4 February 2025 · Shift 3

Number series Write down the difference between one term and the next. The differences are 86-65=21, 114-86=28, 149-114=35 and 191-149=42. Each difference is 7 more than the one before it, so the next difference is 42+7=49. Required term =191+49=240.

Number series Take the difference between consecutive terms. 11-10=1, 13-11=2, 17-13=4, 25-17=8 - every difference is double the previous one. So the next difference is $8\times 2=16$. Required term =25+16=41.

Jumbled letter coding The digits are not written in the order of the letters, so compare the two words as sets. BIGS and SING share the letters I, G and S; their codes 4593 and 9432 share the digits 4, 9 and 3. So B, the letter found only in BIGS, must be the left-over digit 5, and N, the letter found only in SING, must be the left-over digit 2. Hence the code for N is 2.

Letter-cluster analogy Compare the clusters letter by letter, using the position of each letter in the alphabet. M+4=Q, L-4=H, T+4=X, A-4=W, so the rule is +4, -4, +4, -4. It checks out on the second pair: L+4=P, N-4=J, V+4=Z, B-4=X, which is PJZX. Applying it to CWON: C+4=G, W-4=S, O+4=S, N-4=J. Hence CWON is related to GSSJ.

Sign interchange First rewrite every sign the way the question orders. So $8+63-7\times 7÷9$ becomes $8\times 63÷7+7-9$. By BODMAS, do the division and the multiplication first: $8\times 63=504$ and $504÷7=72$. Now 72+7-9=79-9=70. Hence ?=70.

Odd letter-cluster Measure the gap from the first letter to the second, then from the second to the third, counting round the alphabet after Z. NYJ: N+11=Y and Y+11=J. TEP: T+11=E and E+11=P. CNY: C+11=N and N+11=Y. QZK: Q+9=Z, not +11, although Z+11=K is right - so its first gap breaks the pattern. Hence QZK does not belong to the group.

Figure series Count the arrow-arcs in each figure and note which way the arrowheads run. The first figure is a complete circle of 6 arcs; after that one arc is rubbed out every step, giving 5, 4 and 3, so the fifth figure must have 2 arcs. The turning also alternates - clockwise, anticlockwise, clockwise, anticlockwise - so the fifth figure turns clockwise again. Only option (b) shows 2 arcs whose arrowheads run clockwise.

Letter-cluster analogy Compare the two clusters letter by letter. N+2=P, R-5=M, D+2=F, F-5=A, so the rule is +2, -5, +2, -5. It checks out on SFJW: S+2=U, F-5=A, J+2=L, W-5=R, which is UALR. Applying it to GVKQ: G+2=I, V-5=Q, K+2=M, Q-5=L. Hence GVKQ is related to IQML.

Paper folding and punching The first fold brings the top half down and the second brings the left half onto the right, so the punches were made in the bottom-right quarter of the sheet. Unfolding undoes the folds in reverse: first mirror that quarter about the vertical centre line, then mirror the whole bottom half about the horizontal centre line. Each punch therefore appears four times, symmetric about both centre lines. The four cross-shaped punches sit one in each quarter towards the outer corners, and the four solid triangles hang from the top and the bottom edge beside the vertical centre line. Only option (d) shows exactly this double mirror image.

Syllogism Draw the three statements together. All anklets are socks and all socks are flats, so the anklets circle lies completely inside flats - conclusion (I) follows. The statement only says that some flats are trousers; those flats may all lie outside socks, so no trouser need be an anklet - conclusion (II) does not follow. Hence only conclusion (I) follows.

Coded blood relation Read the chain one link at a time. C - D: C is the daughter of D. D ÷ E: D is the brother of E. E × F: E is the son of F. F + G: F is the mother of G. E is F's son and D is E's brother, so D is also a child of F; therefore F is the mother of D. C is the daughter of D, and D is male, so D is the father of C. F is thus the mother of C's father, i.e. F is the father's mother of C.

Letter-cluster series Treat the three positions of the cluster separately. First letters: T, U, V, W - each one step forward, so the next is X. Second letters: K, J, I, H - each one step back, so the next is G. Third letters: F, G, H, I - each one step forward, so the next is J. Hence the missing cluster is XGJ.

Embedded figure Trace the outline of the given figure: a wide inverted U whose two arms each step outwards once near the bottom. In option (c) the top edge of the inner square gives the top of the U, the two inner vertical lines give its arms, the upper line of the middle band gives the two outward steps, and the side edges of the inner square give the two short outer arms. Every part of the outline is found in the same size and the same position, with no turning. Hence the given figure is embedded in option (c).

Letter to number coding Compare each letter with its own number, using the position of the letter in the alphabet. T is the 20th letter and its code is 8, and 20+8=28; K is the 11th letter and its code is 17, and 11+17=28. So the rule is code =28- (position of the letter). Hence HAWP is coded as 20-27-5-12.

Letter series completion There are twenty places, and the letters that are already printed repeat in blocks of five. The 4th, 5th, 6th, 7th and 8th places read u, o, p, q, t, so the block is p q t u o, and the 14th to 17th places u, o, p, q confirm it. Filling every place from the block and reading only the blanks from left to right gives q, t, u, o, p, t, t, o. Hence the series is completed by qtuoptto.

Linear seating Number the seats 1 to 7 from the left. No one sits to the right of M, so M takes seat 7. Three people sit between M and K, so K is in seat 3; two people sit between K and H, so H is in seat 6 (seat 0 does not exist). G is third to the left of L and J is immediately to the right of L; only L in seat 4 fits, which puts G in seat 1 and J in seat 5. I takes the only seat left, seat 2. Between I (seat 2) and H (seat 6) sit K, L and J - three people.

Word rearrangement VARIABLE has the eight letters V, A, R, I, A, B, L, E. Arranged in reverse alphabetical order (Z towards A) they read V, R, L, I, E, B, A, A. Comparing place by place, only the 1st (V), the 4th (I) and the 6th (B) letter stay where they were. Hence the position of three letters remains unchanged.

Dice - opposite faces List the faces that can be seen in each of the two positions. The first position shows A touching N and S; the second shows A touching I and T. So A touches four different faces - N, S, I and T - and every face of a cube touches exactly four faces. The only face left, K, must therefore be the face opposite A. Hence the face opposite K is A.

Circular seating They all face the centre, so a person's right hand points anticlockwise round the table. Fix F. S sits second to F's right, and F sits second to G's right, which places G two seats to F's left. Only F sits between R and E, so R and E are F's two neighbours; E is next to S, so E takes the seat on the S side and R the other one, leaving T for the last seat. Counting from R towards R's right, that is anticlockwise, we pass F, E and S before reaching T. Hence three people sit between R and T.

Venn diagram of classes Cotton and denim are two different cloths - one piece of cloth is not both - so those two classes stay apart. A shirt, however, can be made of cotton and a shirt can be made of denim, so the shirt class cuts across each of them. The correct picture therefore has two separate circles with a third circle overlapping both of them in the middle. Option (d) shows exactly this arrangement.