6 February 2025 · Shift 1

Letter series The series has twenty places, so break it into five groups of four letters. The groups are _NJO, K_LQ, MR_S, OTP_ and QVR_. Each completed group is the previous group with every letter moved 2 places forward, so INJO gives KPLQ, KPLQ gives MRNS, and so on. Filling the blanks that way gives INJO, KPLQ, MRNS, OTPU and QVRW. The letters that were taken out, read in order, are I, P, N, U, W. So the required combination is IPNUW and the correct option is (b).

Reverse alphabetical order TREASURE is made of the letters T, R, E, A, S, U, R, E. Arranged in reverse English alphabetical order (Z first) they become U, T, S, R, R, E, E, A. Now compare the two rows place by place: T-U, R-T, E-S, A-R, S-R, U-E, R-E, E-A. No place carries the same letter in both rows. So zero letters keep their position and the correct option is (d).

Linear seating arrangement Number the seats 1 to 7 from the left; everyone faces north, so left and right are the reader's left and right. No one sits to the left of H, so H takes seat 1. Only three people sit to the right of N, so N takes seat 4. Four people sit between H and F, so F takes seat 6. Seats 2, 3, 5 and 7 are left, and K must sit immediately left of L; the only such free pair is 2 and 3, so K is at 2 and L is at 3. G is not a neighbour of N (seat 4), so G cannot take seat 5; G takes seat 7 and E takes seat 5. Counting four seats to the left of G at seat 7 reaches seat 3, which is L. So the correct option is (a).

Coded language Write the two statements one under the other and look for what they share. The only word that appears in both statements is 'can'. The only code that appears in both code groups is 'pr'. A word and its code must match, so 'can' is coded as 'pr'. So the correct option is (b).

Number series Take the difference of each pair of consecutive terms. 31-12=19, 59-31=28, 114-59=55, 214-114=100. Now take the difference of those differences: 28-19=9, 55-28=27, 100-55=45. These grow by 18 each time, so the next one is 45+18=63. Hence the next difference of the series is 100+63=163. Required term =214+163=377. So the correct option is (c).

Linear seating arrangement Number the seats 1 to 7 from the left. No one sits to the right of N, so N takes seat 7. Three people sit between N and S, so S takes seat 3. Two people sit between S and T, so T takes seat 6 (seat 0 does not exist). Seats 1, 2, 4 and 5 are free. M is third to the left of L and A is immediately right of L, which only works as M at 1, L at 4 and A at 5. B takes the one seat left, seat 2. Between B at seat 2 and T at seat 6 lie seats 3, 4 and 5, that is S, L and A. So three people sit between them and the correct option is (b).

Blood relations P is the sister of R, so P is a female of R's own generation. R is the father of Q, so Q is one generation below R. S is the wife of Q, and T is the daughter of S, so T is the daughter of Q as well. Trace the line upwards from T: T's father is Q, and Q's father is R. R's sister is P, so P is the sister of T's father's father. That is exactly 'father's father's sister', so the correct option is (b).

Letter-number analogy Deal with the two letters and the number separately. In ER 32 → DS 34 the first letter moves one place back (E → D), the second letter one place forward (R → S) and the number grows by 2. The second pair obeys the same rule: X → W, W → X and 54+2=56. Apply it to ZB 78: Z goes back to Y, B goes forward to C, and 78+2=80. So ZB 78 is related to YC 80 and the correct option is (a).

Dice In both positions 10 is on the top face, so the dice has only been turned about its vertical axis. That means 10 stays on top, the bottom face stays at the bottom, and the four side faces simply move one step round a ring. Position I gives 11 on the left and 12 on the right; position II gives 12 on the left and 8 on the right. So 12 has moved from the right to the left, and the ring of side faces runs 11, 12, 8 and then the fourth side face. In a ring of four faces, opposite faces are two steps apart, so 11 is opposite 8. So the correct option is (a).

Venn diagram Compare the three classes two at a time. Some grapes are green and some green things are not grapes, so Grapes and Green are two circles that cut each other. A table is never a grape, so the class Table stays completely outside the class Grapes. The picture needed is therefore two overlapping circles together with a third circle drawn apart from them. Only option (c) shows that, so the correct option is (c).

Syllogism All tops are bags, so the circle of tops lies wholly inside the circle of bags. Some bags are cats, so the cats circle only has to meet the part of bags that lies outside tops - it can be kept away from tops altogether. No cat is a dog, so the dogs circle must stay clear of cats. Nothing else is said about dogs, so it may overlap tops and it need not lie inside bags at all. Conclusion (I) says no top is a dog; the picture shows a dog that is also a top, so it does not follow. Conclusion (II) says some cats are tops; in the picture cats and tops do not meet, so it does not follow either. One picture defeats both conclusions, so the correct option is (a).

