4 February 2025 · Shift 2

Odd letter-cluster out Write the English alphabet position of each letter and look at the gaps between them. PTV gives 16, 20, 22, so the gaps are +4 and +2. CGI gives 3, 7, 9 and EIK gives 5, 9, 11, again gaps +4 and +2. GLM gives 7, 12, 13, so its gaps are +5 and +1. GLM is the only cluster that breaks the +4, +2 pattern, so the answer is (d).

Figure series Count the straight line segments that make up each figure of the series. Figure 1: outer hexagon 6 + six spokes 6 + inner hexagon 6 + triangle 3 = 21 segments. Figure 2: outer hexagon 6 + two spokes 2 + inner hexagon 6 + triangle 3 = 17 segments. Figure 3: outer hexagon 6 + inner hexagon 6 + two sides of the triangle 2 = 14 segments. Figure 4: outer hexagon 6 + inner hexagon 6 = 12 segments. The number of segments lost at each step is 4, then 3, then 2, so the next step must lose 1. Figure 5 therefore has 12-1=11 segments: the outer hexagon complete and the inner hexagon with one side missing. Only option (c) shows exactly this, so the answer is (c).

Letter-cluster analogy Compare the two clusters letter by letter and note the shift at each place. REMY to VHOZ: R +4= V, E +3= H, M +2= O, Y +1= Z. ZFHU to DIJV: Z +4= D (counting on past Z to A), F +3= I, H +2= J, U +1= V. So the shifts are +4, +3, +2, +1 from left to right. Apply them to LBSO: L +4= P, B +3= E, S +2= U, O +1= P. LBSO is related to PEUP, so the answer is (c).

Syllogism Draw the three statements as circles. All wires are cables and all wires are ropes, so the whole wires circle lies inside the cables circle and inside the ropes circle. Some ropes are steels only makes the steels circle touch the ropes circle somewhere; it need not touch the small wires part. Conclusion (I): some steels are wires is therefore a possibility only, not a certainty, so it does not follow. Conclusion (II): every wire is a rope and the same wire is also a cable, so at least those ropes are cables; hence some ropes are cables is certain and it follows. Only conclusion (II) follows, so the answer is (b).

Venn diagram of classes Take the classes two at a time. A bus is a means of transport and an airplane is also a means of transport, so each of them lies completely inside Transport. No bus is an airplane, so the Bus class and the Airplane class have nothing in common and must be drawn apart from each other. The picture needed is therefore one large circle with two separate smaller circles inside it. That is option (a).

Letter-to-number coding Write the alphabet position of each letter and compare it with its code number. P is 16 and its code is 9, and 16+9=25; T is 20 and its code is 5, and 20+5=25. Rule: code =25- (position of the letter). Check on the rest: 25-7=18 for G, 25-13=12 for M, 25-19=6 for S, which gives 9-18-12-6. For KAVO the positions are K =11, A =1, V =22, O =15. So the code is 25-11=14, 25-1=24, 25-22=3, 25-15=10. KAVO is coded as 14-24-3-10, so the answer is (a).

Blood relation coding Break the expression $A÷B+C\times D-E$ one symbol at a time, from left to right. $A÷B$: A is the mother of B, so A is female. B + C: B is the sister of C. $C\times D$: C is the brother of D. D - E: D is the father of E, so D is male. B, C and D are brothers and sisters of one another, so they are children of the same mother. A is the mother of B, therefore A is the mother of C and of D as well. So A is the mother of D, and the answer is (b).

Interchange of signs Rule: after the interchange, read + as -, - as +, $\times$ as $÷$ and $÷$ as $\times$. Given expression: $18÷5+26-45\times 5$. Replacing every sign by its partner: $18\times 5-26+45÷5$. By BODMAS do division and multiplication first: $18\times 5=90$ and $45÷5=9$. The expression becomes 90 - 26 + 9. 90 - 26 = 64 and 64 + 9 = 73. So the required value is 73, and the answer is (b).

Number series Read the series term by term: $79,72,58,37,9,?$ Find the difference between each pair of consecutive terms. 72-79=-7, 58-72=-14, 37-58=-21, 9-37=-28. The numbers subtracted are 7, 14, 21, 28, that is the multiples of 7, so the next number to subtract is 35. Required term =9-35=-26. Hence the answer is (a).

Number coding by length The two code words have no letter in common with their codes, so compare the code with the length of the word instead. TEDI has 4 letters and its code is 11, and 4+7=11. SUCBVO has 6 letters and its code is 13, and 6+7=13. Rule: code = number of letters $+7$. YXJIAZL has 7 letters, so its code is 7+7=14. Hence the answer is (c).

Two-row seating arrangement Row 1 - I, N, K, S face south; row 2 - P, A, R, T face north, and seat 1 of one row faces seat 1 of the other. Number the seats of row 2 from 1 to 4 starting at the left of a north-facing person, so 'immediate right' means the next higher seat number. R is immediate right of A, so (A, R) is (1, 2), (2, 3) or (3, 4). If it were (2, 3), T and P would take seats 1 and 4 and T would sit at an end; if it were (3, 4), P would need a seat to the left of seat 1. So A and R take seats 1 and 2, T takes seat 3 (not an end) and P takes seat 4; row 2 reads A, R, T, P. Only two people sit between K and S, so K and S take seats 1 and 4 of row 1; A faces K, and A is at seat 1, so K is at seat 1 and S at seat 4. The one who faces N sits immediate left of P, that is seat 3, which is T; so N sits at seat 3 and I is left with seat 2. Column by column the facing pairs are K-A, I-R, N-T and S-P, so I and R face each other. Therefore option (a) is correct.

