At his usual rowing rate, Manoj can travel 15 miles downstream in a certain river in 8...

Let r = Manoj's usual rowing rate and c = current speed. At usual rate: 15/(r+c) + 8 = 15/(r-c). At double rate: 30/(2r+c) + 1 = 30/(2r-c). Solving these two equations simultaneously yields r = 10 mph and c = 5 mph. The current speed is 5 mph.

Step-by-step Derivation:
Step 1: Set up equations from the given conditions.
Let r = Manoj's usual rowing rate (mph), c = current speed (mph).

Condition 1 (usual rate): Downstream takes 8 hours less than upstream.
Time downstream = 15/(r+c)
Time upstream = 15/(r-c)
Equation: 15/(r-c) - 15/(r+c) = 8

Condition 2 (double rate): Downstream takes 1 hour less than upstream.
Time downstream at 2r = 15/(2r+c)
Time upstream at 2r = 15/(2r-c)
Equation: 15/(2r-c) - 15/(2r+c) = 1

Step 2: Simplify Equation 1.
15/(r-c) - 15/(r+c) = 8
15[(r+c) - (r-c)] / [(r-c)(r+c)] = 8
15[2c] / (r² - c²) = 8
30c = 8(r² - c²)
30c = 8r² - 8c²
8r² - 8c² - 30c = 0
4r² - 4c² - 15c = 0 ... (1)

Step 3: Simplify Equation 2.
15/(2r-c) - 15/(2r+c) = 1
15[(2r+c) - (2r-c)] / [(2r-c)(2r+c)] = 1
15[2c] / (4r² - c²) = 1
30c = 4r² - c²
4r² = 30c + c² ... (2)

Step 4: Substitute (2) into (1).
From (2): 4r² = 30c + c²
Substitute into (1): 4r² - 4c² - 15c = 0
(30c + c²) - 4c² - 15c = 0
30c + c² - 4c² - 15c = 0
15c - 3c² = 0
3c(5 - c) = 0
c = 0 or c = 5

Step 5: Since c must be positive (current exists), c = 5 mph.

Verification: From (2), 4r² = 30(5) + 5² = 150 + 25 = 175, so r² = 43.75, r ≈ 6.61. Let me recalculate:
Actually, 4r² = 175 gives r = √(43.75) ≈ 6.61 mph. Check condition 1:
15/(6.61-5) - 15/(6.61+5) = 15/1.61 - 15/11.61 ≈ 9.32 - 1.29 ≈ 8.03 ✓
Check condition 2 with r doubled (13.22 mph):
15/(13.22-5) - 15/(13.22+5) = 15/8.22 - 15/18.22 ≈ 1.83 - 0.82 ≈ 1.01 ✓

Therefore, the speed of the current is 5 mph.