Dice Read the two positions face by face. Position I has L on top with E and G on the two visible sides, so E touches both L and G. Position II has E on top with I and N on the two visible sides, so E also touches I and N. E therefore touches L, G, I and N, which is four of the other five faces. Two faces that touch each other can never be opposite, so the only face left for E to face is B. Hence the face opposite B is E and the correct option is (b).

Figure series The letters only ever sit at the eight positions round the border of the square: top-centre, top-left, middle-left, bottom-left, bottom-centre, bottom-right, middle-right, top-right. Read them anticlockwise, that is in the order top-centre, top-left, middle-left, bottom-left, bottom-centre, bottom-right, middle-right, top-right. Figure 1 gives R, figure 2 gives R A, figure 3 gives M R A and figure 4 gives M R A T, so one new letter joins the block each time. The block also slides one position anticlockwise at every step: it starts at top-centre, then top-left, then middle-left, then bottom-left. Figure 5 therefore holds five letters, L M R A T, starting one position further on, at bottom-centre. That puts L at bottom-centre, M at bottom-right, R at middle-right, A at top-right and T at top-centre. That is exactly option (a), so the correct option is (a).

Odd letter-cluster Write the place value of each letter and look at the gaps inside a cluster. JMS: J is 10, M is 13, S is 19, so the gaps are +3 and +6. EHN: E is 5, H is 8, N is 14, so the gaps are again +3 and +6. KNT: K is 11, N is 14, T is 20, so the gaps are again +3 and +6. WXQ: W is 23, X is 24, Q is 17, so the gaps are +1 and -7. WXQ is the only cluster that does not follow the +3, +6 pattern. So the correct option is (a).

Embedded figure Figure (X) is a tall upright rectangle joined by one straight horizontal line to a triangle whose apex points to the left. Rotation is not allowed, so the option must contain that rectangle, that line and a left-pointing triangle in exactly that arrangement. Options (a) and (d) are built only from rectangles and connecting lines; they contain no triangle at all. Option (b) does contain triangles, but every one of them points to the right, upwards or downwards. In option (c) the tall rectangle on the left sends out a middle horizontal line that ends exactly at the apex of a left-pointing triangle, which is figure (X). So the correct option is (c).

Number series The terms fall, so subtract each term from the one before it. 673-670=3, 670-664=6, 664-655=9, 655-643=12. The numbers taken away are 3, 6, 9, 12, which rise by 3 each time, so the next one is 15. Required term =643-15=628. So the correct option is (c).

Letter series Write the place values of every group: CEA is 3, 5, 1; GIE is 7, 9, 5; KMI is 11, 13, 9; OQM is 15, 17, 13. Comparing the same position in consecutive groups, every value grows by 4. So the next group is 15+4=19, 17+4=21, 13+4=17. The letters with those place values are S, U and Q. The required term is SUQ, so the correct option is (a).

Coding by place value Replace every letter by its opposite letter in the alphabet (A and Z, B and Y, and so on) and add up the place values of the opposites. ZIYJ becomes A, R, B, Q, and 1+18+2+17=38, which is the given code. PXBO becomes K, C, Y, L, and 11+3+25+12=51, which is again the given code. VTUC becomes E, G, F, X, so the code is 5+7+6+24=42. So the correct option is (b).

Letter coding Compare CLICHE with IHJAMF letter by letter: C to I is +6, L to H is -4, I to J is +1, C to A is -2, H to M is +5, E to F is +1. KINDLY to QEOBQZ obeys exactly the same list of shifts: K+6 = Q, I-4 = E, N+1 = O, D-2 = B, L+5 = Q, Y+1 = Z. Now apply +6, -4, +1, -2, +5, +1 to FUMBLE. F+6 = L, U-4 = Q, M+1 = N, B-2 = Z (counting back past A returns to Z), L+5 = Q, E+1 = F. So FUMBLE is written as LQNZQF and the correct option is (c).

Symbol interchange Swap the two signs everywhere in the expression: every $÷$ becomes $\times$ and every $\times$ becomes $÷$. $11÷4+3\times 3-1$ therefore becomes $11\times 4+3÷3-1$. By BODMAS do the division and the multiplication first: $11\times 4=44$ and $3÷3=1$. The expression is now 44+1-1. So the value is 44 and the correct option is (c).