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Question 1 Arithmetic & Reasoning
The L.C.M. of two numbers is 2079 and their H.C.F. is 27. If one of the number is 189, find the other:
- A. 299
- B. 197
- C. 295
- D. 297
Correct answer: D. 297
Correct answer (Option D):\nFormula: Product of two numbers = L.C.M. × H.C.F.\nGiven values:\nL.C.M. = 2079\nH.C.F. = 27\nFirst number (A) = 189\nLet the second number be B.\nStep 1: 189 × B = 2079 × 27\nStep 2: B = (2079 × 27) / 189\nStep 3: B = 56133 / 189 = 297\nTherefore, the other number is 297.\nOption D is correct.\n\nWhy others are wrong:\nOptions A (299), B (197), and C (295) do not satisfy the mathematical product relationship between H.C.F. and L.C.M. for the given number 189.\n\nStudy tip:\nAlways remember that the fundamental relationship Product = L.C.M. × H.C.F. holds true for any two positive integers. This is a very frequently asked pattern in Kerala PSC exams.
Question 2 Arithmetic & Reasoning
A can do a piece of work in 6 days and B can do in 4 days. How long will they take if both work together?
- A. 5/12
- B. 2 2/5
- C. 1 1/2
- D. None of these
Correct answer: B. 2 2/5
Correct answer (Option B):\nFormula for combined work rate: Time taken = (A × B) / (A + B)\nGiven values:\nTime taken by A = 6 days\nTime taken by B = 4 days\nStep 1: Total combined days = (6 × 4) / (6 + 4)\nStep 2: Combined days = 24 / 10\nStep 3: Combined days = 12 / 5 = 2 2/5 days\nThus, together they will take 2 2/5 days.\nOption B is correct.\n\nWhy others are wrong:\nOption A represents the combined work done in one day if we add their daily work rates incorrectly. Option C (1 1/2) is too small. Option D is incorrect since the exact value is present in Option B.\n\nStudy tip:\nAlternatively, use the LCM method. Total work = LCM of 6 and 4 = 12 units. A's efficiency = 2 units/day, B's efficiency = 3 units/day. Total efficiency = 5 units/day. Time = 12/5 = 2 2/5 days.
Question 3 Arithmetic & Reasoning
30% of a number is 120. Which is the number?
- A. 300
- B. 400
- C. 500
- D. 450
Correct answer: B. 400
Correct answer (Option B):\nLet the unknown number be x.\nGiven: 30% of x = 120\nStep 1: (30 / 100) × x = 120\nStep 2: 0.3 × x = 120\nStep 3: x = 120 / 0.3\nStep 4: x = 1200 / 3 = 400\nThe required number is 400.\nOption B is correct.\n\nWhy others are wrong:\nOption A (300) would yield 30% as 90. Option C (500) would yield 30% as 150. Option D (450) would yield 30% as 135. None of these equal 120.\n\nStudy tip:\nFor quick mental calculation, if 30% is 120, then 10% is 120 / 3 = 40. Therefore, 100% of the number must be 40 × 10 = 400.
Question 4 Arithmetic & Reasoning
Simplify 0.25 + 0.036 + 0.0075:
- A. 0.2935
- B. 0.02935
- C. 0.00136
- D. 0.0136
Correct answer: A. 0.2935
Correct answer (Option A):\nTo solve decimal addition, align the decimal points vertically:\n 0.2500\n+ 0.0360\n+ 0.0075\n--------\n 0.2935\nThus, the sum is 0.2935.\nOption A is correct.\n\nWhy others are wrong:\nOption B is missing a place value shift, making it 10 times too small. Options C and D are completely mathematically unrelated and arise from incorrect digit placement during alignment.\n\nStudy tip:\nWhen adding decimals with an unequal number of digits after the decimal point, add trailing zeros to make balancing and vertical calculation straightforward.
Question 5 Arithmetic & Reasoning
Simplify (0.2 × 0.2 + 0.01 × 0.01) / 0.401
- A. 1.0
- B. 0.01
- C. 0.1
- D. 0.001
Correct answer: C. 0.1
Correct answer (Option C):\nLet us compute step by step using basic arithmetic operations:\nStep 1: Calculate the numerator parts:\n0.2 × 0.2 = 0.04\n0.01 × 0.01 = 0.0001\nStep 2: Add the numerator values:\n0.04 + 0.0001 = 0.0401\nStep 3: Perform division:\nNumerator / Denominator = 0.0401 / 0.401\nStep 4: Shift decimal points to simplify:\n401 / 4010 = 1 / 10 = 0.1\nOption C is correct.\n\nWhy others are wrong:\nOption A (1.0) is incorrect because the denominator is ten times larger than the numerator. Options B and D represent incorrect decimal placements by factors of 10 and 100 respectively.\n\nStudy tip:\nWhen dividing decimals, multiplying both the numerator and denominator by a suitable power of 10 (like 10000 here) can eliminate decimal places entirely and avoid placement errors.