20 February 2025 · Shift 2

Circular seating For people facing the centre, a person's left is the clockwise side of the table and the right is the anticlockwise side. Three people sit between S and A when counted from the right of S, so A is the fourth seat anticlockwise from S. Only two people sit between A and B, and in a circle of six that is possible only when B sits exactly opposite A. Three seats are left for C, R and T; R must be the seat immediately anticlockwise of C (that is C's immediate right), which fixes C, R and T. Counting from the left of A, that is clockwise, the order is R, C, B, S, T. So four people sit between A and T, and the correct option is (a).

Number series Take the difference between consecutive terms: 451-450=1, 454-451=3, 459-454=5, 466-459=7. The differences are the successive odd numbers 1, 3, 5, 7, so the next difference is 9. Missing term =466+9=475. Hence the correct option is (b).

Letter-to-number coding Replace every letter by its position counted from the END of the alphabet (A=26, B=25, …, Z=1). $XHOS\to 3+19+12+8=42$ and $42\times 2=84$, which is the given code. $CRFY\to 24+9+21+2=56$ and $56\times 2=112$, so the rule is: twice the sum of the reverse positions. $ZIOE\to 1+18+12+22=53$, so the code is $53\times 2=106$. Hence the correct option is (b).

Letter-digit coding 'PAIR' and 'REAP' are built from the same letters except that PAIR has I where REAP has E; P, A and R are common to both. The digits common to 7643 and 7536 are 7, 6 and 3, so those three stand for P, A and R in some order. The digit of 'PAIR' that is left over is 4, so I=4; the digit of 'REAP' that is left over is 5, so E=5. Hence the correct option is (d).

Syllogism All pianists are musicians and all musicians are singers, so the pianists lie completely inside the singers. No singer is a composer, so the whole singer circle stands apart from the composer circle. Conclusion (I): the pianists are inside the singers and no singer is a composer, so no pianist can be a composer - it follows. Conclusion (II): every pianist is a singer, so those pianists are singers who are pianists, which makes 'Some singers are pianists' true - it follows as well. Both conclusions follow, so the correct option is (b).

Paper folding and cutting The first fold takes the bottom half up and the second fold takes the right half onto the left, so the piece that is cut is the top-left quarter and the sheet opens about the horizontal and the vertical centre lines. Every cut therefore appears four times, once in each quarter, mirrored across both centre lines. The small rectangle lies close to the OUTER left edge of the quarter, far from both creases, so its four copies land near the four outer corners of the sheet - not in a group beside the centre line. The slanting cut removes the corner of the quarter that is the centre of the whole sheet; a corner sitting on BOTH creases opens out into one complete four-sided diamond, not a triangle. Only the figure with four rectangles at the four outer corners and one full diamond at the centre matches, so the correct option is (c).

Odd one out (letter-clusters) Compare the second and the third letter of each cluster with its first letter. UVQ: U+1=V and U-4=Q; WXS: W+1=X and W-4=S; IJE: I+1=J and I-4=E. IMH does not obey it: I+4=M and I-1=H, so the two shifts are interchanged. Hence the odd one out is IMH and the correct option is (b).

Letter series (blanks) The series has 12 places; number them 1 to 12 and write down the letters that are already given. T and B are at places 3 and 4, N and A at 6 and 7, B and M at 9 and 10 and A at 12 - the same letters reappear five places later, so the block repeats after every five letters. The repeating block is NATBM, so the series is NATBM / NATBM / NA. The blanks at places 1, 2, 5, 8 and 11 therefore take N, A, M, T and N. Hence the correct option is (a).

Coded blood relations Break the string P<B!S:T/A into single relations and read it from left to right. P<B means P is the son of B; B!S means B is the brother of S; S:T means S is the wife of T; T/A means T is the father of A. T is A's father and S is T's wife, so S is A's mother. B is S's brother, so B is A's mother's brother. P is the son of B, so P is A's mother's brother's son. Hence the correct option is (c).

Linear seating Number the seats 1 to 7 from the left end of the row. Only three people sit to the left of T, so T is at seat 4; only B sits to the right of A, so A is at seat 6 and B at seat 7. Only three people sit between A and S, so S is at seat 2. Seats 1, 3 and 5 are left for M, L and N, and M is to the left of L which is to the left of N, so M = 1, L = 3 and N = 5. Between M (seat 1) and N (seat 5) sit S, L and T - three people, so the correct option is (a).

