21 February 2025 · Shift 2

Syllogism All three statements start from breads: breads lie inside croissants, inside muffins and inside pastries. Nothing at all is said about croissants that are not breads. Conclusion (I): the croissant set only has to hold the breads; the rest of it may lie entirely outside pastries, so "All croissants are pastries" does not follow. Conclusion (II): every bread is a pastry and the same bread is also a muffin, so at least those breads are pastries that are muffins - "Some pastries are muffins" does follow. Only conclusion (II) follows, so the correct option is (d).

Letter series blanks Break the string into five blocks of four letters: _TWR, U_SN, QL_J, MHK_, IDG_. Read the blocks downwards, column by column: the first letters run Y, U, Q, M, I - each one 4 places back, and the other three columns fall 4 places back in exactly the same way. That fills the blanks as Y (block 1), P (block 2), O (block 3), F (block 4) and B (block 5). A check inside any block confirms it: Y $\to$ T is -5, T $\to$ W is +3, W $\to$ R is -5, and I $\to$ D $\to$ G $\to$ B follows the same -5, +3, -5. So the combination is YPOFB and the correct option is (d).

Circular seating Everyone faces the centre, so each person's left hand points clockwise and the right hand points anticlockwise. K sits immediately to F's right, so K is the seat one step anticlockwise from F; K is third to J's right, which puts J two seats clockwise from F. F is third to G's left, so G is three seats anticlockwise from F - that is, four seats clockwise from F. Three seats are still free: the ones one, three and five steps clockwise from F. Counting from L's left, E is two seats clockwise of L, and the only placing that leaves H away from F is L one step clockwise of F, E three steps and H five steps. Now count from H towards H's right (anticlockwise): G, E, J and then L - so three people sit between H and L. Hence the correct option is (a).

Embedded figure Figure (X) is one long straight line with a semicircle hanging below it, and the line sticks out a long way beyond the semicircle on both the left and the right. In option (d) the horizontal double arrow cuts straight through the small inner circle: the part of that circle lying below the arrow is the semicircle, and the arrow carries on far past it on either side - the whole of (X) is present, unrotated. Option (c) also shows half-discs, but there each straight edge is exactly as wide as its own half-disc and stops where the curve stops, so the long overhang that (X) needs is missing. Option (a) has an arc that bulges sideways off a vertical line, and option (b) has only a complete circle, so neither contains a semicircle hanging under a horizontal line. Hence the correct option is (d).

Word-number coding The two words share the letters S, H and O; the two codes share the digits 3, 2 and 1. So S, H, O use up 3, 2, 1 between them in some order, and the letter each word has to itself must take the digit that code has to itself. SHOE keeps E and the digit 4, so E is 4; HOST keeps T and the digit 9, so T is 9. Hence T is coded as 9 and the correct option is (a).

Letter-cluster analogy Compare FNTA with BKRZ letter by letter: F $\to$ B is -4, N $\to$ K is -3, T $\to$ R is -2, A $\to$ Z is -1. JOCY $\to$ FLAX repeats it: J $\to$ F is -4, O $\to$ L is -3, C $\to$ A is -2, Y $\to$ X is -1. Apply the same steps to RWLS: R-4 = N, W-3 = T, L-2 = J, S-1 = R. So RWLS is related to NTJR and the correct option is (b).

Paper folding and cutting The sheet is folded in half upwards, and the half that is left is folded in half upwards again, so the strip that is finally cut is a quarter of the height and carries four layers. Both folds are horizontal, so unfolding only ever mirrors the pattern up and down; the four cut shapes stay in the same four columns and never change left-to-right order. The cut strip reads down-triangle, circle, down-triangle, circle. Opening the second fold puts its mirror image just below it: up-triangle, circle, up-triangle, circle. Opening the first fold copies that pair of strips, again mirrored, into the lower half of the sheet. So from top to bottom the sheet shows down / up / down / up rows of triangles with circles in between - four full strips. Option (c) keeps every triangle pointing down and option (d) keeps every one pointing up, which no mirroring can produce, and option (b) leaves the two middle strips blank although all four layers were cut. Hence the correct option is (a).

Parallel-row seating Draw the two rows one above the other and read the page from left to right. Row 1 faces south, so a row-1 person's right hand points to the left of the page; row 2 faces north, so a row-2 person's left hand points to the left of the page. T is third to K's right, and row 1's right is the left of the page, so T lands at the left end of row 1 and K at the right end: T, _, _, K. The one facing S is a neighbour of K, so S sits in the third place of row 2 from the left of the page; the one facing R is a neighbour of S, which forces R into the second place of row 1 and leaves A third. The one facing T is a neighbour of O, and the person facing T is at the far left, so O takes the second place of row 2; M is somewhere to E's left, so M is first and E is fourth. The end seats are therefore T and K in row 1 and M and E in row 2. Of the four choices only "M and K" names two people who both occupy an end seat - A sits inside row 1, and E is paired with A or R in the other options. Hence the correct option is (b).

