26 November 2025 · Shift 3

Symbol substitution First put the real sign in place of every code symbol. The expression therefore becomes $7\times 6+9÷4$. By BODMAS, division and multiplication are carried out before addition: $7\times 6=42$ and $9÷4=\frac{9}{4}$. Now add the two parts: $42+\frac{9}{4}=\frac{168+9}{4}=\frac{177}{4}$. So the required value is $\frac{177}{4}$, and option (d) is correct.

Position in a row Both positions must first be counted from the same end. Amit is 19th from the back, so from the front his place is 72-19+1=54. Ravi stands at the 27th place from the front and Amit at the 54th place from the front. The people standing strictly between them =54-27-1=26. Hence option (b) is correct.

Position in a row When one object is counted from both ends it gets counted twice, so 1 must be subtracted from the sum of the two positions. Formula: position from the right = total - position from the left $+1$. Position from the right =35-9+1=27. So the tree is 27th from the right end, and option (b) is correct.

Number series Take the difference between consecutive terms: -715-(-701)=-14 and -729-(-715)=-14. The same difference -14 repeats right through the series, so 14 is subtracted at every step. Required term =-785-14=-799. Hence option (b) is correct.

Number series Each term is a little more than double the term before it, so test 'double and add'. $35\times 2+4=74$ and $74\times 2+3=151$. $151\times 2+4=306$ and $306\times 2+3=615$. $615\times 2+4=1234$ and $1234\times 2+3=2471$. So $\times 2+4$ and $\times 2+3$ are used alternately, and the step after $\times 2+3$ is $\times 2+4$. Required term $=2471\times 2+4=4942+4=4946$. Hence option (d) is correct.

Symbol substitution Read the two given equations and find which real signs make them true. Take '' as+and '!' as\div$, the division being done first. First equation: $452+\frac{45}{15}=452+3=455$, which is the value given. Second equation: $578+\frac{84}{12}=578+7=585$, which is also the value given, so the rule is confirmed. Applying the same rule: $317+\frac{65}{13}=317+5=322$. Hence option (b) is correct.

Mirror image The mirror stands along a vertical line, so the reflection is a lateral inversion: the whole row is read back to front and each character is itself flipped left-to-right. The character nearest the mirror becomes the first character of the image and the one farthest from it becomes the last. So the order F, b, 3, e, 5, g becomes g, 5, e, 3, b, F. Each of these is then turned round, so 'b' shows up as 'd', '3' shows up as a shape like 'E' and 'F' faces the other way. Only option (d) has both changes together; the others keep 'b' or 'F' standing the wrong way round. Hence option (d) is correct.

Paper folding and punching The sheet is folded twice, so the paper is four layers thick and every punch will show four times when it is opened. The first crease is the diagonal running from the bottom-left corner to the top-right corner; the second crease joins the centre of the sheet to the bottom-right corner. Undo the second fold first - each punch is reflected in that crease - and then undo the first fold, which reflects everything in the diagonal. The triangular cut touches the centre of the sheet, so its four copies meet there and together make one complete square hole in the middle. The house-shaped cut gives four copies, one on each side of that square, and each of them points inward towards the centre. The round punch gives two holes close together near the bottom-left corner and two near the top-right corner. Only option (b) shows the central square, the four house shapes turned inward and the circles in exactly those two corners. Hence option (b) is correct.

Paper folding and punching The left half is folded onto the right half along the vertical centre line, and then the top-right quarter is folded down along the horizontal centre line. The punching is done on the bottom-right quarter, which is now four layers thick, so each punch will appear four times: $4\times 4=16$ holes in all. Three square punches and one round punch are made, so the opened sheet must carry $3\times 4=12$ squares and $1\times 4=4$ circles. Opening the horizontal fold mirrors the four holes upward about the horizontal centre line; opening the vertical fold then mirrors all eight to the left about the vertical centre line. The finished pattern is therefore symmetric about both centre lines: four squares along the top edge, four along the bottom edge, two squares in each of the two middle rows, and the four circles set inside them. Option (a) has fourteen squares and options (b) and (c) are not symmetric about the two creases, so they cannot be right. Hence option (d) is correct.

