20 November 2025 · Shift 3

Profit and loss Cost price of all the eggs $=20\times 60=1200$, i.e. ₹1,200. Selling price of the first lot $=15\times 70=1050$, i.e. ₹1,050. Eggs left =20-15=5 dozen, sold at ₹55 a dozen, so $5\times 55=275$, i.e. ₹275. Total selling price =1050+275=1325, i.e. ₹1,325. The selling price is more than the cost price, so profit =1325-1200=125, i.e. a profit of ₹125. Hence the correct option is (b).

Combining two ratios Here A:B=3:4 and B:C=5:6; the common term is B, which is 4 in the first ratio and 5 in the second. Make B the same in both — multiply the first ratio by 5 and the second by 4. A:B=15:20 and B:C=20:24. Now B stands for 20 in both, so the three can be written together as A:B:C=15:20:24. Hence the correct option is (b).

Direction sense: distance Let S be the starting point. He goes 6 km north, then 7 km east, and then 6 km south. The 6 km walked north and the 6 km walked south are equal and opposite, so they cancel each other. Only the 7 km walked east is left, and it is a straight line, so his distance from S is 7 km. Hence the correct option is (b).

Direction sense: final direction Let S be the starting point. He walks 12 km west, then 16 km south, and then 12 km east. The 12 km walked west and the 12 km walked east are equal and opposite, so no east–west displacement is left. Only the 16 km walked south remains, so the finishing point lies straight south of S. Hence the correct option is (b).

Letter cluster analogy Compare UPZ with VQA letter by letter: U+1=V, P+1=Q and Z+1=A, because after Z the alphabet begins again at A. So the rule is simply +1 on every letter of the cluster. Apply the same rule to LFZ: L+1=M, F+1=G and Z+1=A. So LFZ : MGA. Hence the correct option is (d).

Series of fractions Simplify each term: $\frac{2}{4}=\frac{1}{2}$, $\frac{3}{6}=\frac{1}{2}$ and $\frac{4}{8}=\frac{1}{2}$. So every term of the series has the same value $\frac{1}{2}$; only the way it is written keeps changing. The numerators run $2,3,4,\dots$ and each denominator is twice its own numerator. The next numerator is 5, so the next denominator is 10 and the missing term is $\frac{5}{10}$. Hence the correct option is (b).

Exact matching of a statement Break the reference into its four parts and test every option against each part: 'Detroit ;' — '15 Augu.st,' — '2024' — '(Tuesday).' Option (a) keeps the place and the month but changes the year to 2025 and the day to (Thursday), so it fails at the last two parts. Option (c) shifts the full stop inside the month — 'Augus.t' in place of 'Augu.st' — and also changes the day to (Thursday). Option (b) repeats every letter, digit, space and punctuation mark of the reference exactly as it stands, including the space before the semicolon and the full stop inside 'Augu.st'. Hence the correct option is (b).

Exact matching of a statement The reference has four things to check: the word 'Vacnaa,' then 'ST:' then the date '6 November' and finally the year '2025.' Option (b) writes 'Vaknaa' with a k in place of the c, so it is wrong at the very first word. Option (c) puts a semicolon after ST instead of a colon and makes the date 16 November instead of 6 November. Option (d) changes the word to 'Vantaa', the date to 21, the month to 'Novembers' and the year to 2026. Option (a) agrees with the reference in every letter, digit and punctuation mark. Hence the correct option is (a).

Missing number in a grid Read the grid column by column, not row by row. In each column the first three numbers are multiplied together and that product is then multiplied by 10 to give the bottom number. First column: $6\times 2\times 6=72$ and $72\times 10=720$. Second column: $5\times 3\times 4=60$ and $60\times 10=600$. Third column: $3\times 4\times 8=96$, so the missing number is $96\times 10=960$. Hence the correct option is (b).

Counting triangles Name the corners A, B, C, D along the top and H, G, F along the bottom, let I–E be the middle horizontal line and C–G the vertical line, and mark the crossings N, O, J, K, L, M and P. The two long slanting lines AG and HC with the sides of the rectangle give AGH, ACH, ACG, ACN, AHN, GHN and CGN — 7 triangles. The middle line IE cuts the same region again and gives AIK, HIJ, CJL, GKL and JKN — 5 more. The vertical CG together with BF gives CGH, BCO, BCP, COP, FGP, FHO, JMO and LMP — 8 more. The short line DM on the right, with BF, DF and IE, gives BDF, BDM, DEM, DFM and EFM — 5 more. Total number of triangles =7+5+8+5=25. Hence the correct option is (a).