Odd letter-cluster Look at the steps inside each cluster. In KOL: K $\to$ O is +4 and O $\to$ L is -3. RVS gives +4 then -3, and HLI gives +4 then -3 as well, so those three match. The same rule read another way: the third letter is one place after the first. LPN starts correctly with L $\to$ P = +4, but P $\to$ N is only -2, and N is two places after L instead of one. LPN therefore breaks the pattern, so the correct option is (d).

Letter-cluster series Compare the clusters column by column instead of letter by letter inside one cluster. First letters: F, K, P, U - each +5. Second letters: T, Y, D, I - also +5, counting on from Z back to A. Third letters: R, W, B, G - again +5. So the next cluster is U+5 = Z, I+5 = N and G+5 = L. The missing term is ZNL, so the correct option is (c).

Coded blood relations Break the string F ! L : A < S / H at each symbol and read it from left to right. F ! L: F is the brother of L. L : A: L is the wife of A. A < S: A is the son of S. S / H: S is the father of H. S is the father of both A and H, so A and H are siblings, and A is male - A is H's brother. L is that brother's wife, and F is L's brother. So F is H's brother's wife's brother, and the correct option is (b).

Interchange of numbers As it stands the left side is $\left(98-45+14\right)+\left(84÷7\right)-36+\left(6\times 2\right)=67+12-36+12=55$, which is 12 more than 43. The left side must fall by 12, so try the swap that changes the two large numbers: interchange 98 and 84. The first bracket becomes 84-45+14=53 and the division becomes $98÷7=14$. Now the left side is 53+14-36+12=43, which equals the right side. Hence the numbers to interchange are 98 and 84 and the correct option is (a).

Number series Take the successive differences: 723-722=1, 722-719=3, 719-714=5, 714-707=7. The amounts taken away are the odd numbers 1, 3, 5, 7, so the next amount is 9. Therefore the missing term is 707-9=698. Hence the correct option is (d).

Number series Take the successive differences: 37-16=21, 79-37=42, 142-79=63, 226-142=84. These are $21\times 1$, $21\times 2$, $21\times 3$ and $21\times 4$, so the next gap is $21\times 5=105$. Therefore the missing term is 226+105=331. Hence the correct option is (d).

Dice - opposite faces 11 is on top in both views, so the die has only been spun about its vertical axis; 11 and the face under it never move. That leaves 12, 8, 10 and one unseen face going round the die as its four side faces. 8 is on the right in view I and on the left in view II, so the die has turned by exactly one quarter turn; the same turn moves 12 out of sight on the left and brings 10 in on the right. Reading the side faces in that turning order gives 12, 8, 10, unseen - and in a ring of four side faces the first and the third are opposite each other. So 12 lies opposite 10, and the correct option is (c).

Alphabetical rearrangement Sort the seven letters of GROUPIE: E, G, I, O, P, R, U. Now compare place by place: G with E, R with G, O with I, U with O, P with P, I with R, E with U. Only the fifth place matches - P is the fifth letter of GROUPIE and also the fifth letter of EGIOPRU. Exactly one letter keeps its position, so the correct option is (a).

Word-number coding The two words share the letters R, I and E; the two codes share the digits 7, 1 and 3. So R, I, E account for 7, 1, 3 between them, and the letter each word has to itself takes the digit that code has to itself. RICE keeps C and the digit 6, so C is 6; FIRE keeps F and the digit 4, so F is 4. Hence C is coded as 6 and the correct option is (c).

Figure series - rotation Every box holds the same four marks - Q, m, η and an arrow - one in each corner, so nothing is added or removed and only the places change. Follow Q: top-left, top-right, bottom-right, bottom-left. It moves one corner clockwise at every step, and each of the other three marks does exactly the same. Four steps make a complete turn, so the fifth box must repeat the first one. That means Q at the top-left, m at the top-right, η at the bottom-left and the arrow at the bottom-right. Only option (a) has that arrangement, so the correct option is (a).

Letter-cluster analogy FCBF $\to$ EBAE moves every letter back by one: F-1 = E, C-1 = B, B-1 = A, F-1 = E. XUTX $\to$ WTSW does exactly the same, one place back at every position. Apply it to NKJN: N-1 = M, K-1 = J, J-1 = I, N-1 = M. Note that the first and last letters are equal in each cluster, so the answer must also end in the letter it starts with - that rules out MKHQ at once. The result is MJIM, so the correct option is (a).

Venn diagram of classes Every rose is a flower, so the rose circle must lie completely inside the flower circle. A bear is neither a rose nor a flower, so its circle must stand apart from both of the others. Option (a) keeps all three apart, which loses the fact that a rose is a flower; option (c) makes all three overlap, which wrongly allows a bear to be a flower; option (d) only lets rose and flower overlap partly instead of one lying inside the other. Only option (b) shows a small circle wholly inside a larger one with a third circle separate from them, so the correct option is (b).