Order and ranking Five different consecutive years starting from 2010 means the years are 2010, 2011, 2012, 2013 and 2014. Riya is the oldest, so Riya must have the earliest year, 2010. Of the five years only 2012 is divisible by 4, so 2012 is the only leap year and Sameer was born in 2012. Tanu was born neither in 2011 nor in 2014, and 2010 and 2012 are already taken, so Tanu was born in 2013. Only 2011 and 2014 are left for Vaibhav and Nikhil; Nikhil is the younger, so Nikhil takes the later year 2014 and Vaibhav 2011. The friend born in 2012 is Sameer, so option (d) is correct.

Odd one out Look at the letters first: P, Q, R, S are four consecutive letters, but the fifth is U, so the letter T has been skipped. The numbers tell the same story: taking alternate terms, P40 and R35 fall by 5, and Q50 and S45 also fall by 5. Carrying the first chain forward, the term after R35 should be T30 - the letter T and the number 30. U25 breaks the letter chain as well as the number chain, so it is the term that does not belong. Hence option (d) is correct.

Number series Write the differences between consecutive terms: 7-3=4, 13-7=6, 21-13=8, 31-21=10, 43-31=12. These differences 4, 6, 8, 10, 12 themselves rise by 2 each time, so the next difference is 12+2=14. Required number =43+14=57. Hence option (d) is correct.

Number analogy The numbers stand in three rows of four; in each row the fourth number is built from the first three. Row 1: $3\times 4\times 5=60$ and $60\times 2=120$, which is the fourth number. Row 2: $6\times 2\times 3=36$ and $36\times 2=72$, so the rule (product of the first three) $\times 2$ holds again. Row 3: $?\times 3\times 4\times 2=96$, that is $?\times 24=96$. So $?=\frac{96}{24}=4$. Hence option (c) is correct.

Counting figures Name the eight meeting points: A, B, C along the top edge, D, E, F along the bottom edge, G where the vertical BE cuts the diagonal DC, and H where AE cuts DC. Eight of the triangles are the smallest pieces, with no line running across them: ABG, AGH, ADH, DEH, EGH, BCG, CEG and CEF. Putting two or more of these pieces together gives thirteen larger triangles: ABE, ACD, ACE, ACG, ACH, ADE, ADG, AEG, BCE, CDE, CDF, CEH and DEG. Total number of triangles =8+13=21. Hence option (a) is correct.

Counting figures Name the meeting points: A is the tall apex, B and L are the two ends of the long horizontal line, O is the foot of the vertical, C, H are the other points on the horizontal line, D and E are the top and bottom corners of the small square, F is the lower end of the vertical, G is the second apex and K is the lowest point on the right. Work from left to right so that nothing is counted twice. Inside the big triangle ABC the vertical AO and the line OD give ABO, AOC, ABC, AOD and OCD - that is 5. The small square OCED and the lines through it give OCE, OEF, OCF and ACF - that is 4 more, making 9. On the right the two small triangles CGH and HKL add 2. Total number of triangles =5+4+2=11. Hence option (a) is correct.

Venn diagram counting The circle stands for books and the triangle for newspapers, so 'books that are newspapers' means the items lying inside the circle and inside the triangle at the same time. Two areas satisfy this: the part of the circle inside the triangle but above the rectangle, which holds 7, and the part common to all three figures, which holds 4. The numbers 21, 6 and 6 lie inside the circle but outside the triangle, so they are not counted. Required number =7+4=11. Hence option (a) is correct.

Circular seating All five face the centre, so for each of them the clockwise side is the left and the anticlockwise side is the right. Fix J's seat. Moving to J's right (anticlockwise), the first seat is skipped and the second one holds I, because exactly one person sits between them on that side. H is second to the left of J, that is two seats clockwise from J. G is second to the right of I; two seats anticlockwise from I is the seat lying immediately clockwise of J, so G sits between J and H. The only seat still empty, the one between J and I, goes to K. Now count two seats anticlockwise from G: the first is J and the second is K. Hence option (c) is correct.