Counting triangles Name the figure as shown: A at the top, B, D, I and J down the left side, C on the slanting side AF, G and H on DF, and L, K and O in the lower rectangle. In the upper part the slanting side AF, the horizontals BC and DF and the lines CG and CH give ABC, ADF, CFG, CFH and CGH — 5 triangles. The wide band between DF and IJ has no slanting line inside it, so it contributes no triangle at all. The lower rectangle ILKJ has both of its diagonals drawn, and such a rectangle always holds 8 triangles. They are the four halves IJL, IJK, JKL and IKL and the four small ones IJO, JKO, KLO and ILO. Total number of triangles =5+0+8=13. Hence the correct option is (a).

Direction sense: turns P is walking towards the west, so his left hand points to the south and his right hand to the north. A left turn therefore makes him face the south. Now facing south, his right hand points to the west, so the right turn brings him back to the west. So one left turn followed by one right turn leaves his direction unchanged, and he is finally facing west. Hence the correct option is (c).

Direction sense: turns Each right turn is a turn of ${90}^{\circ }$, so four right turns make $4\times {90}^{\circ }={360}^{\circ }$ — one complete round, after which a person faces the direction he or she started from. M takes 16 right turns and no left turn, and $16=4\times 4$, so she completes exactly 4 whole rounds. Her facing at the end is therefore the same as her facing at the start. She finishes facing east, so she must have started cycling towards the east. Hence the correct option is (d).

Completing the pattern The question figure is a rectangle cut into four quarters by the two centre lines, and the badge inside it is symmetrical about both of them. In the top-right quarter the double outline runs from the top centre down to the right, so in the missing top-left quarter it must run from that same top-centre point down to the left and then straight down the left side. In the option the piece is shown by itself, so this pair of slanting lines must rise towards the top-right corner and the two short verticals must drop to the bottom edge. The triangle in the top-right quarter points downwards, so by the same symmetry the triangle in the missing quarter must also point downwards. Option (a) alone has both features; (b) and (d) carry an upward triangle and (c) a triangle that points to the right. Hence the correct option is (a).

Odd figure pair In each pair the second figure uses the very same two shapes as the first, but turned inside out: the shape that was outside goes inside and the shape that was inside becomes the outer boundary. In (a) a circle holds a diamond, and then the diamond holds the circle. In (c) a dome holds a parallelogram, and then the parallelogram holds the dome. In (d) a notched circle holds a pentagon, and then the pentagon holds the notched circle. In (b) the first figure is a hexagon holding a triangle, but the second is a rounded square holding a hexagon — the triangle has vanished and a brand-new shape has appeared, so the inside-out rule is broken. Hence the odd pair is option (b).

Figure analogy In each pair the two figures simply turn inside out: the star that forms the outer boundary shrinks and goes inside, and the star that was sitting inside becomes the new outer boundary. The third figure is a six-pointed star carrying a five-pointed star inside it. Interchanging the two, the five-pointed star must become the outer figure and the six-pointed star must sit inside it. Only one option has a five-pointed star as its outer boundary with a six-pointed star inside it; in the other three the exchange has not been made. Hence the correct option is (c).

Figure analogy Compare box A with box B. Every shape stays in its own corner, every shape is turned through a quarter turn, and the shape in the bottom-right corner is additionally shaded. The arrow of A points right and in B it points down, and the plain triangle of A is likewise laid on its side; the right-angled triangle keeps its shape but is filled in. Now treat box C the same way: the upward arrow must become a right-pointing arrow, and the pentagon must tilt through a quarter turn. The six-pointed star has a point every 60°, so a quarter turn makes it stand on a pair of side points — its points now lie to the left and right instead of straight up and down. The downward triangle keeps its shape and is shaded. Option (a) turns the arrow the wrong way and even changes the shaded shape, while (b) and (c) leave the six-pointed star standing upright. Hence the correct option is (d).

