25 February 2025 · Shift 1

Coded blood relation Break the chain $F÷G+H-I\times J$ at every symbol and read each pair on its own. $F÷G$: F is the sister of G. G+H: G is the son of H. H-I: H is the wife of I. $I\times J$: I is the brother of J. H is the wife of I, so I is the husband of H; G is H's son, hence I's son too. F is G's sister, so F is a daughter of the same couple H and I. I is therefore the father of F, and option (d) is correct.

Number series Take the difference of every pair of neighbours: 257-127=130, 407-257=150, 577-407=170, 767-577=190. The differences are 130, 150, 170, 190 - each one is 20 more than the one before it. So the next difference must be 190+20=210. Required term =767+210=977, which is option (a).

Figure series rotation Every box carries the same four things - the letters D and G, a solid star and a hollow triangle - at the top, left, right and bottom. Track the star: box 1 left, box 2 top, box 3 right, box 4 bottom. Left → top → right → bottom is a quarter-turn clockwise each time. D does the same (top → right → bottom → left), and so do G and the triangle, so the whole set turns 90° clockwise at every step. Box 4 holds G top, D left, triangle right, star bottom; one more quarter-turn clockwise sends G to the right, D to the top, the triangle to the bottom and the star to the left. The missing box therefore has D on top, the star on the LEFT, G on the right and the triangle at the BOTTOM - only option (b) does this; option (a) keeps D and G in place but interchanges the star and the triangle.

Letter-cluster analogy Write the alphabet positions of the first pair: $N\left(14\right)\to E\left(5\right)$, $P\left(16\right)\to G\left(7\right)$, $L\left(12\right)\to C\left(3\right)$, $I\left(9\right)\to Z\left(26\right)$. Each letter drops by 9 places; for I, 9-9=0, and on a circular alphabet position 0 means Z. The second pair obeys the same rule: $T\left(20\right)\to K\left(11\right)$, $V\left(22\right)\to M\left(13\right)$, $R\left(18\right)\to I\left(9\right)$, $O\left(15\right)\to F\left(6\right)$. Apply -9 to CEAX: $C\left(3\right)\to 3-9=-6\to 20=T$, $E\left(5\right)\to 22=V$, $A\left(1\right)\to 18=R$, $X\left(24\right)\to 15=O$. CEAX is related to TVRO, so option (d) is correct.

Circular seating All seven face the centre, so for each person LEFT is the clockwise direction and RIGHT is the anticlockwise direction. Three people between I and L counting from I's left puts L in the 4th seat clockwise from I; three people between K and H counting from H's right puts K in the 4th seat anticlockwise from H. G touches both H and L, so G sits between them; taking the seats clockwise as I, _, _, _, L this forces G next to L and H next to G. That fixes H, and counting four seats anticlockwise from H gives K; C then takes the seat immediately anticlockwise of K and D fills the last seat. Counting to D's right (anticlockwise) we meet K, then C, then I - so I sits third to the right of D, option (b).

Syllogism Draw the statements exactly as they stand: rupee lies wholly inside money, paise cuts across rupee, and water is drawn completely away from money. Conclusion (II) 'Some money is water' contradicts the third statement outright - 'No money is water' leaves money and water with nothing in common, so (II) does not follow. Conclusion (I) 'Some rupee is water' is no better: every rupee is money, and no money touches water, so no rupee can be water either. A universal negative can never produce a positive conclusion, so neither follows and option (c) is correct.

Number-letter coding '2590' → 'FACT' and '0129' → 'FITC'. The code letters are not in the order of the digits, so compare the two lines instead of matching them position by position. The digits 2, 9 and 0 occur in both numbers, and the letters F, C and T occur in both codes. So those three digits use up F, C and T between them, and whatever digit is left over in a number must take the letter left over in its code. In '2590' the odd digit is 5 and the odd letter is A, so $5\to A$; in '0129' the odd digit is 1 and the odd letter is I. Hence 1 is coded as I, and option (b) is correct.

Letter-cluster analogy First pair: $S\left(19\right)\to K\left(11\right)$, $O\left(15\right)\to G\left(7\right)$, $M\left(13\right)\to E\left(5\right)$, $R\left(18\right)\to J\left(10\right)$ - every letter falls 8 places. Second pair confirms it: $M\left(13\right)\to E\left(5\right)$, $I\left(9\right)\to A\left(1\right)$, $G\left(7\right)\to Y\left(25\right)$ since 7-8=-1 wraps round, and $L\left(12\right)\to D\left(4\right)$. Now shift OKIN back by 8: $O\left(15\right)\to 7=G$, $K\left(11\right)\to 3=C$, $I\left(9\right)\to 1=A$, $N\left(14\right)\to 6=F$. The result is GCAF, so option (d) is correct; option (a) GCFA has the last two letters interchanged.

Odd letter-cluster Replace each letter by its position and look at the two steps inside every cluster. NKI: 14, 11, 9 gives -3 then -2. AXV: 1, 24, 22 gives -3 then -2, because A steps back to X on a circular alphabet. VSQ: 22, 19, 17 gives -3 then -2. CAX: 3, 1, 24 gives -2 then -3 - the same two steps, but taken in the reverse order. So CAX is the odd one out and option (d) is correct.

