5 February 2025 · Shift 1

Syllogism Draw the three statements together: every lyricist lies inside the singers, the writers cut the singers only partly, and the writers stay completely away from the musicians. Conclusion (I): nothing in the statements keeps the musicians away from the singers, so a musician is free to be a lyricist as well; (I) is only possible, not certain. Conclusion (II): the singers who are writers need not be the singers who are lyricists, so writers and lyricists may have nobody in common; (II) is also only possible. Since neither conclusion is forced by the statements, the answer is option (c).

Venn diagram of classes Take the classes two at a time. Onion is a kind of vegetable, so the whole class Onion lies inside the class Vegetable. A shirt is neither an onion nor a vegetable, so Shirt stays completely apart from both. The required picture therefore shows one small circle drawn inside a larger circle, and one separate circle outside them. That is option (a).

Circular seating All seven face the centre, so for each person the seat on his right is the next seat anticlockwise and the seat on his left is the next seat clockwise. Start with G immediately to the right of N; moving right (anticlockwise) from N past three people places L in the fifth seat. Counting two people to the right of H must land on G, which fixes H two seats clockwise from N. Only one pair of adjacent empty seats is left, so K immediately right of E fits there and F takes the last seat. Now count from the left of F, i.e. clockwise: H, L, K and then E. Three people sit between F and E, so the answer is option (c).

Letter-to-number coding Rule: replace every letter by its opposite letter, add the four positions and double the sum. The opposite of the letter at position p stands at position 27-p. GPZJ: 27-7=20, 27-16=11, 27-26=1, 27-10=17, so the sum is 49 and $49\times 2=98$. OWRS: 27-15=12, 27-23=4, 27-18=9, 27-19=8, so the sum is 33 and $33\times 2=66$. VIXE: V, I, X, E stand at 22, 9, 24, 5, so their opposites stand at 5, 18, 3, 22. The sum is 5+18+3+22=48, and $48\times 2=96$. Hence VIXE is coded as 96, so the answer is option (c).

Letter-cluster series Treat the first, second and third letters of the clusters as three separate series. First letters B, D, ?, H move +2 each step, so the missing first letter is F. Second letters M, N, ?, P move +1 each step, so the missing second letter is O. Third letters Y, W, ?, S move -2 each step, so the missing third letter is U. The missing cluster is FOU, so the answer is option (d).

Odd letter-cluster out Write the alphabet position of each letter and read the two gaps inside every cluster. VSW gives 22, 19, 23, that is -3 and then +4. WTX gives 23, 20, 24 and KHL gives 11, 8, 12, again -3 and then +4. NLO gives 14, 12, 15, that is -2 and then +3, which is a different pair of gaps. So NLO does not belong to the group, and the answer is option (b).

Paper folding and cutting The first fold lifts the bottom half onto the top half, so the horizontal centre line of the sheet is a fold line. The second fold lays the right half onto the left half, so the vertical centre line is the second fold line. The cuts are made in the top-left quarter: a small circle touching the top edge, a bowl-shaped cut below it and another small circle lower down. Opening the second fold copies that column across the vertical line, so two identical columns appear. Opening the first fold copies both columns across the horizontal line, and a bowl reflected downward becomes a dome. Each column now reads circle, bowl, circle, circle, dome, circle from top to bottom, and the two columns are mirror images about the vertical line. This is exactly option (d).

Dice - opposite faces Rule: two faces that are seen together in one view can never be opposite each other. In the first position A is seen with I and E, so A touches I and E. In the second position A is seen with Y and R, so A touches Y and R as well. A therefore touches four faces - I, E, Y and R - and the only letter left over is P. So A and P must be on opposite faces. The face opposite P is A, so the answer is option (b).

Symbol interchange After the interchange every + becomes $÷$, every $÷$ becomes +, every - becomes $\times$ and every $\times$ becomes -. So $16-2\times 104+4÷20$ turns into $16\times 2-104÷4+20$. By BODMAS do the division and the multiplication first: $16\times 2=32$ and $104÷4=26$. The expression becomes 32-26+20. 32-26=6 and 6+20=26. Hence the question mark stands for 26, so the answer is option (c).

Figure series The only thing that changes from box to box is the number of crosses, so count them. The four given boxes hold 15, 14, 12 and 9 crosses. The falls are 1, then 2, then 3, so the next fall must be 4. The fifth box therefore holds 9-4=5 crosses. Option (a) has six, option (b) has seven and option (c) has four; only option (d) shows exactly five. So the answer is option (d).

