25 November 2025 · Shift 3

Symbol substitution First put the real sign in place of every symbol used in the expression. So 24 ! 9 * 4 * 6 becomes $24÷9-4-6$. By BODMAS the division is done first: $24÷9=\frac{24}{9}=\frac{8}{3}$. Now the two subtractions: $\frac{8}{3}-4-6=\frac{8}{3}-10$. $\frac{8}{3}-10=\frac{8-30}{3}=\frac{-22}{3}$. Hence option (a) is correct.

Direction sense Draw the three moves one after the other, starting from the starting point. The 5 km walked south and the 5 km walked north are equal and opposite, so they cancel and no north–south shift is left. Only the 7 km walked towards the west survives. So the finishing point lies 7 km due west of the starting point. Hence option (d) is correct.

Paper folding and punching The first fold brings the right half onto the left half and the second fold brings the bottom half up, so the punched sheet is exactly one quarter of the square. Every punch therefore reappears four times when the paper is opened: once as it is, once mirrored across the vertical fold and once across the horizontal fold, and once across both. The quarter carries a bar near its outer edge, a triangle and a square, so the open sheet must show 4 bars, 4 triangles and 4 squares. The four bars land in the four corners and the four squares form the block of $2\times 2$ squares in the middle. Mirroring across the horizontal fold turns a triangle upside down, so the two upper triangles point up and the two lower ones must point down. Option (b) keeps the lower triangles pointing up, (c) interchanges the squares and the triangles in the lower half, and (d) shows only two bars instead of four. Hence option (a) is correct.

Number series Write down the gap between each pair of consecutive terms. The gaps are 15, 21, 27 and they themselves grow by 6 each time, so the next gap is 27 + 6 = 33. Required term = 75 + 33 = 108. The same series can be read as $3n^2$: $3\times 2^2=12$, $3\times 3^2=27$, $3\times 4^2=48$, $3\times 5^2=75$ and $3\times 6^2=108$. Hence option (b) is correct.

Ordering by birth year The five consecutive years starting from 2002 are 2002, 2003, 2004, 2005 and 2006. Rahul is the oldest, so the earliest year belongs to him: Rahul was born in 2002. Among these five years only 2004 is divisible by 4, so 2004 is the only leap year and Ramesh was born in 2004. Reena is barred from 2003 and 2006, and 2002 and 2004 are already taken, so Reena was born in 2005. Rohan and Ritu must take 2003 and 2006; Ritu is younger, so Ritu got 2006 and Rohan got 2003. Hence option (a) is correct.

Address matching Compare each candidate address with the master address field by field, watching the full stops, hyphens and spaces. In (a) the road is written 'M.G. Road' with two full stops, while the master has the plain 'MG Road'. In (b) the house number is written 'H.No.21' with a full stop, while the master uses a hyphen in 'H.No-21'. In (c) the sector is written 'Sector B 4' with a space, while the master uses a hyphen in 'Sector B-4'. Only (d) keeps every letter, full stop, hyphen and space exactly as in the master address. Hence option (d) is correct.

Number series The differences between consecutive terms are 12, 18, 24, 30, and each difference is 6 more than the one before it. So the next difference is 30 + 6 = 36. Required term = 90 + 36 = 126. The same series is 6 times the triangular numbers: $6\times 1,6\times 3,6\times 6,6\times 10,6\times 15$, and next $6\times 21=126$. Hence option (a) is correct.

Number grid Read the grid row by row: the last entry of a row is the product of the first three entries of that row. Row 1: $4\times 6\times 5=120$, which matches the given entry. Row 2: $3\times 5\times 6=90$, which matches as well. Row 3: $5\times 8\times ?=200$, so $40\times ?=200$. $?=\frac{200}{40}=5$. Hence option (c) is correct.

Number box pattern In each box the bottom-left number is the odd one out, so try to build it from the other three. Rule: bottom-left = (top-left $\times$ top-right) + bottom-right. Box 1: $\left(5\times 4\right)+9=20+9=29$, which is the given entry. Box 2: $\left(5\times 7\right)+5=35+5=40$, which matches as well. Box 3: $\left(3\times 6\right)+?=25$, so 18 + ? = 25 and ? = 7. Hence option (a) is correct.

