13 February 2025 · Shift 3

Circular seating All seven face the centre, so for every one of them the clockwise direction is their left and the anticlockwise direction is their right. Start with M. Counting to the left of M only one person sits between M and I, so clockwise the order runs M, one person, I. N sits third to the left of I, so three seats clockwise from I is N; that fixes M, _, I, _, _, N and leaves one seat after N. Counting to the right of O, two people sit between O and J, so J is three seats anticlockwise from O. Putting J in the seat right after I makes O the last free seat, and H, who must be next to J, takes the seat between J and N. The only seat still empty is the one between M and I, and it goes to G. Reading clockwise the table is M, G, I, J, H, N, O. Third to the right of G means three seats anticlockwise from G: M, then O, then N. So N sits third to the right of G, and the answer is option (c).

Parallel row seating Draw the two rows one above the other and number the four facing pairs 1 to 4 from left to right on the page. Row 1 faces south, so for M, A, R and K the right hand points towards position 1; row 2 faces north, so for P, O, S and T the right hand points towards position 4. Only two people sit between O and P, so O and P take positions 1 and 4 of row 2, and only one person sits between T and P, so T and P are two positions apart. That forces O at 1, T at 2, S at 3 and P at 4. T faces the person immediately to the right of M, and row 1's right runs towards position 1, so M sits at position 3. P at position 4 faces the person second to the left of K, so K sits at position 2; the remaining seats 1 and 4 go to R and A, and R must be to the right of A, that is nearer position 1, so R is at 1 and A at 4. The ends of row 1 are R and A, the ends of row 2 are O and P; only option (b), A and O, names two people who are both sitting at an extreme end. So the answer is option (b).

Figure series rotation The same four letters A, e, T and d appear in every box; only the places they occupy change. In box 1 they sit at the four corners, in box 2 at the four edge midpoints, in box 3 at the corners again, in box 4 at the midpoints, so box 5 must have them back at the corners. Each step turns the whole group through 45 degrees anticlockwise: a corner letter slides to the edge midpoint on its left, and an edge letter slides on to the next corner. Box 4 has T on top, e at the left, d at the right and A at the bottom. Turning that 45 degrees anticlockwise sends T to the top-left, d to the top-right, A to the bottom-right and e to the bottom-left. Option (c) has exactly that: T and d on top, e at the bottom-left and A at the bottom-right. Option (a) carries the same top pair but has A and e interchanged along the bottom, which is what a clockwise turn would give, so it is not the answer.

Dice opposite faces In both positions the letter on the top face is M, so the cube has simply been turned about its vertical axis and M has stayed on top. Position I shows A and P standing next to M, and position II shows G and S standing next to M. A face that is seen together with M in the same view must touch M, so A, P, G and S are all adjacent to M. Four of the six letters therefore touch M, and the only letter never seen with M is I, so I is the face opposite M. If I is opposite M, then M is opposite I, which is option (c).

Venn diagram of classes A train, an aeroplane and a scooter are three different kinds of vehicle. No train is an aeroplane, no aeroplane is a scooter and no scooter is a train, so no two of the three classes share even one member. Three classes with nothing in common are drawn as three separate circles that do not touch, which is option (a). Option (b) makes the circles overlap, option (c) puts one circle inside another and option (d) puts all three one inside the other; each of those claims a common member that does not exist.

Letter-digit coding CAMS is coded as 8137 and MACK is coded as 1238; the letters C, A and M appear in both words. The digits common to the two codes are 1, 3 and 8, so those three digits stand for C, A and M in some order. In CAMS the digit left over is 7, and the letter left over is S, so S is coded 7. In MACK the digit left over is 2, and the letter left over is K, so K is coded 2. The code for K is therefore 2, which is option (a).

Number series Take the difference between one term and the next. 119-109=10, 130-119=11, 142-130=12 and 155-142=13. The difference itself grows by 1 each time, so the next difference is 14. 155+14=169, so the missing term is 169 and the answer is option (a).

Odd letter-cluster First check what all four clusters share: the third letter is two places after the first. U to W, L to N, K to M and H to J are all +2, so that is not the test. Now count from the first letter to the second, moving forward and running past Z back to A when needed. U(21) + 14 = 35, and 35 - 26 = 9 = I; L(12) + 14 = 26 = Z; K(11) + 14 = 25 = Y. H(8) + 14 = 22 = V, but the cluster gives U, which is only +13. HUJ is the cluster whose first step is +13 instead of +14, so it does not belong to the group and the answer is option (d).

Letter-cluster analogy Compare ZBUP with WZSO letter by letter: Z to W is -3, B to Z is -2, U to S is -2 and P to O is -1. The second pair obeys the same shifts: V to S is -3, S to Q is -2, E to C is -2 and X to W is -1. Apply -3, -2, -2, -1 to GMTC: G(7)-3 = D, M(13)-2 = K, T(20)-2 = R and C(3)-1 = B. The required cluster is DKRB, so the answer is option (a).

Alphabetical rearrangement Write the word as it stands: C, R, A, D, L, E, S. Now sort the same seven letters alphabetically: A, C, D, E, L, R, S. Compare the two rows position by position: only the 5th letter L and the 7th letter S sit in the same place in both rows. That is two letters, so the answer is option (a).

