IBPS PO 2025 Prelims Memory-Based Paper 23 August Shift 1 with Answers

Answer: (E) They need far less money to set up and run than ordinary restaurants.

The second paragraph says the biggest advantage is low cost (no costly site, decoration or waiters) and that a cloud kitchen lets young cooks start with far less money and time. The other options are not stated.

Answer: (B) Both (I) and (III)

(I) is stated in the first and third paragraphs. (III) is stated in the last paragraph (a constant rush to meet orders, corners are cut). (II) is not supported: the passage says the sector tripled in 2023, not that it had never grown before.

Answer: (E) A busy cloud kitchen skips cleaning its equipment so that it can send out orders faster.

The concern is that cloud kitchens cut corners on hygiene and safety in the rush to meet demand. Skipping cleaning to send orders faster is exactly this.

Answer: (D) The sector grew to almost three times its earlier size in 2023.

The third paragraph says that in Britain the sector expanded to almost three times its earlier size in 2023.

Answer: (B) Both (I) and (II)

(I) is incorrect: the passage says they increase business for packaging makers and food suppliers. (II) is incorrect: the 27 billion dollars is over 2027 to 2032, not 2027 alone. (III) is correct.

Answer: (D) behind the scenes

The words 'out of the customer's sight' tell us the work is unseen. 'Behind the scenes' means out of public view.

Answer: (E) unattainable

An elusive dream is one that is hard to reach. Of the options, unattainable (out of reach) is closest; achievable is the opposite.

Answer: (E) carried out

Here done means performed or carried out: the work is performed out of sight. It does not mean finished, avoided or cooked.

Answer: (D) touch lightly

A customer needs only to tap a screen once, that is, to touch it lightly. The other options are other senses of tap that do not fit here.

Answer: (D) extract

A rotten tooth is pulled out, that is, extracted. The other verbs do not make sense with a tooth.

Answer: (E) transform

The sentence points to a big positive change, and 'transform the way people...' is the natural phrase. Impede and deteriorate are negative, acclaim means praise, and stagnate does not take an object.

Answer: (B) contributed

Only 'contributed' is followed by 'to' in the sense of helping to cause something. Training and determination contributed to her success.

Answer: (A) promote

A commission that wants fairness would promote (encourage) a level playing field. The other verbs mean breaking, giving up, limiting or breaking up.

Answer: (E) strictly

Rules meant to keep things fair must be followed strictly. Scarcely, hardly and barely mean almost not, and preciously does not fit.

Answer: (C) C

The correct order is C, B, A, D, E. C introduces the problem (exhaustion); B says travel has therefore become an escape; A gives the result (longer holidays); D explains how such breaks refresh the mind; E concludes with what this change shows. First: C.

Answer: (B) B

The correct order is C, B, A, D, E. C introduces the problem (exhaustion); B says travel has therefore become an escape; A gives the result (longer holidays); D explains how such breaks refresh the mind; E concludes with what this change shows. Second: B.

Answer: (A) A

The correct order is C, B, A, D, E. C introduces the problem (exhaustion); B says travel has therefore become an escape; A gives the result (longer holidays); D explains how such breaks refresh the mind; E concludes with what this change shows. Third: A.

Answer: (D) D

The correct order is C, B, A, D, E. C introduces the problem (exhaustion); B says travel has therefore become an escape; A gives the result (longer holidays); D explains how such breaks refresh the mind; E concludes with what this change shows. Fourth: D.

Answer: (E) E

The correct order is C, B, A, D, E. C introduces the problem (exhaustion); B says travel has therefore become an escape; A gives the result (longer holidays); D explains how such breaks refresh the mind; E concludes with what this change shows. Fifth: E.

Answer: (D) B-D

Interchanging oven (B) and touch (D) gives: 'Be careful not to touch the hot dish when you take it out of the oven.'

Answer: (A) A-B and C-D

Interchange A-B and C-D: 'The city is rapidly modernising its bus network, and electric buses offer a great chance to speed up this shift.'

Answer: (E) No interchange required

The sentence is already correct: the temperatures of stars are estimated from measurements of their spectra. Any swap makes it meaningless.

Answer: (A) A-B

Interchanging Growing (A) and involving (B) gives: 'Involving pupils in growing vegetables at school can spark a lasting interest in healthy food.'

