20 February 2025 · Shift 1

Paper folding and punching The dotted line is vertical and the arrow points to the right, so the left half is folded over onto the right half. After that fold only the right half is visible, and a single punch is made a little right of the centre of that half, at mid-height. Unfolding reverses the fold, so the punch reappears on the left half as its mirror image in the vertical fold line. The sheet therefore shows two holes at the SAME height, one on each side of the vertical centre line and equally far from it. Option (b) alone shows that; (a) stacks the two holes one above the other, and (c) and (d) put them at different heights, which a single punch on a vertical fold can never produce.

Figure series Two things change together, so track them separately. Count the strokes in the left column of each figure: 2, 4, 7, 11. The jumps are +2, +3, +4, growing by one each time, so the next jump is +5 and the missing figure needs 11+5=16 strokes. The bracket in the top-right corner turns as well: it opens right, then down (a quarter turn), then up (a half turn), then right (a quarter turn) - quarter and half turns alternate. The next turn is therefore a half turn, and a bracket opening right becomes a bracket opening LEFT. Only option (b) has the bracket opening to the left, and it carries the required 16 strokes, so option (b) is correct.

Letter series The series has 20 places, so break it into five groups of four: _JDI / F_EJ / GL_K / HMG_ / INH_. Now read each position DOWN the five groups instead of across. 1st letters: E, F, G, H, I; 2nd letters: J, K, L, M, N; 3rd letters: D, E, F, G, H; 4th letters: I, J, K, L, M - every column runs in consecutive alphabetical order. The blanks are the 1st letter of group 1, the 2nd of group 2, the 3rd of group 3 and the 4th of groups 4 and 5, giving E, K, F, L, M. The required combination is EKFLM, so option (c) is correct.

Coded blood relations Read the code one link at a time, from left to right. P+Q: P is the daughter of Q. Q-R: Q is the brother of R, so Q is male. $R÷S$: R is the father of S. $S\times T$: S is the wife of T, so S is female. Now walk back from T: S is T's wife, R is S's father, Q is R's brother, and P is Q's daughter. So P is T's wife's father's brother's daughter, and option (c) is correct.

Word-to-number coding 'RUST' → '8264' and 'TURN' → '2843'. The two words share exactly three letters - R, U and T - and the two codes share exactly three digits - 8, 2 and 4. So those three digits stand for R, U and T in some order, and the leftover letter of each word must take the leftover digit. 'RUST' is left with S and the digit 6; 'TURN' is left with N and the digit 3. Hence S is coded as 6, so option (d) is correct.

Letter-cluster analogy Compare MRCS with PUFV letter by letter: M→P, R→U, C→F, S→V. Every letter moves forward by 3 places in the alphabet. The second pair obeys the same rule: V→Y, O→R, L→O, G→J, again +3 each. Apply +3 to SHIM: S→V, H→K, I→L, M→P. SHIM is therefore related to VKLP, so option (c) is correct.

Alphabetical rearrangement Write DELIGHT out with its seven positions: D, E, L, I, G, H, T. Sorting the same seven letters alphabetically gives D, E, G, H, I, L, T. Compare the two rows position by position: D (1st), E (2nd) and T (7th) fall in the same place in both. L, I, G and H all shift, so exactly three letters keep their positions and option (a) is correct.

Embedded figure Fix exactly what figure (X) is made of before looking at the options. (X) is a Y: three identical nodes - each a ring with a SOLID dot at its centre - with a vertical stem from the top node down to a junction and two equal slanting arms from that junction to the two lower nodes. The centre dot is the deciding detail. The circles in options (b) and (d) are plain hollow rings with nothing inside them, and option (c) is built from squares, so none of those three can contain (X). In option (a) the ringed top node, the vertical stem, the junction and the two equal arms running down to the two ringed lower nodes appear exactly as in (X) - same size, same slant, no rotation. So option (a) is correct.

Letter-cluster analogy Compare RFTQ with VKXV position by position: R→V is +4, F→K is +5, T→X is +4, Q→V is +5. DNPJ → HSTO follows the same +4, +5, +4, +5 pattern. Apply it to HLXB: H+4 = L, L+5 = Q, X+4 = B (counting on past Z back to A), B+5 = G. HLXB is therefore related to LQBG, so option (d) is correct.

Syllogism Draw the statements: bins lie wholly inside tins, fins lie wholly inside tins, and pins meet fins in at least one member. Conclusion (I): only those pins that happen to be fins are forced to be tins. The remaining pins may lie entirely outside tins, so 'All pins are tins' does not follow. Conclusion (II): bins and fins are both inside tins, but nothing makes them touch - they can be two separate parts of tins, so 'Some bins are fins' does not follow. A possibility is not a certainty, so neither conclusion follows and option (d) is correct.

