30 November 2025 · Shift 2

Decision making Total revenue = registration fee $\times$ number of participants, so it is the product, not either factor alone, that has to be made as large as possible. A high fee raises the amount per head but shrinks the audience; a low fee fills the hall but earns little from each participant. Either extreme can end with the smaller product. The sound managerial step is therefore to estimate how many participants each fee level would actually attract and then choose the combination whose product is the largest. Option (d) removes the fee altogether and leaves the income to uncertain sponsorship, which does not serve the stated objective. Hence option (c) is correct.

Statement and inference The statement reports only what a study observed: regular breakfast-eaters show lower rates of obesity and other health problems. That is an association between two things. Inference II says regular breakfast 'can prevent' those problems. The hedged word 'can' claims no more than the observed association supports, so II follows. Inference I converts the association into a definite cause — that skipping breakfast causes obesity. A study of this kind shows correlation, not proof of cause, so I does not follow. Inference III advises obese people to stop eating breakfast, which runs against the very finding reported, so it does not follow. Hence only II follows; option (b) is correct.

Statement and assumptions An assumption is what the government must already be taking for granted when it raises the duty. II follows: a higher duty on goods that continue to be imported adds directly to the treasury's collections, which is a standard purpose of raising a duty. III follows: costlier imports push buyers towards Indian fruits and chocolates, so the import bill falls and the trade balance improves — the usual protective purpose of a duty hike. I does not follow: a duty makes such items dearer, not unavailable. Wanting people to 'completely stop' consuming foreign food is far stronger than anything the announcement implies. Hence both II and III follow; option (d) is correct.

Letter series Write each letter by its place in the alphabet: W =23, U =21, S =19, Q =17. Every term is 2 less than the one before it, that is, one letter is skipped each time. Next place =17-2=15, and the 15th letter of the alphabet is O. Hence option (b) is correct.

Coded inequalities First replace every symbol by the relation it stands for: '@' is 'less than', '©' is 'greater than or equal to', '%' is 'equal to', '*' is 'greater than', and the remaining symbol is 'less than or equal to'. The three statements therefore read D<J, $J\le P$ and P=H. Joining them into one chain: $D<J\le P=H$. Conclusion I, H © J, asks for $H\ge J$. The chain gives $H=P\ge J$, so I is true. Conclusion II, P * D, asks for P>D. The chain gives $D<J\le P$, hence D<P, so II is true. Both conclusions are true; hence option (d) is correct.

Number series Take the differences of consecutive terms: 70-55=15, 87-70=17, 106-87=19. The differences 15, 17, 19 themselves increase by 2 each time, so the next difference is 19+2=21. Required term =106+21=127. Hence option (a) is correct.

Number series Each term is one less than twice the term before it. Check: $7\times 2-1=13$, $13\times 2-1=25$, $25\times 2-1=49$. So the required term $=49\times 2-1=98-1=97$. The same pattern shows in the differences 6, 12, 24, which double every step; the next difference is 48 and 49+48=97. Hence option (b) is correct.

Number series The terms grow roughly five times each step, so try 'multiply by 5 and then adjust'. Check: $6\times 5-6=24$, $24\times 5-6=120-6=114$, $114\times 5-6=570-6=564$. The rule $\times 5-6$ holds throughout, so the required term $=564\times 5-6$. $564\times 5=2820$, and 2820-6=2814. Hence option (d) is correct.

Odd one out A hovercraft, a motorboat and a ferry all travel on the surface of the water — the hovercraft on a cushion of air just above it, the other two floating on it. A submersible is built to go below the surface and to work under water, which is exactly what the other three never do. So the submersible is the odd one out; hence option (d) is correct.

Odd one out Anthracite, bituminous and lignite are the ranks of coal, graded by carbon content — lignite the lowest and softest, bituminous in the middle, anthracite the hardest with the highest carbon. Graphite is not a coal at all; it is a crystalline allotrope of carbon, used in pencil leads, as a lubricant and as a moderator in nuclear reactors. So graphite is the odd one out; hence option (d) is correct.

Alternating number series The series moves alternately: first subtract 10, then add 9. Check: 59-10=49, 49+9=58, 58-10=48, 48+9=57. The next step is again -10, so the required term =57-10=47. The same thing is seen in the two interlaced series 59, 58, 57, ... and 49, 48, 47, ..., each falling by 1; the missing term belongs to the second one. Hence option (b) is correct.

Syllogism '40% of Hit are Hop' is a particular statement and means no more than 'some Hit are Hop'. So all three links are of the 'some' type. A chain made only of 'some' statements never carries a definite conclusion from the first class to the last, because the overlapping parts need not be the same parts. Conclusion I, 'Some Hit are Rock', is therefore only a possibility, not a certainty — it does not follow. Conclusion II, 'No Hop is Rock', is equally uncertain, since Hop and Rock are free to overlap through the common Pop portion. I and II are not a complementary pair either (that would need 'Some Hop are Rock' against 'No Hop is Rock' on the same two classes), so no either-or case arises. Hence neither conclusion follows; option (d) is correct.