Interchange of signs Swap the signs first, then work the sum; do not touch the numbers. Every $\times$ is read as -, every - as $\times$, every + as $÷$ and every $÷$ as +. So $88÷7\times 4-21+3$ is read as $88+7-4\times 21÷3$. By BODMAS do multiplication and division first, left to right: $4\times 21=84$ and $84÷3=28$. Then 88+7-28=95-28=67. The value is 67, so the answer is option (a).

Paper folding and cutting The first fold lifts the bottom half on to the top half, so the horizontal centre line of the sheet is a fold line. The second fold lays the right half on to the left half, so the vertical centre line is the second fold line, and the packet left in the hand is the top-left quarter. Two cuts are made in that quarter, side by side and just above its lower edge: on the left a triangle with a flat top and its point downwards, and beside it a triangle whose right angle is at the bottom-right corner. Opening the second fold reflects that pair across the vertical centre line, so the top-right quarter gets the same pair in mirror order. Opening the first fold reflects the whole top half across the horizontal centre line, and a reflection turns every downward point into an upward point. The finished sheet therefore carries eight triangles in two rows of four, clustered round the centre, and the bottom row is the exact mirror image of the top row. Option (b) is the only one in which the lower four triangles are flipped copies of the upper four; option (c) repeats the top row unchanged below, option (a) puts triangles in the four corners and option (d) has only four triangles. So the answer is option (b).

Coded blood relations Read the chain T @ V * I @ O one link at a time. T @ V means T is the brother of V. V * I means V is the father of I. I @ O means I is the brother of O, so I and O are children of the same parents. Since I is V's child and O is I's sibling, V is the father of O as well. T is V's brother, so T is the brother of O's father, that is O's father's brother. The answer is option (b).

Letter series with blanks There are five blanks, so the answer has to be a set of five letters; take option (d), B, K, L, U, Y, and drop them into the blanks in order. The string becomes B G D I F K H M J O L Q N S P U R W T Y. Split it into groups of four: BGDI, FKHM, JOLQ, NSPU, RWTY. Inside every group the steps are the same, +5, -3, +5. For BGDI: B(2) to G(7) is +5, G(7) to D(4) is -3 and D(4) to I(9) is +5. The first letters of the groups, B(2), F(6), J(10), N(14), R(18), also rise by 4 each time. No other option makes all five groups obey this pattern, so the answer is option (d).

Embedded figure Figure (X) is made of two identical arcs. One runs from the top-right corner down to the bottom-right corner and bulges to the left; the other runs from the top-left corner down to the bottom-left corner and bulges to the right. The two arcs cross each other twice, so between the crossings they enclose a tall pointed lens, and outside the crossings four tails flare away to the four corners. Rotation is not allowed, so the answer must show that lens standing upright with all four tails still attached to it. Option (d) contains exactly that: the upright lens sits in the middle of the frame and its four tails run out to the four corners of the inner rectangle, where they meet the two vertical sides. Option (a) is drawn only with straight lines and a circle, option (b) has a single bowl shaped arc and no lens at all, and in option (c) the two arcs bulge towards each other but stop short, so they never cross and no lens is formed. So the answer is option (d).

Letter-cluster series Line the four given terms up and compare the letters that sit in the same place. First letters: B(2), Y(25), V(22), ?, P(16) - each is 3 places before the one before it, so the missing first letter is 22 - 3 = 19 = S. Second letters: T(20), Q(17), N(14), ?, H(8) - again -3, so the missing letter is 14 - 3 = 11 = K. Third letters: D(4), A(1), X(24), ?, R(18) - again -3, so the missing letter is 24 - 3 = 21 = U. The missing term is SKU, which is option (b).

Number series Subtract each term from the one before it. 823-822=1, 822-820=2, 820-817=3 and 817-813=4. The amount taken away grows by 1 each time, so the next subtraction is 5. 813-5=808, so the missing term is 808 and the answer is option (c).

Syllogism Draw the three statements. Some wood are chairs, so the wood circle and the chair circle overlap; all chairs are tables, so the chair circle lies wholly inside the table circle; no table is a bed, so the bed circle stands completely apart. Conclusion (I): the wood that are chairs lie inside the chair circle, and the whole chair circle lies inside the table circle, so that same wood is table as well. 'Some wood are tables' is therefore certain. Conclusion (II): every chair is a table and no table is a bed, so no chair can be a bed. 'All chairs are beds' is not merely uncertain, it is impossible. Only conclusion (I) follows, so the answer is option (d).

Letter-digit coding DAMS is coded as 7803 and MASK is coded as 0386; the letters A, M and S appear in both words. The digits common to the two codes are 0, 3 and 8, so those three digits stand for A, M and S in some order. In DAMS the digit left over is 7 and the letter left over is D, so D is coded 7. In MASK the digit left over is 6 and the letter left over is K, so K is coded 6. The code for K is therefore 6, which is option (d).

Letter-cluster analogy Compare KIFG with HFCD letter by letter: K to H, I to F, F to C and G to D are all 3 places back in the alphabet. The second pair obeys the same rule: A to X, Y to V, V to S and W to T are each -3, with the count running past A back to Z where needed. Apply -3 to CAXY: C(3) becomes Z, A(1) becomes X, X(24) becomes U and Y(25) becomes V. The required cluster is ZXUV, so the answer is option (c).