20 February 2025 · Shift 3

Letter-cluster analogy Put the two clusters one below the other and compare letter by letter. N becomes L and Z becomes X, a step of -2 each; R becomes O and B becomes Y, a step of -3 each (from B count back A, Z, Y). The second pair obeys the same rule: V -2= T, J -2= H, A -3= X and T -3= Q. Now apply -2,-2,-3,-3 to PGXL: P (16)-2= N (14), G (7)-2= E (5), X (24)-3= U (21), L (12)-3= I (9). So PGXL is related to NEUI, i.e. option (c).

Letter-cluster analogy Compare RQAZ with JMSV letter by letter: R (18) to J (10) is -8, Q (17) to M (13) is -4, A (1) to S (19) is -8 counted cyclically, and Z (26) to V (22) is -4. So the steps are -8, -4, -8, -4. LAUJ confirms it: L -8= D, A -4= W (cyclic), U -8= M and J -4= F, which is DWMF. Apply the same steps to NOWX: N (14)-8= F (6), O (15)-4= K (11), W (23)-8= O (15), X (24)-4= T (20). So NOWX is related to FKOT, i.e. option (d).

Embedded figure Figure X is a zig-zag: a long stroke up to a peak, down to a low point, then up again to a second peak, closed at the bottom left by a horizontal base and cut by a short vertical that drops from the descending stroke to that base. Rotation is not allowed, so the peak must still lean to the left and the base must still run along the bottom. In option (b) the left stroke starts exactly at the bottom-left corner and rises to a peak on the top edge; the descending stroke drops to a point from which a vertical reaches the bottom edge, and the short base is part of the square's own bottom edge. From that same point the stroke goes down to the low point and then straight up to the top-right corner - the second peak of X. The other options have no such pair of peaks joined by a low point: (a) is only two diagonals and a centre line, (c) is a wide rectangle whose two slanting strokes spread apart, and (d) has its slanting strokes meeting at a top vertex, not at a bottom one. So the figure is embedded in option (b).

Alphabetical rearrangement Write PUSHING with its seven places numbered, then write the same seven letters in alphabetical order. The letters sorted are G, H, I, N, P, S, U. Compare the two rows place by place: P-G, U-H, S-I, H-N, I-P, N-S, G-U. No place holds the same letter in both rows, so the count is zero, i.e. option (a).

Dice - opposite faces N is on the top face in both views, so the cube has simply been spun about its vertical axis. Spinning about that axis never moves the top face or the bottom face, so the four faces that go round the sides stay the same four in both views. Those side faces are R and I (first view) and T and P (second view) - four different letters. The six letters are N, R, P, I, T and E, so the only letter left for the hidden bottom face is E. The bottom face is opposite the top face, so E is opposite N; therefore N is opposite E, i.e. option (a).

Odd letter-cluster In a three-letter cluster only two gaps matter: first to second, and second to third. LNO: L to N is +2 and N to O is +1. TVW: T to V is +2 and V to W is +1; ACD: A to C is +2 and C to D is +1. JLN: J to L is +2 but L to N is +2 as well, not +1. So JLN breaks the pattern and is the odd cluster, i.e. option (d).

Symbol interchange Two swaps are ordered at once, so four symbols change meaning. Rewrite $44÷9\times 32-8+4$ with the new meanings: $44-9+32÷8\times 4$. Now use BODMAS. Division first: $32÷8=4$. Then multiplication: $4\times 4=16$. Finally left to right: 44-9=35 and 35+16=51. So the question mark stands for 51, i.e. option (d).

Coded blood relation Break the chain at every symbol and read each piece on its own. L * A means L is the mother of A, so A is L's child. A @ Q means A is the husband of Q, so A is male and Q is his wife. Q * M means Q is the mother of M; this only tells us the couple has a child and is not needed for the answer. Q's husband is A, and the mother of A is L. So L is the mother of Q's husband, i.e. Q's husband's mother, option (c). It cannot be (a): that would need L to be the mother of Q's own mother, which the chain never says.

Letter-cluster series Read the groups column by column. First letters: F, G, H, I - each one step forward, so the next is J. Second letters: N, O, P, Q - again one step forward, so the next is R. Third letters: V, W, X, Y - one step forward, so the next is Z. The missing group is JRZ, i.e. option (d).

Number series The terms fall, so take the differences. 153-128=25, 128-105=23, 105-84=21, 84-65=19. The differences are 25, 23, 21, 19 - odd numbers going down by 2, so the next subtraction is 17. 65-17=48. So the missing term is 48, i.e. option (d).

Venn diagram of classes Take the classes two at a time. Every mango is a fruit and no other fruit is a mango, so the class Mango lies completely inside the class Fruit. A table is neither a mango nor a fruit, so the class Table stands completely apart from both. The picture needed is therefore a small circle inside a bigger circle, with a third circle drawn quite separately. That is option (a); (b) makes mango and fruit merely overlap, (c) keeps all three apart, and (d) makes even table overlap the other two.

