27 November 2025 · Shift 3

Ranking in a row Total number of students =80. Rank from the top = total - rank from the bottom +1. Geeta is 35th from the bottom, so her rank from the top =80-35+1=46. Jaishankar is 30th from the top. Students standing between them =46-30-1=15. Hence option (b) is correct.

Ranking in a row Total number of students =40. B is 15th from the bottom, so B's rank from the top =40-15+1=26. A is 12th from the top. Students between them =26-12-1=13. Hence option (c) is correct.

Number series Take the consecutive terms that are printed: 999-987=12, 1035-1023=12 and 1047-1035=12. So the row is an arithmetic progression whose common difference is 12. The missing term stands between 963 and 987, and indeed $987-963=24=2\times 12$. So the missing term =963+12=975. Hence option (a) is correct.

Number series See how each term is built from the one before it: $13\times 3+3=42$. $42\times 3+4=130$ and $130\times 3+3=393$. So the multiplier is always 3, while the number added alternates between 3 and 4. The step after 393 must add 4: $393\times 3+4=1179+4=1183$. Check the rest: $1183\times 3+3=3552$ and $3552\times 3+4=10660$, which are the given terms. Hence option (d) is correct.

Symbol based operation Find the rule from the first pair: $14\times 11=154$ and $154\times 5=770$. Test the same rule on the second pair: $9\times 12=108$ and $108\times 5=540$, which is correct. So the operation is (first number $\times$ second number) $\times 5$. Applying it to 10 and 15: $10\times 15=150$ and $150\times 5=750$. Hence option (b) is correct.

Symbol based operation Read the rule off the first line: $120÷15=8$ and 732-8=724. So the pattern is: first number - (second number $÷$ third number). Check it on the second line: $285÷19=15$ and 845-15=830, which is correct. Apply it to the last line: $270÷18=15$, so the value =512-15=497. Hence option (a) is correct.

Mirror image The mirror AB is a vertical line standing to the right of the figure, so the reflection is a lateral inversion: left and right are exchanged while top and bottom stay as they are. The hatched triangle covers the left half of the top face in the object, so in the image it must cover the right half of the top face. The small hollow triangle points to the left and sits on the left panel, so in the image it points to the right and sits on the right panel. The letter N is on the right panel and is not left-right symmetric, so it shifts to the left panel and appears turned round. The solid arrow inside the bottom arc points to the left, so in the image it points to the right. Option (c) keeps N unreversed and option (a) and (b) place the hatching or the small triangle on the wrong side; only option (d) shows all four changes together. Hence option (d) is correct.

Paper folding and punching First the right half of the square is folded onto the left half along the vertical centre line, and then the top half of that rectangle is folded down along the horizontal centre line. The punched piece is therefore the bottom-left quarter of the sheet, and it is four layers thick, so every punch will show four times when the sheet is opened. Open the horizontal crease first: the two upward triangles are reflected upward across it, and a reflection in a horizontal line turns an upward triangle into a downward one. Open the vertical crease next: everything is reflected left to right, which leaves an upward triangle still upward. So the opened sheet must carry four upward triangles in the lower half and four downward triangles in the upper half, and the pattern must be symmetric about the vertical centre line. The square notch is cut on the vertical crease itself, so it opens into one complete rectangle lying across that line, and its reflection gives a second such rectangle in the other half. In options (a), (b) and (c) at least one triangle points the wrong way, so each of them breaks one of the two symmetries. Hence option (d) is correct.

