SSC CGL Tier II 2024 Paper I — Kerala PSC PYQ Practice with Answers

Correct answer (Option D):\nFirst, let us analyze the side lengths of the given triangle ΔABC:\nAB = 21 cm, BC = 20 cm, AC = 29 cm\nCheck if these sides satisfy the Pythagorean theorem:\n21² + 20² = 441 + 400 = 841\n29² = 841\nSince AB² + BC² = AC², ΔABC is a right-angled triangle with the hypotenuse being AC = 29 cm.\nFor any right-angled triangle inscribed in a circle, the hypotenuse is equal to the diameter of its circumcircle.\nFormula: Circumradius (R) = Hypotenuse / 2\nCalculation: R = 29 / 2 = 14.5 cm\nTherefore, the circumradius length is exactly 14.5 cm.\nOption D is correct.\n\nWhy others are wrong:\nOption A (13.5 cm) is incorrect because 13.5 × 2 = 27 cm, which is not the hypotenuse.\nOption B (21.5 cm) is incorrect because it corresponds to a hypotenuse of 43 cm.\nOption C (32.5 cm) is far too large and does not match half of the longest side.\n\nStudy tip:\nRemember the standard Pythagorean triplets such as (20, 21, 29). For a right triangle, the circumradius is always half the hypotenuse ($R = c/2$), and the inradius is given by $r = (a + b - c) / 2$.

Correct answer (Option D):\nFirst, let us analyze the side lengths of the given triangle ΔABC:\nAB = 21 cm, BC = 20 cm, AC = 29 cm\nCheck if these sides satisfy the Pythagorean theorem:\n21² + 20² = 441 + 400 = 841\n29² = 841\nSince AB² + BC² = AC², ΔABC is a right-angled triangle with the hypotenuse being AC = 29 cm.\nFor any right-angled triangle inscribed in a circle, the hypotenuse is equal to the diameter of its circumcircle.\nFormula: Circumradius (R) = Hypotenuse / 2\nCalculation: R = 29 / 2 = 14.5 cm\nTherefore, the circumradius length is exactly 14.5 cm.\nOption D is correct.\n\nWhy others are wrong:\nOption A (13.5 cm) is incorrect because 13.5 × 2 = 27 cm, which is not the hypotenuse.\nOption B (21.5 cm) is incorrect because it corresponds to a hypotenuse of 43 cm.\nOption C (32.5 cm) is far too large and does not match half of the longest side.\n\nStudy tip:\nRemember the standard Pythagorean triplets such as (20, 21, 29). For a right triangle, the circumradius is always half the hypotenuse ($R = c/2$), and the inradius is given by $r = (a + b - c) / 2$.

Correct answer (Option A):\nWhen two objects move in opposite directions on a track, their relative speed is the sum of their individual speeds.\nFormula: Relative Speed = Speed 1 + Speed 2\nGiven speeds: 20 km/hr and 16 km/hr\nRelative Speed = 20 + 16 = 36 km/hr\nConvert speed to m/s: 36 × (5 / 18) = 10 m/s\nTrack length (Distance) = 2200 m\nFormula: Time = Distance / Relative Speed\nStep 1: Time = 2200 / 10 = 220 seconds\nStep 2: Convert 220 seconds into minutes and seconds\n220 seconds = 3 minutes and 40 seconds\nOption A is correct.\n\nWhy others are wrong:\nOption B (3 mins 20 secs) matches 200 seconds, missing the extra 20 seconds needed.\nOption C (2 mins 50 secs) equals 170 seconds, which represents a calculation breakdown.\nOption D (2 mins 20 secs) equals 140 seconds, which fails to account for the actual combined tracking speed.\n\nStudy tip:\nFor circular tracks, the time taken to meet for the first time anywhere on the track is always Distance / Relative Speed. If they were traveling in the same direction, the relative speed would be the difference: 20 - 16 = 4 km/hr.

Correct answer (Option A):\nWhen two objects move in opposite directions on a track, their relative speed is the sum of their individual speeds.\nFormula: Relative Speed = Speed 1 + Speed 2\nGiven speeds: 20 km/hr and 16 km/hr\nRelative Speed = 20 + 16 = 36 km/hr\nConvert speed to m/s: 36 × (5 / 18) = 10 m/s\nTrack length (Distance) = 2200 m\nFormula: Time = Distance / Relative Speed\nStep 1: Time = 2200 / 10 = 220 seconds\nStep 2: Convert 220 seconds into minutes and seconds\n220 seconds = 3 minutes and 40 seconds\nOption A is correct.\n\nWhy others are wrong:\nOption B (3 mins 20 secs) matches 200 seconds, missing the extra 20 seconds needed.\nOption C (2 mins 50 secs) equals 170 seconds, which represents a calculation breakdown.\nOption D (2 mins 20 secs) equals 140 seconds, which fails to account for the actual combined tracking speed.\n\nStudy tip:\nFor circular tracks, the time taken to meet for the first time anywhere on the track is always Distance / Relative Speed. If they were traveling in the same direction, the relative speed would be the difference: 20 - 16 = 4 km/hr.

Correct answer (Option D):\nLet ∠H = x. In ΔHGK, since HG = GK, the angles opposite to these sides are equal: ∠HKG = ∠H = x.\nBy the exterior angle theorem, ∠KGJ = ∠H + ∠HKG = x + x = 2x.\nIn ΔGKJ, since GK = JK, the angles opposite to these sides are equal: ∠GJK = ∠KGJ = 2x.\nSince G lies on line HJ, triangle HJK is the same as triangle HJK where ∠H = x and ∠J = 2x. We are given HJ = HK, which means the angles opposite to these sides are equal: ∠HKJ = ∠HJK = 2x.\nIn ΔHJK, the sum of angles is 180°:\nx + 2x + 2x = 180°\n5x = 180° → x = 36°\nNow evaluate the components:\n∠HGK is the exterior angle to ΔGKJ, so ∠HGK = 180° - ∠KGJ = 180° - 2x = 180° - 72° = 108°.\nIn ΔGKJ, ∠GKJ = 180° - (∠KGJ + ∠GJK) = 180° - (2x + 2x) = 180° - 4x = 180° - 144° = 36°.\nCalculate the sum inside the brackets: ∠HGK + ∠GKJ = 108° + 36° = 144°.\nFind two-third of this sum: (2 / 3) × 144° = 2 × 48° = 96°.\nOption D is correct.\n\nWhy others are wrong:\nOption A (84°) results from a miscalculation of the base angles.\nOption B (136°) does not scale correctly with the fractional multiplier.\nOption C (90°) is an arbitrary placeholder value.\n\nStudy tip:\nDraw isosceles triangles carefully step-by-step and track identical side notations to systematically map equal variables to unknown angles before evaluating linear constraints.