Answer: (C) B-D

B-D gives: 'Car makers are adding more fast-charging points along highways so that electric-car owners can make long trips without worry.' A needs an infinitive such as 'to report attacks' and C needs a verb; neither E nor F completes them sensibly. So only B-D is correct.

Answer: (C) Both (I) and (II)

A dark horse is a little-known person or team who unexpectedly succeeds. (I) and (II) fit this. In (III) the team was the favourite, which is the opposite of a dark horse.

Answer: (B) B

Promise is followed by the to-infinitive: 'promised to improve'. 'To improving' is wrong, so the error is in part B.

Answer: (C) C

Items in a list should be parallel: cycling, cooking and reading. 'To read' breaks the pattern, so the error is in part C.

Answer: (A) A

'Unexpected' begins with a vowel sound, so the article should be 'an', not 'a'. The error is in part A.

Answer: (B) B

With 'neither ... nor', the verb agrees with the nearer subject, 'assistants', which is plural. So 'were informed' is correct; the error is in part B.

Answer: (E) No error

The sentence is correct: 'will announce' for the future and 'have been checked' (present perfect passive) in the time clause. No error.

Answer: (B) x > y

From I, x = 4, 6. From II, y = 1.2, 2. Every value of x is greater than every value of y, so x > y.

Answer: (C) x < y

From I, x = 2, 2.5. From II, y = 5, 7. Every value of x is less than every value of y, so x < y.

Answer: (A) x = y or no relation can be established

From I, x = 0.5, 2/3. From II, y = 0.5, 0.6. x = 0.5 is less than y = 0.6, while x = 2/3 (about 0.667) is greater than both values of y, so no fixed relation holds.

Answer: (C) x < y

From I, x = -6, -3.5. From II, y = -2.5, -1. Every value of x is less than every value of y, so x < y.

Answer: (B) 147

Round: 780 ÷ 13 + 5 + 23 = 60 + 28 = 88. So ?% of 60 = 88, giving ? = 88 x 100/60 = 146.7, about 147.

Answer: (D) 8

Round: 40% of 250 = 100 and √196 = 14, so 114 = ?² + 50. Then ?² = 64 and ? = 8.

Answer: (A) 255

Round: 21² = 441 and 25% of 220 = 55. 441 - 55 - ? = 131, so ? = 255.

Answer: (C) 21

Round: 24² + 12 = 576 + 12 = 588, and 50% of 56 = 28. ? = 588 ÷ 28 = 21.

Answer: (A) 230%

In bag A, red = 62.5% and blue = 37.5%, so the difference is 25% of the total = 60, giving total 240, red 150, blue 90. In the same way: B total 600 (red 399, blue 201); C total 470 (red 329, blue 141); D total 380 (red 228, blue 152); E total 750 (red 540, blue 210). Red in A and E = 690, blue in A and E = 300. 690/300 x 100 = 230%.

Answer: (E) 10

In bag A, red = 62.5% and blue = 37.5%, so the difference is 25% of the total = 60, giving total 240, red 150, blue 90. In the same way: B total 600 (red 399, blue 201); C total 470 (red 329, blue 141); D total 380 (red 228, blue 152); E total 750 (red 540, blue 210). Blue in B + red in C = 201 + 329 = 530. Red in E = 540. 540 - 530 = 10.

Answer: (B) 23:3

In bag A, red = 62.5% and blue = 37.5%, so the difference is 25% of the total = 60, giving total 240, red 150, blue 90. In the same way: B total 600 (red 399, blue 201); C total 470 (red 329, blue 141); D total 380 (red 228, blue 152); E total 750 (red 540, blue 210). (150 + 540) : 90 = 690 : 90 = 23 : 3.

Answer: (C) 389

In bag A, red = 62.5% and blue = 37.5%, so the difference is 25% of the total = 60, giving total 240, red 150, blue 90. In the same way: B total 600 (red 399, blue 201); C total 470 (red 329, blue 141); D total 380 (red 228, blue 152); E total 750 (red 540, blue 210). (399 + 228 + 540)/3 = 1167/3 = 389.

