22 November 2025 · Shift 1

Time and work Let the number of workers hired at the start be x. Total work = 30x worker-days. Work done in the first 10 days = 10x, so the work still left = 30x - 10x = 20x. After the 10 workers leave, (x - 10) workers finish this remaining work, and they take 30 - 10 + 5 = 25 days. So 25(x - 10) = 20x 25x - 250 = 20x, i.e. 5x = 250, so x = 50 Hence, option (d) is the correct answer.

Direction sense Draw the three moves one after the other from the starting point. The 7 km walked east is exactly undone by the 7 km walked west, so he ends up on the same north-south line on which he started. Only the 15 km walked north is left over, so the finishing point lies 15 km due north of the start. Hence, option (c) is the correct answer.

Direction sense Plot the three moves in order from the starting point. The 12 km walked west and the 12 km walked east are equal and opposite, so they cancel out completely. Only the 12 km walked south survives, so the required distance = 12 km. Hence, option (a) is the correct answer.

Word analogy In the first pair the second word is what the first thing gives to the person who uses it. A mirror gives back a reflection; in exactly the same way a book gives the reader knowledge. 'Words' and 'page' are only parts of a book, and the 'author' is the one who makes the book, not what the book gives. Hence, option (c) is the correct answer.

Number analogy Compare the two numbers of the given pair. $27\times 2=54$ and 54 - 1 = 53 So the rule is: second number = (first number $\times 2)-1$. Applying it to 32: $32\times 2-1=64-1=63$ Hence, option (c) is the correct answer.

Letter cluster analogy Read the first word from its last letter to its first: G-E-R-M-A-N-Y reversed is Y-N-A-M-R-E-G, which is exactly the second word. So the code simply writes the word backwards. BRAZIL reversed letter by letter gives L-I-Z-A-R-B, that is LIZARB. Hence, option (a) is the correct answer.

Number series Take the difference between each pair of consecutive terms. 120 - 20 = 100, 240 - 120 = 120, 380 - 240 = 140 The differences themselves grow by 20 each time, so the next difference must be 140 + 20 = 160. Required term = 380 + 160 = 540 Hence, option (a) is the correct answer.

Number series Use the two differences that can be found at once: 63 - 47 = 16 and 135 - 107 = 28. Between them lie two more differences, and 16, 20, 24, 28 rise by 4 each time. So the missing term = 63 + 20 = 83. Check: 83 + 24 = 107 and 107 + 28 = 135, which agrees with the series. Hence, option (d) is the correct answer.

Alphanumeric series Handle the letters and the numbers as two separate series. The 1st, 3rd and 5th letters run forward from the beginning of the alphabet - C, D, E; the 2nd and 4th run backward from the end - X, W. The fifth term is an odd-numbered term, so its letter is E. The numbers 2, 3, 5, 7 are consecutive prime numbers, so the next number is 11. The missing term is therefore E11. Hence, option (a) is the correct answer.

Matching the exact string Compare each option with the reference 'Washington: July 4th, 1776.' character by character. Option (a) inserts an extra full stop and writes '4.th'; option (c) spells the name as 'Washingtan'; option (d) writes the year as 1767 instead of 1776. Option (b) repeats every letter, digit, space and punctuation mark of the reference without a single change. Hence, option (b) is the correct answer.

Number grid In each row, multiply the first three numbers and then add 5 to that product. Row 1: $5\times 3\times 4=60$ and 60 + 5 = 65 Row 2: $3\times 7\times 8=168$ and 168 + 5 = 173 Row 3: $7\times 5\times 3=105$, so ? = 105 + 5 = 110 Hence, option (b) is the correct answer.

Number grid In each row, add the first three numbers and multiply the sum by 3 to get the fourth number. Row 1: $\left(8+7+2\right)\times 3=17\times 3=51$ Row 3: $\left(6+3+1\right)\times 3=10\times 3=30$ Row 2: $60÷3=20$, so 9 + 5 + ? = 20, which gives ? = 6 Hence, option (c) is the correct answer.

Counting triangles Name the points of the figure as shown; the two ovals and the two rhombuses carry no triangle at all, so only the straight lines matter. Take the triangles group by group so that none is counted twice. With the apex A as a vertex: ABC, ABD, ACD, AEF, APQ, ABM and the eight smaller ones that the median AD cuts out with PQ, EF, HK, EL and BM - 14 in all. With B as a vertex but not A: BEL, BCM and the three pieces that AD and EL cut from them - 5 in all. Lying wholly inside the square EFGH and in the strip below it: FGK, HKM, EHW, GHY, DLT and six more formed where EL, BM, HK and AD cross - 11 in all. Total number of triangles = 14 + 5 + 11 = 30 Hence, option (c) is the correct answer.

