13 February 2025 · Shift 2

Linear seating, two rows No one sits to the right of B, so B takes the east end of row 2, and as exactly two people sit between N and B, N takes the west end. Row 1 faces south, so for that row 'right' means towards the west. The person facing B sits opposite him at the east end, and the person facing O sits two seats to the west of that person, which puts O in the second seat of row 2 and leaves D in the third. I sits immediately right of F and G immediately right of I, so row 1 runs G, I, F, T from west to east; the other possible order T, G, I, F would place G opposite O, which is not allowed. The ends are G and T in row 1 and N and B in row 2, and the only option that names two of these four is T and N, so option (b) is correct.

Paper folding and punching The sheet is folded once along the horizontal crease and once along the vertical crease, so the single punch lies on one quarter and must reappear once in each of the four quarters. Opening the vertical fold mirrors the punch left to right, and opening the horizontal fold then mirrors both marks top to bottom. So the two upper marks must be the upside-down images of the two lower ones, and the left marks the left-right images of the right ones: every comma points outwards, away from the centre. Option (a) merely repeats the lower pair at the top instead of turning it over, option (b) shows only two punches instead of four and option (c) leaves every comma on the left, so option (d) is correct.

Coded blood relations Read the chain one link at a time: O / M means O is the father of M, M : I means M is the wife of I, I ! T means I is the brother of T, and T < S means T is the son of S. T is the son of S and I is T's brother, so I is also a son of S. M is the wife of I, so M is the wife of S's son, and O is M's father. Hence O is the father of S's son's wife, i.e. son's wife's father, so option (c) is correct.

Alphabetical rearrangement Arranged in English alphabetical order, the letters of FROCKED become C, D, E, F, K, O, R. The third letter from the left end is E and the second letter from the right end is O. E is the 5th letter and O the 15th, so the letters strictly between them number 15-5-1=9 (F, G, H, I, J, K, L, M, N). Hence option (a) is correct.

Number series Split the digits into the terms 991, 1000, 1010, 1021, 1033. The gap between consecutive terms grows by one each time: +9, +10, +11, +12. So the next gap is +13, giving 1033+13=1046. Hence option (d) is correct.

Odd one out, letter-clusters In each cluster count the step from the first letter to the second and from the second to the third, moving forward and starting again at A after Z. AJS, OXG and KTC all move +9 and then +9 ($O\to X\to G$ and $K\to T\to C$ simply run past Z). VEM moves $V\to E$, which is +9, but then $E\to M$, which is only +8. So VEM breaks the pattern and option (b) is correct.

Letter-cluster analogy Compare the two clusters letter by letter: the first letter moves +4 and each of the other three moves +2. The second pair obeys the same rule: D+4=H, W+2=Y, Q+2=S and I+2=K. Applying it to VAPD gives V+4=Z, A+2=C, P+2=R and D+2=F. So VAPD is related to ZCRF and option (a) is correct.

Number series Split the digits into the terms 712, 711, 713, 710, 714, 709, 715. The differences change sign each time and grow by one: -1, +2, -3, +4, -5, +6. So the next difference is -7, giving 715-7=708. Hence option (a) is correct.

Embedded figure Figure (X) is a horizontal bar carrying a short arm that hangs down from its left end, a short arm that rises from its right end, a long stem dropping from a point near the middle of the bar, and a rectangle attached to the RIGHT of the foot of that stem. Option (c) contains all five parts in exactly that arrangement, and without turning the figure: the wide horizontal line, the line running down from its left end, the short line rising from its right end, the central vertical, and the small rectangle sitting to the right of the foot of that vertical. Option (a) is built only from a chord and two triangles, option (b) has no straight stem ending in a rectangle at its side, and option (d) is only a diagonal with lines parallel to the sides, so option (c) is correct.

Letter coding with place values Replace every letter by its opposite letter (A↔Z, B↔Y, C↔X and so on), add the place values of the four opposites and double the total. Check with JRZQ: the opposites are Q, I, A, J with place values 17+9+1+10=37, and $37\times 2=74$. LOBT gives O, L, Y, G, i.e. 15+12+25+7=59, and $59\times 2=118$. For XGCF the opposites are C, T, X, U, so the sum is 3+20+24+21=68 and the code is $68\times 2=136$, i.e. option (a).

