28 November 2025 · Shift 1

Ranking after interchange When two girls exchange places, each of them simply takes the seat the other was standing on. Reena moves from 9th to 18th from the left, so the seat she now occupies — Pooja's original seat — is the 18th from the left. That very seat is also 14th from the right, so the number of girls in the row =18+14-1=31. Pooja now stands on Reena's old seat, which is 9th from the left. Her position from the right =31-9+1=23. Hence Pooja's new position is 23rd from the right.

Position in a list Shyam is counted once while counting from the top and once again while counting from the bottom, so his own place gets counted twice. Total students = rank from the top + rank from the bottom -1. =12+18-1 =29 Hence there are 29 students in the list.

Number analogy Look for a rule that ties the pair on the left of the equality to the number on its right. $18\times 4=72$ and $72\times 3=216$, so the product of the pair has been tripled. Test it on the second pair: $51\times 27=1377$ and $1377\times 3=4131$, which is the given number. Apply the same rule to the third pair: $31\times 52=1612$. $1612\times 3=4836$.

Number series Write down the gaps between consecutive terms. 130-109=21, 153-130=23 and 178-153=25. The gaps themselves increase by 2 each time, so the next gap must be 27. Required term =178+27=205. The same series can be read as ${10}^2+9,{11}^2+9,{12}^2+9,{13}^2+9,{14}^2+9$.

Number series Take the differences of consecutive terms. 54-41=13, 69-54=15 and 86-69=17. These differences rise by 2 every time, so the next difference is 19. Required term =86+19=105. Read another way, the terms are $6^2+5,7^2+5,8^2+5,9^2+5,{10}^2+5$.

Paper folding and punching In the first step the top right and the bottom left corners are folded inwards, and in the second step the sheet is folded along the diagonal that joins the top right corner to the bottom left corner. Both punches are made in the lower right part of the packet, well clear of the two small corner flaps, so at each punch only two layers are cut — the lower right half of the sheet and the upper left half lying on it. Two punches through two layers give $2\times 2=4$ holes when the sheet is opened. Opening the diagonal fold reflects each hole across that diagonal, so the pair near the bottom right corner is copied as a matching pair near the top left corner. The corner flaps carry no punch, so they add nothing further; the opened sheet shows exactly two holes near the bottom right corner and their mirror images near the top left corner. Option (a) strings the four holes along the main diagonal, option (c) shows six holes and option (d) keeps all four holes in the lower half, so only option (b) is correct.

Paper folding and punching The sheet is folded twice — the lower half upwards and then the right half on to the left half — so the punched packet is one quarter of the square and every punch cuts four layers. Three punches through four layers give $3\times 4=12$ holes in the opened sheet. Undo the vertical fold first: the three holes are reflected in the vertical centre line, so an identical mirrored group appears near the top right corner. Now undo the horizontal fold: all six holes are reflected in the horizontal centre line, giving a mirrored group near each of the two lower corners. So the opened sheet must carry the same group of three holes near all four corners, each group being the mirror image of the group beside it. Only option (b) has this double mirror symmetry: (a) repeats the group by sliding it instead of mirroring it, (c) breaks the symmetry in the lower half and (d) scatters the holes with no symmetry at all.

Number puzzle rows In every row the last number is far larger than the other three, so a product must be involved. Row 1: $3\times 5\times 4=60$ and $60\times 2=120$, which is the fourth number. Row 2: $6\times 2\times 7=84$ and $84\times 2=168$, again the fourth number. So the rule is: the product of the first three numbers, doubled, gives the fourth. Row 3: let the missing number be x, then $\left(5\times 3\times x\right)\times 2=90$. 30x=90 x=3

Number puzzle boxes Each box carries three small numbers and one larger one, so try a product followed by an addition. Box 1: $5\times 4+3=23$, the large number of that box. Box 2: $2\times 7+6=20$, again the large number. So the rule is: (second number $\times$ third number) + first number = the large number. Box 3: $6\times x+4=58$ 6x=54 x=9

Counting triangles First number the smallest triangles that the lines cut out of the figure. The two crossing diagonals of the left half give the three small triangles 1, 2 and 3; the long diagonal and the vertical line make triangle 4 below the rectangle; the right half is cut into triangles 5 and 6. That is 6 simple triangles. Joining neighbours gives three more: 1+2, 1+3 and 5+6 are each a triangle. One large triangle spans the whole figure — the top edge, the long diagonal from the top left corner and the vertical line close it, and it contains parts 2 and 4. Total =6+4=10 triangles.

