7 February 2025 · Shift 1

Coded blood relations Fix the four symbols first: + means mother of, - means brother of, $\times$ means wife of, $÷$ means father of. Now read $K÷L\times M-N+O$ one link at a time. $K÷L$: K is the father of L, and $L\times M$: L is the wife of M. M-N: M is the brother of N, and N+O: N is the mother of O. N is O's mother and M is N's brother, so M is O's mother's brother.

Letter cluster analogy Compare the first pair letter by letter using alphabet positions. I(9)+12=U(21), L(12)+12=X(24), O(15)+12=27-26=A(1), R(18)+12=30-26=D(4). The second pair obeys the same rule: C+12=O, F+12=R, I+12=U, L+12=X. Apply +12 to WZCF: $W\left(23\right)\to I\left(9\right)$, $Z\left(26\right)\to L\left(12\right)$, $C\left(3\right)\to O\left(15\right)$, $F\left(6\right)\to R\left(18\right)$. So WZCF is related to ILOR.

Letter rearrangement ARMCHAIR uses the letters A, R, M, C, H, A, I, R. Written in reverse alphabetical order they become R, R, M, I, H, C, A, A. Now match the two rows place by place. Place 2 keeps R, place 3 keeps M and place 5 keeps H; every other place changes. So three letters remain unchanged.

Letter cluster analogy In the first pair every letter moves 9 places back: J(10)-9=A(1), I(9)-9=Z(26), L(12)-9=C(3), T(20)-9=K(11). The second pair confirms it: $Z\to Q$, $Y\to P$, $B\to S$, $J\to A$. Apply -9 to YXAI: $Y\left(25\right)\to P\left(16\right)$, $X\left(24\right)\to O\left(15\right)$, $A\left(1\right)\to R\left(18\right)$, $I\left(9\right)\to Z\left(26\right)$. So YXAI is related to PORZ.

Letter to digit coding 'AGED' is coded 7613 and 'EARG' is coded 1236. The letters common to both words are A, G and E. The digits common to both codes are 6, 1 and 3, so A, G and E use those three digits. The digit of 7613 that is left over is 7, which must stand for the left over letter D. The digit of 1236 that is left over is 2, so R is coded as 2.

Figure series Each frame holds the same four marks q, ★, s and t, only their corners change. Frame 1: q top left, ★ middle left, s middle right, t bottom right. Frame 2: q middle left, ★ bottom left, s top right, t middle right. Frames 3 and 4 repeat frames 1 and 2, so the two patterns simply alternate. The 5th frame must therefore repeat frame 1, which is option (c).

Interchange of numbers Work out the left hand side as it stands, using BODMAS. $9\times \left(34-15\right)+\left(105÷7\right)+28+\left(13\times 4\right)=9\times 19+15+28+52=266$, not 218. The value must fall by 48, so a large number inside the bracket has to shrink. Interchange 34 and 28: $9\times \left(28-15\right)+\left(105÷7\right)+34+\left(13\times 4\right)=9\times 13+15+34+52$. =117+15+34+52=218, which matches the right hand side. So 34 and 28 must be interchanged.

Letter series with blanks The series has 10 places, four of them blank: _ D _ L _ F _ O L J. Places 8, 9 and 10 read O, L, J and places 2 and 4 read D and L, which hints at a block of five letters used twice. Taking the block as F D O L J the whole series reads F D O L J F D O L J. Every given letter then agrees, and the blanks at places 1, 3, 5 and 7 need F, O, J and D. So the missing letters in order are FOJD.

Number series Take the difference between consecutive terms. 1001-1000=1, 1000-998=2, 998-995=3, 995-991=4. The number subtracted grows by 1 each time, so the next step is -5. 991-5=986.

Number series Take the difference between consecutive terms. 621-620=1, 624-621=3, 629-624=5, 636-629=7. The differences are consecutive odd numbers, so the next difference is 9. 636+9=645.

Venn diagram of classes Every parrot is a bird, so the parrot circle must lie wholly inside the bird circle. A cat is neither a parrot nor a bird, so its circle stays completely apart from both. The picture needed is a small circle inside a larger one, with a third circle standing alone. Only option (b) shows that arrangement.

Dice opposite faces View I shows A on top with U and L on the sides; view II shows A on top with R and I on the sides. A stays on top in both views, so the cube has only been turned about its vertical axis. That makes U, L, R and I the four side faces of the dice. Only Q is left for the hidden bottom face, so Q lies opposite A. Hence the face opposite Q is A.

Word to number coding The codes are single digits, so count the letters of each word. 'ZXFGA' has 5 letters and is coded 3, and 5-2=3. 'BTGHASQR' has 8 letters and is coded 6, and 8-2=6. So the rule is code = number of letters $-2$. 'YWNERI' has 6 letters, so its code is 6-2=4.

Linear seating All six face north, so number the seats 1 to 6 from the reader's left. No one sits to the right of H, so H takes seat 6. Two people sit between H and G, so G takes seat 3, and D, being immediately right of G, takes seat 4. E sits immediately left of G, so E takes seat 2, and W then takes seat 1 because two people sit between W and D. The only seat left for F is seat 5. G and D occupy the seats between E (2) and F (5), so two people sit between them.

Syllogism Draw only what the three statements force: taps overlap soaps, taps overlap creams, soaps overlap oils. None of the statements links oils with taps, nor creams with soaps. The oils can sit entirely in the part of soaps that lies outside taps, so conclusion (I) need not be true. The creams can sit entirely in the part of taps that lies outside soaps, so conclusion (II) need not be true. A conclusion follows only if it holds in every possible diagram, so neither conclusion follows.

Linear seating All six face north, so number the seats 1 to 6 from the reader's left. Only two people sit to the right of R, so R takes seat 4, and F, being immediately left of R, takes seat 3. Two people sit between R and E, so E takes seat 1; this also leaves exactly one person between E and F. S sits immediately right of Q, and seats 5 and 6 are the only adjacent pair still free, so Q takes seat 5 and S seat 6. The only seat left for G is seat 2. F and R occupy the seats between G (2) and Q (5), so two people sit between them.

Odd letter cluster Turn each cluster into alphabet positions and read the two gaps. TOJ: 20,15,10, so the gaps are -5 and -5; MHC: 13,8,3, again -5 and -5. GBW: 7,2,23; from B the count of 5 backwards runs past A and lands on W, so the gaps are -5 and -5 as well. QJE: 17,10,5, so the gaps are -7 and -5. QJE is the only cluster whose first gap is not 5, so it does not belong to the group.

Dice opposite faces Two faces that appear together in one view must be adjacent, never opposite. View I shows U with E and I, so U is adjacent to E and to I. View II shows U with L and S, so U is adjacent to L and to S as well. U therefore touches four faces - E, I, L and S - and a cube face touches exactly four others. The only face U does not touch is R, so R lies opposite U.

Embedded figure Figure (X) is one straight line rising to the right, carrying an arrowhead part way along it whose two barbs point left and down. Since rotation is not allowed, look for that same slant with the head in the middle of the stroke, not at its end. Option (b) has no arrowheads at all and option (c) is built only from horizontal and vertical strokes. In option (a) the rising line carries its heads at the two ends, so the middle arrowhead of (X) is missing. Option (d) contains the rising line with the arrowhead sitting on it exactly as in (X), so (X) is embedded in option (d).

Letter cluster series Split every term into its three letters and follow each column on its own. First letters A, B, C, D rise by 1, so the next one is E. Second letters F, H, J, L rise by 2, so the next one is N. Third letters Q, R, S, T rise by 1, so the next one is U. Putting the three together, the missing term is ENU.