A motorboat travels 118km in 2 hours going upstream. It travels190 going downstream in the same amount of time. What is the rate of the boat in still water and what is the rate of the current?

Question

lgbasallote · Answer

upstream speed = speed of motorboat (in still water) - current
downstream speed = speed of motorboat (in still water) + current

let b = speed of boat in still; c = current

you are given a distance and some time...you're looking for the rate of boat in still and rate of current...

Hint: first thing you'll do here is to solve for the downstream speed and the upstream speed..do you know how?

lgbasallote · Answer

to find downstream speed...you  use rate = $\large \frac{distance}{time}$

dumbcow · Answer

side note: i have seen so many of these problems, they need to come up with different rate problems

anonymous · Answer

thnx Igbasallote, that helped

lgbasallote · Answer

well..i'll continue anyway :DDD so downstream speed = $\Large \frac{190}{2}$ = 95 km/h

upstream speed = $\Large \frac{118}{2}$ = 59 km/h

we put this back in the system...

95 km/h = b + c <----where b is boat speed; c is current speed
59 km/h = b - c

we can use substitution method here by letting c = 95 - b..we put htat in the second equation...

59 = b -(95 - b)
59 = b -95 + b
95 + 59 = 2b
154 = 2b
b = 77 km/h

knowing this... we can find c..

95 = 77 + c
95 - 77 = c
18 km/h = c

anonymous · Answer

wow ur good. thnx!

lgbasallote · Answer

heh thanks ^_^ <tips hat> hope that explanation was clear enough :DDD

anonymous · Answer

does (-2v^2 +7v-6)+(v^2 +5v+4) equal -v^2 +12v-2?

lgbasallote · Answer

$\Huge \mathbb{YEP!}$ $\Huge \checkmark$

anonymous · Answer

lol. ty

anonymous · Answer

$$\sqrt{-40}/\sqrt{-2}=?  and \sqrt{-5}*\sqrt{-20}=?$$

lgbasallote · Answer

$\huge \frac{\sqrt{40}}{\sqrt{-2}}$ = $\huge \sqrt{\frac{40}{-2}}$

as for multiplication...

$\huge \sqrt{-5} * \sqrt{-20} = \sqrt{(-5)(-20)}$

does that give you any hints?

anonymous · Answer

rectangle length is 5 times twice its width, and the area is88ft^2. wat are the length and width of the rectangle?

lgbasallote · Answer

let L = length; w = width
L = 5(2w)
a = 88
a = lw
a = 5(2w)(w)

substitute values..

anonymous · Answer

gtg