19 November 2025 · Shift 1

Ages: sum and difference Let the age of B be x years; then A, being 15 years older, is x+15 years. Their sum is 45 years, so x+(x+15)=45. 2x+15=45 2x=30, hence x=15. Check: B is 15 years and A is 15+15=30 years, and 15+30=45. So B is 15 years old and option (a) is correct.

Simple interest Formula: $\mathrm{S.I.}=\frac{P\times R\times T}{100}$ Here the principal P=6000, the rate $R=10\%$ per annum and the time T=4 years. $\mathrm{S.I.}=\frac{6000\times 10\times 4}{100}$ $=\frac{240000}{100}=2400$ Hence the simple interest is ₹2,400 and option (b) is correct.

Direction sense: turns Start facing north and walk 7 km. A right turn always moves the face one quarter-turn clockwise, so from north he now faces east; he walks 5 km in that direction. A second right turn takes him one more quarter-turn clockwise, from east to south, and the last 7 km are walked facing south. Walking does not change the direction faced, so he is finally facing south. Hence, option (b) is correct.

Direction sense: distance Take the starting point as the origin: 20 km east, then 20 km north, then 20 km west. The 20 km walked east and the 20 km walked west are equal and opposite, so together they bring him back to the vertical line through the start. Only the 20 km walked north is left over, so the finishing point lies straight north of the starting point. Therefore the required distance is 20 km. Hence, option (c) is correct.

Decision making at work A mistake in a report that has already reached the client can damage the client's trust, so the first duty is to limit the damage honestly and quickly. Informing the senior at once and offering a correction does exactly that: the error is owned up to, and a remedy is placed on the table before the client discovers it. Ignoring the mistake or waiting for the client to point it out only lets a small error grow into a serious loss of credibility. Shifting the blame on to a colleague is unethical and does not correct the report at all. Hence, option (c) is correct.

Letter analogy Write the alphabet positions: P = 16, R = 18, S = 19 and C = 3, E = 5, F = 6. Each letter has moved back 13 places: 16-13=3, 18-13=5, 19-13=6. Apply the same rule to H = 8, S = 19, T = 20, treating the alphabet as a cycle of 26. 8-13=-5, and -5+26=21, which is U; 19-13=6 is F; 20-13=7 is G. So HST becomes UFG, and option (b) is correct.

Vowel analogy All the letters involved are vowels, so read them in the vowel order A, E, I, O, U. In A : E, the second letter is simply the vowel that comes immediately after the first. The vowel immediately after O in the same series is U, so O : U completes the analogy. Hence, option (b) is correct.

Number series in blocks Read the long digit string in blocks of four digits: 2002, 2012, 2022. Each block is 10 more than the block before it: 2012-2002=10 and 2022-2012=10. So the next block must be 2022+10=2032. Hence, option (b) is correct.

Exact replication check Compare each option with the reference 'Mexico City: 14th February, 1929 CE.' character by character. In (b) the era is written 'C.E.' with two full stops, while the reference has plain 'CE'. In (c) the city is spelt 'Mexico Citi' instead of 'Mexico City'. In (d) the month is spelt 'Februry' — the letter 'a' of 'February' is missing. Only (a) agrees with the reference in every letter, number and punctuation mark, so option (a) is correct.

Exact replication check Compare each option with the reference 'Accra (PG): Jun 5, 1952.' character by character. In (b) the city name carries a stray underscore — 'Accr_a' instead of 'Accra'. In (c) the month is written 'Jun.' with a full stop, which the reference does not have. In (d) the year has been altered from 1952 to 1972. Option (a) alone repeats the city, the code in brackets, the month, the date and the year exactly as given, so option (a) is correct.

Embedded figure First fix the shape to be found: a four-sided figure with a sharp peak near the top, a long side falling from the peak to a point on the right, a long base running back to the far left and a short side closing it on the left. Since rotation is not allowed, this quadrilateral must sit in the answer figure at exactly the same tilt. In option (a) those four strokes are all present and in that very position — the sharp peak, the long right side, the long base sloping down to the left and the short left side. In options (b), (c) and (d) the corresponding inner quadrilateral is far more even-sided: the peak is blunt and the long sloping base is missing, so the given shape cannot be traced in them without turning it. Hence, option (a) is correct.

