RRB Assistant Loco Pilot CBT 1 — 16 Feb 2026 — Shift 3 — Official Paper — Kerala PSC PYQ Practice with Answers

Correct answer (Option A):\nLet us evaluate the statements to determine the tallest pole among the five poles (I, L, O, M, and D):\nFrom Statement (I): O is shorter than D (D > O). This does not tell us about the other three poles.\nFrom Statement (II): M is taller than D (M > D).\nCombining both statements: We get the partial order M > D > O. However, we have absolutely no information regarding the remaining two poles, I and L. Either I or L could still be taller than M. Therefore, the data in both statements together are not sufficient to answer the question.\nOption A is correct.\n\nWhy others are wrong:\nOption B is incorrect because even after combining both statements, the positions of I and L remain completely unknown.\nOptions C and D are incorrect because individual statements provide even less information about the remaining four poles.\n\nStudy tip:\nIn data sufficiency problems involving ordering or ranking, ensure that you can uniquely identify the extreme position (tallest/shortest) relative to all elements before declaring the data sufficient.

Correct answer (Option C):\nA convex mirror is a diverging mirror that always forms a virtual, erect, and diminished image regardless of the position of the object in front of it. When an object is placed anywhere between infinity and the pole (P) of a convex mirror, the image is formed behind the mirror between the pole (P) and the principal focus (F). This image is always smaller than the object (diminished), upright (erect), and cannot be caught on a screen (virtual).\nOption C is correct.\n\nWhy others are wrong:\nOptions A, B, and D are incorrect because a convex mirror never produces real or inverted images for real objects, nor does it produce enlarged or same-sized images when the object is outside the mirror.\n\nStudy tip:\nMemorize mirror behaviors: Convex mirrors always form virtual, erect, and diminished images. Concave mirrors can form both real/inverted and virtual/erect images depending on the object distance relative to the focal length.

Correct answer (Option B):\nLet us analyze the statements using standard set logic:\nStatement 1: 'All kings are bombs' means the set of Kings is entirely inside the set of Bombs.\nStatement 2: 'No parrots are kings' means the set of Parrots has no overlap with the set of Kings.\nNow evaluate Conclusion (I): 'Some bombs are not parrots.' Since all kings are bombs, those bombs that are kings cannot be parrots. Therefore, this conclusion is definitely true.\nEvaluate Conclusion (II): 'Some bombs are kings.' Since the set of Kings is completely contained inside Bombs, the overlapping region represents bombs that are kings. Thus, this conclusion also follows.\nTherefore, both conclusions logically follow.\nOption B is correct.\n\nWhy others are wrong:\nOptions A, C, and D are incorrect because they fail to acknowledge that both statements satisfy both conditions simultaneously under logical deductions.\n\nStudy tip:\nUse Venn diagrams to quickly visualize syllogisms. If a sub-set is completely inside a main set, any element excluded from the sub-set does not automatically clear the main set, but the existing intersection guarantees both 'some' and non-overlap restrictions.

Correct answer (Option B):\nThe nucleus of a eukaryotic cell is surrounded by a double-layered membrane known as the nuclear membrane or nuclear envelope. This membrane separates the genetic material inside the nucleus from the surrounding cytoplasm. It contains small pores (nuclear pores) that regulate the exchange of materials, such as RNA and proteins, between the nucleus and the rest of the cell.\nOption B is correct.\n\nWhy others are wrong:\nOption A is incorrect because the cell wall is the outer protective layer found in plant cells, fungi, and bacteria.\nOption C is incorrect because cytoplasm is the jelly-like fluid filling the cell outside the nucleus.\nOption D is incorrect because the plasma membrane is the outer boundary layer around the entire cell.\n\nStudy tip:\nRemember that the nucleus has a double membrane (nuclear envelope), while organelles like mitochondria and chloroplasts also possess double membranes. Ribosomes and centrioles lack membranes entirely.

Correct answer (Option B):\nLet us compute the total unengraved surface area step-by-step:\n\nFormula: Total Surface Area of a Cube = 6 × side²\nGiven: side = 50 cm\nStep 1: Total Surface Area = 6 × 50²\nStep 2: Total Surface Area = 6 × 2500 = 15000 cm²\n\nNow, let us calculate the total area of the engraved circular spots. A standard die has numbers from 1 to 6.\nTotal number of circular spots = 1 + 2 + 3 + 4 + 5 + 6 = 21 spots\nGiven: Diameter of each spot = 10 cm, so Radius (r) = 5 cm\nFormula: Area of one circle = πr²\nStep 3: Area of 1 spot = π × 5² = 25π cm²\nStep 4: Total Engraved Area = 21 × 25π = 525π cm²\nStep 5: Total Engraved Area ≈ 525 × 3.14159265 ≈ 1649.34 cm²\n\nStep 6: Unengraved Surface Area = Total Area - Engraved Area\nStep 7: Unengraved Area = 15000 - 1649.34 = 13350.66 cm²\nRounding off to the nearest integer gives 13351 cm².\nOption B is correct.\n\nWhy others are wrong:\nOptions A, C, and D do not match the correct numerical derivation or involve incorrect calculation of the number of total dots on a standard 6-sided die.\n\nStudy tip:\nAlways calculate the total sum of elements carefully (here the sum of dots from 1 to 6 equals 21) rather than treating each face as having only one circle.