Answer: (D) 600

In bag A, red = 62.5% and blue = 37.5%, so the difference is 25% of the total = 60, giving total 240, red 150, blue 90. In the same way: B total 600 (red 399, blue 201); C total 470 (red 329, blue 141); D total 380 (red 228, blue 152); E total 750 (red 540, blue 210). Balls in F = 125% of 540 = 675. Blue in F = 150/2 = 75. Red in F = 675 - 75 = 600.

Answer: (A) 18

In bag A, red = 62.5% and blue = 37.5%, so the difference is 25% of the total = 60, giving total 240, red 150, blue 90. In the same way: B total 600 (red 399, blue 201); C total 470 (red 329, blue 141); D total 380 (red 228, blue 152); E total 750 (red 540, blue 210). Balls moved = 20% of 240 = 48. Red balls moved = 359 - 329 = 30. Blue balls moved = 48 - 30 = 18.

Answer: (E) 31.58%

Cell phones = TVs + difference: A 520, B 880, C 1080, D 1280, E 440. Items: A 840, B 1480, C 1760, D 2040, E 520. Items by E = 520, TVs by D = 760. (760 - 520)/760 x 100 = 31.58% less.

Answer: (D) 672

Cell phones = TVs + difference: A 520, B 880, C 1080, D 1280, E 440. Items: A 840, B 1480, C 1760, D 2040, E 520. Items by F = 120% of 1760 = 2112. TVs = 7/22 x 2112 = 672.

Answer: (C) 680

Cell phones = TVs + difference: A 520, B 880, C 1080, D 1280, E 440. Items: A 840, B 1480, C 1760, D 2040, E 520. Cell phones by C and E = 1080 + 440 = 1520. Items by A = 840. 1520 - 840 = 680.

Answer: (B) 19:22

Cell phones = TVs + difference: A 520, B 880, C 1080, D 1280, E 440. Items: A 840, B 1480, C 1760, D 2040, E 520. 760 : 880 = 19 : 22.

Answer: (A) 185

Cell phones = TVs + difference: A 520, B 880, C 1080, D 1280, E 440. Items: A 840, B 1480, C 1760, D 2040, E 520. Defective TVs = 12.5% of 760 = 95. Defective cell phones = 280 - 95 = 185.

Answer: (E) 800

Cell phones = TVs + difference: A 520, B 880, C 1080, D 1280, E 440. Items: A 840, B 1480, C 1760, D 2040, E 520. (880 + 1080 + 440)/3 = 2400/3 = 800.

Answer: (B) 90

Let lab assistants in C = x; professors in C = x + 6, so staff in C = 2x + 6 = 82 + 24 = 106 and x = 50. Lab assistants in B = 54, so professors in A = 54 - 44 = 10 and lab assistants in A = 72. Professors in B = 102 - 10 - 56 = 36, so staff in B = 54 + 36 = 90. Staff in B = 90.

Answer: (D) 1:2

Let lab assistants in C = x; professors in C = x + 6, so staff in C = 2x + 6 = 82 + 24 = 106 and x = 50. Lab assistants in B = 54, so professors in A = 54 - 44 = 10 and lab assistants in A = 72. Professors in B = 102 - 10 - 56 = 36, so staff in B = 54 + 36 = 90. 36 : 72 = 1 : 2.

Answer: (C) 24

Let lab assistants in C = x; professors in C = x + 6, so staff in C = 2x + 6 = 82 + 24 = 106 and x = 50. Lab assistants in B = 54, so professors in A = 54 - 44 = 10 and lab assistants in A = 72. Professors in B = 102 - 10 - 56 = 36, so staff in B = 54 + 36 = 90. Staff in D = 3/4 x 72 = 54. Lab assistants in D = 50 - 20 = 30. Professors in D = 54 - 30 = 24.

Answer: (B) 11.11%

Let lab assistants in C = x; professors in C = x + 6, so staff in C = 2x + 6 = 82 + 24 = 106 and x = 50. Lab assistants in B = 54, so professors in A = 54 - 44 = 10 and lab assistants in A = 72. Professors in B = 102 - 10 - 56 = 36, so staff in B = 54 + 36 = 90. 10/90 x 100 = 11.11%.

Answer: (D) 61

Let lab assistants in C = x; professors in C = x + 6, so staff in C = 2x + 6 = 82 + 24 = 106 and x = 50. Lab assistants in B = 54, so professors in A = 54 - 44 = 10 and lab assistants in A = 72. Professors in B = 102 - 10 - 56 = 36, so staff in B = 54 + 36 = 90. (72 + 50)/2 = 61.

