# Math word problems

• Dec 8th 2009, 01:27 PM
brownjovi
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.
• Dec 8th 2009, 03:03 PM
skeeter
Quote:

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 $r$ = row rate in km/hr

$t$ = time in hrs upstream

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

$(r-3)t = 3$

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

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

sub this into the 2nd equation ...

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

solve for $r$
• Dec 9th 2009, 04:18 AM
HallsofIvy
Since we do not know the time taken to row upstream and downstream separately, let " $t_1$" be the time taken to row upstream and let " $t_2$" be the time taken to row downstream. Let "v" be their speed in still water.

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

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

Those, together with $t_1+ t_2= 80$ give three equations to solve for the three values $t_1$, $t_2$, and v. Of course, you are only asked for v so if you could find a way to eliminate $t_1$ and $t_2$, reducing to a single equation in v, that would be excellent.
• Dec 9th 2009, 12:04 PM
brownjovi
Quote:

Originally Posted by skeeter

sub this into the 2nd equation ...

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

How would I remove the r-3 from

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

?
• Dec 9th 2009, 02:28 PM
skeeter
Quote:

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

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

?

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

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

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

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

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

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

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