18 February 2025 · Shift 3

Embedded figure Figure (X) is a right-angled triangle: its slanting side runs corner to corner, from the bottom-left up to the top-right. The mid-point of that slanting side is joined by three lines - one going right to the vertical side, one going straight down to the base and one going down to the bottom-right corner. In option (b) the diagonal of the square is exactly that slanting side, the base and the right side of the square are the other two sides of the triangle, and the half-horizontal, the half-vertical and the short diagonal all meet at the centre of the square, which is the mid-point of that diagonal. In options (a), (c) and (d) no line at all runs corner to corner across the figure, so the slanting side of (X) cannot be found in any of them. So figure (X) is embedded in option (b).

Letter series with blanks There are five blanks, so break the completed series into five groups of four letters, each group starting with a blank. Read the groups column by column: the 1st letters give T, R, P, N, L; the 2nd give Y, W, U, S, Q; the 3rd give S, Q, O, M, K; the 4th give X, V, T, R, P. Every column steps two places back in the alphabet, so the five groups are TYSX, RWQV, PUOT, NSMR and LQKP. The blanks are the first letter of each group, so in order they are T, W, O, R and P. So the required combination is TWORP, i.e. option (d).

Venn diagram Hands are a part of the body and eyes are a part of the body, so both of these classes lie completely inside the class Body. No hand is an eye and no eye is a hand, so the two smaller classes do not touch each other at all. The picture needed is therefore one big circle carrying two separate smaller circles inside it, which is option (d). Option (a) leaves one class completely outside the body, option (b) makes all three classes overlap one another, and option (c) lets a small class cross the boundary of the big one.

Linear seating Number the seats 1 to 7 from the left end of the row. "Only L sits to the right of N" means exactly one person is to N's right, so N takes seat 6 and L seat 7. Only one person sits between M and N, so M takes seat 4. Only two people sit between T and M, so T is at seat 1 or seat 7; seat 7 is already L, so T takes seat 1. Seats 2, 3 and 5 are left for S, B and A, and S is before B while B is before A, so S sits at 2, B at 3 and A at 5. Between A (seat 5) and S (seat 2) sit the people at seats 3 and 4, that is B and M. So two people sit between A and S, i.e. option (b).

Alphabetical rearrangement The given word is GLORIFY. Its seven letters in English alphabetical order are F, G, I, L, O, R, Y, which gives FGILORY. Now compare the two words place by place: G against F, L against G, O against I, R against L, I against O and F against R - all six differ. Only the seventh place matches, because Y stands there in both words. So exactly 1 letter keeps its position, i.e. option (c).

Sign interchange Rule: wherever $\times$ appears read $÷$ and wherever $÷$ appears read $\times$; wherever + appears read - and wherever - appears read +. So $72+5-42\times 7÷1$ has to be read as $72-5+42÷7\times 1$. By BODMAS do the division and the multiplication first: $42÷7=6$ and $6\times 1=6$. Now work from the left: 72-5=67 and 67+6=73. So the required value is 73, i.e. option (c).

Number series Take the difference of every pair of neighbouring terms: 320-310=10, 331-320=11, 343-331=12 and 356-343=13. The differences themselves are 10, 11, 12, 13 - each one bigger than the last by 1, so the next difference must be 14. Therefore the missing term is 356+14=370. So the answer is 370, i.e. option (d).

Number series Use the gaps that can be seen: 729-719=10, 739-729=10 and 769-759=10. So the series simply adds 10 to each term to get the next one. The missing term is therefore 739+10=749, and it fits on the other side too, because 759-749=10. So the answer is 749, i.e. option (d).

Syllogism All three statements are particular ("some"), and two particular statements can never force a definite conclusion. For (I): the actors who are producers need not be the same actors who are superstars, so no link between superstars and producers is fixed. For (II): "some superstars are actors" and "some superstars are writers" only describe two parts of the superstars, and those two parts may have nobody in common. The arrangement drawn above satisfies all three statements, yet in it no superstar is a producer and no actor is a writer. Since a single such arrangement makes both conclusions false, neither of them follows. So the answer is "Neither conclusion (I) nor (II) follows", i.e. option (b).

Circular seating All seven face the centre, so for each of them the right-hand side is the anticlockwise direction; build the circle by walking to the right. Start at F. J sits immediately to F's right, and counting on from F to the right exactly three people lie between F and H. G sits immediately to L's right, and L and G are the only pair that can fill those three places along with J, so to the right of F the order runs F, J, L, G, H. Counting to the right from K, exactly two people lie between K and J; putting K immediately to the right of H leaves E between K and F, which gives exactly two (E and F) - so the full order to the right is F, J, L, G, H, K, E. Now walk to the right from E: F is 1st, J is 2nd, L is 3rd and G is 4th. So G sits fourth to the right of E, i.e. option (b).

