28 November 2025 · Shift 3

Ranking after exchange When two students exchange places, each simply takes the seat the other was standing on. Ravi moves from 10th to 24th from the left, so the seat he now occupies — Sumit's original seat — is 24th from the left. That same seat is 15th from the right, so the total number of students =24+15-1=38. Sumit now stands on Ravi's old seat, which is 10th from the left. His position from the right =38-10+1=29. Hence option (b) is correct.

Ranking in a row Bring both ranks to the same end of the row before comparing them. Neha is 24th from the right, so from the left her place is 50-24+1=27. Arjun stands at the 18th place from the left. The students standing between them occupy places 19 to 26, i.e. 27-18-1=8 students. Hence option (c) is correct.

Symbol based operation In each given pair, subtract the second number from the first and halve the difference. Test it on the first pair: $\frac{56-20}{2}=\frac{36}{2}=18$, which is the value given. Test it on the second pair: $\frac{74-30}{2}=\frac{44}{2}=22$, again the value given. Apply the same rule to 63 and 18: $\frac{63-18}{2}=\frac{45}{2}=22.5$. Hence option (c) is correct.

Symbol based operation In each given pair, multiply the two numbers and then multiply the product by 3. First pair: $12\times 20=240$ and $240\times 3=720$, the value given. Second pair: $15\times 20=300$ and $300\times 3=900$, again the value given. For 6 and 25: $6\times 25=150$ and $150\times 3=450$. Hence option (b) is correct.

Number series Take the difference between consecutive terms: 28-19=9, 39-28=11, 52-39=13. The differences 9, 11, 13 are consecutive odd numbers, each 2 more than the one before, so the next difference is 15. Required term =52+15=67. Hence option (b) is correct.

Number series Take the difference between consecutive terms: 18-11=7, 27-18=9, 38-27=11. The differences 7, 9, 11 rise by 2 each time, so the next difference is 13. Required term =38+13=51. Hence option (d) is correct.

Paper folding and punching The first fold brings the left half of the sheet on to the right half; the second fold brings the top half down on to the bottom half, so only the bottom-right quarter is left in hand. Four cuts are made in that quarter: a small square near its upper-left, a circle near its middle, and two small squares along its lower edge. Opening the second fold mirrors all four cuts upwards across the horizontal centre line, and opening the first fold mirrors everything sideways across the vertical centre line. So each cut appears four times and the opened sheet carries $4\times 3=12$ squares and $4\times 1=4$ circles. The squares form a ring of four along the top edge, four along the bottom edge and a block of four at the centre, with the four circles set in between. Hence option (d) is correct.

Paper folding and punching The first fold is along the diagonal running from the top-left corner to the bottom-right corner, so the upper-right triangle falls on the lower-left triangle. The second fold turns the right part of that triangle on to its left part along the vertical centre line. Two cuts are then made: a triangular bite on the vertical folded edge and another triangular bite on the bottom edge. Opening the second fold doubles them — the bite that sat on the fold line becomes a complete diamond hole, and the bite on the bottom edge becomes two notches placed symmetrically about the centre of that edge. Opening the diagonal fold copies the whole bottom pattern on to the right edge, because a mirror in that diagonal turns the bottom edge into the right edge. The opened sheet therefore shows two notches on the bottom edge with a wide diamond hole between them, and two notches on the right edge with a tall diamond hole between them. Hence option (c) is correct.

Missing number in grid Read the grid column by column: the number in the bottom row is the square of the sum of the three numbers above it. First column: 11+5+3=19 and ${19}^2=361$. Second column: 17+3+7=27 and ${27}^2=729$. Third column: $1225={35}^2$, so the three numbers of that column must add up to 35. Therefore 8+15+?=35, which gives ?=35-23=12. Hence option (c) is correct.

Missing number in figures In each figure the bottom-left number equals the product of the two top numbers increased by the bottom-right number. Figure (i): $4\times 7+2=28+2=30$. Figure (iii): $3\times 5+4=15+4=19$. Applying the same rule to figure (ii): $4\times 5+?=22$, i.e. 20+?=22. Therefore ?=22-20=2. Hence option (a) is correct.

Counting triangles The line joining the mid-points of the top and the bottom side cuts the rectangle into two equal halves, and each half carries both of its diagonals. In a rectangle with both diagonals drawn there are 8 triangles: 4 small ones around the crossing point and 4 halves of the rectangle itself. So the two halves give 8+8=16 triangles. Two more triangles use both halves together: the one standing on the whole top side with its apex at the bottom mid-point, and the one standing on the whole bottom side with its apex at the top mid-point. Total number of triangles =16+2=18. Hence option (c) is correct.

Venn diagram regions The triangle stands for bags, the circle for leather and the rectangle for brown, so read every number by the shapes that enclose it. Leather items that are not brown lie inside the circle but outside the rectangle — that is only the region marked 5. So their number is 5 and not 14, which makes statement I false. Brown items which are only bags lie inside the rectangle and the triangle but outside the circle; no number is written in that region, so the count there is 0 and not 4. Statement II is therefore false as well. Hence option (d) is correct.