Venn diagram reading Fix the three shapes first: the rectangle is buses, the circle is black things and the triangle is bikes. Something that is black, a bike and a bus at the same time must lie inside all three shapes, and that is the single region marked 3. So black bikes that are also buses number 3, not 9 — statement I is not correct. Buses that are not bikes are the parts of the rectangle lying outside the triangle, i.e. the regions 8, 9 and 5. 8+9+5=22, not 24 — statement II is not correct either. Since both the statements are wrong, the answer is 'Both I and II'.

Circular seating All five face the centre, so for each person the right hand side is the anticlockwise side of the table. 'Only one person sits between O and P towards the right of P' places O two seats to the right of P. 'N is second to the left of P' places N two seats to the left of P, which in a circle of five is three seats to the right of P. 'M is second to the right of O' places M four seats to the right of P. Seats one, two, three and four to the right of P are thus taken by nobody, O, N and M respectively, so the only seat left — the first to the right of P — must belong to Q. Hence Q sits to the immediate right of P.

Comparison ranking Arrange the six toys from the costliest downwards. 'The price of T is less than only R' means R is the costliest and T comes second. Q is stated to be third. The remaining three places go to S, U and P, and since S>U>P they fall in that order. The full order is R>T>Q>S>U>P. Cheaper than Q are S, U and P, and the only option naming two of these three is 'U and P'.

Symbol substitution Put back the real sign in place of every symbol used in the expression. So 6 * 4 # 12 % 8 becomes $6-4\times 12+8$. By BODMAS the multiplication is done first: $4\times 12=48$. 6-48+8=14-48 =-34

Figure analogy Compare the first pair: everything inside the square — the arrow, the small rectangle on its stem and the tiny triangle in the corner — has been turned through ${180}^{\circ }$. In the first figure the arrow points up, the rectangle sits on the lower part of the stem and the triangle is in the top left corner; in the second the arrow points down, the rectangle has moved to the upper part and the triangle to the bottom right corner. So the rule is a half turn of the whole figure. Apply it to the third figure: the block arrow pointing up becomes a block arrow pointing down, and the letter P in the top left corner moves to the bottom right corner. A capital P given a half turn reads like a small 'd', and that is exactly what option (c) shows. Option (a) only flips the figure top to bottom and leaves P upright, option (b) mirrors it left to right so the arrow is not turned at all, and option (d) turns the arrow through ${90}^{\circ }$ instead of ${180}^{\circ }$.

Figure analogy Every box carries the same seven symbols — B, S, T, U, A, a star and a small circle — in seven places: the three of the top row, the centre, and the three of the bottom row. Compare the 2nd box with the 3rd: the top row slides one place to the right and so does the bottom row (the last symbol coming round to the front), while the centre symbol stays where it is. The 1st box becomes the 2nd by a different move: the two end symbols of the top row change places, the two end symbols of the bottom row change places, and the middle column slides down one step (top-middle goes to the centre, the centre goes to bottom-middle, bottom-middle goes to top-middle). The 3rd box becomes the 4th by that same second move, so the two moves act turn by turn and the step from the 4th box to the missing one must be the sliding move again. The 4th box has U, star, B on top and A, S, circle below, so sliding each row one place right gives B, U, star on top and circle, A, S below, with T still in the centre. That is exactly the figure of option (b), so option (b) is the correct answer.

Figure series Each box holds six vertical rods: the first three carry their symbol at the top and the last three carry it at the bottom. Box 1 to box 2: the first top symbol changes from a square to a star. Box 2 to box 3: the last bottom symbol changes from a star to a square. Box 3 to box 4: the second top symbol changes from a square to a star — so one change happens on top from the left and the next at the bottom from the right, turn by turn. The next change is therefore at the bottom: the second bottom symbol must turn from a star into a square. The missing figure thus shows star, star, square on top and star, square, square below, which is option (b), so option (b) is the correct answer.