Answer: (E) 72

Let the boat's speed in still water be b. 2(b + 3) = 3(b - 3), so b = 15 km/h. X = 2 x 18 = 36 km, so 2X = 72.

Answer: (A) 44494

A: 6400 x 4 + 4480 x 8 = 61440. B: 7200 x 12 = 86400. C: 2880 x 8 = 23040. Total = 170880. Share of A and C = 84480/170880 x 90000 = Rs 44494 (approx.).

Answer: (C) 3

Let the parts be k, so 8k + xk = 55. After adding: 8k + 5 = xk + 30, so (8 - x)k = 25. Dividing: (8 + x)/(8 - x) = 55/25 = 11/5, so 40 + 5x = 88 - 11x, x = 3. (Check: milk 40, water 15; after adding 45 and 45.)

Answer: (C) 56

Let breadth = b, length = 2b. Path area = (2b + 4)(b + 4) - 2b² = 12b + 16 = 352, so b = 28 and length = 56 m.

Answer: (E) 350

Train speed = 20 m/s, athlete = 5 m/s. Train length = (20 - 5) x 10 = 150 m. Train + platform = 20 x 25 = 500 m, so platform = 350 m.

Answer: (B) 15%

CP = 624/0.78 = Rs 800. MP = 1040. SP = 800 + 84 = 884. Discount = 156/1040 x 100 = 15%.

Answer: (A) Rs 1200

Interest from A = 0.30X; amount = 1.30X. Interest from B = 1.30X x (1.1² - 1) = 0.273X. Difference = 0.30X - 0.273X = 0.027X = 32.40, so X = 1200.

Answer: (D) 12

(7m + 15w) x 15 = (18m + 20w) x 8 gives 39m = 65w, so m : w = 5 : 3. Taking m = 5, w = 3, work = 80 x 15 = 1200 units. (5X + 60) x 10 = 1200, so X = 12.

Answer: (D) 36

The multiples are 24, 28, 32, 36 (average 30). The second smallest is 28, so the even numbers are 28, 30, 32, 34, 36. The largest is 36.

Answer: (C) 3

R = 32 - 12 = 20, Q = 56 - 20 = 36, P = 64 - 36 = 28. (28 + 12) : (36 - 6) = 40 : 30 = 4 : 3, so X = 3.

Answer: (D) Four

F is at the left end, so no one is to the left of F; then no one is to the right of C, so C is at the right end. D is fifth to the left of C, so four persons sit between D and C; hence three sit between D and F, and D is 5th from the left. C is then 10th, at the right end, so there are 10 persons. B is second to the left of C (8th), A is four places from B (4th, since 12th does not exist), E is three places from D (2nd, as 8th is B). A is 3rd to the right of F, so G is 3rd to the left of C (7th). Order from left: F, E, _, A, D, _, G, B, _, C. E is 2nd and G is 7th, so four persons sit between them.

Answer: (B) A

F is at the left end, so no one is to the left of F; then no one is to the right of C, so C is at the right end. D is fifth to the left of C, so four persons sit between D and C; hence three sit between D and F, and D is 5th from the left. C is then 10th, at the right end, so there are 10 persons. B is second to the left of C (8th), A is four places from B (4th, since 12th does not exist), E is three places from D (2nd, as 8th is B). A is 3rd to the right of F, so G is 3rd to the left of C (7th). Order from left: F, E, _, A, D, _, G, B, _, C. Third to the left of G (7th) is the 4th position, A.

Answer: (E) 10

F is at the left end, so no one is to the left of F; then no one is to the right of C, so C is at the right end. D is fifth to the left of C, so four persons sit between D and C; hence three sit between D and F, and D is 5th from the left. C is then 10th, at the right end, so there are 10 persons. B is second to the left of C (8th), A is four places from B (4th, since 12th does not exist), E is three places from D (2nd, as 8th is B). A is 3rd to the right of F, so G is 3rd to the left of C (7th). Order from left: F, E, _, A, D, _, G, B, _, C. There are 10 persons.

