Question 1

What is the position of Radha from the shorter end?

Accepted Answer

According to the question, Sunita is taller than Seema and Renu. 1. Sunita > Seema/Renu. Renu is shorter than Radha and Gauri 2. Renu < Radha/Gauri. Bina is taller than Radha. 3. Bina > Radha. Bina shorter than Sunita. 4. Bina < Sunita. Sunita is not the tallest and Renu is not the shortest. 5. Sunita (not tallest) 6. Renu (not Shortest) On arranging the above data, we get from 3 and 4, 7. Sunita > Bina > Radha from 7 and 1, 8. Sunita > Bina > Radha > Renu/Seema. from 8 , 5 and 6. ? Gauri > Sunita > Bina > Radha > Renu > Seema. Third

Question 2

Who is the shortest?

Accepted Answer

According to the question, Sunita is taller than Seema and Renu. 1. Sunita > Seema/Renu. Renu is shorter than Radha and Gauri 2. Renu < Radha/Gauri. Bina is taller than Radha. 3. Bina > Radha. Bina shorter than Sunita. 4. Bina < Sunita. Sunita is not the tallest and Renu is not the shortest. 5. Sunita (not tallest) 6. Renu (not Shortest) On arranging the above data, we get from 3 and 4, 7. Sunita > Bina > Radha from 7 and 1, 8. Sunita > Bina > Radha > Renu/Seema. from 8 , 5 and 6. ? Gauri > Sunita > Bina > Radha > Renu > Seema. Seema is shortest.

Question 3

Who is the tallest?

Accepted Answer

According to the question, Sunita is taller than Seema and Renu. 1. Sunita > Seema/Renu. Renu is shorter than Radha and Gauri 2. Renu < Radha/Gauri. Bina is taller than Radha. 3. Bina > Radha. Bina shorter than Sunita. 4. Bina < Sunita. Sunita is not the tallest and Renu is not the shortest. 5. Sunita (not tallest) 6. Renu (not Shortest) On arranging the above data, we get from 3 and 4, 7. Sunita > Bina > Radha from 7 and 1, 8. Sunita > Bina > Radha > Renu/Seema. from 8 , 5 and 6. ? Gauri > Sunita > Bina > Radha > Renu > Seema. Gauri is tallest among them.

Question 4

Which of the following represents the descending order of weights of the articles?

Accepted Answer

According to question, A's weight is twice as much as of B. 1. A = 2B. B's weight is four and half times as much as of C. 2. B = 4.5C C's weight is half times as much as of D. 3. C = D/2. D's weight is half as much as of E. 4. D = E/2. E weight is less than A but more than C. 5. A > E > C from 1 and 2 6. A = 2B = 9C from 6 and 3 7. A = 2B = 9C = 4.5D from 7 and 4 8. A = 2B = 9C = 4.5D = 2.25E so from 8 and 5 A > B > E > D > C Descending order of weights of the articles will be A, B, E, D, C

Question 5

Which of the following is the heaviest in weight?

Accepted Answer

According to question, A's weight is twice as much as of B. 1. A = 2B. B's weight is four and half times as much as of C. 2. B = 4.5C C's weight is half times as much as of D. 3. C = D/2. D's weight is half as much as of E. 4. D = E/2. E weight is less than A but more than C. 5. A > E > C from 1 and 2 6. A = 2B = 9C from 6 and 3 7. A = 2B = 9C = 4.5D from 7 and 4 8. A = 2B = 9C = 4.5D = 2.25E so from 8 and 5 A > B > E > D > C So A is the heaviest in the weight.

Question 6

E is heavier than which of the following two articles?

Accepted Answer

According to question, A's weight is twice as much as of B. 1. A = 2B. B's weight is four and half times as much as of C. 2. B = 4.5C C's weight is half times as much as of D. 3. C = D/2. D's weight is half as much as of E. 4. D = E/2. E weight is less than A but more than C. 5. A > E > C from 1 and 2 6. A = 2B = 9C from 6 and 3 7. A = 2B = 9C = 4.5D from 7 and 4 8. A = 2B = 9C = 4.5D = 2.25E so from 8 and 5 A > B > E > D > C E is heavier than D and C.

