18 February 2025 · Shift 2

Odd letter-cluster Write each letter's place in the alphabet and look at the two jumps inside the cluster. LIM gives $L\left(12\right)\to I\left(9\right)=-3$ and $I\left(9\right)\to M\left(13\right)=+4$; BYC and QNR repeat the same pair (from B, three steps back runs A, Z, Y). IGJ gives $I\left(9\right)\to G\left(7\right)=-2$ and $G\left(7\right)\to J\left(10\right)=+3$, so both of its jumps are one short. Hence IGJ does not belong to the group - option (d).

Figure series Read the four corners of every box; the strip simply alternates between two arrangements. Boxes 1 and 3 carry the filled dot at the top-left, h at the top-right, the hollow square at the bottom-left and x at the bottom-right. Boxes 2 and 4 are the other arrangement, x and the hollow square on top with h and the filled dot below. The fifth box continues the alternation, so it must repeat boxes 1 and 3 exactly. Option (b) also starts with the filled dot, but it puts the hollow square beside it; only option (d) has the filled dot AND h on the top row with the hollow square and x below - option (d).

Number coding Put the two words beside their codes and find what the pair shares. CALF and FACE share the letters C, A and F, and their codes 7904 and 0489 share the digits 9, 0 and 4. So C, A and F use up 9, 0 and 4 in some order, which leaves each word's unshared letter tied to its unshared digit. L is the extra letter of CALF and 7 the extra digit, so L = 7; E is the extra letter of FACE and 8 the extra digit, so E = 8. Hence the code for E is 8 - option (d).

Letter analogy Compare the given pair letter by letter instead of guessing a rule. $X\to W$, $U\to T$, $Y\to X$, $Z\to Y$: every letter of XUYZ moves one step BACK, and EBFG $\to$ DAEF does exactly the same. Apply -1 to DAEF: $D\to C$, $A\to Z$ (A wraps round to the end of the alphabet), $E\to D$, $F\to E$. That gives CZDE - option (d).

Letter analogy One common shift does not work here, so test the four positions separately. USTD $\to$ VUWH uses +1, +2, +3, +4, and PAVG $\to$ QCYK uses the very same four shifts. Apply them to ZJOU: Z+1 = A (Z wraps round to the start), J+2 = L, O+3 = R, U+4 = Y. So ZJOU is related to ALRY - option (d).

Blood relations Fix the generations first: P is the father of Q and also the father of R, so Q and R are sisters. R is the wife of S, and T is the son of S, so T is the son of R as well. R is therefore T's mother, and Q is R's sister. So Q is T's mother's sister - option (b).

Circular seating Everybody faces the centre, so each person's RIGHT runs anticlockwise round the table; walk the seats in that direction. Seat C, and put I on C's immediate right. Three people between H and C counted from C's right makes H the fourth seat to C's right. Two people between K and I counted from K's right makes I the third seat to K's right, so K sits three seats before I - that is the fifth seat to C's right. The two seats still empty are the second and third to C's right, and G on L's immediate right fills them as L then G; D takes the last seat, sixth from C. Now count to D's right: C is first, I second, L third and G fourth. So G sits fourth to the right of D - option (a).

Letter series completion The series is twenty places long and every option supplies eight letters, so twelve letters are already printed - try a block of five repeated four times. The printed letters fit the block PMSTV perfectly: PMS at places 1-3, P at 6, TV at 9-10, M at 12, TV at 14-15, PM at 16-17 and V at 20. The blanks are therefore places 4, 5, 7, 8, 11, 13, 18 and 19, which the block fills with T, V, M, S, P, S, S, T. That combination is TVMSPSST - option (b).

Syllogism A conclusion follows only if it is true in EVERY picture the statements allow, so draw the least restricted one. 'All cages are nests' puts Cages wholly inside Nests; 'Some nests are houses' only needs Nests and Houses to overlap somewhere; 'No house is a den' keeps Dens clear of Houses. Nothing forces Dens away from Cages, so the picture may let dens sit inside the Cages circle - 'No cage is a den' need not be true, and (I) does not follow. The Nests-Houses overlap need not reach the Cages circle at all, so 'Some houses are cages' need not be true either, and (II) does not follow. Hence neither conclusion follows - option (d).

Embedded figure List the three parts of (X): one straight UPRIGHT stroke, ONE teardrop loop hanging from its LOWER-LEFT, and ONE long shallow curve leaving the stroke to the right. Option (b) carries all three exactly so - the stroke is vertical, the single loop hangs on its lower-left and one curve runs off to the right; the square and the small triangle are merely drawn on top of it. Option (a) has the very same shape turned through about ${45}^{\circ }$, and the question forbids rotation. In option (c) the loop hangs on the RIGHT of a slanted stroke, and option (d) has TWO loops instead of one. So (X) is embedded only in option (b).