Answer: (A) 1

F E R M E N T A T I O N: the 7th letter is T, 10th I, 11th O and 12th N. The only meaningful word from T, I, O, N is INTO, so the answer is 1.

Answer: (D) More than three

Qualifying pairs: 7 and 9 (one digit between in both), 7 and 2 (four between in both), 3 and 1 (one between), 1 and 4 (two between), 4 and 3 (adjacent, the last two digits). That is five pairs, which is more than three.

Answer: (C) Sister-in-law

Every child's parents are in the family, and the only couple who can be parents of the brothers E and F is B and D, so E and F are both sons of B and D (B has no daughter). A's father must be a son of B, and A's mother his wife. E has no child. If C were A's father, his wife would have to be A's mother, but G is C's sister-in-law and H is A's aunt, so neither can be C's wife and mother of A. So F is A's father. F's wife cannot be H (H is A's aunt), so F is married to G, who is D's daughter-in-law. The third couple must include C; C cannot marry E, because then H would be unmarried and could be A's aunt only as a sister of A's parents, which is impossible (B has no daughter, and G's parents are not in the family). So C is married to H. Since H is female, C is male and A is female. G being C's sister-in-law makes C a son of B. So B and D have sons C, E and F; C is married to H, F to G, and A is the daughter of F and G. G is the wife of E's brother F, so G is E's sister-in-law.

Answer: (A) Four

Every child's parents are in the family, and the only couple who can be parents of the brothers E and F is B and D, so E and F are both sons of B and D (B has no daughter). A's father must be a son of B, and A's mother his wife. E has no child. If C were A's father, his wife would have to be A's mother, but G is C's sister-in-law and H is A's aunt, so neither can be C's wife and mother of A. So F is A's father. F's wife cannot be H (H is A's aunt), so F is married to G, who is D's daughter-in-law. The third couple must include C; C cannot marry E, because then H would be unmarried and could be A's aunt only as a sister of A's parents, which is impossible (B has no daughter, and G's parents are not in the family). So C is married to H. Since H is female, C is male and A is female. G being C's sister-in-law makes C a son of B. So B and D have sons C, E and F; C is married to H, F to G, and A is the daughter of F and G. Males: B, C, E and F, so four.

Answer: (C) D

Every child's parents are in the family, and the only couple who can be parents of the brothers E and F is B and D, so E and F are both sons of B and D (B has no daughter). A's father must be a son of B, and A's mother his wife. E has no child. If C were A's father, his wife would have to be A's mother, but G is C's sister-in-law and H is A's aunt, so neither can be C's wife and mother of A. So F is A's father. F's wife cannot be H (H is A's aunt), so F is married to G, who is D's daughter-in-law. The third couple must include C; C cannot marry E, because then H would be unmarried and could be A's aunt only as a sister of A's parents, which is impossible (B has no daughter, and G's parents are not in the family). So C is married to H. Since H is female, C is male and A is female. G being C's sister-in-law makes C a son of B. So B and D have sons C, E and F; C is married to H, F to G, and A is the daughter of F and G. E is a son of B and D, so D is his mother.

Answer: (B) Only conclusion II follows

J = P = R ≥ G, so J ≥ G and II follows. S = G ≤ R and L ≤ R give no relation between S and L, so I does not follow.

Answer: (A) Only conclusion I follows

M > D > E = T ≥ A, so A < M and I follows. F is linked to the rest only through H < C < T, and Q only through M, so the relation between F and Q is unknown; II does not follow.

Answer: (C) Either conclusion I or II follows

X and R (and so Z and Q) are not connected by a definite relation, so neither conclusion follows on its own. But Z = X and Q = R, so II says X ≤ R; I (X > R) and II (X ≤ R) together cover every case and exactly one must be true. Hence either I or II follows.

Answer: (C) The one who likes Pink

Final arrangement: 9 April D (Yellow), 26 April F (White), 9 July A (Pink), 26 July E (Green), 9 September C (Red), 26 September B (Blue). A, who likes Pink, was born on 9th July.

Answer: (E) F - White - 26th April

Final arrangement: 9 April D (Yellow), 26 April F (White), 9 July A (Pink), 26 July E (Green), 9 September C (Red), 26 September B (Blue). Only F - White - 26th April matches.