Odd letter-cluster Write the alphabet number of each letter and find the step from the 1st letter to the 2nd and from the 2nd to the 3rd. KAQ: K(11) to A(1) is 10 places back, A(1) to Q(17) is 16 places forward. RHX: R(18) to H(8) is 10 back, H(8) to X(24) is 16 forward. WMC: W(23) to M(13) is 10 back, M(13) to C(3) is 16 forward, counting on past Z to A again. QEU: Q(17) to E(5) is 12 places back - not 10 - even though E(5) to U(21) is the usual 16 forward. So the cluster that breaks the pattern −10, +16 is QEU, i.e. option (d).

Letter-cluster analogy Compare ADGI with BFJM letter by letter: A+1=B, D+2=F, G+3=J and I+4=M. The second pair uses the very same steps: E+1=F, H+2=J, L+3=O and N+4=R. Apply +1, +2, +3, +4 to GKPQ: G+1=H, K+2=M, P+3=S and Q+4=U. So GKPQ is related to HMSU, i.e. option (b).

Number coding Count the letters, not their values: 'DSBQCA' has 6 letters and is coded 4; 'SJTB' has 4 letters and is coded 2. In both cases the code is 2 less than the number of letters: 6-2=4 and 4-2=2. 'XTPLIKAE' has 8 letters, so its code is 8-2=6. So the code is 6, i.e. option (c).

Paper folding and cutting The square sheet is first folded along its horizontal middle line, so the lower half comes up on to the upper half. The half-sheet is then folded along its vertical middle line from right to left, so what is left is the top-left quarter with four thicknesses of paper. Every cut therefore goes through all four thicknesses and must show up four times, once in each quarter of the opened sheet. The small circle is cut at the outer corner of the folded quarter, so on opening it gives one small circle at each of the four corners of the square. The arch is cut in the body of the quarter, so it gives four arches; opening the horizontal fold flips the lower two, so their notch opens upwards while the upper two keep it opening downwards. The opened sheet must therefore carry four corner circles, two arches with the notch opening downwards in the top half and two with the notch opening upwards in the bottom half - that is option (a). Option (b) shows only two circles and a single arch, option (d) shows only two arches instead of four, and option (c) has the arches the wrong way up in each half.

Letter-cluster series Handle the three letters of a term in separate columns. First letters: D, B, Z, X - each is 2 places behind the one before it (D−2=B, B−2=Z counting back past A, Z−2=X), so the next first letter is V. In every term the second letter is the same as the first, so it is V as well. Third letters: I, D, Y, T - each is 5 places behind the one before it (I−5=D, D−5=Y, Y−5=T), so the next third letter is O. So the missing term is VVO, i.e. option (b).

Letter-number analogy Take the letters first: in KB → SK, K+8=S and B+9=K; in BP → JY, B+8=J and P+9=Y. So the first letter always moves 8 places forward and the second letter 9 places forward. Now the numbers: 50-23=27 and 39-12=27, so 27 is added every time. Apply all three steps to MH 31: M+8=U, H+9=Q and 31+27=58. So MH 31 is related to UQ 58, i.e. option (d).

Dice A die carries each number only once, so the 3 on top of the first view and the 3 on top of the second view are the same face. The die has therefore only been turned about its vertical axis; the top face 3 never moved. The faces 1 and 2 in the first view and 4 and 5 in the second view are all side faces, so all four of them touch the 3 and none of them can be opposite it. Only the sixth face, 6, is left for that place, so 6 lies opposite 3. Hence the face opposite 6 carries the number 3, i.e. option (a).

Number coding Count the letters: 'NFTBZ' has 5 letters and is coded 20; 'ZOQ' has 3 letters and is coded 12. In both cases the code is four times the number of letters: $5\times 4=20$ and $3\times 4=12$. 'GZFESITY' has 8 letters, so its code is $8\times 4=32$. So the code is 32, i.e. option (d).

Blood relation coding Read the expression $P+Q\times R÷S-T$ from the left, one sign at a time. P+Q: P is the daughter of Q, so P is a girl and Q is her parent. $Q\times R$: Q is the wife of R, so Q is the mother and R is the father of P. $R÷S$: R is the father of S, so S is another child of R and Q. S-T: S is the brother of T, so T is a child of the same parents R and Q. P and T therefore have the same parents, and P is a girl. So P is the sister of T, i.e. option (c).

Figure series Every box of the series holds exactly two marks: a small triangle and one other symbol beside or below it. Follow the triangle alone: it is at the top in figure 1, at the left in figure 2, at the bottom in figure 3 and at the right in figure 4. That is a quarter turn anticlockwise at every step, so in figure 5 the triangle must come back to the top and the second mark must sit below it. The marks themselves never turn - the triangle points up in all four boxes - so the second mark must also be shown standing upright, and no symbol is used twice in the strip. Only option (d) has the triangle on top with a fresh upright symbol, a small circle, below it. Options (b) and (c) put the triangle at the bottom instead of the top, and option (a) reuses figure 4's pin and lays it on its side.