Counting triangles Break the figure into the four blocks it is built from and count each block separately. A square with both of its diagonals drawn always holds 8 triangles - 4 quarters and 4 halves - so the big square gives 8. The tilted square on the right also has both of its diagonals drawn (the right side of the big square and the horizontal line to the far corner), so it gives 8 more. The triangle standing on top has its median drawn down to the centre; with the two diagonals of the square it accounts for 7 more. The two crossing lines in the bottom-left corner add 3 more. Total = 8 + 8 + 7 + 3 = 26 Hence, option (c) is the correct answer.

Direction sense Each right turn is a quarter turn, so four right turns make $4\times {90}^{\circ }={360}^{\circ }$ and leave the direction unchanged. Six right turns $=6\times {90}^{\circ }={540}^{\circ }={360}^{\circ }+{180}^{\circ }$, so their whole effect is a half turn. A half turn simply reverses the direction, so the final direction is exactly opposite to the starting one. He finally faces east, and the opposite of east is west, so he started walking towards the west. Hence, option (d) is the correct answer.

Direction sense A turn to the right swings a walker ${90}^{\circ }$ clockwise and a turn to the left swings him ${90}^{\circ }$ anticlockwise. 8 turns to the right $=8\times {90}^{\circ }={720}^{\circ }$ clockwise. 8 turns to the left $=8\times {90}^{\circ }={720}^{\circ }$ anticlockwise. Net turning $={720}^{\circ }-{720}^{\circ }=0^{\circ }$, and since rotations simply add up, the order of the turns does not matter. So the final facing is the same as the starting facing, which was south. Hence, option (c) is the most appropriate answer.

Figure completion The question figure is a square cut into four quarters by its centre lines, and the quarter that has been removed is the bottom-right one. Read what the three surviving quarters contain: two rectangles drawn concentrically about the centre — a tall narrow one and a short wide one. Each rectangle is symmetrical about the centre, so the missing quarter must carry the right-hand side and the right half of the lower edge of both of them, giving two nested corners. Above the horizontal centre line a complete X is drawn inside the tall rectangle; below it only the left half of a second X is visible. The missing quarter must therefore also supply that second X's right half — a wedge like '<' whose tip touches the centre of the square. Only figure (a) has both nested corners together with the wedge opening towards the right; (b) turns the wedge the other way round, (c) pushes the two corners up towards the top edge and (d) stretches one arm of the wedge to the far corner. Hence, option (a) is the most appropriate answer.

Figure completion Here the piece cut away (shown by the dashed lines) is the top-left quarter of the square. The three quarters that remain hold two rectangles drawn concentrically about the centre, one large and one small. By that symmetry the missing quarter must show the top-left corner of the large rectangle and, closer to the centre, the top-left corner of the small one — two nested corners and nothing else along the edges. The small rectangle also carries both of its diagonals. Three of the four half-diagonals are already drawn, so the fourth one, running from the small rectangle's top-left corner down to the centre, has to be supplied too. Figure (c) shows exactly two nested corners plus that single short diagonal; (a) adds a stray diagonal at the outer corner, (b) draws the diagonal right across the tile from corner to corner and (d) has three nested corners instead of two. Hence, option (c) is the most appropriate answer.

Figure analogy Compare the first pair. The rectangle itself does not change, and none of the four small elements moves to a new corner. The only change is that every element crosses the boundary of the main figure: the triangle and the plus go from inside to outside, while the circle and the pentagon go from outside to inside. So the rule is: keep each element at its own corner and send it to the other side of the boundary. Apply this to the third figure, where the star and the hexagon lie inside the ellipse and the diamond and the square lie outside it. The answer figure must therefore show the star outside at the top-left, the diamond inside at the top-right, the square inside at the bottom-left and the hexagon outside at the bottom-right. Option (b) leaves the diamond and the square outside, option (a) leaves the diamond sitting on the ellipse instead of fully inside it, and option (d) keeps the star inside. Hence, option (c) is the correct answer.