Linear seating R is the immediate neighbour of both P and T, and T sits immediately to the left of Q, so these four form one fixed block: P R T Q. If the block took seats 1 to 4, the last two seats would go to S and U; S may not sit at an end, so S would be at seat 5 and U at seat 6 — but then nobody is to the right of U. If the block took seats 2 to 5, only the two ends would be left and S cannot sit at an end. So the block occupies seats 3 to 6, and seats 1 and 2 are left for U and S; since S is not at an end, S takes seat 2 and U seat 1. The final row is U, S, P, R, T, Q, so the two extreme ends are held by U and Q. Hence option (d) is correct.

Linear seating R sits immediately to the left of M and M is at an extreme end, so M must be at the boys' right end with R next to him. Number the seats 1 to 8 from the boys' own left to their own right: M is at seat 8, R at seat 7 and the other end, seat 1, is T's. N has both Q and R as immediate neighbours; the only seat free beside R is 6, so N sits at seat 6 and Q at seat 5. O is fourth to the left of N, that is at seat 6-4=2. P is not a neighbour of O, so P cannot take seat 3; therefore P sits at seat 4 and S at seat 3. Counting three seats to the right of Q (seat 5) gives seats 6, 7 and 8 — N, R and M — so M is third to the right of Q. Hence option (c) is correct.

Symbol substitution Replace every letter by the sign it stands for: P by $÷$, Q by $\times$, R by + and S by -. The expression becomes $24\times 10÷5+6-8$. By BODMAS, division and multiplication are done first: $24\times 10÷5=\frac{240}{5}=48$. Now 48 + 6 - 8 = 46. Hence option (a) is correct.

Balancing an equation Put the signs of option (a) — $÷$, +, $\times$, - — into the four stars in that order. The equation becomes $96÷8+7\times 5-11$. By BODMAS, $96÷8=12$ and $7\times 5=35$. So LHS =12+35-11=36, which is exactly the RHS. Option (b) would give $96÷8\times 7-5+11=90$, which is not 36. Hence option (a) is correct.

Odd letter-number pair In each pair the first letter moves one step forward, the second letter three steps forward, and the number is doubled. BW78 → CZ156: B+1=C, W+3=Z and $78\times 2=156$. ME108 → NH216: M+1=N, E+3=H and $108\times 2=216$. NJ136 → OM272: N+1=O, J+3=M and $136\times 2=272$. UD94 → VG198: the letters obey the rule, but $94\times 2=188$ and not 198. Hence option (b) is correct.

Letter coding Write the word backwards first and then shift its letters. TABLE written backwards is ELBAT; its first three letters move 5 steps back and the last two move 4 steps back: E-5=Z, L-5=G, B-5=W, A-4=W, T-4=P, giving ZGWWP (counting back from A continues at Z). The same rule fits the second word: CHAIR backwards is RIAHC, and R-5=M, I-5=D, A-5=V, H-4=D, C-4=Y give MDVDY. PIANO backwards is ONAIP, so O-5=J, N-5=I, A-5=V, I-4=E, P-4=L. The code is JIVEL, so option (b) is correct.

Letter coding Reverse the word first and then move every letter 3 steps back in the alphabet. FRIEND reversed is DNEIRF; D-3=A, N-3=K, E-3=B, I-3=F, R-3=O, F-3=C, which gives AKBFOC. The rule also fits GLOBAL: reversed it is LABOLG, and L-3=I, A-3=X, B-3=Y, O-3=L, L-3=I, G-3=D give IXYLID (counting back from A continues at Z). MODERN reversed is NREDOM, so N-3=K, R-3=O, E-3=B, D-3=A, O-3=L, M-3=J. The code is KOBALJ, so option (b) is correct.

Alphanumeric series Split every term into its first letter, its second letter and its number. First letters M, N, O, ?, Q move one step forward each time, so the missing letter is P. Second letters V, X, Z, ?, D move two steps forward each time, so after Z comes B. The numbers 186, 206, 226, ?, 266 rise by 20 each time, so the missing number is 226+20=246. The missing term is PB246, so option (a) is correct.