A speedboat takes 1 hour longer to go 24 miles up a river th

A speedboat takes 1 hour longer to go 24 miles up a river than to return. If the boat cruises at 10 miles per hour in still water, what is the rate of the current?

If R*T=D, and R is speed + or - current, how can you find the rate without the time, and how can you find the time without the rate?

Solution

Let x=rate of current
distance(d)=rate(r) times time(t) or d=rt or t=d/r
distance upstream=distance downstream=24 mi
rate upstream=10-x
rate downstream=10+x
time upstream=(distance upstream)/(rate upstream)=24/(10-x)
time downstream=(distance downstream)/(rate downstream)=24/(10+x)
Now, we are told that time upstream minus one hour equals time downstream. So our equation to solve is:
24/(10-x)-1=24/(10+x) Multiply both sides by (10-x)(10+x)
24(10+x)(10-x)/(10-x)-1(10-x)(10+x)=24(10-x)/(10+x)/(10+x) clear fractions
24(10+x)-(10-x)(10+x)=24(10-x) clear parens
240+24x-100-10x+10x+x^2=240-24x subtract 240 from and add 24x to both sides
240-240+24x+24x-100+x^2=240-240-24x+24x collect like terms
x^2+48x-100=0 ----------------quadratic equation in standard form
This equation can be factored:
(x+50)(x-2)=0
x=-50 mph----------------discount the negative value for speed
and
x=2 mph----------------------speed of the current
CK
10-x=10-2= 8 mph -----------------------speed upstream
10+x=10+2=12 mph--------------------------speed downstream
time upstream=24/8=3 hours
time downstream=24/12=2 hours
so 3-2=1----takes 1 hour longer upstream