26 November 2025 · Shift 2

Position in a row When a person's place is counted from both ends, that person gets counted twice, so 1 must be subtracted. Formula: total = position from left + position from right -1. Total =15+18-1=32. So there are 32 people in the row, and option (b) is correct.

Blood relation 'The only daughter of my father' is the woman herself, since she is her father's one and only daughter. So the sentence really says: his wife is this very woman. If the woman is the man's wife, the man is her husband, so option (b) is correct.

Syllogism Statement 1 is a particular affirmative: at least one book is a notebook. Statement 2 is a universal affirmative: every notebook lies inside the class of files. Those books that are notebooks must therefore be files as well, so 'Some books are files' is certain - conclusion I follows. Conclusion II claims that no book is a file, which directly contradicts what has just been proved, so it does not follow. Only conclusion I follows, and option (a) is correct.

Position in a row First bring both positions to the same end. Position of Q from the left =35-15+1=21. P is 12th from the left and Q is 21st from the left, so the bicycles strictly between them number 21-12-1=8. Hence option (c) is correct.

Position in a row Convert B's position to the left end: position of B from the left =70-21+1=50. A is 24th from the left and B is 50th from the left. Books lying between them =50-24-1=25. Hence option (c) is correct.

Number series Work with the terms that are given: $75=36\times 2+3$, $154=75\times 2+4$ and $313=154\times 2+5$. So the rule is: double the term and add a number that grows by 1 at every step. Running the rule backwards from 36, the missing term $=\frac{36-2}{2}=17$. Check the first step: $8\times 2+1=17$, which fits perfectly. Hence the missing term is 17 and option (c) is correct.

Number series Take the differences between consecutive terms: -54-(-50)=-4, -64-(-54)=-10, -80-(-64)=-16 and -102-(-80)=-22. The differences -4,-10,-16,-22 each fall by 6. So the next difference is -22-6=-28, which gives the missing term =-102-28=-130. Check onwards: -130-34=-164 and -164-40=-204, exactly the terms that follow. Hence option (b) is correct.

Number series Test the rule 'double and add': $12\times 2=24$ and 29-24=5. $29\times 2=58$ and 64-58=6; $64\times 2=128$ and 135-128=7. $135\times 2=270$ and 275-270=5; $275\times 2=550$ and 556-550=6. So the number added repeats in the cycle 5, 6, 7. The next addition is therefore 7, giving the missing term $=556\times 2+7=1112+7=1119$. Hence option (b) is correct.

Symbol operation Find what the symbol # does to the two numbers standing on either side of it. $15\times 6=90$ and $450÷90=5$, so the product has been multiplied by 5. Check with the second pair: $12\times 8\times 5=480$, exactly the value given. So $a\#b=a\times b\times 5$, which gives $20\#9=20\times 9\times 5=900$. Hence option (b) is correct.

Symbol operation Look for a rule that links the three numbers to the value on the right. In the first case $300÷15=20$ and 500-20=480, which is the value given. In the second case $450÷25=18$ and 920-18=902, so the same rule holds. The rule is therefore: first number - (second number $÷$ third number). So the required value $=745-\frac{336}{14}=745-24=721$, and option (b) is correct.

Symbol substitution Put the real sign in place of each symbol: # becomes $\times$, % becomes $÷$ and ! becomes +. The expression becomes $16\times 4÷2+7$. By BODMAS, $\times$ and $÷$ are done from left to right: $16\times 4=64$, then $64÷2=32$. Finally add: 32+7=39. Hence option (d) is correct.

Syllogism 'Some Flowers are Red' guarantees at least one thing that is a Flower and Red at the same time. 'All Flowers are Beautiful' places that thing inside Beautiful as well, so it is Beautiful and Red together - conclusion I follows. 'No Red is Artificial' keeps every Red thing outside Artificial, so those Red flowers cannot be Artificial. That means at least some Flowers are not Artificial - conclusion II follows too. Both conclusions follow, so option (c) is correct.

Stack arrangement Number the places 1 (bottom) to 6 (top). Three boxes between A and B means their places differ by 4, so the pair of places is (1, 5) or (2, 6). B is immediately below C, that is C=B+1, and C is not at place 6. If B were at 5, C would be at 6, which is not allowed; if B were at 1, then A is at 5 and C at 2, leaving 3, 4, 6 for D, E, F - and D=E+1 forces E at 3, D at 4 and F at 6, which contradicts 'F is below E'. So B is at 2, C at 3 and A at 6. Places 1, 4, 5 are left; D=E+1 puts E at 4 and D at 5, and F takes place 1, correctly below E. A stands at place 6 and C at place 3, so the boxes at places 4 and 5 lie between them - that is 2 boxes. Hence option (b) is correct.

Letter cluster series Split each cluster into its three letter positions and track the positions separately. First letters G, J, ?, P, S move +3 each time, so the missing first letter is J+3=M. Second letters Z, X, ?, T, R move -2 each time, so the missing second letter is X-2=V. Third letters M, R, ?, B, G move +5 each time (after Z the count starts again at A), so the missing third letter is R+5=W. The cluster is therefore MVW, so option (a) is correct.

Ranking by weight 'C is heavier than only 3 boxes' means exactly 3 boxes are lighter than C, so C is the 2nd heaviest of the five. A and B are lighter than D and E is heavier than D, so E>D>A,B. The only box that can stand above C is E, which gives the full order E>C>D>A,B. So D is lighter than C, that is, D weighs less than 50 kg. Among the options only 42 kg is below 50 kg, so option (c) is correct.

