10 February 2025 · Shift 1

Figure series Compare the two Greek symbols first: in boxes 1 and 3 the symbol μ stands on the left and θ on the right, while in boxes 2 and 4 they are the other way round, so the pair changes sides at every step. Box 5 is again an odd-numbered box, so it must show μ on the left and θ on the right; option (b) has θ on the left and is ruled out. Now the letters: every box carries one capital on top and one small letter below, and no letter is ever used twice - F-k, R-d, D-m, I-b. Option (a) brings back R and k and option (c) brings back R and m, both already used in the strip; only option (d) supplies a fresh pair, J above and a below, with μ on the left and θ on the right. Hence the figure that replaces the question mark is (d).

Letter-cluster analogy Take the first pair letter by letter: R→P, M→K, D→B, H→F - every letter moves 2 steps backward in the alphabet. Check the rule on the second pair: S→Q, U→S, G→E, V→T - again -2 on each letter. Apply the same shift to MXKD: M→K, X→V, K→I, D→B. So MXKD is related to KVIB, and the correct option is (c).

Number coding SKIP = 5269 and SILK = 2576; the two words are built from the same letters except for one letter each. S, K and I occur in both words, and the digits 5, 2 and 6 occur in both codes, so those three digits stand for S, K and I in some order. P is the only letter left over in SKIP and 9 the only digit left over in its code, so P = 9. L is the only letter left over in SILK and 7 the only digit left over in its code, so L = 7, and the correct option is (d).

Parallel-row seating Number the four seats of each row 1 to 4 as we look at the picture. Row 1 faces south, so for those four people 'right' points towards seat 1; row 2 faces north, so for them 'left' points towards seat 1. Only two people sit between L and D, so L and D take the two ends of row 2, i.e. seats 1 and 4. D faces the person second to the left of T, so seat of D = seat of T + 2. Seat 1 would leave no room for T, so D is in seat 4, T in seat 2 and L in seat 1. Only one person sits between O and D, so O takes seat 2, and M is left with seat 3. O faces the person immediately to the right of P, that is the person in seat 2, so P sits in seat 3. Seats 1 and 4 of row 1 are left for E and S, and E must sit to the right of S - towards seat 1 - so E is in seat 1 and S in seat 4. The four end seats are held by E and S in row 1 and by L and D in row 2, and the only option naming two of these four is E and D. Hence the correct option is (c).

Letter series with blanks There are five blanks and twenty places in all, so break the string into five groups of four letters. The groups are _URV | Q_SW | RW_X | SXU_ | TYV_ , and one letter of the option goes into each group. Read the groups column-wise: the first letters run P, Q, R, S, T; the second U, V, W, X, Y; the third R, S, T, U, V; the fourth V, W, X, Y, Z - every column moves one step forward. So the blanks are P, V, T, Y and Z in that order, which is option (a).

Letter-cluster series Treat the three places of the cluster separately. First letters: T → U → V → W, one step forward each time, so the next one is X. Second letters: E → D → C → B, one step backward each time, so the next one is A. Third letters: J → K → L → M, one step forward each time, so the next one is N. The cluster that follows WBM is therefore XAN, and the correct option is (a).

Coding by place values Replace each letter by its place value in the English alphabet and add the four values. A+C+E+G=1+3+5+7=16, which is the code given for ACEG. B+O+Q+T=2+15+17+20=54, so the rule holds for the second word too. U+W+I+F=21+23+9+6=59. So UWIF is coded as 59, and the correct option is (c).

Paper folding and punching The sheet is folded twice: first the lower half is turned up over the upper half, then the left half is turned over on to the right half, so what is left is the top-right quarter. The punch is made in that quarter, so on opening the mark must appear once in each of the four quarters. Opening the vertical fold copies the mark into the top-left quarter as its left-right mirror image. Opening the horizontal fold carries both of those marks down into the lower half as top-bottom mirror images. So the four marks must form mirror pairs across both fold lines, which only option (d) shows.

Venn diagram of classes Every frog is an amphibian but not every amphibian is a frog, so the circle for Frog lies completely inside the circle for Amphibian. A crow is a bird, so no crow is an amphibian and no crow is a frog; its circle must stand entirely apart from the other two. The diagram needed is therefore a small circle inside a bigger one with a third circle outside both, which is option (a).

Number series Write down the gaps between consecutive terms: 86-77=9, 104-86=18, 140-104=36 and 212-140=72. The gaps themselves double every time: $9,18,36,72$, so the next gap must be $72\times 2=144$. Required term =212+144=356. Hence the correct option is (b).

