# Help with a word problem

• Feb 16th 2014, 10:10 PM
Earl6282
Help with a word problem
It's embarrassing to say but I'm basically having to re-learn algebra after not touching the subject for about 5 years.
I'm going through ACT practice tests and this problem came up. The guide book helps you put the word problem into equation form but just assumes you know what to do from there to solve for X. And this one has been driving me crazy.

"Amy drove 200 miles at an average speed of 10 mph faster than her usual average speed. If she completed the trip in 1 hour less than usual, what is her usual driving speed?"
(r) is her usual speed.
Using the "time= distance/rate" formula, it guides me to come up with an equation of:

(200/r+10) = (200/r)+1
parentheses only there to distinguish fractions from integers

I know the answer, I've seen the key. It's 40. It's the process of getting to the answer that I can't figure out for the life of me, and it looks like it should be very obvious.
Best I could do was fill up half a page of scrap paper of breaking it out and breaking it back down again until I could "simplify" it down to r2+10r=2000, but even from there I don't know what to do. It took way more time than anyone should spend on one problem during a timed ACT test, so I was wondering if someone could walk me through this, preferably with shortcuts. It would be greatly appreciated.
• Feb 16th 2014, 11:18 PM
romsek
Re: Help with a word problem
Quote:

Originally Posted by Earl6282
It's embarrassing to say but I'm basically having to re-learn algebra after not touching the subject for about 5 years.
I'm going through ACT practice tests and this problem came up. The guide book helps you put the word problem into equation form but just assumes you know what to do from there to solve for X. And this one has been driving me crazy.

"Amy drove 200 miles at an average speed of 10 mph faster than her usual average speed. If she completed the trip in 1 hour less than usual, what is her usual driving speed?"
(r) is her usual speed.

Let her usual speed be s mph.

Her usual time is then D/s = Ts = 200/s hrs

Now at speed s+10 she finishes 1 hr faster than usual i.e.

Tf = 200/s - 1

Now Tf is how long it takes to drive 200 miles at the faster speed (s+10)

Tf = 200/(s+10) = 200/s - 1

Now it's just grind through some algebra

$$\frac{200}{s+10}=\frac{200}{s}-1$$

$$1=\frac{200}{s}-\frac{200}{s+10}$$

$$1=\frac{200s+2000-200s}{s(s+10)}$$

$$1=\frac{2000}{s^2+10s}$$

$$s^2+10s-2000=0$$

$$(s+50)(s-40)=0$$

$$s=-50\mbox{ or }s=40$$

since we assume a positive speed only s=40 makes sense
• Feb 17th 2014, 06:54 AM
Earl6282
Re: Help with a word problem
That second to last step, (s+50)(s-40) = 0

Is there a trick to coming up with those two numbers?
• Feb 17th 2014, 07:13 AM
romsek
Re: Help with a word problem
Quote:

Originally Posted by Earl6282
That second to last step, (s+50)(s-40) = 0

Is there a trick to coming up with those two numbers?

$$(x-r_1)(x-r_2) = x^2+(r_1+r_2)x+ r_1 r_2$$

so if you've got an equation in the form $$x^2+ bx + c$$ you need to find $$r_1 \mbox{ and }r_2$$ such that

$$r_1+r_2=b \mbox{ and } r_1 r_2 = c$$

in this case 2 numbers such that

$$r_1+r_2=10 \mbox{ and } r_1 r_2 = 1000$$

2000 only has a handful of factors that are even close matching the first part so a little trial and error quickly gets you the right answer.

A bit more systematic way of solving this is using the quadratic formula. If you have

$$a x^2 + b x + c=0$$

then the roots of the equation are given by

$$r_1,\;r_2=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

here a=1, b=10, c=-2000

$$r_1,\;r_2=\frac{-10 \pm \sqrt{100+8000}}{2} =$$

$$\frac{-10 \pm 90}{2} = 40\mbox{ and }-50$$