Question 7

E is lighter in weight than which of the other two articles?

Accepted Answer

According to question, A's weight is twice as much as of B. 1. A = 2B. B's weight is four and half times as much as of C. 2. B = 4.5C C's weight is half times as much as of D. 3. C = D/2. D's weight is half as much as of E. 4. D = E/2. E weight is less than A but more than C. 5. A > E > C from 1 and 2 6. A = 2B = 9C from 6 and 3 7. A = 2B = 9C = 4.5D from 7 and 4 8. A = 2B = 9C = 4.5D = 2.25E so from 8 and 5 A > B > E > D > C So, E is lighter in weight than A and B.

Question 8

Which of the following is the lightest in weight?

Accepted Answer

According to question, A's weight is twice as much as of B. 1. A = 2B. B's weight is four and half times as much as of C. 2. B = 4.5C C's weight is half times as much as of D. 3. C = D/2. D's weight is half as much as of E. 4. D = E/2. E weight is less than A but more than C. 5. A > E > C from 1 and 2 6. A = 2B = 9C from 6 and 3 7. A = 2B = 9C = 4.5D from 7 and 4 8. A = 2B = 9C = 4.5D = 2.25E so from 8 and 5 A > B > E > D > C So lightest in weight is C.

Question 9

Raja walks slower then Raghu and Raghu walks as fast as Guru and Krishna walks faster than Guru. Who walks the fastest?

Accepted Answer

Arrangement in descending order, Krishna > Guru > Raghu > Raja Hence, Krishna is the fastest from all of them.

Question 10

If the 5th data of a month is Tuesday, what data will be 3 days after the 3rd Friday in the month?

Accepted Answer

If the 5th day of the month is Tuesday. Then, first day of the month will be Friday. The third Friday will fall on 15th of the month. So, the date 3 days after 3rd Friday will be 19th.

Question 11

If 1st October is a Sunday, then 1st November will be

Accepted Answer

As 1st October is a Sunday. So, 29th October is also a Sunday. Now, 31st October is Tuesday. Hence, 1st November will be Wednesday.

Question 12

Samant remembers that his brother's birthday is after 15th but before 18th of February' while his sister remembers that her brothers' birthday is after 16th but before 19th of February, is Samant's brother's birthday?

Accepted Answer

According to Birth Date Samant 16, 17 Sister 17, 18 Here, 17th is common to both, Hence, confirm data is 17th February.

Question 13

Praveen correctly remembers that his father's birthday is after 21th May but before 27th May whereas his sister correctly remembers that their father's birthday is after 24th May but before 30th May. On which day in May was definitely their father's…

Accepted Answer

According to Praveen - 22, 23, 24, ,25, 26 According to his sister - 25, 26, 27, 28, 29 So, Common data is 25 or 26.

Question 14

Interval reasoning – Two siblings recall their father’s birthday within overlapping date ranges in December. Sangeeta remembers it was certainly after 8th but before 13th (i.e., 9th–12th). Natasha remembers it was definitely after 9th but before 14t…

Accepted Answer

Introduction / Context: Many reasoning problems ask you to intersect two people’s partial memories to find a unique date. If the overlap still leaves multiple possibilities, the information is insufficient. Given Data / Assumptions: Sangeeta: after 8th and before 13th → {9, 10, 11, 12}. Natasha: after 9th and before 14th → {10, 11, 12, 13}. We assume “after” and “before” are strict. Concept / Approach: Compute the intersection of both sets. Only if the intersection has exactly one date can we answer uniquely; otherwise, the data are insufficient. Step-by-Step Solution: Intersection = {9,10,11,12} ∩ {10,11,12,13} = {10, 11, 12}.There are three feasible dates; no rule picks just one. Verification / Alternative check: If any added constraint (e.g., “even date only”) existed, the set might reduce; here, none exists. Why Other Options Are Wrong: 10th/11th/12th are each possible but not uniquely determined. “None of these” is invalid because there are feasible dates; the issue is non-uniqueness. Common Pitfalls: Misreading “after/before” as inclusive; that would alter the bounds. Final Answer: Data inadequate

Question 15

Intersecting memories with parity clue – Sneha: father’s birthday is after 11 June but before 16 June; younger brother: after 13 June but before 18 June; elder brother remembers it is an even date. What is the definite date?