Missing letters series Count the string with its blanks: it is 12 letters long, and 12 divides neatly into three blocks of 4. The letters that are already there — x y _ z at the start — suggest the block x y y z. Testing xyyz / xyyz / xyyz against the given string fits every letter that is printed. The gaps are the 3rd, 8th, 9th and 12th places, which the pattern fills with y, z, x and z. So the missing letters in order are y z x z; hence option (b) is correct.

Alpha-numeric series Each term has a letter and a number, so handle the two parts separately. Letters: Y =25, W =23, U =21, S =19 — each falls by 2, so the next place is 19-2=17, which is Q. Numbers: $8=2^3$, $64=4^3$, $216=6^3$, $512=8^3$ — cubes of consecutive even numbers, so the next is ${10}^3=1000$. Putting the two together, the missing term is Q1000. Hence option (c) is correct.

Letter series completion Ten letters are printed and each option supplies eight, so the completed string has 18 letters — three repetitions of a six-letter block. The printed letters F _ _ K J _ _ R O _ _ T F R _ _ J T match the block F R O K J T repeated three times. Reading off the four gaps, two letters at a time: R O, then T F, then K J, then O K. So the required sequence is R, O, T, F, K, J, O, K. Hence option (c) is correct.

Number series Instead of the terms themselves, look at the gaps between two consecutive terms. 8-3=5, 15-8=7, 24-15=9, 35-24=11 — the gaps are consecutive odd numbers. So the next gap must be 13, giving the missing term =35+13=48. The same series can be read as $2^2-1,3^2-1,4^2-1,5^2-1,6^2-1$, whose next term is $7^2-1=48$. Hence option (b) is correct.

Number analogy Two complete pairs are given, so first find the rule from them. 7756-6802=954 and 7309-6355=954, so the second number is always 954 more than the first. Applying the rule backwards, the missing first number =10889-954=9935. Check: 9935+954=10889, exactly as required. Hence option (d) is correct.

Number analogy The second number is about seven times the first, so test (first number $\times 7$) with a small adjustment. $927\times 7=6489$ and 6489-3=6486. $691\times 7=4837$ and 4837-3=4834, so the rule is (first number $\times 7$) -3. For 514: $514\times 7=3598$, and 3598-3=3595. Hence option (d), 3595, is correct.

Number analogy The second number is roughly fourteen times the first, so try (first number $\times 14$) with a small addition. $410\times 14=5740$ and 5740+8=5748. $373\times 14=5222$ and 5222+8=5230, so the rule is (first number $\times 14$) +8. Reversing it for 1898: 1898-8=1890 and $1890÷14=135$. Hence option (b), 135, is correct.

Odd letter-number pair Treat the letters and the number separately. The letters behave alike in all four pairs: the first letter moves 1 step forward and the second letter 3 steps forward — D→E, T→W; M→N, R→U; N→O, O→R; B→C, Y→B (counting on past Z). So the number must decide the odd one; here the number is simply doubled. $244\times 2=488$, $196\times 2=392$ and $148\times 2=296$ — three pairs obey the rule. But $256\times 2=512$, whereas the pair shows 502. Hence option (d) is correct.

Odd letter-number pair Check the letters first: in every pair both letters move 6 steps forward — V→B, F→L; O→U, Y→E; H→N, G→M; T→Z, K→Q. Since the letters are alike everywhere, compare the rise in the numbers. 630-440=190, 550-360=190 and 710-520=190 — three pairs rise by the same 190. But 460-280=180, which breaks the pattern. Hence option (b) is correct.

Odd number pair The second number of each pair is much larger than the first, so test whether it is the square of the first. ${14}^2=196$, ${16}^2=256$ and ${15}^2=225$ — three pairs match exactly. But ${17}^2=289$, while the fourth pair shows 279, which is 10 short. Hence option (d) is correct.

Letter-number cluster analogy Deal with the letters and the number one at a time. In TC760 : QZ530 both letters go 3 steps back — T→Q and C→Z — and the number falls by 760-530=230. BS1040 : YP810 confirms it — B→Y, S→P, and 1040-810=230. For GL840: G→D, L→I, and 840-230=610. So the required cluster is DI610. Hence option (c) is correct.

Letter-number cluster analogy The letters do not shift in place — they are first interchanged and each is then moved 2 steps back. KV72 → TI215: V goes back 2 to T and takes the first position, K goes back 2 to I and takes the second position. BZ93 → XZ278 agrees: Z goes back 2 to X, and B goes back 2 to Z (counting back past A). The number is multiplied by 3 and 1 is subtracted: $72\times 3=216$, 216-1=215; $93\times 3=279$, 279-1=278. For EW103: W goes back 2 to U, E goes back 2 to C, and $103\times 3=309$, 309-1=308. So the required cluster is UC308. Hence option (b) is correct.

Letter-number cluster analogy The first cluster is QO112, the two letters being Q and O. The letters are interchanged and each is then moved 7 steps back: O→H takes the first place and Q→J takes the second, giving HJ. GC84 → VZ294 confirms it: C→V and G→Z, counting back past A. The number is multiplied by $\frac{7}{2}$: $112\times \frac{7}{2}=392$ and $84\times \frac{7}{2}=294$. For BS56: S→L takes the first place, B→U takes the second, and $56\times \frac{7}{2}=196$. So the required cluster is LU196. Hence option (b) is correct.