pra1785 wrote:
can someone explain what is wrong with my explanation below and why I am not getting the right answer?
Upstream
D= 90 miles
S= v-3 mph
T=H
Downstream
d= 90 miles
s= v+3
t=h
H=h+1/2
distance=speed x time
Upstream
90=(v-3) (h+1/2).....(i)
Downstream
90=(v+3)h ....(ii)
with equation i and ii
(v-3)(h+1/2)=(v+3)h
After this, I am stuck. Is the above logic right at all?
pra1785, in a different kind of question, the logic would work more efficiently. Your logic is not wrong here. It just makes things a lot harder, I think.
Distances are equal, so it does seem like a natural step to set each leg's (r * t) equal, but . . .
The problem is that you have defined one unknown variable in terms of another. You have two unknown variables and one equation. The answer choices are values, not variable expressions (in which case your equation, without more, might work, see below).
You will have to use a known value at some point. IMO, it's easier to do so at the beginning.
Your equation yields \(\frac{1}{2}v - 6h = \frac{3}{2}\)
"Solve" for time, which you have called h:
h = \(\frac{(v - 3)}{12}\)
Now you must "go back" and use the known value of 90.
h\(_2\) = \(\frac{(v-3)}{12}\)
h\(_1\) = \(\frac{90}{(v+3)}\)
\(\frac{(v-3)}{12}\) = \(\frac{90}{(v+3)}\)
\(v^2 - 9 = 1080\)
\(v^2 = 1089\)
\(v = 33\)
Finally (whew!), if h = \(\frac{(v - 3)}{12}\), then
\(\frac{(33 - 3)}{12}\) = h
h = \(\frac{30}{12}\) = 2.5 hrs
If answer choices WERE variable expressions, depending on the prompt, the variable-variable solution ("h = ...") to your equation as-is (i.e., not taken further as I did, with D = 90) would be among them.
But the answer choices have actual values. So you need one variable and one known in your equation. Known D / unknown rates = known time.
When distances are equal, it often works efficiently to write \(r_1*t_1 = r_2*t_2\). I don't think it's efficient here. But if you are very quick and very accurate with algebra, it's probably fine.
Does that help?
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