Letter series Track the three letter positions separately instead of the clusters as a whole. First letters E, F, G, H rise by +1, so the next is I; second letters B, C, D, E rise by +1, so the next is F; third letters F, G, H, I rise by +1, so the next is J. Putting the three together gives IFJ; option (b) has the same letters but in the order I, J, F, which does not match position by position. So the missing term is IFJ - option (d).

Circular seating All seven face the centre, so a person's RIGHT is the anticlockwise direction; build the circle in that direction. Seat I and put J on I's immediate right. Three people between H and I counted from I's right makes H the fourth seat to I's right. Two people between O and J counted from O's right makes J the third seat to O's right, so O sits three seats before J - the fifth seat to I's right. The second and third seats to I's right are still free, and G on N's immediate right fills them as N then G; M takes the last seat, sixth from I. Count to M's right: I is first, J second, N third and G fourth. So G sits fourth to the right of M - option (c).

Number coding Pair each letter with its own code number and add the two. FWOD: F(6)+23 = 29, W(23)+6 = 29, O(15)+14 = 29, D(4)+25 = 29; SINH does the same, 19+10 = 29, 9+20 = 29, 14+15 = 29, 8+21 = 29. So the code of a letter is 29 - (its place in the alphabet). For TYLB: 29-20 = 9, 29-25 = 4, 29-12 = 17, 29-2 = 27. That is 9-4-17-27 - option (c).

Paper folding and punching The bottom half is folded up and then the left half is folded to the right, so the cut is made in the TOP-RIGHT quarter of the sheet. Open the vertical fold first: that quarter is copied to the top-left as a LEFT-RIGHT MIRROR, so its double arrow-head turns round and points right. Open the horizontal fold next: the whole top half is copied downwards as an UP-DOWN FLIP, so in the lower half the bar and the arrow-head sit BELOW the pair of curves. Options (a) and (b) point all four arrow-heads the same way instead of mirroring the left column, and option (d) also leaves the lower half the right way up. Only option (c) points the two left-hand arrow-heads right and the two right-hand ones left, with the lower half turned upside down - option (c).

Venn diagram of classes Bangalore is a city inside India, and India is a country inside Asia. So the three classes are nested one within the other, each one completely inside the next. Three circles drawn one inside the other is exactly option (a). Option (c) offers only two circles for three classes, while (b) and (d) leave part of India outside Asia, which is false. Hence option (a) is correct.

Number series Write the differences between consecutive terms. 155-105 = 50, 196-155 = 41, 228-196 = 32, 251-228 = 23. The differences themselves fall by 9 each time, so the next difference is 23-9 = 14. 251+14 = 265. So the missing term is 265 - option (b).

Number series Take the differences between consecutive terms. 22-18 = 4, 27-22 = 5, 33-27 = 6, 40-33 = 7. The differences rise by 1 each time, so the next one is 8. 40+8 = 48. So the missing term is 48 - option (b).

Rearranging letters The eight letters of ALTERING are A, E, G, I, L, N, R, T; written from Z downwards they become T, R, N, L, I, G, E, A. Now compare the two rows place by place: A/T, L/R, T/N, E/L, R/I, I/G, N/E, G/A. Not a single position holds the same letter in both rows. So the number of letters whose position stays unchanged is zero - option (b).

Symbol substitution Replace P by $\times$, Q by $÷$, R by - and S by +, then simplify each option by BODMAS. Option (b) becomes $\left(160÷10\right)+\left(30\times 2\right)-\left(16\times 4\right)+45$. = 16 + 60 - 64 + 45 = 121 - 64 = 57 The other three simplify to 124, 20 and -59, none of which is 57. So the required expression is option (b).

Paper folding and punching The bottom half is folded up and then the left half is folded to the right, so the cut is made in the TOP-RIGHT quarter of the whole sheet. Opening the vertical fold copies that quarter to the top-left as a LEFT-RIGHT MIRROR: the F is reversed and moves to the other side of the wavy stem. Opening the horizontal fold copies the whole top half downwards as an UP-DOWN FLIP: in the lower half the arc opens upward and the stem rises from it. Option (b) simply repeats one mark four times, option (a) leaves the lower half the right way up, and option (c) keeps the F facing the same way in both columns. Only option (d) carries BOTH changes - a mirrored pair across the top and the same pair turned upside down below - option (d).