Answer: (D) Three

Final arrangement: 9 April D (Yellow), 26 April F (White), 9 July A (Pink), 26 July E (Green), 9 September C (Red), 26 September B (Blue). F (26 April) and B, the Blue-liker (26 September): A, E and C are between them, so three.

Answer: (B) F

Final arrangement: 9 April D (Yellow), 26 April F (White), 9 July A (Pink), 26 July E (Green), 9 September C (Red), 26 September B (Blue). F likes White.

Answer: (B) Only I

Final arrangement: 9 April D (Yellow), 26 April F (White), 9 July A (Pink), 26 July E (Green), 9 September C (Red), 26 September B (Blue). I is incorrect (D was born first). II is correct (F on 26 April, Pink-liker A on 9 July, with no one between). III is correct (July has 31 days and E likes Green). So only I is incorrect.

Answer: (A) A

Final order of buying: 1st E (Sedan), 2nd A (Limousine), 3rd D (Hatchback), 4th F (Coupe), 5th C (Convertible), 6th B (SUV). A bought the Limousine.

Answer: (A) C

Final order of buying: 1st E (Sedan), 2nd A (Limousine), 3rd D (Hatchback), 4th F (Coupe), 5th C (Convertible), 6th B (SUV). The person just before D is A (2nd); three turns later is the 5th buyer, C.

Answer: (B) SUV

Final order of buying: 1st E (Sedan), 2nd A (Limousine), 3rd D (Hatchback), 4th F (Coupe), 5th C (Convertible), 6th B (SUV). Each person is related to the car bought immediately after his own turn: A (2nd) to the Hatchback (3rd), E (1st) to the Limousine (2nd). C (5th) relates to the SUV (6th).

Answer: (C) SUV

Final order of buying: 1st E (Sedan), 2nd A (Limousine), 3rd D (Hatchback), 4th F (Coupe), 5th C (Convertible), 6th B (SUV). Between E (1st) and the Hatchback buyer (3rd) there is one person. One more than the count after X means zero persons after X, so X is the last car bought, the SUV.

Answer: (D) E

Final order of buying: 1st E (Sedan), 2nd A (Limousine), 3rd D (Hatchback), 4th F (Coupe), 5th C (Convertible), 6th B (SUV). E bought the Sedan.

Answer: (E) Both conclusions I and II follow

I: nothing prevents clubs from overlapping allowances, so the possibility holds. II: the courts that are allowances can never be revenue (no allowance is revenue), so some courts are definitely not revenue. Both follow.

Answer: (D) Neither conclusion I nor II follows

I: editions that are not investors may or may not be affects, so I is not certain. II: all investors are verdicts and no verdict is an affect, so no investor can be an affect; the possibility is ruled out. Neither follows.

Answer: (A) Only conclusion I follows

I: nothing stops all skippers from also being hears, so the possibility holds. II: some players are hears, and those players may be wins, so 'No win is a hear' is not certain. Only I follows.

Answer: (E) QSTU

JLMP, ACDG, EGHK and LNOR each follow +2, +1, +3 (for example J10 L12 M13 P16). QSTU is Q17 S19 T20 U21, which is +2, +1, +1, so it is the odd one.

Answer: (E) E

Ages: 1947 - 78, 1959 - 66, 1964 - 61, 1973 - 52, 1978 - 47, 1988 - 37, 2001 - 24. The only pair summing to another age is 24 + 37 = 61, so C and F are 24 and 37 and D is 61. A is just older than F, so F cannot be 24 (then A would be 37); F = 37 and A = 47, C = 24. G is 66 or 78; A + G is prime only for 47 + 66 = 113, so G = 66. B + 47 must be divisible by 5, so B = 78. E = 52. Final: B 1947, G 1959, D 1964, E 1973, A 1978, F 1988, C 2001. E was born in 1973.

Answer: (B) Three

Ages: 1947 - 78, 1959 - 66, 1964 - 61, 1973 - 52, 1978 - 47, 1988 - 37, 2001 - 24. The only pair summing to another age is 24 + 37 = 61, so C and F are 24 and 37 and D is 61. A is just older than F, so F cannot be 24 (then A would be 37); F = 37 and A = 47, C = 24. G is 66 or 78; A + G is prime only for 47 + 66 = 113, so G = 66. B + 47 must be divisible by 5, so B = 78. E = 52. Final: B 1947, G 1959, D 1964, E 1973, A 1978, F 1988, C 2001. Between B (1947) and A (1978): G, D and E, so three.

