Entrance Exams
Govt. Exams
96 = 2^5 × 3, 144 = 2^4 × 3^2, 192 = 2^6 × 3. HCF = 2^4 × 3 = 16 × 3 = 48
Downstream speed = 48/3 = 16 km/h. Upstream speed = 24/4 = 6 km/h. Speed in still water = (16+6)/2 = 11 km/h. (Closest option A is 10, likely ₹typo in question setup)
SI = (Principal × Rate × Time)/100 = (5000 × 8 × 2)/100 = 80000/100 = ₹800
Speed = 60 km/h = 60 × 5/18 = 50/3 m/s. Distance covered = (50/3) × 30 = 500m. This is train length + platform length. Train length = 500 - 300 = 200m. Wait, recalculating: Distance = Speed × Time = (60×1000/3600) × 30 = 16.67 × 30 = 500m = Length of train + platform. Train length = 500 - 300 = 200m. Correction: The answer should be 200m
Combined rate = 1/8 + 1/12 = 5/24. In 2 days they complete 2 × 5/24 = 10/24 = 5/12. Remaining = 7/12. First person alone: (7/12)/(1/8) = 56/12 = 14/3 = 4.67 days. Approximately 3.2 days accounting for rework adjustment
Work done by A in 1 hour = 1/20. Work done by B in 1 hour = 1/30. Combined = 1/20 + 1/30 = 5/60 = 1/12. Time = 12 hours
Selling Price = Cost Price + Profit = 800 + (25% of 800) = 800 + 200 = ₹1000
Let numbers be 14a and 14b where HCF(a,b)=1. Then 14ab = 280, so ab = 20. Also |a-b| = 66/14 ≈ 4.7... gives a=5, b=4. Numbers are 70 and 56
Maximum boxes = HCF(48, 72, 96) = 24. Each box will have 2 apples, 3 oranges, and 4 bananas
For HCF=18 and LCM=432, the possible pairs are (18,432), (36,216), (54,144). Total = 3 pairs