Dice - opposite faces The face X is common to both positions and it stands in the same place, so the second cube is only the first one turned about the axis through X. Position I - A on top, X on the left, D on the right. Position II - E on top, X on the left, N on the right. A quarter turn about that axis would bring the top face A round to the right, so Position II would show A on the right; it shows N, so the cube has been given a half turn. A half turn sends the top face to the bottom and the right face to the back, so the new top E is the face opposite A and the new right N is the face opposite D. That fixes A-E and D-N, and the two faces left over, X and R, must pair up. Therefore option (c) is correct: E lies opposite A.

Alphabet series Inside every group the second letter is 2 places before the first letter and the third letter is 2 places after it. WUY is $23,21,25$; KIM is $11,9,13$; YWA is $25,23,1$; MKO is $13,11,15$. So the whole series is carried by the first letters W, K, Y, M, that is $23,11,25,13$. The jumps alternate -12 and +14: 23-12=11, 11+14=25, 25-12=13. The next jump is +14, so 13+14=27, and counting round the 26 letters 27-26=1, which is A. With A as the first letter, 2 places before A is Y and 2 places after A is C, so the group is AYC. Therefore option (c) is correct.

Dice - opposite faces The face F is drawn in the same place (the front-left face) in both positions, so the second cube is only the first one turned about the axis through F. Position I - A on top, F on the left, I on the right. Position II - L on top, F on the left, V on the right. A quarter turn about that axis would bring A or I back into view on the top or on the right, but Position II shows L and V, so the cube has been given a half turn. A half turn sends the top face to the bottom and the right face to the back, so the new top L is the face opposite A and the new right V is the face opposite I. That fixes the pairs A-L and I-V; the only faces left, F and O, must therefore lie opposite each other. Therefore option (b) is correct: F lies opposite O.

Number series The series reads $14,23,42,71,110,?$ Take the differences: 23-14=9, 42-23=19, 71-42=29, 110-71=39. Every difference is 10 more than the one before it, so the next difference is 39+10=49. Required term =110+49=159. Therefore option (d) is correct.

Linear seating arrangement All seven face north, so fix one direction as 'right' and keep it the same throughout the line. V is immediate right of Q and immediate left of R, which gives the block Q, V, R. P is immediate right of R and immediate left of S, which gives the block R, P, S; the common letter R joins the two blocks into Q, V, R, P, S. U is second to the right of S, so exactly one seat lies between S and U: Q, V, R, P, S, _, U. T is the only person left, so T fills that seat and the line is Q, V, R, P, S, T, U, with U at an end as required. The person sitting between S and U is T. Therefore option (a) is correct.

Mirror image The mirror MN is a vertical line, so the image is the figure reversed left to right; nothing moves up or down. In the given figure the arrow head is at the top left and the solid dot is at the bottom right, so in the image the arrow head must point right at the top and the dot must sit at the bottom left. A left-right flip also keeps every stroke at its own height, so the two-line stroke stays on top and the single line carrying the dot stays at the bottom. Option (a) turns the arrow the right way but shifts the two-line stroke to the bottom; options (b) and (c) are not reflected at all, their arrow still points left. Therefore option (d) is correct.

Letter series completion Eight letters have to be filled in, so write the series out with all eight blanks: _ K _ U T O _ L _ T O _ _ U T O K L _ _, twenty places in all. Twenty places built on a repeating pattern means one block of five letters used four times, so set the four blocks one under the other. Reading the columns of the four blocks, position 1 shows O, position 2 shows K, position 3 shows L, position 4 shows U and position 5 shows T, so the repeating block is OKLUT. Now complete each block: block 1 needs O and L, block 2 needs K and U, block 3 needs K and L, block 4 needs U and T. Reading the fillers from left to right gives O, L, K, U, K, L, U, T. Therefore option (b) is correct.

Alphabetical rearrangement Arrange the seven letters of JOSTLED in alphabetical order: D, E, J, L, O, S, T. Now write the original word and the sorted word one under the other and compare position by position. The pairs are J/D, O/E, S/J, T/L, L/O, E/S and D/T, so not one position carries the same letter. Therefore option (c) is correct: 0 letters keep their position.

Letter cluster analogy Compare the given pairs letter by letter using the alphabet positions. PSPN to OROM: $P\left(16\right)\to O\left(15\right)$, $S\left(19\right)\to R\left(18\right)$, $P\left(16\right)\to O\left(15\right)$, $N\left(14\right)\to M\left(13\right)$. YBYW to XAXV: $Y\left(25\right)\to X\left(24\right)$, $B\left(2\right)\to A\left(1\right)$, $Y\left(25\right)\to X\left(24\right)$, $W\left(23\right)\to V\left(22\right)$. So the rule is the same throughout: every letter moves one step backward in the alphabet. Apply -1 to QTQO: $Q\left(17\right)\to P\left(16\right)$, $T\left(20\right)\to S\left(19\right)$, $Q\left(17\right)\to P\left(16\right)$, $O\left(15\right)\to N\left(14\right)$, giving PSPN. Therefore option (d) is correct.