Math and Puzzles

Let the youngest child's age be x. The ages are x, x+3, x+6, x+9, x+12. Sum = 5x + 30 = 50, so 5x = 20 and x = 4. The arithmetic progression with common difference 3 is the key structure. Checking: 4 + 7 + 10 + 13 + 16 = 50.

Let the number be 10a + b where a is tens digit and b is units digit. Difference: (10a + b) - (10b + a) = 9a - 9b = 9(a - b) = 36. Therefore a - b = 4. The digit difference is always 4, regardless of which digits (as long as a > b to keep the difference positive). The key is that the difference is always a multiple of 9.

Since the lily doubles every day, if the pool is fully covered on day 20, then on day 19 it must have been exactly half-covered. Doubling half gives full coverage. The common mistake is thinking it's a linear growth problem, but it's exponential. Working backwards: day 20 = full, day 19 = half, day 18 = quarter, etc.

The snail climbs net 1 meter per day (5 up, 4 down). After 15 days: 15 meters reached. On day 16, climbs 5 meters from 15 to reach 20 meters and exits - doesn't slide back. The trick is that on the final day, the snail reaches the top before nightfall. Common trap: dividing 20/1 = 20, but this ignores the final ascent.

The largest circle that fits inside a square is the inscribed circle, whose diameter equals the side of the square. With side 14m, radius = 7m. Area = πr² = π × 49 ≈ 154 sq m.

To find 30% of 1986, multiply 1986 by 0.30. 1986 * 0.3 = 595.8. The calculation is straightforward and matches option 381919.

50 times 5 paise is 250 paise per day. Over 50 days, the total is 250 * 50 = 12,500 paise. Since 100 paise equals 1 rupee, 12,500 paise equals 125 rupees.

Total distance = 120km, total time = 20min + 40min = 60min = 1 hour. Average speed = total distance / total time = 120 km / 1 hour = 120 km/h. The speeds of individual segments (144 km/h and 108 km/h) don't affect the overall average.

To find 1% of any number, divide the number by 100 or move the decimal point two places to the left. 1354 ÷ 100 = 13.54. This is a fundamental percentage calculation.

When two trains approach each other, their relative speed equals the sum of individual speeds. Distance = 55 km, time = 15 min = 0.25 hours. Relative speed = 55/0.25 = 220 kmph. If one train moves at 120 kmph, the other must be 220 - 120 = 100 kmph for them to meet in 15 minutes.

This is a divisibility problem involving a 101-digit number with 50 twos, then one middle digit, then 50 threes. For such a number to be divisible by 7, the middle digit must be 5. This follows from the mathematical property that 222...25333... (with equal counts of 2s and 3s) is divisible by 7.

The square root of 3.6 is approximately 1.897, not 0.6. Additionally, if you square 0.6, you get 0.36, not 3.6. The statement incorrectly suggests that 0.6 x 0.6 = 3.6, when actually 0.6 x 6 = 3.6. The correct relationship is 1.897 x 1.897 = 3.6.

Let CP of 1 article be 1. CP of 12 articles = 12. Selling 12 at 80% profit gives SP = 12 x 1.8 = 21.6. But for every 12 bought, customer gets 15 (12 paid + 3 free). Effective SP per article = 21.6/12 = 1.8. Total SP for 15 articles = 21.6. CP of 15 articles = 15. Profit = 21.6 - 15 = 6.6. Profit% = 6.6/15 x 100 = 44%.

For equal distances, average speed is the harmonic mean, not arithmetic mean. Total distance = 2d. Time for first half = d/40, time for second half = d/60. Total time = d/40 + d/60 = 5d/120. Average speed = 2d / (5d/120) = 240/5 = 48 kmph. The arithmetic mean (50 kmph) is incorrect.

The series is: 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, ... We sum digits sequentially: 1 (sum=1), 2+2=4 (sum=5), 3+3+3=9 (sum=14), 4x4=16 (sum=30), 5x5=25 (sum=55), 6x6=36 (sum=91), 7x7=49 (sum=140), 8x8=64 (sum=204), 9x9=81 (sum=285), then we need 15 more. The next digits are 1, 1, 1,... and 1+1+1+1+1+1+1+1+1+1+1+1+1+1+1 = 15. So the 300th digit is 1.

Let the father's age be F and son's age be S. 3 years ago: F-3 = 7(S-3), so F = 7S - 18. Present: F = 5S. Equating: 5S = 7S - 18, giving 2S = 18, so S = 9. Then F = 5×9 = 45. Checking: 3 years ago, father was 42 and son was 6, and 42 = 7×6. Option C (45 & 9) is correct.

Total age of 30 boys = 30 × 14 = 420 years. When teacher is included, 31 people have average 15, so total = 31 × 15 = 465 years. Teacher's age = 465 - 420 = 45 years. This is a standard average problem where you work backwards from totals.

Total runs after 10 innings = 10 × 21.5 = 215. After 11 innings at average 24, total needed = 11 × 24 = 264. Runs required in 11th inning = 264 - 215 = 49. This tests understanding of how averages change with new data points.

Profit sharing ratio is based on investment × time. A: 1000×4=4000, B: 1500×8=12000, C: 2000×10=20000. Total = 36000. A's share = (4000/36000)×585 = 65, B's share = (12000/36000)×585 = 195, C's share = (20000/36000)×585 = 325.