Accepted Answer

Introduction / Context: This problem combines overlapping date ranges with one extra parity clue (“even date”). Intersect ranges first; then apply the even/odd filter to isolate a single date. Given Data / Assumptions: Sneha: 12–15 June (after 11th, before 16th). Younger brother: 14–17 June (after 13th, before 18th). Elder brother: the date is even. Concept / Approach: Step 1: Compute intersection of the two ranges. Step 2: From the intersection, keep only even dates because of the final clue. Step-by-Step Solution: Intersection of {12,13,14,15} and {14,15,16,17} = {14,15}.Apply “even” → {14}. Verification / Alternative check: Check both original bounds: 14 is after 11 and before 16; also after 13 and before 18; and 14 is even. All conditions satisfied. Why Other Options Are Wrong: 12 is not in the second range; 16 is not in Sneha’s range; “Data inadequate” is incorrect because the even-date clue resolves the tie. Common Pitfalls: Treating “before” as inclusive; mixing up the parity condition. Final Answer: Fourteenth

Question 16

Bird order on a branch – Five birds (Crow, Pigeon, Little Pigeon, Big Crow, Eagle) fly one after another. Big Crow flies after Crow but ahead of Eagle; Pigeon is between Crow and Big Crow. Which bird is last?

Accepted Answer

Introduction / Context: We translate relational clues into a partial ordering, then extend it to determine the last position. Such problems hinge on recognizing that some elements must be to the right of all others due to constraints. Given Data / Assumptions: Big Crow is after Crow, and ahead of Eagle. Pigeon is strictly between Crow and Big Crow. Little Pigeon has no positional constraint except being one of the five. Concept / Approach: From “Pigeon between Crow and Big Crow,” the triplet order is Crow < Pigeon < Big Crow. From “Big Crow ahead of Eagle,” we extend to Crow < Pigeon < Big Crow < Eagle. Little Pigeon can be anywhere relative to these, but cannot force anyone after Eagle. Step-by-Step Solution: Establish base chain: Crow < Pigeon < Big Crow.Append Eagle to the right: … < Big Crow < Eagle.Little Pigeon is free; placing it anywhere does not move Eagle leftward of Big Crow. Verification / Alternative check: Try placements of Little Pigeon at the extreme left, middle, or before Eagle; in all valid placements, Eagle remains to the right of Big Crow, hence is last. Why Other Options Are Wrong: Pigeon/Big Crow cannot be last due to the given chain; “None of these” and “Little Pigeon” conflict with the forced position of Eagle. Common Pitfalls: Reversing “ahead of” vs “after,” or ignoring the “between” constraint. Final Answer: Eagle

Question 17

Digit selection → perfect square – Using the 1st, 4th, and 6th digits of 531697 (i.e., 5, 6, 7), form a 3-digit perfect square of a two-digit even number. What is the second digit of that two-digit even number?

