Direction : Read the given information carefully and answer the given…
2021
Direction : Read the given information carefully and answer the given questions based on it.:
Seven persons S, E, V, B, A, D and U sit in a row facing north and they all work in a company with different working experience of 15yr, 7yr, 2yr, 23yr, 19yr, 6yr and 12yr. All the information is not necessarily in the same order as given. The persons with vowel name have experience of even number year and neither sits at the end nor at exactly middle of the row. Three persons sit between S and the one having 23yr experience. One person sits between S and A. The number of persons sits to the right of A is same as the number of persons sits to the left of the one having experience 2.5 times of A. B neither have highest year experience nor sits adjacent to S. The person having least experience sits second to the left of D. U does not have least experience. The neighbors of B have more experience than B.
Who among the following sits fourth to the left of the one having 7yr experience?
- A.
V
- B.
A
- C.
E
- D.
D
- E.
None of these
Attempted by 3 students.
Show answer & explanation
Correct answer: B
Concept: In a linear (row) arrangement, fix the seats first, then translate every hint into a position relation. The key opening insight here is the parity-and-placement rule: vowel-named persons (E, A, U) must each have an even number of years, and even values among 15, 7, 2, 23, 19, 6, 12 are only 2, 6 and 12. Since there are exactly three vowel names and three even values, the vowels take {2, 6, 12} in some order. "2.5 times of A" must land on a value present in the list, which forces A = 6 (because 2.5 × 6 = 15).
How to solve, step by step:
Vowels E, A, U carry the even experiences {2, 6, 12}; each vowel sits off the two ends and off the exact middle seat.
"2.5 × A" must exist in the list: only A = 6 works (2.5 × 6 = 15), so the comparison person has 15 years.
U is not the least, so U ≠ 2; with A = 6 this leaves U = 12 and E = 2. The remaining {15, 7, 23, 19} go to S, V, B, D, with B not the highest (B ≠ 23).
Place S and the 23-year person with exactly three seats between them, and S and A with exactly one seat between them; combined with "persons right of A = persons left of the 15-year person," A is pinned near the left and S two seats to A's right.
Finally, the least value (2 years = E) must sit second to the left of D, and every neighbour of B must out-rank B in experience — which forces B (7 years) to the right end beside the 12-year person. Only one seating satisfies all of these simultaneously, left to right:
V(23), E(2), A(6), D(19), S(15), U(12), B(7).
Answer to the question: The person with 7 years is at the extreme right of the row. Counting four seats to that person's left lands on the seat holding A. Hence A sits fourth to the left of the 7-year person.
Cross-check: From the 7-year seat, the four seats to its left in order hold 12, 15, 19 and 6 years; the fourth of these is the 6-year person, who is A — consistent. Every clue (vowel parity, the 23-year gap of three, the one-person gap between S and A, the second-to-left-of-D least value, and B's richer neighbours) holds in this single arrangement.