11 February 2025 · Shift 2

Dice - opposite faces Both pictures show the same die, so every letter that shares a view with another letter must touch it - two faces that meet cannot be opposite. View 1 shows N together with A and T, so N touches A and T. View 2 shows N together with S and H, so N also touches S and H. N therefore touches four of the other five letters: A, T, S and H. The only letter left for the face opposite N is I. So the face opposite I is N, i.e. option (c).

Letter-cluster series Look inside a cluster first: Z, X, V fall by 2 each time, and so do P, N, L and F, D, B. Now look across the clusters, at their first letters only: Z =26, P =16, F =6, V =22. Each first letter is 10 less than the one before it, counting back round the alphabet: 6-10=-4, which is 26-4=22, i.e. V. So the next cluster starts at 22-10=12, i.e. L. Falling by 2 twice from L gives L, J, H. So the missing term is LJH, i.e. option (b).

Letter-digit coding In this kind of code the digits are not in the order of the letters, so compare the two words as sets. SOUL and LOST have the letters S, O and L in common; their codes 6415 and 6548 have the digits 6, 4 and 5 in common. So those three letters use up exactly those three digits. In SOUL the letter left over is U, and in 6415 the digit left over is 1. So U is coded as 1, i.e. option (b).

Interchange of signs Write down what each sign turns into before touching the numbers. So $51÷30\times 36-6+6$ becomes $51-30+36÷6\times 6$. Now apply BODMAS. Division first: $36÷6=6$. Then multiplication: $6\times 6=36$. Finally addition and subtraction from the left: 51-30=21 and 21+36=57. So the question mark stands for 57, i.e. option (b).

Odd letter-cluster out Measure the two gaps inside each cluster. RWD: R $\to$ W is +5, and W $\to$ D is +7 (X, Y, Z, A, B, C, D). FKR: F $\to$ K is +5 and K $\to$ R is +7; DIP: D $\to$ I is +5 and I $\to$ P is +7. JOD: J $\to$ O is +5, but O +7 gives V, not D. So JOD is the cluster that breaks the +5, +7 rule, i.e. option (a).

Figure series Two things change from box to box: the three corner letters and the arrow. The letters read B, N, K then K, B, N then N, K, B then B, N, K - the same three letters keep sliding round, so the fifth box must read K, B, N (top-left, bottom-left, bottom-right). The arrow turns ${45}^{\circ }$ anticlockwise each time: it points east, then north-east, then north, then north-west. The next turn takes it to west, i.e. the arrow lies flat with its head on the left. Only one option has K, B, N with a flat arrow pointing left. So the missing figure is option (c).

Linear seating All seven face north, so left and right are the same as on the page; number the seats 1 to 7 from the left. Only three people sit to the left of L, so L takes seat 4. Only F sits to the right of G, so G takes seat 6 and F seat 7. Only three people sit between G and K, so K takes seat 2 - seats 3, 4 and 5 then lie between them. E, H and N are left for seats 1, 3 and 5, and E is left of N which is left of H, so E =1, N =3, H =5. Between E (seat 1) and N (seat 3) sits only K. So one person sits between E and N, i.e. option (b).

Number series Take the differences between consecutive terms. 460-450=10, 471-460=11, 483-471=12, 496-483=13. The difference itself grows by 1 each step, so the next difference is 14. Hence the missing term is 496+14=510. So the answer is 510, i.e. option (c).

Mirror image MN is a vertical line drawn to the right of the figure, so the mirror flips the figure from left to right; top and bottom stay exactly where they are. In the given figure the ball is on top and the stem hangs below - both lie on the centre line, so neither of them changes. The only feature that is not symmetric is the pair of wings: the left wing is hollow and the right wing is solid. A left-right flip must therefore put the solid wing on the left and the hollow wing on the right. Option (b) keeps the solid wing on the right, so it is the figure itself, not its image; options (c) and (d) make both wings the same, which loses the difference altogether. So the mirror image is option (a).

Syllogism Draw the three statements on one picture: beats inside music, music inside rhythms, and the tune circle kept completely apart from rhythms. Conclusion (I): every beat lies inside rhythms, and rhythms share nothing with tunes, so no beat can be a tune - (I) follows. Conclusion (II): all beats are rhythms, and that at once gives rhythms which are beats - (II) follows. So both conclusions follow, i.e. option (a).