Accepted Answer

Introduction / Context: We must permute selected digits to get a 3-digit square whose root is a two-digit even number; then report the second (unit) digit of that even root. The special outputs ‘@’ (no such number) or ‘c’ (more than one) are sentinels. Given Data / Assumptions: Digits available: positions 1,4,6 of 531697 → {5,6,7}. Seek a 3-digit perfect square of an even two-digit number. Concept / Approach: List permutations of {5,6,7} and check against known 3-digit squares. The only candidate matching is 576 = 24^2. Step-by-Step Solution: Permutations: 567, 576, 657, 675, 756, 765.Perfect square match: 576 = 24^2.Even root = 24 → second digit (units) = 4. Verification / Alternative check: Other permutations are not perfect squares (quick check against the list of 3-digit squares). Why Other Options Are Wrong: ‘2’ or ‘6’ are not the unit digit of the unique valid even root; ‘@’ (none) and ‘c’ (more than one) do not apply. Common Pitfalls: Overlooking 576; assuming squares must end with 5 or 6 only for certain roots without checking. Final Answer: 4

Question 18

District size comparison – In five districts: Akbarpur smaller than Fatehpur; Dhanbad bigger than Palamu; Bara Banki bigger than Fatehpur but smaller than Palamu. Which district is the biggest?

Accepted Answer

Introduction / Context: Translate comparative statements into a partial order and identify the maximal element. Watch for chains that force one element above all others. Given Data / Assumptions: Akbarpur < Fatehpur. Dhanbad > Palamu. Bara Banki > Fatehpur but < Palamu. Concept / Approach: From Bara Banki > Fatehpur and Bara Banki < Palamu, we have Fatehpur < Bara Banki < Palamu. Since Dhanbad > Palamu, Dhanbad exceeds all three in that chain. Also, Akbarpur < Fatehpur, so Dhanbad is higher than Akbarpur as well. Step-by-Step Solution: Chain: Akbarpur < Fatehpur < Bara Banki < Palamu.Given Dhanbad > Palamu → Dhanbad is above every district in the chain. Verification / Alternative check: No other statement contradicts Dhanbad’s superiority; hence Dhanbad is uniquely the largest. Why Other Options Are Wrong: Each of Akbarpur, Fatehpur, Bara Banki, and Palamu is strictly below another district; none can be maximal. Common Pitfalls: Misreading “bigger than” for two different pairs; ignoring the transitive implications. Final Answer: Dhanbad

Question 19

Single-student two-sided rank – Neha is 21st from the top and 30th from the bottom in her class. What is the total number of students?

Accepted Answer

Introduction / Context: When both top-rank and bottom-rank of the same student are given, the class size is the sum minus one: N = top + bottom − 1. Given Data / Assumptions: Neha’s ranks: 21st from top, 30th from bottom. Concept / Approach: For a single lineup, positions counted from both ends overlap on the student herself once; hence subtract one to avoid double counting. Step-by-Step Solution: N = 21 + 30 − 1 = 50. Verification / Alternative check: Check: If there are 50 students, the 21st from top is the 30th from bottom (since 50 − 21 + 1 = 30). Consistent. Why Other Options Are Wrong: 48/49/51 would break the complement relation between the two given ranks. Common Pitfalls: Forgetting to subtract one. Final Answer: 50

Question 20

Paired-position swaps – Number 53146872: swap (1↔5), (2↔6), (3↔7), (4↔8). After rearrangement, what is the 5th digit from the right end?

Accepted Answer

Introduction / Context: Index-swapping exercises require careful mapping of original positions to new ones, then translating a right-based index to a left-based position. Given Data / Assumptions: Original digits (1-based): [5,3,1,4,6,8,7,2]. Swap pairs: (1,5), (2,6), (3,7), (4,8). Concept / Approach: Perform each swap pair once on the original indices. After swaps, convert “5th from right” in an 8-digit number to the left index: 8 − 5 + 1 = 4. Step-by-Step Solution: After swaps → [6,8,7,2,5,3,1,4].5th from right = position 4 = digit 2. Verification / Alternative check: Rightmost four digits are 4 (1st), 1 (2nd), 3 (3rd), 2 (4th); the next (5th from right) is indeed 2 at left index 4. Why Other Options Are Wrong: 1/4/6/8 occur elsewhere but not at the queried position. Common Pitfalls: Applying swaps sequentially and reusing modified positions incorrectly; mis-converting right index to left. Final Answer: 2

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