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kristina0812 · Dec 21, 2011, 10:18 PM

John traveled 400 km from Manila to a nearby province. If in returning home he reduced his speed by 9 km and able to reach Manila 2 hours more time than going to the province, what was his original speed?

corrigan · Dec 22, 2011, 06:11 PM

okay, I had a typo, and it messed everything up.

here we go again:

let $s$ be the original speed and let $t$ be the original time.
So we have $s \cdot t =400$ and $(t+2)(s-9)=400$

They both equal 400 because that is the distance he traveled both to and from Manilla.
So we set them equal to each other and get:
$t \cdot s = (t+2)(s-9)$
[math] \Rightarrow /, t cdot s = t \cdot s -9t + 2s -18 [\math]
[math] \Rightarrow /, 0 = -9t + 2s -18 [\math]
[math] \Rightarrow /, 9t = 2s -18 [\math]
[math] \Rightarrow /, t = \frac{1}{9}(2s -18) [\math]

So now we have $t$ in terms of $s$ , and we just plug that into the original equation and get:

[math] \frac{1}{9}(2s -18) \cdot s = 400 [\math]
[math] \Rightarrow \, (2s -18) \cdot s = 3600 [\math]
[math] \Rightarrow \, 2s^2 -18 s = 3600 [\math]
[math] \Rightarrow \, s^2 -9 s = 1800 [\math]
[math] \Rightarrow \, s^2 -9 s - 1800 = 0 [\math]

Now we apply the quadratic formula and get:
$A=1, B=-9, C=-1800$

$s = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 1 \cdot (-1800)}}{2 \cdot 1}$

$\, = \frac{9 \pm \sqrt{7200}}{2}$

$\, = \frac{9 \pm 60\sqrt{2}}{2}$

$\, = \frac{9]{2} \pm 30\sqrt{2}$

Now since $\frac{9]{2} - 30\sqrt{2}$ is negative and we can't go a negative speed, are only viable answer is

$s = \frac{9]{2} + 30\sqrt{2}$

I hope this helps.

corrigan · Dec 22, 2011, 06:16 PM

wow, I just can't get anything right today :(

your answer is $\frac{9}{2} + 30 \sqrt{2}$ , If you can wade through my horrible proofreading skills above, I explained it.