Missing letters series The series has 20 places and each option supplies 9 letters, so look for one block repeated four times. The first five places read C R _ S _, which can only be the word CROSS. Writing CROSS four times under the series matches every letter that is already printed. Reading only the blank positions from left to right gives O, S, C, S, R, O, C, O, S. That is OSCSROCOS, so the answer is option (a).

Alphabetical rearrangement OPERATOR is made of the letters O, P, E, R, A, T, O, R. Arranged in alphabetical order they read A, E, O, O, P, R, R, T. Now compare the two words place by place: O with A, P with E, E with O, R with O, A with P, T with R, O with R and R with T. Not a single place carries the same letter in both words. Hence zero letters keep their position, so the answer is option (a).

Letter-cluster analogy Compare the two clusters of each pair letter by letter. CGJK to IMPQ: C+6=I, G+6=M, J+6=P and K+6=Q. HLQR to NRWX: H+6=N, L+6=R, Q+6=W and R+6=X, so the rule is +6 on every letter. Applying +6 to AEMO: A+6=G, E+6=K, M+6=S and O+6=U. So AEMO is related to GKSU, and the answer is option (c).

Coded blood relations Break the chain into its three links and read each one with the given code. E * L means E is the husband of L, so E and L are a married couple. L # B means L is the mother of B, so B is a child of L and therefore also a child of E. B @ P means B is the father of P, and only a male can be a father, so B is male. A male child of E is his son, so B is the son of E, and the answer is option (d).

Number series The terms fall steadily, so take the difference between consecutive terms. 173-151=22, 151-130=21, 130-110=20 and 110-91=19. The number that is subtracted drops by 1 each time, so the next one to subtract is 18. Hence the missing term is 91-18=73, so the answer is option (c).

Letter-cluster analogy Compare the clusters letter by letter, counting backwards and wrapping round from A to Z when needed. FROM to WJFE: F moves back 9 to W, R moves back 8 to J, O moves back 9 to F and M moves back 8 to E. VLEG to MDVY: V back 9 is M, L back 8 is D, E back 9 is V and G back 8 is Y, the same -9, -8, -9, -8 rule. Applying it to UNDI: U back 9 is L, N back 8 is F, D back 9 is U and I back 8 is A. So UNDI is related to LFUA, and the answer is option (b).

Number series The terms rise steadily, so take the difference between consecutive terms. 676-666=10, 687-676=11, 699-687=12 and 712-699=13. The number that is added grows by 1 each time, so the next one to add is 14. Hence the missing term is 712+14=726, so the answer is option (d).

Letter-digit coding The digits of a code are jumbled, so treat each code as a set of four digits rather than a letter-by-letter match. BASE uses the digits 8, 9, 7, 0 and ASET uses the digits 0, 6, 9, 7. The letters A, S and E occur in both words, so the three digits common to both codes, namely 9, 7 and 0, must belong to A, S and E. The digit left over in the code of BASE is 8, so 8 stands for B, the extra letter of BASE. The digit left over in the code of ASET is 6, so 6 stands for T, the extra letter of ASET. Hence the code for T is 6, so the answer is option (b).

Circular seating All five face the centre, so the person sitting to someone's right occupies the next seat anticlockwise. Put Anjali first; Hansraj sits to her right and Roshan sits to the right of Hansraj, which fixes three seats in a row. Panjiri must touch Roshan, so Panjiri takes the next seat, and the last seat goes to Kunti. Kunti now sits between Panjiri and Anjali, so the neighbours of Anjali are Kunti and Hansraj. Roshan sits between Hansraj and Panjiri, and Panjiri sits between Roshan and Kunti, so options (b), (c) and (d) are all false. Only statement (a) fits the arrangement, so the answer is option (a).

Embedded figure Figure (X) is a crescent: a strongly curved outer arc and a flatter inner arc that meet at a point above and a point below, with the hollow facing left. So look for an option in which one circle is cut by the arc of a second, flatter circle, leaving a sliver of exactly that shape. In option (a) the plain large circle is overlapped on its left by the spoked circle, and the part of the large circle left outside the spoked one is precisely the crescent (X), in the same position and without any rotation. Option (b) gives a ring, option (c) gives a bowl with straight lines and option (d) gives a dome cut by a band, and none of them contains the crescent. So the answer is option (a).