Figure series — rotation Every square is cut into eight triangles by its two diagonals and its two centre lines, and the same four symbols — J, R, 7 and T — occupy four of those triangles. Track one symbol at a time. R stands in the upper right triangle in figure 1, in the lower bottom-right triangle in figure 2, in the lower left triangle in figure 3 and in the upper top-left triangle in figure 4. That is a move of two triangles clockwise at every step, i.e. the whole square is turned through ${90}^{\circ }$ clockwise each time; the other three symbols move in exactly the same way. So figure 2 is figure 1 turned through ${90}^{\circ }$, figure 3 through ${180}^{\circ }$ and figure 4 through ${270}^{\circ }$. One more quarter turn completes ${360}^{\circ }$, so the fifth figure must be figure 1 itself: J in the upper left triangle, R in the upper right, 7 in the lower left and T at the bottom, every symbol standing the right way up. Option (b) is that figure. (a) and (c) turn the T upside down, (c) also reverses the R, and (d) replaces the J by an L. Hence, option (b) is the most appropriate answer.

Figure series — rotation Each box holds the same four objects: an arrow with a white head, a half-shaded hexagon, a half-shaded pentagon and an arrow with a white tail. Follow the hexagon. It is at the top right of box 1, at the top left of box 2 and at the bottom left of box 3 — one quarter turn anticlockwise about the centre of the box at every step. The other three objects move the same way, and each object is itself turned through ${90}^{\circ }$ anticlockwise, so the shaded half swings round with it and every arrow points a quarter turn further round. Box 3 is therefore box 1 turned through ${180}^{\circ }$, and the missing box is box 1 turned through ${270}^{\circ }$ anticlockwise, which is the same as turning box 1 through ${90}^{\circ }$ clockwise. That places the pentagon at the top left pointing right with its shaded half below, the white-headed arrow at the top right pointing right, the arrow with the white tail at the bottom left pointing left and the hexagon at the bottom right with its shaded half on top. Hence, option (b) is the most appropriate answer.

Odd one out — letter-number Test the two halves of each pair separately: first how the letters move, then how the number changes. (a) $B\to G$ and $K\to P$, each letter +5; the number goes $650\to 980$, i.e. +330. (b) $D\to I$ and $N\to S$, each letter +5; the number goes $720\to 1050$, i.e. +330. (c) $P\to U$ and $R\to W$, each letter +5; the number goes $810\to 1140$, i.e. +330. (d) $F\to K$ and $L\to Q$, each letter +5; but the number goes $540\to 920$, i.e. +380. The letter rule holds everywhere, so it cannot decide the question; the number rule +330 is broken only in (d). Hence, option (d) is the most appropriate answer.

Letter-number coding Replace each letter by its opposite letter (A↔Z, B↔Y, C↔X, …), whose position is $27-\left(\mathrm{position}\mathrm{of}\mathrm{the}\mathrm{letter}\right)$, add those four values and then add the number of letters in the word. RING: $R\to 27-18=9$, $I\to 27-9=18$, $N\to 27-14=13$, $G\to 27-7=20$. Sum = 9+18+13+20 = 60, and 60+4 = 64, which is the given code. BOOK: 25+12+12+16 = 65, and 65+4 = 69, again the given code. FAIL: $F\to 27-6=21$, $A\to 27-1=26$, $I\to 27-9=18$, $L\to 27-12=15$. Sum = 21+26+18+15 = 80, so the code is 80+4 = 84. Hence, option (b) is the most appropriate answer.

Letter coding Line the two words up letter by letter and treat consonants and vowels by separate rules. Consonants move two places back: $T\left(20\right)\to R\left(18\right)$, $G\left(7\right)\to E\left(5\right)$, $R\left(18\right)\to P\left(16\right)$. Vowels are replaced by the next vowel in the order A, E, I, O, U: $I\to O$ and $E\to I$. Now apply the same rules to LION: $L\left(12\right)\to J\left(10\right)$, $I\to O$, $O\to U$, $N\left(14\right)\to L\left(12\right)$. So LION is written as JOUL. Hence, option (a) is the most appropriate answer.

Word formation from letters LANTERN is built from the letters A, E, L, N, N, R, T — the letter N occurs twice and every other letter once. A word can be formed from an option only if that option supplies each of these letters at least as many times as LANTERN needs it. RENTALN, ENTRALN and NERTALN are all made of exactly those seven letters in a different order, so LANTERN can be formed from each of them. ANTLELN, however, is made of A, E, L, L, N, N, T — it carries a second L but no R at all, and LANTERN cannot be written without an R. Hence, option (b) is the most appropriate answer.