Linear seating arrangement U is at the extreme right end, so seat 6 is fixed for U. R is an immediate neighbour of both Q and S, so R sits between them: the block is either Q – R – S or S – R – Q. S is immediately to the right of T, so T must sit just left of S; in Q – R – S the seat left of S is taken by R, so only S – R – Q works, giving the block T – S – R – Q. P is not at either end, so P cannot take seat 1; the four-seat block must therefore fill seats 1 to 4 and P takes seat 5. The row is T, S, R, Q, P, U, so T is at the extreme left end and option (a) is correct.

Linear seating arrangement Number the eight seats 1 to 8 in the direction of the boys' own right, so 'immediate right' always means the next higher seat number. S is to the immediate right of W, so W cannot be in seat 8; W takes seat 1 and S seat 2, and P, who is at the other end, takes seat 8. R and S are both immediate neighbours of Q, so Q sits between them; S is in seat 2, hence Q is in seat 3 and R in seat 4. V is third to the right of Q, so V is in seat 3+3=6. Only seats 5 and 7 are left, for T and U; T is not an immediate neighbour of R (seat 4), so T cannot take seat 5 and must take seat 7, leaving seat 5 for U. Between S (seat 2) and V (seat 6) sit Q, R and U, so 3 boys are sitting between them and option (c) is correct.

Logical inference The statement is a universal negative: failing the physical fitness test rules an athlete out of the National Games. Turn it round (contrapositive): if a person did take part in the National Games, that person cannot have failed the test. Rohit did participate, so the only thing that follows with certainty is that he did not fail the physical fitness test. Topping the test, practising for months and receiving a special exemption are extra details that the statement never supplies, so none of them can be inferred; hence option (b) is correct.

Interchange of numbers The given equation is $3\times 5+8-6÷1$. Only the two whole numbers 3 and 8 are swapped, so every 3 becomes 8 and every 8 becomes 3: $8\times 5+3-6÷1$. Now apply BODMAS — division first, $6÷1=6$, then multiplication, $8\times 5=40$. The value is 40+3-6=37. Hence option (c) is correct.

Interchange of numbers Swap the whole numbers 4 and 5 everywhere in each equation and then test it with BODMAS. I. $6\times 4+5-10÷2=29$ becomes $6\times 5+4-10÷2$. Its value is 30+4-5=29, which is exactly the right-hand side, so equation I stays correct. II. $8+4\times 3-5÷1=25$ becomes $8+5\times 3-4÷1$. Its value is 8+15-4=19, which is not 25, so equation II is not correct. Only equation II fails after the interchange, so option (b) is correct.

Letter coding-decoding Write the alphabetical positions of GLOW and its code CIKS: G = 7, L = 12, O = 15, W = 23 and C = 3, I = 9, K = 11, S = 19. The shifts are 7-3=4, 12-9=3, 15-11=4 and 23-19=4, i.e. the pattern −4, −3, −4, −4. FISH → BFOD confirms the pattern: 6-4=2 (B), 9-3=6 (F), 19-4=15 (O), 8-4=4 (D). Apply the same shifts to DARK: 4-4=0, which wraps round to 26, i.e. Z, and 1-3=-2, which wraps round to 24, i.e. X. Then 18-4=14 gives N and 11-4=7 gives G, so DARK is coded as ZXNG and option (a) is correct.

Letter coding-decoding Positions of HEAD and its code CYVX: H = 8, E = 5, A = 1, D = 4 and C = 3, Y = 25, V = 22, X = 24. H → C is 8-5=3; E → Y is 5-6=-1, which wraps round to 25; A → V is 1-5=-4, which wraps round to 22; D → X is 4-6=-2, which wraps round to 24. So the shifts run −5, −6, −5, −6, and MILK → HCGE confirms it: 13-5=8 (H), 9-6=3 (C), 12-5=7 (G), 11-6=5 (E). For PENT, 16-5=11 gives K and 5-6=-1 wraps round to 25, i.e. Y. Then 14-5=9 gives I and 20-6=14 gives N, so PENT is coded as KYIN and option (c) is correct.

Alphanumeric series Each term is two letters followed by a three-digit number, so treat the three parts separately. The first letters are J, G, D, A — every step is three places back, so the next is 1-3=-2, which wraps round to 24, i.e. X. The second letters are S, P, M, J — again three places back, so the next is 10-3=7, i.e. G. The numbers 400, 440, 480, 520 rise by 40 each time, so the next is 520+40=560. The missing term is therefore XG560, and option (a) is correct.

Alphanumeric series Split every term into its two letters and its number. The first letters A, G, M, S each move six places forward, so the next is 19+6=25, i.e. Y. The second letters D, J, P, V also move six places forward, so the next is 22+6=28, which wraps round to 28-26=2, i.e. B. The numbers 10, 20, 40, 80 double each time, so the next is $80\times 2=160$. The missing term is YB160, so option (a) is correct.

Alphanumeric series The first letters run B, D, F, ?, J and the second letters run E, G, I, ?, M — both move two places forward each time. So the missing term begins with 6+2=8, i.e. H, followed by 9+2=11, i.e. K. For the numbers, $15\times 3+2=47$ and $47\times 3+2=143$, so each number is 3 times the previous one plus 2. Hence the missing number is $143\times 3+2=431$, and the check $431\times 3+2=1295$ matches the last term. The missing term is HK431, so option (d) is correct.