25 February 2025 · Shift 2

Blood relation coding Read 'T < I / M : E ! S' one link at a time, using the code given. 'T < I' means T is the son of I, and 'I / M' means I is the father of M, so T and M are children of the same father - T is M's brother. 'M : E' means M is the wife of E, so E is the husband of T's sister M. 'E ! S' means E is the brother of S. Now start from S: E is S's brother, M is that brother's wife, and T is that wife's brother. Hence T is S's brother's wife's brother, so the correct option is (b).

Odd letter-cluster In each cluster compare the first letter with the second and then with the third. UWT, OQN and MOL all follow the same rule: second letter = first +2, third letter = first -1. In GHF the second letter is only one step ahead (G+1=H), not two, so the second step is broken. Therefore GHF does not belong to the group and the correct option is (c).

Paper folding and punching Always open the folds in the reverse order in which they were made. The last fold brought the left half onto the right half, so opening it puts a second punch in the top-left quarter - a left-right mirror image of the first. The first fold brought the bottom half up onto the top half, so opening that copies both punches into the bottom half as top-bottom mirror images. The answer must therefore carry four punches whose left pair mirrors the right pair AND whose lower pair mirrors the upper pair. Option (a) shows only two punches, option (b) repeats the top row unchanged at the bottom (no mirroring about the horizontal fold) and option (c) repeats the left punch unchanged on the right (no mirroring about the vertical fold). Only option (d) is mirrored about both fold lines, so the correct option is (d).

Parallel row seating Number the four seats of each row 1 to 4 from the reader's left. 'T sits second to the right of A' gives A-1 and T-3, or A-2 and T-4. 'Only two people sit between A and P' needs A and P at the two ends, so A takes seat 1, T seat 3, P seat 4 and S the remaining seat 2. 'K sits second to the left of R' gives R-1 and K-3, or R-2 and K-4. O faces S, so O must take seat 2 of row 1; that rules out R-2, leaving R-1, K-3 and W-4. The ends are R and W in row 1 and A and P in row 2; the only pair offered that is made of two end seats is R and P. Hence the correct option is (a).

Letter-cluster analogy Compare the two given pairs letter by letter. ZLER to BJBT: Z+2=B, L-2=J, E-3=B, R+2=T; NDSG to PBPI repeats exactly the same shifts. Apply +2, -2, -3, +2 to YIHB: Y+2=A, I-2=G, H-3=E, B+2=D. This gives AGED, so the correct option is (d).

Number series Write down the difference between consecutive terms. 45-45=0, 46-45=1, 50-46=4, 59-50=9. The differences are $0^2$, $1^2$, $2^2$, $3^2$, so the next difference is $4^2=16$. Missing term =59+16=75, so the correct option is (d).

Word coding Compare the two statements and find what they share. 'words' is the only word common to both sentences, and 'kj' is the only code common to both codes. A word that appears in both must carry the code that appears in both, so 'words' is coded 'kj'. Hence the correct option is (b).

Syllogism All three statements are of the form 'All P are cloth', so dress, shirts and ties are three groups lying inside cloth. Nothing in the statements forces any two of those three groups to overlap - they can be drawn completely apart inside cloth. Conclusion (I) 'Some ties are dress' therefore need not be true, and conclusion (II) 'Some shirts are tie' need not be true either. A universal affirmative shared parent never guarantees an overlap between its children, so neither conclusion follows and the correct option is (a).

Embedded figure Figure (X) is one arrow pointing up, one arrow leaving the middle of its stem towards the left, and one arrow leaving the foot of its stem towards the right. Because rotation is not allowed, the answer must contain those three arrows in exactly those directions and joined at exactly those two points. In option (a) the upward arrow, the left arrow from the middle of its stem and the right arrow from its foot are all present, unturned and unbroken. Options (b), (c) and (d) are built only of plain line segments with no arrowheads at all, so figure (X) cannot sit inside any of them. Hence the correct option is (a).

Letter series Every term is three consecutive letters, so only the first letter of each term matters. The first letters are A(1), F(6), L(12), S(19), whose gaps are +5, +6, +7. The next gap is +8, so the first letter is 19+8=27; the alphabet has 26 letters, so 27-26=1, that is A. The term is therefore ABC and the correct option is (a).

