12 February 2025 · Shift 1

Odd one out: letter-cluster Compare the gap between consecutive letters of each cluster, using their positions in the alphabet. In LPJ, OSM and MQK the first letter moves +4 and then -6, so the third letter is always 2 less than the first. In OIP the gaps are -6 and then +7, which breaks that pattern. So OIP does not belong to the group, and the correct option is (a).

Letter series with blanks The string is twenty letters long, so break it into five blocks of four letters with one blank in each block. Inside every block the letters move -5, then +4, then -5. The first letters of the five blocks are T, R, P, N, L - each one 2 less than the block before it. Reading the five blanks in order gives T, M, O, H, F. So the correct option is (c).

Number coding of letters JIVE → 1597 and FIVE → 5781; the letters I, V and E appear in both words. The digits common to both codes are 1, 5 and 7, so I, V, E use 1, 5, 7 in some order. That leaves 9 for J in the first code and 8 for F in the second. So the code for F is 8, and the correct option is (c).

Interchange of signs Interchange the signs first: - becomes $÷$, $÷$ becomes -, $\times$ becomes + and + becomes $\times$. So $7+45-15\times 5÷12$ becomes $7\times 45÷15+5-12$. $7\times 45=315$ $315÷15=21$ 21+5-12=14 So the required value is 14, and the correct option is (a).

Number coding of letters CLAP → 6452 and PALM → 5428; the letters L, A and P appear in both words. The digits common to both codes are 4, 5 and 2, so L, A, P use 4, 5, 2 in some order. That leaves 6 for C in the first code and 8 for M in the second. So M is coded as 8, and the correct option is (d).

Letter-cluster analogy Compare DEATH with GLITZ letter by letter, using alphabet positions. D→G is +3, E→L is +7, A→I is +8, T→T is unchanged and H→Z is +18. GRAVE→JYIVW obeys exactly the same five shifts, so the rule is confirmed. Applying the same shifts to BENCH: B+3=E, E+7=L, N+8=V, C stays C and H+18=Z. So BENCH is related to ELVCZ, and the correct option is (d).

Coded blood relations Read 'F ! L : A < S / H' one link at a time, from left to right. F ! L means F is the brother of L, and L : A means L is the wife of A. A < S means A is the son of S, and S / H means S is the father of H. S is therefore the father of both A and H, so A and H are siblings. Now trace back from H: A is H's brother, L is A's wife and F is L's brother. So F is H's brother's wife's brother, and the correct option is (b).

Paper folding and cutting The first fold brings the top half down, so its crease is the horizontal centre line and the packet is the bottom half. The second fold brings the right half onto the left, so its crease is the vertical centre line and the packet is the bottom-left quarter. The cut is a rectangle removed from the TOP edge of that quarter, and that top edge is the first crease, not a free edge of the paper. Opening the vertical fold mirrors the cut across the vertical centre line, giving two cuts side by side. Opening the horizontal fold mirrors each of them across the horizontal centre line, so each cut closes into a complete rectangular hole sitting astride that line. The sheet therefore shows exactly two rectangular holes, both cut through by the horizontal centre line and placed symmetrically about the vertical centre line, and neither hole touches an edge. So the correct option is (a).

Letter-cluster series Take the three letter positions of the clusters separately. The first letters run B, C, D, E, so the next one is F. The second letters run F, G, H, I, so the next one is J. The third letter is A in every term, so it stays A. The missing term is therefore FJA, and the correct option is (a).

Circular seating arrangement All seven face the centre, so each person's left is the clockwise direction and their right is anticlockwise. Two people sit between D and C counting from C's left, so D is three seats clockwise from C. H is second to the left of D, so H is two seats clockwise from D; H is on I's immediate right, so I is one seat clockwise from H. K is third to the left of I, i.e. three seats clockwise from I, which places K between C and D. Only the seat between C and K and the seat between D and H are left for G and L; L cannot be next to K, so L sits between D and H and G sits between C and K. Counting three seats to G's right, i.e. anticlockwise, gives C, then I, then H. So H sits third to the right of G, and the correct option is (c).

Venn diagram of classes Every whale is a mammal, so the class Whale lies completely inside the class Mammal. No reptile is a mammal and no reptile is a whale, so Reptile stands completely apart from both. The diagram needed is therefore one small circle wholly inside a larger circle, with a third circle separate from both. Option (a) keeps all three apart, (c) overlaps all three, and (d) only overlaps Whale with Mammal instead of enclosing it. So the correct option is (b).