Symbol coding STAIR and TEARS are five-letter words built from almost the same letters: S, T, A and R occur in both. Their codes also share four symbols, and only one symbol of each code is missing from the other. So those four shared symbols stand for S, T, A and R in some order. I is the only letter of STAIR that TEARS does not contain, and the dollar sign is the only symbol of the first code that the second code does not contain. Therefore I is coded as the dollar sign, and the correct option is (d).

Number series The terms grow roughly three times and then twice, so test a multiply-and-adjust rule on the first step: $9\times 3-2=25$. The second step follows the other rule: $25\times 2+3=53$. Check that the two rules alternate: $53\times 3-2=157$ and $157\times 2+3=317$. The next step uses $\times 3-2$ again: $317\times 3-2=951-2=949$. Hence the correct option is (c).

Paper folding and cutting The sheet is folded once, the left half being turned over on to the right half, so the paper is now a double-thick right half. Both cuts are made on the folded edge itself: the small rectangle near the top and the larger rectangle at the middle each have one side lying on the fold. A cut that lies on the fold opens out into a single hole of double the width, centred on the fold line. So the opened sheet carries two rectangular holes, a small one above and a larger one below, both straddling the vertical centre line, which is option (b).

Odd letter-cluster out Measure the two steps inside each cluster with the place values of the letters. KOI: $K\left(11\right)\to O\left(15\right)$ is +4 and $O\left(15\right)\to I\left(9\right)$ is -6. LPJ: $L\left(12\right)\to P\left(16\right)$ is +4 and $P\left(16\right)\to J\left(10\right)$ is -6; OSM: $O\left(15\right)\to S\left(19\right)$ is +4 and $S\left(19\right)\to M\left(13\right)$ is -6. MTW: $M\left(13\right)\to T\left(20\right)$ is +7 and $T\left(20\right)\to W\left(23\right)$ is +3, a completely different pattern. Three clusters share the pattern $+4,-6$ while MTW does not, so the odd one out is option (b).

Alphabetical rearrangement Write the word and then its letters in alphabetical order, one row under the other. GRISTLE rearranged in alphabetical order becomes EGILRST. Comparing place by place, only the third place holds the same letter I in both rows. So exactly one letter keeps its position, and the correct option is (b).

Operations on digits Work through 5163748 digit by digit: an odd digit gains 2 and an even digit loses 2. $5\to 7$, $1\to 3$, $6\to 4$, $3\to 5$, $7\to 9$, $4\to 2$, $8\to 6$. The new number is 7345926, whose highest digit is 9 and lowest digit is 2. Required sum =9+2=11, so the correct option is (c).

Embedded figure Figure (X) is made of three parts: an upright box, a curved cap resting on its top edge, and two straight lines running from the two top corners of the box down to its centre. Since rotation is not allowed, look for exactly that upright capped box with its top corners joined to the middle. Options (a) and (b) are built only of straight shafts and triangular heads, so they hold no capped box at all, and in (d) the cap sits over a free-standing hourglass with no box around it. Option (c) carries the capped box in the middle of the figure, and the upper half of the cross drawn inside it gives the two lines from the top corners to the centre. Hence figure (X) is embedded in option (c).

Syllogism The three statements form one chain: the crows lie inside the rocks, the rocks inside the plants and the plants inside the flowers. Conclusion (I): following the chain, the whole crow circle lies inside the flower circle, so all crows are flowers - it follows. Conclusion (II): all rocks are plants and all plants are flowers, so every rock is a flower; a class lying wholly inside another always gives the 'some' relation back, so some flowers are rocks - it follows. Both conclusions follow, so the correct option is (d).

Coded blood relations Break the chain into its four links and read each sign separately. F+G: F is the mother of G. $G\times H$: G is the wife of H. $H÷I$: H is the father of I. I-J: I is the brother of J. I is a child of H and G, and J is I's brother or sister, so J is a child of H and G as well. That makes G the mother of J. F is the mother of G, and G is the mother of J. So F is the mother of J's mother, and the correct option is (b).

Circular seating All seven face the centre, so for each of them the left hand points clockwise and the right hand anticlockwise. Fix K anywhere and count three seats clockwise to seat G, which leaves two empty seats between K and G. Only K sits between L and N, so L and N are the two immediate neighbours of K. N sits second to the left of H, i.e. two seats clockwise from H. Putting N on the clockwise side of K would push H on to the seat L must take, so N is K's anticlockwise neighbour and L its clockwise neighbour; H then sits two seats anticlockwise from N. Two seats are still empty, and F cannot be next to G, so F takes the seat between H and N while E takes the seat between L and G. Counting from E towards its right, that is anticlockwise, we pass L and then K before reaching N. So two people sit between them, and the correct option is (c).