21 February 2025 · Shift 3

Alphabetical rearrangement Put the letters of DUMPERS in alphabetical order: D, E, M, P, R, S, U. The new letter-cluster is DEMPRSU. Fourth from the left end = P; second from the right end = S. Between P and S the alphabet has only Q and R, so 2 letters lie between them. So the correct option is (a).

Figure series rotation Nothing is added or removed: the pentagon, the circle inside it and the forked stalk stay the same throughout. Only the orientation changes - the stalk points up, then up-right, then right, then down-right, so the figure turns 45° clockwise at every step. One more turn of 45° puts the figure 180° from the first frame: the pentagon on top with its apex pointing up and the stalk hanging below it. The stalk must still end in an OPEN fork whose two arms spread apart - option (d) closes them into an arrowhead and option (b) ends in a straight cross-bar, while option (c) is the first frame unturned. So the correct option is (a).

Letter series The twelve places are built from one block of five letters repeated over and over. The printed letters fix that block: C at place 1, B at place 5 and A, X, B at places 8, 9, 10 fit the block CZAXB. Writing CZAXB / CZAXB / CZ gives C Z A X B C Z A X B C Z, and the Z already printed at place 12 confirms it. Reading only the six blank places, from left to right, gives Z, A, X, C, Z, C. So the correct option is (c).

Letter-cluster series Write each cluster as alphabet positions: BCG = 2, 3, 7; EFJ = 5, 6, 10; HIM = 8, 9, 13; KLP = 11, 12, 16. Inside a cluster the second letter is (first + 1) and the third is (second + 4). From one cluster to the next the first letter jumps by 3: 2, 5, 8, 11, so the next is 14. Next cluster: first =14=N, second =15=O, third =15+4=19=S, i.e. NOS. So the correct option is (a).

Paper folding and punching First fold: the top half comes down on the bottom half, so the horizontal centre line is a crease. Second fold: the right half comes across on to the left half, so the vertical centre line is a second crease. The punch is made through the folded bottom-left quarter, and two folds mean four thicknesses of paper - so the cut shows up once in every quarter, four cuts in all. Each crease acts as a mirror, so every cut is a reflection of its neighbour: in all four the solid bar stays on the OUTER side and the curve faces the centre of the sheet. Options (a) and (c) show only two cuts, and (d) turns the two upper cuts round so their bars point inwards. So the correct option is (b).

Letter-digit coding TIDE → 2986 and KITE → 8216. The two words share the letters T, I and E; the two codes share the digits 2, 8 and 6. So T, I and E use up 2, 8 and 6, and the only letter left in TIDE is D against the only digit left, 9. So the correct option is (a).

Circular seating All seven face the centre, so for each of them the left hand runs clockwise and the right hand anticlockwise. L is immediately to the right of F, so L takes the seat one step anticlockwise from F. Three people between L and H counted from L's left puts H four seats clockwise from L; three people between F and J counted from F's left puts J four seats clockwise from F. E is immediately to the right of K, so E must be one seat anticlockwise of K; of the three seats still empty only one adjacent pair is left, which fixes K and E, and G takes the last seat. Now count from K towards K's right, i.e. anticlockwise: E, then F. Only one person, E, sits between them, so the answer is One. So the correct option is (c).

Letter-number analogy AR 18 → HO 31: A + 7 = H, R − 3 = O and 18 + 13 = 31. KN 05 → RK 18: K + 7 = R, N − 3 = K and 05 + 13 = 18, so the rule holds. Apply it to QL 15: Q + 7 = X, L − 3 = I and 15+13=28. That gives XI 28; option (a) keeps the letters but adds only 12 to the number. So the correct option is (b).

Sign interchange Change every sign first, then calculate - never the other way round. $41+8÷6-24\times 2$ becomes $41-8\times 6+24÷2$. By BODMAS do × and ÷ first: $8\times 6=48$ and $24÷2=12$. 41-48+12=5. So the correct option is (a).

Linear seating Number the six seats 1 to 6 from the left end to the right end of the row. No one sits to the right of F, so F takes seat 6. Three people sit between F and J, so J is three seats away plus one: seat 2. Two people sit between J and G, so G is seat 5 (seat −1 does not exist). H sits second to the left of I, and only seats 1, 3 and 4 are free; H = 1 with I = 3 is the only fit, so S takes seat 4. Between H (seat 1) and S (seat 4) sit J and I - two people. So the correct option is (a).