Figure analogy Leave the outer frame of each box aside and count the sides of the shapes drawn inside it. The first figure holds a square and a pentagon, that is 4+5=9 sides, and its partner carries exactly 9 small symbols, set out in three rows of three. The third figure holds a pentagon and a square, again 5+4=9 sides, so the missing figure must carry 9 symbols. The symbol itself is not the point — what must be matched is the count, three rows of three. Options (a), (c) and (d) show only 5, 7 and 6 crosses, so they cannot stand for 9 sides. Hence the correct option is (b).

Figure series The circle is cut into five equal sectors and two marks travel round it — a shaded dot and a hollow circle. Follow the shaded dot: upper-left, right, lower-left, upper-right. Each step it jumps two sectors clockwise. Follow the hollow circle: upper-left, lower-left, bottom, right. Each step it slips one sector anticlockwise. In the fourth figure the shaded dot is in the upper-right sector, so two sectors clockwise brings it to the bottom sector. The hollow circle is in the right sector, so one sector anticlockwise brings it to the upper-right sector. The missing figure therefore has the shaded dot at the bottom and the hollow circle at the upper right. Hence the correct option is (b).

Odd number pair Look at what has to be done to the left number to reach the right number. $14\to {\left(14+2\right)}^2={16}^2=256$. $12\to {\left(12+2\right)}^2={14}^2=196$. $16\to {\left(16+2\right)}^2={18}^2=324$. So the rule is: add 2 and square. $18\to {\left(18+2\right)}^2={20}^2=400$, but the pair printed is 18-420. Hence the odd pair is option (d).

Letter coding Match the code letter by letter against the word. In 'SPPU' → 'NJKO': $S\to N$ is -5, $P\to J$ is -6, $P\to K$ is -5, $U\to O$ is -6. 'GBOT' → 'BVJN' obeys the very same pattern: $G\to B$ is -5, $B\to V$ is -6 (counting back past A to Z), $O\to J$ is -5, $T\to N$ is -6. So the shifts are -5,-6,-5,-6 in that order. For 'KBNT': K-5=F, B-6=V, N-5=I, T-6=N. The code is 'FVIN'. Hence the correct option is (d).

Coded language Take the statements two at a time: the word they share must carry the code they share. Statements (ii) 'light bring happiness' and (iii) 'happiness produced joy' have only 'happiness' in common, and only the code 'mj' in common, so happiness = mj. Statements (i) 'bring the light' and (ii) 'light bring happiness' have 'bring' and 'light' in common, and 'eb' is the code that runs through both, so 'eb' is the code of 'bring'. That leaves 'db' for 'the' in statement (i), and the remaining code of statement (ii) for 'light'. Therefore 'bring happiness' = 'eb' + 'mj', which is printed as 'mj eb'; the codes need not be written in the order of the words. Hence the correct option is (b).

Coded language Statements (i) and (iii) share 'apple' and 'very'; the codes they share are 'kp' and 'st'. Statements (i) and (iv) share 'is' and 'very'; the codes they share are 'lm' and 'st'. 'very' is the word common to both these comparisons, so very = st, and then apple = kp and is = lm. From (i) the only code left is 'qr', so sweet = qr. In (iv) 'fruit very is tasty' = 'mn st lm ab', so fruit and tasty take 'mn' and 'ab'; 'fruit' also occurs in (ii) and (iii), where 'mn' appears but 'ab' does not, so fruit = mn. Therefore 'very sweet fruit' = 'st qr mn', which is printed as 'qr mn st'. Hence the correct option is (c).

Word formation LIGHT needs the five letters L, I, G, H and T, each of them present in the option word. (a) LITHOTOMY has L, I, T and H but no G. (b) HOLLOWING has L, I, G and H but no T. (d) HOLOGRAPH has L, G and H but neither I nor T. (c) LIGHTER supplies L, I, G, H and T, so LIGHT can be picked out of it. Hence the correct option is (c).

Letter cluster analogy Do not compare the clusters letter by letter in the order given — write the first cluster backwards first. QPSU written backwards is USPQ; moving each of these letters one step forward gives V, T, Q, R, that is VTQR. GXOE written backwards is EOXG; one step forward gives F, P, Y, H, that is FPYH. So the rule is: reverse the cluster, then add 1 to every letter. HCSJ written backwards is JSCH; one step forward gives K, T, D, I. The required cluster is KTDI. Hence the correct option is (b).