## Tuesday, April 3, 2012

### this week's MGRE Math Beast Challenge

From here:

A predator is 80 meters behind its prey, which is running away at a rate of 40 kilometers per hour. If the predator chases at 48 kilometers per hour and both animals run along a straight-line path at their respective constant rates, how long will it take, in seconds, for the predator to catch the prey? (1 kilometer = 1,000 meters)

(A) 10

(B) 24

(C) 36

(D) 80

(E) 100

Go to it! My answer will appear in the comments.

_

#### 5 comments:

Kevin Kim said...

When dealing with "work/rate" problems, remember this basic formula, for which I use the mnemonic "DIRT":

d = rt

i.e., Distance equals rate times time.

[NB: for work, just change "d" to "w" and remember "WART":
w = rt.]

In the problem we've been given, we know the rates. We also know that the times have to be equal, because the predator catches up to its prey. (I'm imagining a really slow cheetah for some reason.) To find the distance, we can set the times equal to each other.

If d = rt, then t = d/r.

The prey is going to run x distance before it gets caught. The predator is 80m behind, so it's going to have to run x + 80. Since the speed is given in km/h, let's change the distance to kilometers, and rewrite the above as x + .08 (because 80m = .08km).

We can now set the respective times for predator and prey to be equal to each other:

Predator: t = (x + .08)/48

Prey: t = x/40

So--

(x + .08)/48 = x/40

Thus...

x + .08 = (48/40)x

x + .08 = (6/5)x = 6x/5

.08 = 6x/5 - x = x/5

x = .08(5) = .4 (km)

Knowing this, we can now plug .4km into the original "d = rt" formula for the predator:

(.4 + .08) = 48t

t = .48/48 = 1/100 hr = 36 seconds.

I'd say the answer is (C).

Charles said...

I arrived at C as well, but here was my reasoning:

1) The difference in speed between the two animals is 8 kph. Since we're dealing with meters and looking for a time in second, I converted that to meters/second: 8000 meters / 3600 seconds = 2.2 (repeating), or 2 2/9.

2) Divide the distance that we need to cover (80 m) by that and you get 36 seconds.

Is that also a valid solution?

Kevin Kim said...

C,

Interesting.

I get the first step, but how did you arrive at 80m as the distance to cover? The predator starts 80m behind the prey, and actually has to cover a total of 480m to catch up to it, since both predator and prey are running forward the whole time. (The prey will run only 400m before being caught.)

Your method is probably perfectly valid, but I'm just having trouble seeing how that second part works.

OK.... I just tried the math myself, and your solution does work just fine.

d = rt

.08km = 8kph • t

t = .08/8 = 1/100 of an hour = 36 sec.

Now I'm left to wonder just how this more economical method works. I guess the idea is that the 80m gap needs to be closed, and even though the gap is moving forward with the pursuers, it's narrowing at a rate of 8kph, so it zeroes out in 36 seconds.

OK, I think I get the logic. My approach focuses on absolute distance; yours focuses on relative distance. Bravo!

I suspect MGRE will publish a solution that's similar to yours next week.

Charles said...

Yep, that's pretty much it--thinking relatively as opposed to absolutely.

I got the inspiration for this from driving, actually. If I'm driving along at 100 kph and someone drives past me at 20 kph, to me it looks like they are going 20 kph. So I guess I just automatically thought in relative terms

Charles said...

Doh... obvious gaffe in previous post:

"...and someone drives past me at 120 kph..."

(I blame this on Blogger's new captcha system, which I find so painful that I no longer preview my comments because I don't want to have to puzzle out the words twice.)