Mirror image MN is a vertical mirror standing to the right of the figure, so the image is the figure turned over left-to-right; top stays top and bottom stays bottom. List what the figure contains: the outer square, both mid-lines, the diamond joining the four mid-points, and ONE corner-to-corner diagonal running from the top-left corner to the bottom-right corner. The square, the two mid-lines and the diamond are each already symmetric about a vertical line, so a left-right flip leaves all of them exactly as they were. Only the single diagonal moves: top-left-bottom-right becomes top-right-bottom-left, which puts the crossings in the top-right and bottom-left quarters instead of the top-left and bottom-right ones. Option (a) is the only figure whose lone diagonal runs top-right to bottom-left; (b) leaves it top-left to bottom-right (an unflipped copy), (c) has no corner diagonal at all and (d) draws both diagonals but drops the diamond.

Letter series Read the first letters down the series: A, C, E, G, ?, K - each is two places after the one before it. The second letters do the same (B, D, F, H, ?, L) and so do the third letters (C, E, G, I, ?, M). So the missing cluster begins two places after G, that is at I, and each cluster is simply three consecutive letters. The missing term is therefore IJK, and it is indeed followed by KLM, so option (a) is correct.

Letter series blanks The string has 20 places and five blanks, so break it into five groups of four: _EIM / J_KO / LI_Q / NKO_ / PMQ_. Now read the groups column by column. Column 2 is E, _, I, K, M and column 3 is I, K, _, O, Q - both rise by two letters a step. Column 1 is _, J, L, N, P, so it must start at H, and column 4 is M, O, Q, _, _, so its last two entries are S and U. Filling every column at the rate of +2 gives HEIM, JGKO, LIMQ, NKOS, PMQU. Reading the blanks in order gives H, G, M, S, U, so option (c) is correct.

Paper folding and cutting Fold 1 brings the top half down on the bottom half, so the first crease is the horizontal centre line of the sheet. Fold 2 brings the right half across to the left, so the second crease is the vertical centre line; the packet is now the bottom-left quarter, four layers thick. The cut takes a small triangle off the packet's top-right corner - and that corner is exactly the centre of the whole sheet. Opening fold 2 mirrors that triangle across the vertical centre line; opening fold 1 mirrors both triangles across the horizontal centre line. The four triangles all meet at the centre and join into a single diamond-shaped hole there, which is option (a); (b) shows an upright square, (c) puts two holes in opposite corners and (d) shows two diamonds away from the centre.

Number series Take the differences: 52-28=24, 82-52=30, 118-82=36, 160-118=42. These differences are 24, 30, 36, 42 - they themselves grow by 6 each time. So the next difference is 42+6=48. Required term =160+48=208, which is option (c).

Paper folding and punching Fold 1 brings the right half onto the left half and fold 2 brings the bottom half up onto the top half, so the packet is a quarter of the sheet, four layers thick. One punch through four layers must therefore leave four marks - one in each quarter of the opened sheet, which already rules out option (d) with its two marks. Opening the second fold reflects the mark across the horizontal centre line, so the lower marks are the upper ones turned over top-to-bottom. Opening the first fold reflects both marks across the vertical centre line, so the right-hand marks are the left-hand ones turned over left-to-right. Only option (a) shows both mirrorings - the open side of the curve faces inward on each side; (b) repeats the same mark left and right without flipping it, and (c) repeats it unchanged in all four quarters.

Circular seating All six face the centre, so LEFT means the clockwise direction and RIGHT means the anticlockwise direction. One person between X and D counting from D's left puts X two seats clockwise from D; 'Z sits second to the left of C' puts Z two seats clockwise from C. Z must also touch both E and D, so with the seats taken clockwise the only consistent order is C, E, Z, D, Y, X. Check it: from D the seats clockwise are Y then X, so exactly one person (Y) sits between X and D; from C they are E then Z, so Z is second to C's left; Z's neighbours are E and D. Counting to Y's right, that is anticlockwise, we pass D and reach Z - exactly one person sits between them, so option (b) is correct.

Venn diagram of classes Take the three classes two at a time and ask whether any member of one can also be a member of the other. A car is a machine, a horse is an animal and chocolate is a food: no car is a horse, no horse is chocolate and no chocolate is a car. None of the three is contained in another either, so there is no room for an overlap or for one circle inside another. The three classes must therefore be drawn as three circles that neither touch nor overlap - option (d). Options (a) and (b) make the classes share members, and (c) puts one class wholly inside another.

Symbol substitution (BODMAS) Replace every letter by the sign it stands for. The equation becomes $\left(59-38\right)\times 4+\left(12+8\right)÷2+42-26$. Brackets first: 59-38=21 and 12+8=20, giving $21\times 4+20÷2+42-26$. Then multiplication and division: $21\times 4=84$ and $20÷2=10$, giving 84+10+42-26. Finally addition and subtraction: 84+10=94, 94+42=136, 136-26=110. So the question mark stands for 110, option (c).

Letters between two letters EXCITED is made of E, X, C, I, T, E, D; sorted alphabetically these become C, D, E, E, I, T, X. The second letter from the left end is D and the first letter from the right end is X. D is the 4th letter of the alphabet and X the 24th, so the count strictly between them is 24-4-1=19. The answer is 19, option (d).

Letter-number coding 'GALS' → '7610' and 'LADS' → '0168'. The digits are jumbled, so compare the two words instead of reading them position by position. The letters A, L and S appear in both words, and the digits 6, 1 and 0 appear in both codes. So A, L and S use up 6, 1 and 0 between them, and the odd letter of each word must take the odd digit of its code. In GALS the odd letter is G and the odd digit is 7, so $G\to 7$; in LADS the odd letter is D and the odd digit is 8. Hence D is coded as 8, option (b).