Parallel-row seating Number the four places 1 to 4 as printed, from the left of the page, for both rows. Row 2 faces north, so for them left is the page's left; row 1 faces south, so for them left is the page's right. S sits third to the left of R, so in row 2 S takes place 1 and R place 4. R's only immediate neighbour is place 3, and that person faces M, so M takes place 3 of row 1. L sits third to the right of N; for row 1 right means towards place 1, so N takes place 4 and L place 1. That leaves place 2 of row 1 for O, which is indeed to the right of M as required. O is not facing P, so P takes place 3 of row 2 and Q takes place 2. Q is at place 2 of row 2 and O is at place 2 of row 1, so the two face each other. So O faces Q, i.e. option (c).

Letter-cluster analogy Compare the pair letter by letter: U $\to$ Z is +5, M $\to$ O is +2, C $\to$ E is +2, P $\to$ U is +5. Check it on the second pair: Q $\to$ V is +5, B $\to$ D is +2, X $\to$ Z is +2, K $\to$ P is +5. So the rule is +5, +2, +2, +5. Apply it to EJAH: E +5= J, J +2= L, A +2= C, H +5= M. So EJAH is related to JLCM, i.e. option (c).

Letter-cluster analogy Take the first pair: B =2 and P =16, a jump of +14; Y =25 and M =13, which is also +14 once you go past Z (25+14=39, and 39-26=13). V $\to$ J and T $\to$ H are +14 as well, and the second pair QNKI $\to$ EBYW obeys the same +14. Apply +14 to NKHF: N =14 gives 28-26=2= B; K =11 gives 25= Y. H =8 gives 22= V; F =6 gives 20= T. So NKHF is related to BYVT, i.e. option (c).

Letter-digit coding The digits are jumbled, so compare the two words as sets of letters and sets of digits. BASH and HASE share the letters A, S and H; the codes 5431 and 1234 share the digits 1, 3 and 4. So those three letters take up exactly those three digits. The letter left in HASE is E and the digit left in 1234 is 2. So E is coded as 2, i.e. option (c).

Letter series completion Count the places first: there are 21 places in all, and 7 of them are blank. The letters already printed - s t o l b, then o l b r i s t, then r i - all fit the block s-t-o-l-b-r-i written three times. The first block is s t o l b r i, so the four blanks after stolb must be r, i, s, t. The printed o l b r i s t then completes the second block and starts nothing new, so the three blanks that follow are the start of the third block: o, l, b. Reading the blanks from left to right gives r i s t o l b. So the answer is ristolb, i.e. option (b).

Dice - opposite faces Two faces seen together in one view must touch each other, so they cannot be opposite. A appears in both views: with L and D in view 1, and with Z and I in view 2. So A touches L, D, Z and I - four of the five other letters. The only letter that can lie opposite A is the one never seen with it, namely N. So the face opposite N is A, i.e. option (c).

Number series Consecutive differences (10, 8, 28, 8) show no pattern, so compare every second term instead. Terms in the odd places: 7, 25, 61. The gaps are 25-7=18 and 61-25=36, so the gap doubles. Terms in the even places: 17, 53 and then the missing one. The first gap is 53-17=36, so the next gap is $36\times 2=72$. Hence the missing term is 53+72=125. So the answer is 125, i.e. option (d).

Alphabetical rearrangement Write LIGHTER as it stands: L, I, G, H, T, E, R. Sort the same seven letters alphabetically: E, G, H, I, L, R, T. Now compare place by place: L-E, I-G, G-H, H-I, T-L, E-R, R-T. Not a single place holds the same letter in both rows. So the answer is zero, i.e. option (b).

Coded blood relation Break the expression at each sign and read it from the left. M+N means M is the brother of N, so M is male. N-O means N is the sister of O, so M, N and O are children of the same parents. $O\times P$ means O is the mother of P. $P÷Q$ means P is the father of Q, which is not needed for this question. P's mother is O, and M is a brother of O. So M is P's mother's brother, i.e. option (d).

Venn diagram of classes Take the three classes two at a time. Every sister is a female, so the whole of sister lies inside female. A doctor may be a man or a woman, so doctor only partly overlaps female - it can neither contain it nor sit inside it. A sister can also be a doctor, so doctor must cut into the sister region too. The diagram needed is therefore a small ellipse (sister) inside a large one (female), with a third ellipse (doctor) crossing both and still sticking out. So the correct Venn diagram is option (c).