Figure series: rotation Every figure is a double-headed arrow carrying a short stem that leaves the arrow's midpoint at a right angle and ends in a small hollow circle. From each figure to the next the whole shape turns ${45}^{\circ }$ clockwise, the stem and circle turning with the arrow. In the fourth figure the stem points right and the arrow is vertical, so in the fifth the stem must point down-right and the arrow must lie along the bottom-left to top-right diagonal. Only option (a) shows that: the arrow on that diagonal, the stem still meeting it at a right angle, and the circle still hollow. Option (d) does put the arrow on the right diagonal, but its stem is horizontal, so it no longer meets the arrow at a right angle; (b) and (c) have not turned at all and (c) also fills the circle in. So the correct option is (a).

Embedded figure Figure (X) is a single straight stroke running from bottom-left to top-right, with one arc that springs from the lower end of that stroke, bulges to the upper left and rejoins the stroke about halfway along it. Rotation is not allowed, so look for that same stroke at that same slant, with that same arc on its upper-left side. In option (b) the lower-left diamond's bottom-right edge runs straight on into the upper-right diamond's edge, giving one unbroken bottom-left to top-right stroke, and the curve inside the left diamond leaves the lower end of that stroke and rejoins it exactly as in (X). In (a) the long slanting stroke carries an arrowhead, in (c) the strokes all meet at one centre and end in circles and half-discs, and in (d) the long stroke is horizontal, so none of them holds (X) as drawn. So the correct option is (b).

Number series: two alternating rules The series moves by two rules used turn and turn about. 17-8=9 and $9\times 3+2=29$. 29-8=21 and $21\times 3+2=65$. 65-8=57, so the next step must be $\times 3+2$. $57\times 3+2=171+2=173$ So the missing term is 173, and the correct option is (c).

Linear seating arrangement Number the seven seats 1 to 7 from the left; all face north, so left means a smaller number. Only two people sit to the left of Y, so Y is in seat 3. Three people sit between Y and W, so W is in seat 7. X is second to the right of Y, so X is in seat 5, and two people sit between X and M, so M is in seat 2. Seats 1, 4 and 6 are left for O, L and N, and O is left of L which is left of N, so O is in 1, L in 4 and N in 6. To the right of L (seat 4) sit X, N and W. So three people sit to the right of L, and the correct option is (b).

Alphabetical rearrangement Arrange the letters of GUILDER in English alphabetical order. GUILDER becomes D, E, G, I, L, R, U. The fourth letter from the left end is I and the first from the right end is U. I is the 9th letter and U is the 21st, so the letters strictly between them number 21-9-1=11. So the answer is 11, and the correct option is (c).

Letter-cluster analogy Compare YFOT with CIRX letter by letter. Y→C is +4, F→I is +3, O→R is +3 and T→X is +4. UBHW→YEKA follows the very same shifts +4, +3, +3, +4, so the rule holds. Applying it to SAEX: S+4=W, A+3=D, E+3=H and X+4=B (counting on past Z). So SAEX is related to WDHB, and the correct option is (d).

Paper folding and cutting The first fold brings the bottom half up, so its crease is the horizontal centre line; the second brings the right half onto the left, so its crease is the vertical centre line. The packet left is the top-left quarter: its bottom edge and its right edge are the two creases, while its top and left edges are free edges of the sheet. The round hole sits in the middle of the quarter, so it is clear of both creases and simply repeats - one circle in each of the four quadrants. The small square is cut at the corner where the two creases meet, so its four copies join up into a single square hole at the centre of the sheet. The corner cut crosses the free TOP edge and the vertical crease, so it opens into a V-notch at the middle of the top edge and, after the horizontal fold opens, an identical V-notch at the middle of the bottom edge. It never touches the free left edge, so the left and right edges must stay perfectly straight - that is what rules out (d), which notches them as well. Option (a) turns the four circles into squares and the centre square into a circle, and (b) puts the cuts at the four corners instead of the edge midpoints. So the correct option is (c).

Number series: even differences Take the difference between consecutive terms. The differences are +2, +4, +6, +8 - consecutive even numbers, so the next difference is +10. 730+10=740 So the missing term is 740, and the correct option is (c).

Syllogism: statements and conclusions A conclusion follows only if it is true for every arrangement allowed by the statements. Conclusion (I): 'Some chargers are batteries' and 'Some batteries are energies' are both particular statements, and two particular premises never give a definite conclusion; the chargers that are batteries may all lie outside energies. Conclusion (II): batteries and powers are linked only through chargers, and the chargers that are batteries need not be the chargers that are powers, so no link is forced. Neither conclusion is therefore compelled by the statements. So the correct option is (d).