (i) is the same conflict reordered.Common Pitfalls:Reading "only when" as "if and only if." It is one-way, not biconditional.Final Answer:(ii) (iii)

Question 17

Logical deduction – "only toppers get IIT admission": Only the toppers in Class 12 get admitted to IITs. Which pair is logically consistent/consequential? (i) Sid got admission in IIT. (ii) Sid is a topper in Class 12. (iii) Sid did not get admissio…

Accepted Answer

Introduction / Context:"Only toppers get admitted" means admission implies topper (Admission → Topper). It does not say every topper must get admission.Given Data / Assumptions:Rule: If someone is admitted to an IIT, they are a topper.No claim that every topper is admitted.Concept / Approach:We check which pair aligns with Admission → Topper and avoids invalid converses.Step-by-Step Solution:(i) Admission true ⇒ (ii) Topper must be true; hence (i) (ii) is a valid consequence.Pairs like (iii) (iv): "no admission ⇒ not topper" is the inverse fallacy and is not entailed.Verification / Alternative check:A Venn view: IIT-admitted is a subset of toppers. Anyone inside admitted is inside topper.Why Other Options Are Wrong:(iii) (iv) and (iv) (iii) commit the inverse error. (ii) (i) lists the same two truths but the consequence is specifically Admission ⇒ Topper; (i) (ii) cleanly expresses it.Common Pitfalls:Assuming the converse (Topper ⇒ Admission) which is not stated.Final Answer:(i) (ii)

Question 18

Logical deduction – biconditional ("if and only if"): "Arya takes fries if and only if she gets it free with a cheese burger." Which pair must hold together? (i) Arya got a burger. (ii) Arya had fries. (iii) Arya got a cheese burger. (iv) Arya did n…

Accepted Answer

Introduction / Context:An "if and only if" (iff) statement creates a two-way linkage: Fries ⇔ (Free-with-cheese-burger). Either side being true forces the other true; either side being false forces the other false.Given Data / Assumptions:Fries iff Cheese-burger-free (with fries) condition.Plain "got a burger" in (i) is irrelevant unless it is a cheese burger given free with fries.Concept / Approach:From the biconditional, (ii) ↔ (iii). Thus whenever (iii) is true, (ii) is true, and vice versa.Step-by-Step Solution:(iii) ⇒ (ii) by "only if" direction (if CB-free then fries).(ii) ⇒ (iii) by "if" direction (if fries then CB-free).Therefore both pairs (iii) (ii) and (ii) (iii) are correct.Verification / Alternative check:Truth-table of p ⇔ q shows p and q always share the same truth value.Why Other Options Are Wrong:(i) (ii) does not follow; (i) says nothing about cheese or free. (iv) contradicts (ii).Common Pitfalls:Confusing "only if" (one-way) with "if and only if" (two-way).Final Answer:Both (a) and (b)

Question 19

Logical deduction – "only if" sleep condition: "Arjun falls asleep only if it is a soft bed." Which pair is consistent/consequential? (i) Arjun fell asleep. (ii) Arjun remains awake. (iii) It was a soft bed. (iv) It was a hard bed.

Accepted Answer

Introduction / Context:"Only if soft bed" translates to Sleep ⇒ SoftBed. Its contrapositive is HardBed ⇒ NotSleep (i.e., remain awake). The converse (SoftBed ⇒ Sleep) is not guaranteed.Given Data / Assumptions:Rule: If Arjun fell asleep then the bed was soft.If the bed was hard, he could not have slept.Concept / Approach:Look for pairs that respect Sleep ⇒ Soft and Hard ⇒ Awake. Pair (ii) (iv) asserts awake and hard bed, which aligns with the contrapositive.Step-by-Step Solution:From "only if": Sleep → Soft.Contrapositive: Hard → NotSleep (Awake).Thus (ii) (iv) is a valid, consistent pair.Verification / Alternative check:(i) together with (iii) would also be consistent, but that exact pairing is not offered; (iii) (ii) and (ii) (iii) invert cause/effect without necessity.Why Other Options Are Wrong:(i) (ii) is contradictory. (ii) (iii) and (iii) (ii) have no necessary link from (iii) to (ii).Common Pitfalls:Assuming Soft ⇒ Sleep. The rule is one-way.Final Answer:(ii) (iv)

Question 20

Logical deduction – "only those in Tilli are unfriendly": Interpret the implication and choose the pair that follows. (i) Rohan lives in Tilli. (ii) Rohan lives in Rampur. (iii) Rohan is unfriendly. (iv) Rohan may be unfriendly.

Accepted Answer

Introduction / Context:"Only those living in Tilli are unfriendly" means Unfriendly ⇒ LivesInTilli. It does not say everyone in Tilli is unfriendly; it restricts the set of unfriendly people to Tilli residents.Given Data / Assumptions:Rule: If someone is unfriendly, they must live in Tilli.People outside Tilli are not unfriendly (contrapositive).Concept / Approach:We seek a pair where (iii) implies (i). That is exactly (iii) (i): if Rohan is unfriendly, then he lives in Tilli.Step-by-Step Solution:Unfriendly ⇒ LivesInTilli.Thus the pair (iii) (i) is a valid consequence.(i) alone does not imply (iii); living in Tilli does not force unfriendliness.Verification / Alternative check:Set-theory view: Unfriendly ⊆ TilliResidents.Why Other Options Are Wrong:(i) (iii) asserts Tilli ⇒ Unfriendly (converse). (ii) (iv) allows unfriendly in Rampur, which contradicts the rule.Common Pitfalls:Equating "only those" with "if and only if."Final Answer:(iii) (i)

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