Letter series with blanks Twenty places and five blanks split the series into five groups of four letters. The groups are _GIM, K_JN, LI_O, MJL_ and NKM_. Reading down each column - G, H, I, J, K and I, J, K, L, M and M, N, O, P, Q - every letter is one step ahead of the letter above it, and the first column must run J, K, L, M, N. So the blanks are J, H, K, P and Q in that order. That combination is JHKPQ, so the correct option is (d).

Sign interchange First rewrite the expression with the signs interchanged. $11-9÷2+16\times 8$ becomes $11+9\times 2-16÷8$. By BODMAS do division and multiplication first: $9\times 2=18$ and $16÷8=2$. =11+18-2 =29-2=27 Hence the correct option is (a).

Paper folding and punching Two holes are punched in the folded quarter, which is the bottom-right quarter of the sheet. Opening the last fold (left half onto right half) mirrors both holes across the vertical centre line, giving 4 holes in the bottom half. Opening the first fold (top half down onto the bottom half) mirrors all four across the horizontal centre line, giving 8 holes in all. So the answer must show 8 holes in two columns and four rows, symmetric about both centre lines. Option (a) has 3 holes and option (b) has 4, both in a single column, and option (d) has only 4 holes placed unsymmetrically. Only option (c) shows the required 8 holes, so the correct option is (c).

Figure series Count the straight lines that make up each figure of the series. The open chevron uses 2 lines, the triangle 3 lines and the square 4 lines. The count rises by one each time, so the missing figure must be drawn with exactly 5 straight lines. Option (a) is a square whose bottom side is replaced by two slanting lines - two sides, the top and two slants, that is exactly 5 lines. Options (c) and (d) are a trapezium and a parallelogram, 4 lines each - the same count as the previous figure - and option (b) needs 6 lines (a five-line star plus the separate line beneath it). Hence the correct option is (a).

Number series The differences 9, 8, 26, 8 have no pattern, so look for two operations used turn by turn. $5\times 2+4=14$, then 14+8=22, then $22\times 2+4=48$, then 48+8=56. The next step is the '$\times 2+4$' step: $56\times 2+4=112+4=116$. Hence the correct option is (a).

Letter-cluster coding Add the alphabetical positions of the four letters of each cluster. LQAB: 12+17+1+2=32, which is the given code. CMGP: 3+13+7+16=39, again the given code, so the rule is confirmed. SJZC: 19+10+26+3=58. Hence the correct option is (a).

Linear seating All six face north, so their left is the reader's left; number the seats 1 to 6 from the left. 'No one sits to the right of C' puts C in seat 6, and 'only two people sit between C and B' puts B in seat 3. B is immediately right of L, so L takes seat 2, and B is immediately left of M, so M takes seat 4. 'Only two people sit between A and M' now leaves seat 1 for A, and K takes the one seat left, seat 5. A, L, B and M all sit to the left of K, that is four people. Hence the correct option is (c).

Letter-cluster analogy Compare each given pair letter by letter, keeping the order of the letters. MEGA to OFIB: M+2=O, E+1=F, G+2=I, A+1=B; UIOE to WJQF uses the same +2, +1, +2, +1. Apply it to ALUH: A+2=C, L+1=M, U+2=W, H+1=I, giving CMWI. Option (a) CMIW and options (c) and (d) hold the same four letters but in a shuffled order, which the rule never allows. Hence the correct option is (b).

Venn diagram of classes Every consonant is a letter of the alphabet and every vowel is a letter of the alphabet, so both lie completely inside 'Alphabet'. No letter is a consonant and a vowel at the same time, so those two circles must not touch each other. The diagram needed is therefore one large circle holding two smaller circles that are separate from each other. Option (a) makes all three overlap, option (b) leaves one class outside the alphabet, and option (d) puts the vowels inside the consonants. Hence the correct option is (c).

Alphabetical rearrangement Arrange the letters of DAUGHTER in English alphabetical order: A, D, E, G, H, R, T, U. Now set the new word ADEGHRTU under the original word and compare position by position. Only the 4th letter G and the 5th letter H stand in the same place in both words. So two positions remain unchanged and the correct option is (c).