Answer: (A) 37 years

Ages: 1947 - 78, 1959 - 66, 1964 - 61, 1973 - 52, 1978 - 47, 1988 - 37, 2001 - 24. The only pair summing to another age is 24 + 37 = 61, so C and F are 24 and 37 and D is 61. A is just older than F, so F cannot be 24 (then A would be 37); F = 37 and A = 47, C = 24. G is 66 or 78; A + G is prime only for 47 + 66 = 113, so G = 66. B + 47 must be divisible by 5, so B = 78. E = 52. Final: B 1947, G 1959, D 1964, E 1973, A 1978, F 1988, C 2001. D is 61 and C is 24, so the difference is 37 years.

Answer: (D) B

Ages: 1947 - 78, 1959 - 66, 1964 - 61, 1973 - 52, 1978 - 47, 1988 - 37, 2001 - 24. The only pair summing to another age is 24 + 37 = 61, so C and F are 24 and 37 and D is 61. A is just older than F, so F cannot be 24 (then A would be 37); F = 37 and A = 47, C = 24. G is 66 or 78; A + G is prime only for 47 + 66 = 113, so G = 66. B + 47 must be divisible by 5, so B = 78. E = 52. Final: B 1947, G 1959, D 1964, E 1973, A 1978, F 1988, C 2001. B, born in 1947, is the oldest.

Answer: (C) 66 years

Ages: 1947 - 78, 1959 - 66, 1964 - 61, 1973 - 52, 1978 - 47, 1988 - 37, 2001 - 24. The only pair summing to another age is 24 + 37 = 61, so C and F are 24 and 37 and D is 61. A is just older than F, so F cannot be 24 (then A would be 37); F = 37 and A = 47, C = 24. G is 66 or 78; A + G is prime only for 47 + 66 = 113, so G = 66. B + 47 must be divisible by 5, so B = 78. E = 52. Final: B 1947, G 1959, D 1964, E 1973, A 1978, F 1988, C 2001. G was born in 1959, so G is 66.

Answer: (C) C

Final arrangement, written from west to east. Row 1 (facing north): A, B, C, G, F, E, D. Row 2 (facing south): P, S, R, Q, V, U, T. So A faces P, B faces S, C faces R, G faces Q, F faces V, E faces U and D faces T. (For Row 1 the right is east; for Row 2, facing south, the right is west.) C faces R.

Answer: (A) A sits at an end of Row 1

Final arrangement, written from west to east. Row 1 (facing north): A, B, C, G, F, E, D. Row 2 (facing south): P, S, R, Q, V, U, T. So A faces P, B faces S, C faces R, G faces Q, F faces V, E faces U and D faces T. (For Row 1 the right is east; for Row 2, facing south, the right is west.) A is at the west end of Row 1. U faces E, Q is fourth in Row 2, T faces D and three persons (C, G, F) sit between B and E.

Answer: (D) F - U

Final arrangement, written from west to east. Row 1 (facing north): A, B, C, G, F, E, D. Row 2 (facing south): P, S, R, Q, V, U, T. So A faces P, B faces S, C faces R, G faces Q, F faces V, E faces U and D faces T. (For Row 1 the right is east; for Row 2, facing south, the right is west.) A-P, B-S, C-R and G-Q face each other. F faces V, not U, so F - U is the odd pair.

Answer: (E) One

Final arrangement, written from west to east. Row 1 (facing north): A, B, C, G, F, E, D. Row 2 (facing south): P, S, R, Q, V, U, T. So A faces P, B faces S, C faces R, G faces Q, F faces V, E faces U and D faces T. (For Row 1 the right is east; for Row 2, facing south, the right is west.) B faces S. B faces north, so B's left is west, where only A sits.

Answer: (B) R

Final arrangement, written from west to east. Row 1 (facing north): A, B, C, G, F, E, D. Row 2 (facing south): P, S, R, Q, V, U, T. So A faces P, B faces S, C faces R, G faces Q, F faces V, E faces U and D faces T. (For Row 1 the right is east; for Row 2, facing south, the right is west.) V faces F. V faces south, so V's right is west: Q, then R. R is second to the right of V.