Paper folding and punching The circular sheet is first folded upward along the horizontal diameter and then folded to the left along the vertical crease, so the punched piece is the upper-left quarter and it is four layers thick. Four layers means each of the three punches - the square, the ellipse and the small circle - must reappear exactly four times, giving $3\times 4=12$ holes in all. Open the vertical crease first: the three holes are reflected left to right, so the upper half carries a square, an ellipse and a circle on each side of that crease. Open the horizontal crease next: the whole upper half is reflected downward, so the lower half repeats the upper half turned upside down. The opened sheet is therefore symmetric about both creases, and moving inward from the rim the order of the punches must stay square, then ellipse, then circle - exactly as they lie in the punched quarter. Option (a) shows only 10 holes, so it cannot come from a four-layer fold. Option (b) has 12 holes but its lower half is a copy of the upper half instead of its mirror image, so it fails the horizontal crease. Option (c) is symmetric yet puts the circles at the rim and the squares near the centre, reversing the order in which the quarter was punched. Hence option (d) is correct.

Conditional selection A valid selection must contain exactly 2 fruits and 3 snacks and must satisfy all five conditions. Combination I gives the fruits Mango and Banana and the snacks Sandwich, Biscuit and Chocolate; Mango comes with Sandwich, Cake is absent and Orange is absent, so every condition holds. Combination II carries three fruits (Apple, Orange, Mango) and only two snacks, so the 2-and-3 requirement itself fails; it also puts Biscuit with Cake and takes Mango without Sandwich. Combination III takes Sandwich without Mango, but condition 3 says the two must always go together, so it is rejected. Only combination I is possible. Hence option (b) is correct.

Number series Every term is a perfect square: $25=5^2$, $64=8^2$, $169={13}^2$, $441={21}^2$ and $1156={34}^2$. The bases $5,8,13,21,34$ form a Fibonacci-type chain, because 5+8=13, 8+13=21 and 13+21=34. So the next base is 21+34=55. The required term $={55}^2=3025$. Hence option (b) is correct.

Missing number in a grid Work down each column and compare the bottom entry with the three above it. First column: $7\times 3\times 2=42$ and 42-1=41. Second column: $3\times 6\times 2=36$ and 36-1=35. So the rule is (product of the three numbers) $-1$. Third column: $6\times ?\times 3=71+1=72$, so $?=\frac{72}{18}=4$. Hence option (b) is correct.

Counting triangles The narrow strip down the middle is bounded by two parallel lines, so no triangle can cross it; the figure splits into a left part and a right part which are mirror images of each other. In the left part let L be the left end of the horizontal diameter, let A,B,P,C,D be the points of the left chord from top to bottom, and let T,M,S be the top, middle and bottom points of the strip's left edge. Smallest triangles: LBP, LPC, BPM, CPM, BTM, CSM - that is 6. Triangles made of two of these: LBM, LCM, LBC, BMC - that is 4. Triangles made of three of these: LTM and LSM - that is 2. One triangle uses the whole left part: LTS - that is 1. So the left part contains 6+4+2+1=13 triangles, and by symmetry the right part contains 13 as well. Total number of triangles =13+13=26. Hence option (a) is correct.

Counting quadrilaterals Every line in the figure is horizontal or vertical, so each quadrilateral is a rectangle, fixed by choosing two horizontal lines and two vertical lines. Call the horizontals $H_1$ (top), $H_2$, $H_3$ and $H_4$ (bottom); all four run the full width, so only the vertical lines restrict the choice. The verticals are: the two side edges (full height), one line from $H_1$ to $H_3$, a short line from $H_1$ to $H_2$, and one line from $H_2$ to $H_4$. Between $H_1$ and $H_2$ four verticals run right across, giving ${}^4{C}_2=6$ rectangles, and between $H_2$ and $H_3$ four verticals again give 6. Between $H_1{H}_3$, $H_2{H}_4$ and $H_3{H}_4$ only three verticals run across each time, giving ${}^3{C}_2=3$ rectangles in each case. Between $H_1$ and $H_4$ only the two side edges run the whole height, giving ${}^2{C}_2=1$ rectangle. Total =6+6+3+3+3+1=22. Hence option (d) is correct.

Venn diagram counting 'Cars that are buses' means everything lying inside both the square and the triangle, whether or not it also lies inside the circle. The part of the square inside the triangle but outside the circle holds 9. The part common to the square, the triangle and the circle holds 7. Required number =9+7=16. Hence option (b) is correct.

