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

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?

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

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

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