Mirror image MN is a vertical mirror standing beside the figure, so the image is the figure turned left-to-right; top stays top. Check the figure itself: the two corner diagonals, the horizontal bar, the V and the three lines running down to the base are all placed symmetrically about its own vertical centre line. A left-to-right flip therefore leaves it looking exactly as it was. So the image must keep every line of the original - and the deciding detail is the count of lines reaching the base: two side verticals AND the short centre vertical below the V. Option (a) drops that centre vertical, option (c) drops the two side verticals and option (b) moves the fan to the top and adds a line from the top edge. So the correct option is (d).

Odd letter-cluster Compare the two steps inside each cluster, using alphabet positions. LOI: 12 → 15 is +3, and 15 → 9 is −6. PSM: 16 → 19 is +3, and 19 → 13 is −6. RUO: 18 → 21 is +3, and 21 → 15 is −6. TWS: 20 → 23 is +3, but 23 → 19 is only −4. The second step breaks the pattern only in TWS, so it is the odd one out. So the correct option is (d).

Word-number coding The code has nothing to do with which letters are used - only with how many there are. TNRA has 4 letters and its code is 28, and $4\times 7=28$. LOPGVO has 6 letters and its code is 42, and $6\times 7=42$, so the rule is (number of letters) × 7. JDSHKQB has 7 letters, so its code is $7\times 7=49$. So the correct option is (d).

Number series Series: 54, 55, 58, 63, 70, ? Take the differences: 55-54=1, 58-55=3, 63-58=5, 70-63=7. They are the consecutive odd numbers 1, 3, 5, 7, so the next difference is 9. Missing term =70+9=79. So the correct option is (d).

Venn diagram of classes Every cheetah is a wild animal, so the class Cheetah lies wholly inside the class Wild animals - one circle completely within another. A cow is a domestic animal, so it is neither a cheetah nor a wild animal; its circle must touch neither of the other two. So the diagram needs exactly one small circle inside a big one, plus a third circle standing apart. Option (a) puts the Cow circle inside Wild animals as well, (c) lets Cow and Cheetah overlap, and (d) leaves Cheetah outside Wild animals. So the correct option is (b).

Syllogism All three statements are of the form 'All X are rugs', so each of sacks, mugs and pugs sits inside rugs. Nothing at all is said about how sacks, mugs and pugs stand to one another. They can therefore be drawn as three separate circles inside rugs, and that picture breaks both conclusions at once. Since a conclusion must hold in EVERY possible picture, 'Some pugs are sacks' does not follow and neither does 'Some mugs are pugs'. So the correct option is (c).

Number series Series: 53, 63, 76, 92, 111, ? Differences: 63-53=10, 76-63=13, 92-76=16, 111-92=19. The differences themselves rise by 3 each time, so the next one is 19+3=22. Missing term =111+22=133. So the correct option is (b).

Coded blood relations Break the chain P − Q + R × S ÷ T into its four links and read each one with the code. P − Q: P is the brother of Q. Q + R: Q is the daughter of R. R × S: R is the wife of S. S ÷ T: S is the father of T. So R and S are a married couple; Q is their daughter, and P, being Q's brother, is their son. S is also the father of T, so T is another child of the same couple. P is male and shares his parents with T, so P is the brother of T. So the correct option is (b).

Letter-number analogy MH 45 → KF 40: M − 2 = K, H − 2 = F and 45 − 5 = 40. TN 35 → RL 30: T − 2 = R, N − 2 = L and 35 − 5 = 30, so both letters fall by 2 and the number by 5. Apply it to ZI 25: Z − 2 = X, I − 2 = G and 25-5=20. That gives XG 20; option (c) moves only the first letter and leaves the number untouched. So the correct option is (b).

Paper folding and cutting First fold brings the top half down, so the horizontal centre line is a crease; the second fold brings the left half across, so the vertical centre line is a second crease. Both cuts are made through the folded bottom-right quarter, and two folds give four thicknesses, so each cut reappears once in every quarter of the opened sheet. The small square sits in the OUTER corner of that quarter, so the four squares land in the four corners of the sheet. The circle sits near the INNER corner of the quarter, i.e. beside the point where the two creases cross, so the four circles gather round the centre of the sheet. Only the option that keeps the squares in the corners and the circles round the centre - four of each - can be right. So the correct option is (a).