Number series The series is 21, 75, 120, 156, 183, ? Take the differences: 75-21=54, 120-75=45, 156-120=36, 183-156=27. The differences themselves fall by 9 each time, so the next difference is 27-9=18. Required term =183+18=201, so option (a) is correct.

Number series The terms grow roughly three times each step, so test $\times 3$ with a correction. $120\times 3+1=361$ and $361\times 3+1=1084$. $1084\times 3+1=3253$ and $3253\times 3+1=9760$, so the rule $\times 3+1$ holds throughout. Next term $=9760\times 3+1=29280+1=29281$, so option (d) is correct.

Symbol interchange The interchanges are $-\leftrightarrow ÷$ and $\times \leftrightarrow +$, so replace every sign in the expression. $9+56-8\times 4÷9$ becomes $9\times 56÷8+4-9$. By BODMAS do $\times$ and $÷$ first, from left to right: $9\times 56=504$, then $504÷8=63$. Now the remaining additions and subtractions: 63+4-9=58. The value is 58, so option (c) is correct.

Letter-cluster series Treat the three positions of the clusters ICD, MGH, QKL, UOP separately. 1st letters: I, M, Q, U - each is 4 places after the previous one, so the next is U+4=Y. 2nd letters: C, G, K, O - again +4 each, so the next is O+4=S. 3rd letters: D, H, L, P - again +4 each, so the next is P+4=T. The missing cluster is YST, so option (b) is correct.

Venn diagram of classes Test each pair of classes in turn. A king can also be a father, but not every king is a father and not every father is a king, so Kings and Fathers must be two circles that cross without either lying inside the other. A queen is a woman, so no queen is a king and no queen is a father; the Queens circle has to stand completely clear of both. So the picture needed is two intersecting circles plus a third circle that touches neither of them. Option (c) is exactly that. In (a) and (b) the Queens circle cuts Kings or Fathers, and in (d) one class is drawn wholly inside another, which is true of no pair here.

Odd one out Turn each cluster into alphabet positions and measure the two gaps inside it. NVD: N(14) +8 = V(22), and V(22) +8 = D, counting on past Z back to the start. JRZ: J(10) +8 = R(18), R(18) +8 = Z(26). PXF: P(16) +8 = X(24), X(24) +8 = F. IOY breaks it: I(9) +6 = O(15) and O(15) +10 = Y(25) - the two gaps are 6 and 10, not 8 and 8. IOY does not belong to the group, so option (b) is correct.

Coding by letter count The two given words share no letters and no obvious alphabet shift, so look at their length instead. 'KIXCUP' has 6 letters and its code is 10; 'DZYM' has 4 letters and its code is 8. In both cases the code is the number of letters plus 4: 6+4=10 and 4+4=8. 'RSAWYQI' has 7 letters, so its code is 7+4=11, and option (b) is correct.

Paper folding and cutting First fold: the arrow points up, so the bottom half is folded onto the top half along the horizontal centre line. Second fold: the arrow points left, so the right half is folded onto the left half along the vertical centre line; the top-left quarter is left facing us. The cuts are made on that quarter - a circle, then a bar, then a circle, one below the other. Opening the vertical fold mirrors the quarter to the right of the vertical centre line, so the top half carries TWO identical columns. Opening the horizontal fold mirrors both columns below the horizontal centre line, so each column reads circle, bar, circle above the line and circle, bar, circle below it. That gives 8 circles and 4 bars, symmetric about both centre lines, which is option (c). Option (b) puts only three circles and two bars in each column (it forgets one mirror), and (a) and (d) scatter the cuts round the border instead of in two columns.

Linear seating arrangement Number the seven seats 1 to 7 from left to right; everybody faces north, so 'right' means a higher seat number. 'Only K sits to the right of J' means exactly one person is to J's right, so J takes seat 6 and K seat 7. Only one person sits between I and J, so I takes seat 4. Only two people sit between M and I, so M takes seat 1, with seats 2 and 3 in between. Seats 2, 3 and 5 are left for L, G and H, and G must be right of L but left of H, so L = 2, G = 3, H = 5. Between H (seat 5) and L (seat 2) sit G and I - two people - so option (a) is correct.

Linear seating arrangement Number the seven seats 1 to 7 from left to right; everybody faces north, so 'right' means a higher seat number. Only two people sit to the right of K, so K takes seat 5, and H, sitting immediately to K's left, takes seat 4. Only two people sit between K and F, so F takes seat 2 (seats 3 and 4 lie between). Only two people sit between E and H puts E at seat 1 or seat 7. N sits immediately to the right of L, and the only free adjacent pair left is seats 6 and 7, so L = 6 and N = 7; that forces E = 1. The one seat still empty, seat 3, goes to G. Between G (seat 3) and L (seat 6) sit H and K - two people - so option (c) is correct.