Ranking by length Only two rectangles are longer than Q, so Q stands third from the top among the five. No rectangle is shorter than S, so S is the shortest of all and takes the fifth place. For the remaining three the given chain is T>R>P, so the two rectangles above Q must be T and R, and P must sit between Q and S. The complete order therefore is T>R>Q>P>S. T stands at the top, so option (a) is correct.

Ranking by price Q is the second highest, so exactly one toy is costlier than Q. The other clues join into a single chain R>T>S>U>P, in which R stands above every one of the four. Since Q occupies the second place, the toy above Q can only be R, giving R>Q>T>S>U>P. The highest price belongs to R, so option (b) is correct.

Circular seating All eight face the centre, so for each woman 'left' runs clockwise and 'right' runs anticlockwise. Fix W. Q is second to the left of W, and R is the immediate neighbour of both W and Q, so R fills the single seat between them. P is third to the left of Q, S is immediately to the right of W, U is fourth to the left of S and V is second to the right of W. Placing all of them gives the clockwise order W, R, Q, U, T, P, V, S, with T taking the only seat left over. The seat to T's right is the anticlockwise neighbour, which is U, so option (a) is correct.

Symbol substitution Replace every letter by the sign it stands for. The expression 48 B 8 C 2 A 12 D 3 becomes $48-8\times 2+12÷3$. By BODMAS, division and multiplication are done first: 48-16+4. 48-16+4=36, so option (b) is correct.

Balancing an equation Put the four signs of an option into $18*6*54*9*12=114$ in the given order and simplify by BODMAS. Option (a) gives $18\times 6-54÷9+12$. Doing multiplication and division first: 108-6+12. 108-6+12=114, which is exactly the right-hand side. Hence option (a) is correct.

Odd letter-number pair In every pair the letters move forward by fixed steps and, more decisively, the number on the left is doubled to give the number on the right. In (a) $156\times 2=312$, in (b) $204\times 2=408$ and in (d) $198\times 2=396$, so all three obey the rule. In (c) doubling gives $175\times 2=350$, but the pair shows 360. Hence H175R - I360T breaks the pattern and option (c) is correct.

Letter coding Write the given word backwards first and then shift the letters. GHOST written backwards is TSOHG, and the shifts that turn it into QOLDD are -3,-4,-3,-4,-3. The same test works on the second pair: BRAIN backwards is NIARB, and -3,-4,-3,-4,-3 gives KEXNY. STORM backwards is MROTS; applying -3,-4,-3,-4,-3 gives J, N, L, P, P. The code is JNLPP, so option (b) is correct.

Letter coding Keep the first and the last letter in place and reverse the four middle letters. MANTRA becomes M + RTNA + A, that is MRTNAA; adding +4 to every letter gives QVXREE. The rule holds for the second word too: POTION becomes P + OITO + N, that is POITON, and +4 on each letter gives TSMXSR. For CAMERA the middle letters AMER reverse to REMA, so the word becomes CREMAA. Adding +4 to each letter gives GVIQEE, so option (b) is correct.

Alphanumeric series Handle the two letters and the number separately. The first letters run T, S, R, Q, P — one step back each time — so the missing term begins with R. The second letters run I, G, E, C, A — two steps back each time — so the second letter is E. The numbers fall by 32 each time: 278-32=246 and 246-32=214; the check 214-32=182 matches the fourth term. The missing term is RE214, so option (b) is correct.

Wrong term in series The first letters D, K, R, Y, F each move 7 steps forward, and in every term the second letter is 7 steps ahead of the first. So the letter part of all five terms is correct and the fault must lie in the numbers. The numbers should be multiplied by 5 each time: $12\times 5=60$, $60\times 5=300$, $300\times 5=1500$. The last number should therefore be $1500\times 5=7500$, but the series prints 7600. Hence FM7600 is the wrong term and option (d) is correct.