Ranking by marks 'Marks of Q are less than only P' means exactly one student, namely P, is above Q, so P is first and Q is second. Then R, S and T all stand below Q. 'S is less than R but more than T' gives R>S>T. Putting the two facts together, the full order from the highest is P>Q>R>S>T. Only S and T score less than R, so the required number is 2. Hence option (b) is correct.

Ranking by weight V is the second heaviest, so exactly one friend is heavier than V. U is not the heaviest and U is not V, so the heaviest friend has to be W, because Z, X and Y are all lighter than W. That fixes the top of the order as W>V, and since U comes below V, W>V>U. Z is lighter than U, and X and Y are lighter than Z, giving W>V>U>Z>X and Y (X and Y in either order). Below U stand Z, X and Y — three friends in all. Hence option (c) is correct.

Figure analogy Count the sides of the figure drawn inside each box. The first box holds a triangle, which has 3 sides, and the second box holds a rectangle, which has 4 sides. So the rule is simply: the second figure has one side more than the first. The third box holds a house-shaped figure — a triangle sitting on a rectangle — whose outline is a pentagon with 5 sides. The missing figure must therefore have 5+1=6 sides, that is, a hexagon. Hence option (b) is correct.

Figure analogy In every figure the eight entries sit on the border of a 3×3 grid, so read them clockwise starting from the top-left corner. First figure: 2, 0, 4, 5, 6, 7, 8, 9 and second figure: 4, 5, 6, 7, 8, 9, 2, 0. The same eight entries appear, but each one has moved two places anticlockwise along the border. Third figure, read the same way: 6, ↑, 5, M, ↑, P, D, S. Moving each entry two places anticlockwise gives 5, M, ↑, P, D, S, 6, ↑. Placed back on the border this reads 5, M, ↑ in the top row, ↑ and P in the middle row and 6, S, D in the bottom row. Hence the correct option is (b).

Figure series Two things change from one figure to the next: the number of triangles and the position of the shaded cell in the small 2×2 square. The triangles run 2, 3, 4, 5, so the fifth figure must carry 6 triangles, and every triangle in the series points upwards. The shaded cell goes top-right → top-left → bottom-left → bottom-right, i.e. one corner anticlockwise each time. The corner that follows bottom-right in the anticlockwise round is the top-right corner again. So the missing figure has 6 upward triangles with the top-right cell shaded, which is option (c).

Symbol substitution First replace every sign by the one the question gives it, and only then apply BODMAS. So $56-8+4÷15\times 5$ becomes $56÷8\times 4-15+5$. $56÷8\times 4=7\times 4=28$ 28-15+5=13+5=18 Hence the required value is 18, option (c).

Balancing an equation Put each set of signs into $12*4*8*2*5=23$ in the given order and simplify by BODMAS. Option (b) gives $12+4\times 8÷2-5$. $8÷2=4$, so the expression is $12+4\times 4-5=12+16-5$. 12+16-5=23, which is exactly the right-hand side. Hence the correct combination is $+,\times ,÷,-$, option (b).

Odd letter-number cluster In each cluster the number should be the sum of the alphabetical positions of the four letters. J+K+M+N=10+11+13+14=48 U+V+X+Y=21+22+24+25=92 F+G+I+J=6+7+9+10=32 C+D+G+H=3+4+7+8=22, while the cluster carries 21. So CDGH21 does not obey the rule, option (c).

Odd letter-number pair In every pair both letters move 2 steps forward, so the letters give no clue about the odd one; check the numbers. 500-350=150, 725-575=150 and 890-740=150. 660-500=160, which is 10 more than the common fall. So NO660 – PQ500 is the odd pair, option (d).

Odd number pair In each pair the first number should be the perfect square of the second number. ${21}^2=441$, ${23}^2=529$ and ${26}^2=676$ — all three match. ${24}^2=576$, not 600. So 600 – 24 breaks the rule, option (c).

Odd number group In each triplet the middle number should be the first number plus the cube of the third number. $142+7^3=142+343=485$ $215+{10}^3=215+1000=1215$ $310+{11}^3=310+1331=1641$ $180+{13}^3=180+2197=2377$, but the group shows 2370. So (180, 2370, 13) does not follow the rule, option (d).

Letter-number analogy Compare each cluster with its partner: the first letter moves one step forward, the second letter one step backward. The number is halved: $500÷2=250$ and $120÷2=60$. The missing cluster stands first, so work backwards from DG200: D-1=C and G+1=H. For the number, 200 is the half, so the original is $200\times 2=400$. So the missing cluster is CH400, option (b).

Letter-number analogy Both letters of the second cluster are 4 steps ahead of the letters of the first cluster. The new number is the square of the old number increased by 5. ${10}^2+5=100+5=105$ and ${12}^2+5=144+5=149$. For GK14: G+4=K and K+4=O, while ${14}^2+5=196+5=201$. So the required cluster is KO201, option (a).