Coded operations Replace every code letter by the sign it stands for. The expression 48 A 12 B 3 D 5 C 2 becomes $48+12÷3-5\times 2$. By BODMAS, division and multiplication come first: $12÷3=4$ and $5\times 2=10$. So the value =48+4-10=42. Hence option (a) is the correct answer.

Signs in equation Put each set of signs into $42*18*36*3*2=36$ and simplify by BODMAS. Option (b) puts +, -, $÷$, $\times$ in order, giving $42+18-36÷3\times 2$. Division and multiplication first: $36÷3=12$ and then $12\times 2=24$. So the left side =42+18-24=60-24=36, which matches the right side. No other set of signs reaches 36, so option (b) is the correct answer.

Odd letter-number pair In every pair the number is simply doubled: $108\times 2=216$, $132\times 2=264$, $244\times 2=488$ and $296\times 2=592$. So the numbers agree everywhere and the letters must settle the question. In ES–FV, BI–CL and QV–RY the first letter moves +1 and the second letter moves +3: S(19)+3=V(22), I(9)+3=L(12), V(22)+3=Y(25). In DY–EC the first letter moves +1 as usual, but the second moves +4, because Y(25)+4=29-26=3=C. That pair alone breaks the rule, so option (a) is the correct answer.

Odd letter-number pair In every pair the number is multiplied by 1.5: $120\times 1.5=180$, $246\times 1.5=369$, $122\times 1.5=183$ and $310\times 1.5=465$. In AX–CC, BC–DH and KV–MA the first letter moves +2 and the second moves +5: X(24)+5=29-26=3=C, C(3)+5=H(8), V(22)+5=27-26=1=A. In SQ–UW the first letter moves +2 as usual, but the second moves +6, because Q(17)+6=23=W. That pair is therefore the odd one out, so option (a) is the correct answer.

Odd number group Check the two gaps inside each triple. 315-285=30 and 285-255=30; 492-462=30 and 462-432=30; 678-648=30 and 648-618=30. So in three of the groups the numbers fall by a constant 30 each time. In (754,724,700) the first gap is 754-724=30, but the second is only 724-700=24. That group breaks the pattern, so option (d) is the correct answer.

Odd number group Read each triple as (first, second, third) and look for one rule that ties all three numbers together. (120,243,363): $120\times 2+3=243$, and 120+243=363. (155,313,468): $155\times 2+3=313$, and 155+313=468. (182,367,549): $182\times 2+3=367$, and 182+367=549. So the rule is: second $=2\times$ first $+3$, and third = first + second. Test the remaining group (145,292,437): $145\times 2+3=293$, but the second number is 292. The third number fails as well, since 145+293=438, not 437. This group alone breaks the rule, so option (c) is the correct answer.

Letter-number analogy Letters first: T(20)-3=Q(17) and C(3)-3=Z(26), and again B(2)-3=Y(25) and S(19)-3=P(16). So both letters step back 3 places, C rolling round to Z and B to Y. Numbers next: 760-210=550 and 1040-210=830, so every number falls by 210. For GL840, G(7)-3=D(4) and L(12)-3=I(9), which gives DI. And 840-210=630, so the required cluster is DI630. Hence option (c) is the correct answer.

Letter-number analogy Take the numbers first: $72\times 3-1=215$ and $93\times 3-1=278$, so the number rule is $\times 3-1$. Hence the missing number $=103\times 3-1=309-1=308$. For the letters, a plain forward or backward shift does not work, so try moving each letter and then exchanging the two. MX: M(13)-2=K(11) and X(24)-2=V(22), and writing them the other way round gives VK. DB: D(4)-2=B(2) and B(2)-2=Z(26), and exchanging gives ZB — the rule holds. GY: G(7)-2=E(5) and Y(25)-2=W(23), and exchanging gives WE. So the required cluster is WE308, and option (b) is the correct answer.