Letter-cluster series Compare the clusters position by position: every letter of one cluster moves +4 to give the letter below it in the next cluster. First letters: $E\to I\to M\to Q$; second letters: $F\to J\to N\to R$; third letters: $L\to P\to T\to X$. So the missing cluster is Q+4=U, R+4=V and X+4=B (counting on past Z), i.e. UVB. Hence option (a) is correct.

Figure series The three symbols keep their fixed triangular arrangement, and at every step the whole group turns a quarter-turn clockwise inside the box. So @ travels middle-left → top-middle → middle-right → bottom-middle, and the triangle and the star travel with it, keeping the same positions relative to @. Four quarter-turns complete a full turn, so the fifth figure repeats the first: @ at the middle of the left edge, the triangle in the bottom-left corner and the star beside it in the bottom row. Option (b) puts all three symbols in the bottom row, option (c) interchanges the triangle and the star, and option (d) shifts @ to the right edge, so option (a) is correct.

Symbol interchange After the interchange, $\times$ is read as +, + as $\times$, - as $÷$ and $÷$ as -. So $57÷9\times 50-5+2$ becomes $57-9+50÷5\times 2$. By BODMAS, divide first: $50÷5=10$, then multiply: $10\times 2=20$. This leaves 57-9+20=48+20=68, so option (a) is correct.

Linear seating Everyone faces north, so a person's immediate right is the next seat towards the east. M is at an end, P is immediately right of M, N immediately right of P and L immediately right of N, which fixes the block M, P, N, L starting from the left end. O, Q, R must also stay together in that order, so they fill the remaining three seats and the line reads M, P, N, L, O, Q, R. The only person between M and N is P, so option (d) is correct.

Syllogism Draw the statements: pumps lie inside motors and also inside machines, while engines lie inside machines only. Every pump is a machine as well as a motor, so at least the pumps are machines that are motors; hence 'Some machines are motors' definitely follows. Nothing links engines to motors, and the diagram shows the engines can sit entirely in the part of 'machines' lying outside 'motors', so 'All engines are motors' need not be true. Only conclusion (II) follows, so option (c) is correct.

Dice, opposite faces The two views show the same cube, and the faces O and S appear in both of them. When two views of a cube share two faces, the two faces that differ must be opposite each other, so C (seen only in view I) is opposite V (seen only in view II). The remaining pair then splits as O opposite E and S opposite L. So the face opposite V is C, and option (a) is correct.

Letter series with blanks Break the string into five groups of four letters: _BDG / I_GJ / LH_M / OKM_ / RNP_. Reading the groups downwards, each column advances by +3: the second letters are B, E, H, K, N, the third are D, G, J, M, P and the fourth are G, J, M, P, S. The first letters must therefore be F, I, L, O, R, so the first blank is F, and the other blanks are E, J, P and S. The blanks in order give FEJPS, so option (c) is correct.

Letter-digit coding ALSO and LOAN are made of the same letters except that ALSO has S where LOAN has N; the letters A, L and O are common to both. Their codes behave the same way: 4381 and 1436 share the digits 4, 3 and 1. So the leftover digit of ALSO, 8, stands for S, and the leftover digit of LOAN, 6, stands for N. Hence N is coded as 6 and option (b) is correct.

Venn diagram of classes Every eagle is a bird, so the circle for eagles must lie completely inside the circle for birds. A house is neither an eagle nor a bird, so its circle must stay entirely apart from both of them. Only option (b) shows a small circle drawn inside a larger one with a third circle standing separately; (a) and (d) make the classes overlap and (c) offers a single circle for three different classes. Hence option (b) is correct.

Letter-cluster analogy Compare SOMI with HDBX letter by letter: every letter moves -11 places, counting backwards and starting again at Z after A. Check: S-11=H, O-11=D, M-11=B and I-11=X (since 9-11=-2, which is 24, i.e. X). The second pair obeys the same rule: N-11=C, J-11=Y, H-11=W and D-11=S. For YUSO: Y-11=N, U-11=J, S-11=H and O-11=D, giving NJHD, so option (a) is correct.