This math problem really has me puzzled, please help?

Question

Debbie traveled by boat 5 miles upstream to fish in her favorite spot. Because of the 4-mph current, it took her 20 minutes longer to get there than to return. How fast will her boat go in still water?

Answer 1

let speed of boat in still water be w
let the time taken to return be t
distance = speed x time
Then going there will be (w-4)(t+[1/3]) = 5
Also returning (w+4)(t) = 5
2 equations 2 unknowns

From 2nd eqn
t = 5/(w+4)

Put into first eqn
(w-4)[(19+w)/(3w+12)]=5
(w-4)(19+w)=5(3)(w+4)
w^2 +15w - 76 = 15w + 60
w^2 = 136
w = 11.66 mph

Speed in still water is 11.66 mph

Answer 2

Let still water speed = x mph
Slow speed (against current) = (x - 4) mph
Fast speed (with current) = (x + 4) mph
t1 = 5 / (x - 4) is time against current.
t2 = 5 / (x + 4) is time with current.
5 / (x - 4) - 5 / (x + 4) = 20
5.(x + 4) - 5 .(x - 4) = 20.(x - 4).(x + 4)
40 = 20x² - 320
20x² = 360
x² = 18
x = 3√2
Speed in still water = 3√2 mph

Answer 3

you must use distance = rate * time

Start by labeling all variables and converting units.
distance = d = 5 miles
let b = boat rate
current c = 4 mph
time = t down
20 minutes = 1/3 hour

(rate up* time up) = distance up
(rate down * time down) = dist down

(b-4) (t + 1/3) = 5
(b+4)(t) = 5

Two equations with two unknowns, solve by method of your choice.

I will solve for t in eqn II and sub into eqn I

t = 5/ (b+4) ==> put in other eqn

(b-4) (5 / (b+4) + 1/3) = 5 ===> solve for b, multiply by 3(b+4)

15(b-4) +(b-4)(b+4) = 15(b+4)
15b - 60 + b^2 - 16 = 15b + 60
b^2 = 136
b = 2sqrt34 miles per hour

ANSWER: Boat goes 2sqrt34 miles per hour
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check:
time = dist/rate
time up = 5 / (2sqrt34 - 4) about=
time down = 5 / 2sqrt34 about =

note these are about 1/3 of an hour (20 min) apart.

Answer 4

unknown because u ddnt put the time of traveling without the current