Question 9
1, 2, 9, 4, 25, ___, 7
(a) 36 (b) 33 (c) 27 (d) 35
Solution
Note: the small “²” on 2, 4 and 7 doesn’t show in the print, so I’ve read the sequence as 1, 2², 9, 4², 25, ___, 7².
- Write each term as a square: 1², 2², 3², 4², 5², ___, 7².
- The bases go 1, 2, 3, 4, 5, ?, 7, so the missing base is 6.
- Missing term = 6² = 36.
Final Answer: (a) 36
Question 35
BDAC : FHEG :: NPMO : ___
(a) QTRS (b) RQTS (c) TRQS (d) RTQS
Solution
- In alphabet positions, B D A C = 2, 4, 1, 3 and F H E G = 6, 8, 5, 7.
- Each letter moves forward 4 (2→6, 4→8, 1→5, 3→7).
- N P M O = 14, 16, 13, 15.
- Add 4 to each: 18, 20, 17, 19 = R, T, Q, S.
Final Answer: (d) RTQS
Question 36
FxyZ : YyxF :: AefM : ___
(a) RefL (b) AefL (c) LefL (d) LfeA
Solution
- Reverse FxyZ to get ZyxF.
- Compared with YyxF, only the first letter changes: Z → Y, one letter back.
- So the rule is: reverse the word, then move the first letter back by 1.
- Reverse AefM to get MfeA.
- Move M back one to L, giving LfeA.
Final Answer: (d) LfeA
Questions 17 to 22
A Travel Agent is arranging sight seeing tours in the following locations L, M, N, O, P, Q, R, S, and T. Each tour has the following conditions:
- If M is included, both Q and R must be included.
- P can be included only if O is included.
- If Q is included, M must be included along with either N or T or both.
- P and Q cannot be combined together.
- O, R and T cannot be combined together.
- N, S and R cannot be combined together.
- L and R cannot be combined.
Rules 5 and 6 mean all three can’t be in the same tour. Any two of them together is fine.
Question 17
If M is included in a tour, the minimum number of other cities that must be included is
(a) 3 (b) 4 (c) 5 (d) 6
Solution
- Rule 1: M is in, so Q and R must be in.
- Rule 3: Q is in, so N or T must be in.
- No other rule forces a city in.
- The minimum is Q + R + (N or T) = 3 cities.
Final Answer: (a) 3
Question 18
Which of the following tours cannot be included with P?
(a) L (b) M (c) N (d) O
Solution
- Rule 4: P is in, so Q is out.
- Rule 1: M needs Q, so M is out.
- L, N and O don’t clash with P. O is actually required with P.
Final Answer: (b) M
Question 19
Which of the following is an acceptable tour group?
(a) M N O P (b) M N Q R (c) M N Q S (d) L M Q R
Solution
- (a) M needs Q and R, and both are missing ✗
- (b) M has Q and R, Q has M and N, and nothing clashes ✓
- (c) M needs R, which is missing ✗
- (d) L is with R ✗ (and Q has neither N nor T)
Final Answer: (b) M N Q R
Question 20
Which of the following would have to be removed from a group to form an acceptable tour?
(a) M (b) Q (c) O (d) R
Solution
Note: the question doesn’t name the group, so I’ve taken it as all nine locations.
- With all nine included, rules 4, 5, 6 and 7 are broken.
- R is in three of those clashes (rules 5, 6 and 7), so remove R.
- M needs R, so M goes. Q needs M, so Q goes too, which also fixes rule 4.
- That leaves L, N, O, P, S, T, which is an acceptable tour.
- Removing M, Q or O instead still leaves either O, R, T or N, S, R together.
Final Answer: (d) R
Question 21
Which of the following can be an acceptable tour if one more destination is added?
(a) M, P, Q (b) M, Q, R (c) N, S, R (d) R, T, P
Solution
- (a) P with Q is a clash, and adding a city can’t remove it ✗
- (b) The only thing missing is N or T. Adding N gives M, N, Q, R ✓
- (c) N, S and R together is a clash ✗
- (d) P needs O, but adding O puts O, R and T together ✗
Final Answer: (b) M, Q, R
Question 22
How many destinations can be used for a tour consisting of one city only?
(a) 3 (b) 4 (c) 5 (d) 6
Solution
- M needs Q and R, P needs O, and Q needs M, so these 3 can’t be alone.
- L, N, O, R, S and T don’t need any other city, so each can be alone.
- 9 − 3 = 6.
Final Answer: (d) 6