Embedded figure Break the question figure into its six strokes: the short upper-left edge rising to the peak, the long edge from the peak down to the right point, the long line from the left point down to that same right point, the long edge from the peak down to the bottom-left corner, the vertical from the peak to the base, and the short base itself. The fingerprint of the figure is the crossing: the line from the left point and the line from the peak to the bottom-left corner must cut each other on the way, with the vertical running the full height between them. Option (c) carries all six strokes in exactly that position, so the whole figure can be traced on it without any turning. In option (a) the stroke from the peak stops on the left edge and never reaches the bottom-left corner; in option (b) the vertical stops half-way down and does not touch the base; in option (d) there is no line running from the left point across to the right point. Hence, option (c) is correct.

Number grid pattern Work down each column: the first three numbers must produce the fourth. First column: $15\times 3\times 6=270$ and $270\times 10=2700$. Second column: $7\times 9\times 5=315$ and $315\times 10=3150$. So the rule is (first) $\times$ (second) $\times$ (third) $\times 10$. Third column: $9\times 6\times 5=270$, hence $270\times 10=2700$. Hence, option (c) is correct.

Counting triangles Count the triangles that stand on the upper horizontal line first; five points lie on it — the two outer corners, the two lower vertices of the top petal and the point where the vertical crosses it. The centre is joined to all five of those points, so choosing any two of them gives a triangle: ${}^5{C}_2=10$ triangles. The top tip is joined to three of those points, which gives ${}^3{C}_2=3$ more triangles. Two further triangles use the vertical as a side: the top tip, the centre and one lower vertex of the petal, on the left and on the right. So the upper half yields 10+3+2=15 triangles, and by symmetry the lower half yields 15 more. The left tip forms 2 triangles with the centre — one with each of the two left corners — and the right tip forms 2 in the same way. Total =15+15+2+2=34. Hence, option (d) is correct.

Counting quadrilaterals The figure is one big rectangle cut by two full horizontal lines and three part-length vertical lines. A rectangle is fixed by a top line, a bottom line and two verticals that both run the whole way between them, so count the verticals available for each pair of horizontals. Top strip: the two sides and the short vertical inside it give 3 verticals, so ${}^3{C}_2=3$ rectangles. Middle strip: the two sides and the two verticals inside it give 4, so ${}^4{C}_2=6$ rectangles. Bottom strip: the two sides and the vertical that runs through it give 3, so ${}^3{C}_2=3$ rectangles. Taller ones: top + middle has only the two sides, giving 1; middle + bottom has 3 verticals, giving 3; the whole figure gives 1. Total =3+6+3+1+3+1=17. Hence, option (c) is correct.

Direction sense — turns A right turn is a quarter turn, that is a turn of ${90}^{\circ }$ in the clockwise sense. She starts facing East; the first right turn makes her face South, the second West and the third North. The fourth right turn brings her back to East, because $4\times {90}^{\circ }={360}^{\circ }$ is one full turn. Any smaller number of right turns leaves her facing South, West or North, so 4 is the least number that works. Hence, option (a) is correct.

Completing the pattern The big square is cut into four quarters and the three completed quarters carry the very same motif, each time turned over like a mirror image. In the top-left quarter the long line runs from the outer corner of the square to the centre, a short line cuts off the corner that touches the vertical middle line, and the small circle sits on the far side of the long line, near the outer edge. The bottom-left quarter is the top-left quarter turned over about the horizontal middle line, and the bottom-right quarter is the bottom-left quarter turned over about the vertical middle line — that is why the two short lines meet in a V at the bottom. So the missing top-right quarter must be the top-left quarter turned over about the vertical middle line: the long line from the top-right corner of the square down to the centre, the short line cutting the corner at the top of the middle line, and the circle in the lower-right part. Option (a) is exactly that; in (b) an extra parallel line appears, in (c) the circle lies on the wrong side of the long line, and in (d) the long line stops short of the corner. Hence, option (a) is correct.

Figure analogy — water image Compare the first two figures: the round dot that was at the upper end of the slanting line is at the lower end afterwards, while the diamond has moved from the lower end to the upper end. Nothing has changed from left to right — only top and bottom — so the second figure is the first figure turned over about a horizontal line, that is its water image. The shading agrees: the shaded half lies below the slanting line in the first figure and above it in the second. Turning the third figure over in the same way makes the triangle point downwards with its dotted tip at the bottom, lifts the stalk and its dot above the (now upper) base, and moves the band of stripes into the lower half. That is option (b); (a) and (c) leave the triangle standing upright and (d) pushes the stalk away from the middle. Hence, option (b) is correct.