Letter-to-number coding Each letter is first replaced by its opposite letter, the one as far from Z as it is from A; its place value is 27- (place value of the letter). Add the four opposite place values and then add 6. Check on WGOA: the opposites of W, G, O, A are D, T, L, Z with values 4, 20, 12, 26; 4+20+12+26=62 and 62+6=68. Check on ISLT: the opposites of I, S, L, T are R, H, O, G with values 18, 8, 15, 7; 18+8+15+7=48 and 48+6=54. For DXRI the opposites are W, C, I, R with values 23, 3, 9, 18. 23+3+9+18=53, and 53+6=59. So DXRI is coded as 59, i.e. option (d).

Circular seating All seven face the centre, so for each of them the left hand side is the clockwise direction. Fix F anywhere. E sits immediately to F's left, so E is one seat clockwise from F. G sits third to the left of F, so G is three seats clockwise from F; that leaves exactly one empty seat between E and G. K must touch both L and G. If K took the single seat between E and G, both its neighbours would already be E and G and L could not sit beside it, so K is G's other neighbour and L sits next to K. H is not a neighbour of G, so H cannot take the seat between E and G; J takes it and H fills the last seat, next to L and F. Counting three seats clockwise from J: G, then K, then L. So L sits third to the left of J, i.e. option (c).

Figure series Two things change from box to box: the arrow and the circles. The arrow points up, then up-right, then down-right, then down - clockwise turns of ${45}^{\circ }$, ${90}^{\circ }$, ${45}^{\circ }$. The turns alternate, so the next turn is ${90}^{\circ }$ clockwise and the arrow must point to the left. The circles march clockwise round the corners: a plain circle appears at the top-right, then it gets a bar and a plain circle appears at the bottom-right, then that one gets a bar as well. So the fifth box keeps barred circles at the top-right and bottom-right and gains a fresh plain circle at the bottom-left. Options (a) and (d) fail on the arrow - down-left is a turn of only ${45}^{\circ }$ and up is a turn of ${180}^{\circ }$. Option (b) has exactly the right circles but its arrow also points down-left, so it too breaks the turning rule. Only option (c) has both a left-pointing arrow and an unbarred circle at the bottom-left, so the answer is (c).

Linear seating Number the six seats 1 to 6 from the left end. Everyone faces north, so their left is the left of the row as we read it. No one sits to the left of M, so M is in seat 1. Exactly three people sit between M and K, so K is in seat 5. Only three people sit to the right of O, so seats 4, 5 and 6 lie to his right and O is in seat 3. Z sits immediately to the left of O, so Z is in seat 2. O's neighbours are seats 2 and 4; N is not a neighbour of O, so N cannot take seat 4 and L does, leaving seat 6 for N. Seat 6 is the rightmost end, so N sits there, i.e. option (c).

Number series Each term is a little more than twice the one before it, so test the rule 'double and add 1'. $110\times 2+1=221$ and $221\times 2+1=443$. $443\times 2+1=887$ and $887\times 2+1=1775$, so the rule holds throughout. Therefore the next term is $1775\times 2+1=3551$. So the answer is 3551, i.e. option (c).

Paper folding and punching The first fold brings the top half down on to the bottom half; the second fold brings the right half of that strip on to its left half. The punch is therefore made through four thicknesses, and on the folded packet it lies in the bottom-left quarter. Undo the vertical fold first: the mark is copied into the bottom-right quarter as a left-right mirror image, so the two bottom marks face each other. Now undo the horizontal fold: both bottom marks are copied into the top half as up-down mirror images. The opened sheet therefore carries four marks, one per quarter, and each is a mirror image of its neighbour across the fold line - never a plain repeat. Option (a) shows only two marks, so two thicknesses were missed. Option (b) repeats the left-hand mark on the right instead of mirroring it, and option (c) repeats the bottom row at the top instead of mirroring it. Only option (d) mirrors across both fold lines, so the answer is (d).

Syllogism Draw the three statements: Treks lies wholly inside Hikes, Hikes and Climbs merely cross, and Walks lies wholly inside Climbs. Walks is inside Climbs, but nothing forces it into the small part where Climbs meets Hikes; it can be drawn in the far side of Climbs, clear of Hikes. In that legal picture no walk is a hike, so conclusion (II) is only possible, not certain. Every trek is inside Hikes, so a walk could be a trek only by being a hike first; since that is not certain, conclusion (I) is not certain either. Neither conclusion follows, i.e. option (b).

Letter-to-number coding Add the place values of the four letters and see how far the code is from that sum. XIAP: 24+9+1+16=50, and the code is 60, which is 10 more. NGQC: 14+7+17+3=41, and the code is 51, again 10 more. So the rule is: sum of the place values, plus 10. TFIZ: 20+6+9+26=61, so the code is 61+10=71. So the answer is 71, i.e. option (c).

Letter series with blanks The series is five groups of four letters, one blank in each group. The known first letters are D, G, J and M - a chain of +3 - so the first group starts at A and the first blank is A. Inside a complete group such as GI_H the letters are G (7), I (9), blank, H (8), i.e. first, first +2, blank, first +1. So the third letter of a group is first +5: for GI_H it is 7+5=12, which is L, the third blank. D_IE: the second letter is 4+2=6, which is F, the second blank. JLO_: the fourth letter is 10+1=11, which is K, the fourth blank. MOR_: the fourth letter is 13+1=14, which is N, the fifth blank. Reading the blanks in order gives A, F, L, K, N, i.e. option (a).