Alice rows a boat downstream for 273 miles. The return trip upstream took 18 hours longer. If the current flows at 3 mph, how fast does Alice row in still water?

Answer 1

Hi Candice, to solve this problem use the formula: D(istance) = R(ate) * T(ime). We know the speed of the stream, 3 miles per hour. We know the distance, 273 miles. We know it took 18 hours longer for Alice to return to her starting point.

Downstream:

D = R * T

273 = (r + 3) * T, the speed down stream is Alice's rowing speed, r, plus the speed of the water.

Solve for T by dividing both sides by r + 3:

273 / (r + 3) = T

Upstream:

D = R * T

273 = (r - 3) * (T + 18), the speed up stream is Alice's rowing speed, r, minus the speed of the water. In addition, it took Alice 18 more hours to row up stream.

Solve for T:

Divide both sides by r - 3

273 / (r - 3) = T + 18

Subtract 18 from both sides:

273 / (r - 3) - 18 = T

The T from the first equation must equal the T from the second equation, therefore:

273 / (r + 3) = 273 / (r - 3) - 18

Clear the fractions by dividing the entire equation by (r + 3) * (r - 3):

273 * (r - 3) = 273 * ( r + 3) - 18 * (r - 3) * (r - 3)

273r - 819 = 273r + 819 - 18 * (r² - 9)

The 273r terms cancel out, subtract 819 from both sides:

-1638 = -18 * r² + 162

Subtract 162 from both sides:

-1800 = -18 * r²

Divide both sides by -18:

r² = 100

Take the square root of both sides:

r = +/- 10, we can discard the solution r = -10 since the rowing speed cannot be negative

Alice's speed in still water is 10 miles per hour.

Questions?

Answer 2

273 = (R+3)*t = Rt + 3t

273 = (R-3)*(t+18) = Rt + 18R - 3t - 54

Subtracts Equation #2 MINUS Equation #1:

0 = 18R - 6t - 54

0 = 3R - t - 9

t = 3r-9

273 = (r+3)(3r-9)

273 = 3(r+3)(r-3)

91 = (r+3)(r-3)

91 = r^2 - 9

100 = r^2

10 = r <--- only positives allowed

check:

273 = (10+3)*t

273 = 13*t

t = 273/13 = 21

(10+3)*21 = 13*21 = 273

(10-3)*39 = 7*39 = 273

Her speed is 10.

It takes her 21 hours to go downstream and 39 hours to go upstream