Mathematics

The effective speed downstream is the sum of the boat's speed in still water and the stream speed: 20 kmph + 6 kmph = 26 kmph. The time taken to cover 60 km downstream is distance divided by speed, which is 60/26 = 30/13 hours. This matches option D. Therefore, the correct answer is D.

Upstream speed = 3/3 = 1 km/hr and downstream speed = 15/3 = 5 km/hr. Let speed in still water be b and speed of current be c. Then b − c = 1 and b + c = 5. Solving gives b = 3 and c = 2 km/hr. Hence, the speed of the current is 2 km/hr.

The man's effective speed downstream is the sum of his rowing speed in still water and the speed of the current: 15 km/h + 3 km/h = 18 km/h. Converting this speed to meters per second: (18 * 1000) / 3600 = 5 m/s. To cover 60 meters at 5 m/s, the time taken is distance/speed = 60 / 5 = 12 seconds. Therefore, the correct answer is D (12 seconds).

The man rows at 7.5 km/hr in still water. It takes him twice as long to row upstream as downstream, meaning the time upstream is double the time downstream for the same distance. Let the speed of the stream be s km/hr. Then upstream speed = 7.5 - s and downstream speed = 7.5 + s. Since time is distance/speed, and time upstream is twice time downstream, we have: (distance / (7.5 - s)) = 2 × (distance / (7.5 + s)). Simplifying, (7.5 + s) = 2(7.5 - s), which leads to 7.5 + s = 15 - 2s, or 3s = 7.5, so s = 2.5 km/hr. Therefore, the rate of the stream is 2.5 km/hr, which corresponds to option B.

Compound interest formula: CI = P[(1 + r)^n − 1]. Here CI = 33, r = 10% = 0.1, n = 3. So, (1.1)^3 = 1.331, hence CI = P(1.331 − 1) = 0.331P. Therefore, P = 33 / 0.331 ≈ 100. Hence, the correct principal is Rs. 100.

The speed downstream is the sum of the boat's speed in still water and the current speed: 60 + 20 = 80 kmph. The speed upstream is the difference: 60 - 20 = 40 kmph. Therefore, the correct answer is 80 kmph downstream and 40 kmph upstream.

The effective speed downstream is 9 + 1 = 10 kmph, and upstream is 9 - 1 = 8 kmph. Let the distance be d km. Total time = d/10 + d/8 = 4.5 hours. Solving gives d = 32 km.

Let the amounts with P and Q be x. According to the problem, R has two-thirds of the total amount with P and Q combined, so R = (2/3)x. The total amount among P, Q, and R is Rs. 6000, so we have x + R = 6000. Substituting R, we get x + (2/3)x = 6000, which simplifies to (5/3)x = 6000. Solving for x gives x = (6000 * 3)/5 = 3600. Therefore, R's amount is (2/3)*3600 = 2400. Hence, the amount with R is Rs. 2400, which corresponds to option C.

Let the total capacity be x litres. Initially, milk and water are in ratio 4:5, so milk = 4k and water = 5k with 9k = x. Adding 8 litres of milk fills the can, so new milk = 4k + 8, water = 5k, total = x. The new ratio is 6:5, so (4k + 8)/5k = 6/5. Solving gives k = 8, so x = 9k = 72 litres. However, 72 is not an option, so re-checking: The ratio after addition is 6:5, so (4k + 8)/(5k) = 6/5 => 5(4k + 8) = 6*5k => 20k + 40 = 30k => 40 = 10k => k = 4. So capacity x = 9k = 36 litres. But 36 is not an option either. Re-examining the problem: The can is filled after adding 8 litres milk, so total capacity = initial volume + 8 litres milk. Initial volume = 9k, so total capacity = 9k + 8. But the problem states the can is full after adding 8 litres milk, so total capacity = 9k + 8. The new ratio is (4k + 8) : 5k = 6 : 5. Cross-multiplied: 5(4k + 8) = 6*5k => 20k + 40 = 30k => 40 = 10k => k = 4. So initial volume = 9*4 = 36 litres, total capacity = 36 + 8 = 44 litres. Therefore, the capacity of the can is 44 litres, which corresponds to option B. Hence, the correct answer is B.

The word TIME has 4 distinct letters. The number of three-letter words that can be formed without repetition is the number of permutations of 4 letters taken 3 at a time, which is calculated as 4P3 = 4 × 3 × 2 = 24. Therefore, 24 distinct three-letter words can be formed from the letters of TIME. This matches option D.