1. ## Math word problems

Hi, new here. I got a question here that I do not understand how to do at all. It'd be nice if someone explained how to solve it.

It took a crew 80 minutes to row 3km upstream and back again. If the rate of the flow of the stream was 3km/h, what was the rowing rate of the crew.

I did:

3=(1.33)(x+3)
3=(1.33)(x-3)

I was going to solve by substitution but I realized that 1.33 cannot be the time for both going up and downstream.

2. Originally Posted by brownjovi
Hi, new here. I got a question here that I do not understand how to do at all. It'd be nice if someone explained how to solve it.

It took a crew 80 minutes to row 3km upstream and back again. If the rate of the flow of the stream was 3km/h, what was the rowing rate of the crew.

I did:

3=(1.33)(x+3)
3=(1.33)(x-3)

I was going to solve by substitution but I realized that 1.33 cannot be the time for both going up and downstream.
let $\displaystyle r$ = row rate in km/hr

$\displaystyle t$ = time in hrs upstream

$\displaystyle \frac{4}{3} - t$ = time in hrs downstream

$\displaystyle (r-3)t = 3$

$\displaystyle (r+3)\left(\frac{4}{3} - t\right) = 3$

from the 1st equation ... $\displaystyle t = \frac{3}{r-3}$

sub this into the 2nd equation ...

$\displaystyle (r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3$

solve for $\displaystyle r$

3. Since we do not know the time taken to row upstream and downstream separately, let "$\displaystyle t_1$" be the time taken to row upstream and let "$\displaystyle t_2$" be the time taken to row downstream. Let "v" be their speed in still water.

Then their speed upstream is $\displaystyle v- \frac{4}{3}$ and that equals "distance divided by time: $\displaystyle v- \frac{4}{3}= \frac{3}{t_1}$.

Their speed downstream is $\displaystyle v+ \frac{4}{3}$ and so $\displaystyle v+ \frac{4}{3}= \frac{3}{t_2}$.

Those, together with $\displaystyle t_1+ t_2= 80$ give three equations to solve for the three values $\displaystyle t_1$, $\displaystyle t_2$, and v. Of course, you are only asked for v so if you could find a way to eliminate $\displaystyle t_1$ and $\displaystyle t_2$, reducing to a single equation in v, that would be excellent.

4. Originally Posted by skeeter

sub this into the 2nd equation ...

$\displaystyle (r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3$
How would I remove the r-3 from

$\displaystyle (\frac{4}{3} - \frac{3}{r-3})$

?

5. Originally Posted by brownjovi
How would I remove the r-3 from

$\displaystyle (\frac{4}{3} - \frac{3}{r-3})$

?
$\displaystyle (r+3)\left(\frac{4}{3} - \frac{3}{r-3}\right) = 3$

$\displaystyle (r+3)\left(\frac{4(r-3)}{3(r-3)} - \frac{9}{3(r-3)}\right) = 3$

$\displaystyle (r+3)\left(\frac{4(r-3)-9}{3(r-3)}\right) = 3$

$\displaystyle (r+3)\left(\frac{4r-21}{3(r-3)}\right) = 3$

$\displaystyle \frac{(r+3)(4r-21)}{3(r-3)} = 3$

$\displaystyle (r+3)(4r-21) = 9(r-3)$

expand both sides, combine like terms and solve the resulting quadratic equation.