Letter-cluster analogy Compare the two clusters letter by letter, position by position. The shifts are -3, +2, +2, -3, and the same four shifts turn RBUX into ODWU, which confirms the rule. Apply them to NCTV: N-3=K, C+2=E, T+2=V, V-3=S. So NCTV is related to KEVS and the correct option is (b).

Letter series Compare the terms letter by letter: $A\to V$, $V\to Q$, $Q\to L$ - each first letter moves 5 places back. The second and third letters do the same: $T\to O\to J\to E$ and $E\to Z\to U\to P$. So from LEP: L-5=G, E-5=Z (from E, five steps back runs A, then wraps to Z) and P-5=K. The missing term is GZK, so the correct option is (b).

Alphabetical rearrangement Write CLIMBER and, under it, its seven letters arranged alphabetically: B, C, E, I, L, M, R. Compare the two rows place by place: C-B, L-C, I-E, M-I, B-L, E-M all differ. Only the 7th place holds the same letter R in both rows. So exactly one letter keeps its position and the correct option is (d).

Letter-cluster analogy Compare the pairs position by position. The shifts are +2, +3, +2, +3; they also turn UELO into WHNR, which confirms the rule. Apply them to JDVB: J+2=L, D+3=G, V+2=X, B+3=E. So JDVB is related to LGXE and the correct option is (c).

Embedded figure Figure (X) is a square with a right-angled triangle standing on its top-right corner: one leg goes straight up from that corner and the other goes straight to the right, and a slanting line closes them. Both legs of the triangle are as long as the side of the square, so the option must contain a square whose top-right corner carries exactly that triangle, in exactly that orientation. In option (c) the lower-left quarter of the big square is that square; the upper half of the vertical centre line is the upward leg, the right half of the horizontal centre line is the rightward leg, and the slant from the top of the centre line to the middle of the right edge closes the triangle. In (a) and (b) no square has a right-angled triangle of the same size sitting on its top-right corner, and in (d) the horizontal lines do not pass through the corner of any such square. Hence the correct option is (c).

Paper folding and cutting The bottom half is folded up and then the right half is folded onto the left, so the cut piece is the top-left quarter and the sheet opens about the horizontal and the vertical centre lines. Each cut therefore appears four times, so 1 star and 2 circles become 4 stars and 8 circles - that alone rules out (b), which has one star, and (c), which has only four circles. In the folded quarter the star sits at its centre, one circle sits near the top close to the vertical crease and the other sits near the outer left edge close to the horizontal crease. So the circles open into a pair side by side at the top centre, a matching pair at the bottom centre and a pair one above the other at the middle of each side - none of them lands in a corner or at the centre of the sheet, which rules out (d). Hence the correct option is (a).

Figure series Every box carries the same four signs on the diagonal from its bottom-left corner to its top-right corner; only their order changes. Reading each box from the bottom-left upwards: box 1 is $+,-,÷,\times$; box 2 is $-,+,\times ,÷$; box 3 is $\times ,÷,-,+$; box 4 is $÷,\times ,+,-$. From box 1 to box 2 the signs swap inside each pair (1st with 2nd, 3rd with 4th); from box 2 to box 3 the first pair changes places with the second pair; from box 3 to box 4 they swap inside each pair again. So box 4 to box 5 must exchange the two pairs once more: $÷,\times ,+,-$ becomes $+,-,÷,\times$. The missing box has + at the bottom-left, then -, then $÷$ and $\times$ at the top-right, so the correct option is (b).

Interchange of signs Write down what each sign becomes before touching the numbers. The given equation $32+49-7\times 4÷3$ therefore becomes $32\times 49÷7+4-3$. By BODMAS, do $\times$ and $÷$ from left to right: $32\times 49=1568$ and $1568÷7=224$. Then 224+4-3=225. Hence the correct option is (a).

Venn diagram of classes Bread is one kind of food, so every bread is a food and the whole 'Bread' region must lie inside 'Food'. Wool is not a food and no food is wool, so 'Wool' must be drawn completely apart from 'Food'. The diagram must therefore show one small closed figure inside a bigger one, plus a third figure that touches neither. Option (a) is the only one of that shape - (b) makes wool partly food, (c) makes all three nested and (d) puts two separate kinds inside food - so the correct option is (a).

Number series Take the difference between consecutive terms: 40-36=4, 48-40=8, 64-48=16, 96-64=32. Each difference is double the one before it, so the next difference is $32\times 2=64$. Missing term =96+64=160. Hence the correct option is (d).