Figure analogy — water image In the first pair the circle that stood at the top is at the bottom afterwards and the pentagon that stood at the bottom is at the top, while the left and right positions do not change at all. The triangle also turns over — it points up in the first figure and down in the second — so the second figure is the water image of the first, its mirror image in a horizontal line. Apply the same to the third figure: the '=' at the bottom rises to the top, and the '+' stays on the left, because a mirror in a horizontal line never moves anything sideways. The 'p' at the top turns over into a 'b' at the bottom, and the 'g' on the right turns over into a '6' on the right. So the answer must show '=' at the top, '+' on the left, '6' on the right and 'b' at the bottom. Hence, option (d) is correct.

Figure analogy — shapes swapped Read the four cells of figure A: a triangle inside a square, a diamond inside a hexagon, a circle inside a pentagon and a triangle inside a circle. Figure B holds the very same four pairs, but in each pair the inner shape has become the outer one and the outer shape has become the inner one — the triangle-inside-a-square becomes a square-inside-a-triangle. Each pair also crosses over to the diagonally opposite cell: the top-left pair of A appears at the bottom-right of B, and the top-right pair of A appears at the bottom-left of B. Figure C holds a circle inside a diamond (top-left), a square inside a circle (top-right), a triangle inside a hexagon (bottom-left) and a diamond inside a pentagon (bottom-right). Swapping each pair and sending it to the opposite cell gives a pentagon inside a diamond at the top-left, a hexagon inside a triangle at the top-right, a circle inside a square at the bottom-left and a diamond inside a circle at the bottom-right. Hence, option (a) is correct.

Figure series — marks rotating The frame of lines is exactly the same in every figure; only the four marks sitting at its four ends change place. Follow the ringed dot: it is at the bottom-left, then the top-left, then the top-right, then the bottom-right — it moves one corner clockwise each time. One more clockwise step brings it back to the bottom-left, so the missing figure must carry the ringed dot at the bottom-left; that rules out options (a) and (c). Now follow the plus: it is at the top-left in the third figure and at the top-right in the fourth, again one corner clockwise, so in the fifth figure it must be at the bottom-right. Option (b) has a cross there, not a plus, while option (d) has the ringed dot at the bottom-left and the plus at the bottom-right, exactly as required. Hence, option (d) is correct.

Letter coding — fixed shift Match 'CSJP' with 'GWNT' letter by letter: $C\to G$, $S\to W$, $J\to N$ and $P\to T$ — every letter moves 4 places forward. The second pair confirms the rule: $E\to I$, $S\to W$, $B\to F$ and $C\to G$ are all +4 as well. Apply +4 to 'BUPQ': B+4=F, U+4=Y, P+4=T and Q+4=U. So the code is 'FYTU'. Hence, option (d) is correct.

Letter coding Look at the first letters only: H (8th) becomes C (3rd) and D (4th) becomes Y (25th) — both move 5 places back, counting backwards past A into Z. Now notice that 'MONEY' and 'HONEY' differ in the first letter alone; the block 'ONEY' is common to both. So that block must be coded exactly as it was in 'HONEY': $O\to Y$, $N\to I$, $E\to I$ and $Y\to T$. Only the first letter is new: M is the 13th letter and 13-5=8, which is H. So 'MONEY' becomes 'HYIIT'. Hence, option (b) is correct.

Letter coding — reverse and shift A straight letter-by-letter comparison of 'GHTMIK' with 'PMPYLJ' gives no steady pattern, so write the word backwards first: 'GHTMIK' reversed is 'KIMTHG'. Now compare with 'PMPYLJ': $K\to P$ is +5, $I\to M$ is +4, $M\to P$ is +3, $T\to Y$ is +5, $H\to L$ is +4 and $G\to J$ is +3. The second pair obeys the same rule: 'NEATLY' reversed is 'YLTAEN', and +5,+4,+3,+5,+4,+3 gives D,P,W,F,I,Q — exactly 'DPWFIQ'. 'DACBYP' reversed is 'PYBCAD', so P+5=U, Y+4=C, B+3=E, C+5=H, A+4=E and D+3=G. The code is 'UCEHEG'. Hence, option (c) is correct.

Word formation 'SHOES' is made of S, H, O, E and S — the important point is that the letter S is needed twice. 'SCHOOL' has only one S, so 'SHOES' cannot be built from it. 'STATEMENT' has no H and no O, and 'BATSMAN' has no H, no O and no E. 'SCHOOLERS' is made of S, C, H, O, O, L, E, R and S — it carries two S's, an H, an O and an E, which is everything 'SHOES' needs. Hence, option (b) is correct.