Counting triangles Number the smallest closed triangular regions of the figure. Simple triangles: 1, 2 and 3 inside the left square, 4 and 5 on the right, and 6 hanging below the rectangle — that is 6 triangles. Two regions join to make a bigger triangle wherever their common edge lies on a longer straight line. In the left square 1 + 2 and 1 + 3 each give a half-square triangle, and on the right 4 + 5 gives one more — 3 triangles. One large triangle is still left: the top edge of the rectangle, the long slanting line and the vertical line enclose it. Total = 6 + 3 + 1 = 10. Hence option (b) is correct.

Counting triangles Name the points as shown and then count in groups so that nothing is repeated. With the short line BC: ABC, ADB and ADC — 3 triangles. The whole figure and its two halves cut by the axis AP: AOQ, APO and APQ — 3 triangles. The two long lines OF and QE cross the axis at I and give AIE, AIF, AIO, AIQ, EOI, FIQ, AEQ, AFO, OEQ and OFQ — 10 triangles. Below I, the line LN and the base OQ give ILM, IMN, ILN, IPO, IPQ and IOQ — 6 triangles. The two side flaps add GJO and HKQ — 2 triangles. Total = 3 + 3 + 10 + 6 + 2 = 24. Hence option (c) is correct.

Circular seating Everybody faces the centre, so a person's right hand points anticlockwise; 'second to the right' therefore means two seats anticlockwise. Fix S first. Q sits two seats anticlockwise from S, and P sits two seats anticlockwise from Q. Reading anticlockwise the five seats are S, _, Q, _, P. R and T must fill the two empty seats, but R cannot be next to P, and the seat between Q and P touches P. So R takes the seat between S and Q, and T takes the seat between Q and P. One seat anticlockwise from P brings us back to S, so S is immediately to the right of P. Hence option (d) is correct.

Circular seating Since all five face the centre, counting to the right means counting anticlockwise. Place S, then Q two seats anticlockwise of S and P two seats anticlockwise of Q, which gives S, _, Q, _, P anticlockwise. R may not be an immediate neighbour of P, so R cannot take the seat between Q and P; R sits between S and Q and T sits between Q and P. The complete anticlockwise order is therefore S, R, Q, T, P. Counting to the right from R: first Q, second T, third P. Hence option (b) is correct.

Direction after turns Each turn is a quarter turn, so start from North and apply the turns of option (b) one at a time. North, turn right → East; turn left → North; turn right → East; turn left → North; turn left → West. So this combination leaves him facing West. Options (a) and (c) contain three right turns against only two left turns, which brings him to East instead. Hence option (b) is correct.

Completing the pattern The two diagonals cut the rounded square into four equal triangles, and each triangle carries the same design read from its own base towards the centre. That design is: one thick bar close to the base, then two thin lines parallel to it, and finally a pair of small arcs at the tip that meets the centre. The missing piece is the bottom triangle, so its base is the bottom edge and its tip points up to the centre. Rotation is not allowed, so the required figure must show the arcs at the top tip and the thick bar at the bottom, exactly as in option (c). In (a) the upper thin line is broken into two dashes, in (b) the tip carries a closed oval and an extra line, and in (d) the tip carries one large arc instead of the pair of small ones. Hence option (c) is correct.

Figure series: rotation Three things change from figure to figure: the four-pointed star, the big dark blob at a corner and the small dot inside the circle. The star turns through ${45}^{\circ }$ each step, so it is slanted, like the letter X, in figures 1 and 3 and upright, like a plus sign, in figure 2; the missing figure must therefore show the upright star. The big blob moves ${90}^{\circ }$ clockwise round the circle: top-left, then top-right, then bottom-right, so next it reaches the bottom-left corner. The small dot turns ${45}^{\circ }$ anticlockwise about the centre: it lies on the upper-right arm, then on the top arm, then on the upper-left arm, so in the fourth figure it must lie on the left arm — directly left of the centre. Only one figure has the upright star, the blob at the bottom-left and the dot to the left of the centre, so option (d) is correct.

Water image series In every figure the symbol at the bottom-left is the water image (an up-down mirror) of the symbol that stood at the top-right of the figure just before it, and a brand-new symbol is drawn at the top-right. Test it on the first pair: the Y at the top-right of figure 1 comes back upside down at the bottom-left of figure 2, and the same happens at every later step. So the answer figure must carry, at its bottom-left, the water image of the symbol at the top-right of figure 5. That symbol has a short flag at the top-left, a stem with two small cross-bars, and a long bar at its foot. Turning it upside down puts the long bar on top and leaves the short flag at the bottom-left, because a water image never exchanges left and right. Option (b) is the symbol mirrored left to right, option (c) is the symbol unchanged and option (d) is the symbol turned through ${180}^{\circ }$, so option (a) is correct.

Symbol substitution Replace each letter by the sign it stands for. The expression becomes $36÷4\times 2+8-5$. By BODMAS, division and multiplication come first, working from left to right: $36÷4=9$ and $9\times 2=18$. Now 18 + 8 - 5 = 21. So the value is 21 and option (b) is correct.

Interchanging numbers Interchanging the numbers 5 and 9 means every 5 becomes 9 and every 9 becomes 5; all the other numbers stay where they are. Equation I becomes $5\times 4+9-6÷2=20+9-3=26$, which is not 20. Equation II becomes $3\times 9+5-12÷4=27+5-3=29$, which is not 34. Both equations fail after the interchange, so option (c) is correct.

Number pair: perfect square In each pair check whether the number on the right is the square of the number on the left. ${17}^2=289$, ${16}^2=256$ and ${15}^2=225$ — these three pairs follow the rule exactly. ${14}^2=196$, but the pair offers 197, which is one more than the square. Hence 14 - 197 is the odd pair out, so option (d) is correct.

Letter coding Compare the code with the word letter by letter, using the positions A = 1 … Z = 26. G(7) → B(2) is -5, O(15) → I(9) is -6, L(12) → G(7) is -5 and D(4) → X(24) is -6, counting back past A into Z. FISH → ACNB follows the very same -5, -6, -5, -6 pattern, which confirms the rule. Apply it to TASK: T(20) -5 = O, A(1) -6 = U, S(19) -5 = N, K(11) -6 = E. So TASK is coded as OUNE and option (b) is correct.

Word code language Take the statements two at a time and pick out the word they share and the code they share. Statements 1 and 2 have only 'hard' in common and only 'ma' in common, so hard → ma. Statements 1 and 3 have only 'work' in common and only 'ka' in common, so work → ka. Statements 2 and 3 have only 'growth' in common and only 'pa' in common, so growth → pa. Hence 'work growth' is coded as 'ka pa' and option (b) is correct.

Word formation The given word DEMOCRATIC supplies the letters A, C, C, D, E, I, M, O, R, T. COMRADE, CREDIT and METRIC use only these letters, and the repeated C is available because DEMOCRATIC contains C twice. ACTION needs the letter N, and there is no N anywhere in DEMOCRATIC. So ACTION cannot be formed and option (c) is correct.

Word formation COMMUNICATION supplies the letters A, C, C, I, I, M, M, N, N, O, O, T, U. ACTION, UNION and MOTION can all be built from these — UNION needs two N's and MOTION needs two O's, and both are available. MUSIC needs the letter S, which does not occur in COMMUNICATION at all. So MUSIC cannot be formed and option (c) is correct.

Alphanumeric series Split every term into its three parts and follow each part on its own. The first letters are M, K, I, G — each two steps back in the alphabet, so the next one is E. The second letters are Z, V, R, N — each four steps back, so the next one is J. The numbers grow by 45 every time. So the number is 335 + 45 = 380, the term is EJ380 and option (c) is correct.