In Part I of Organizing Your Information, we talked about rate time distance problems. In this one, we’ll tackle the Grizzly Peak problem from the Veritas Prep Arithmetic homework. I can’t tell you how many questions have been asked about this one in office hours, or via email support.

Here’s the problem:

The bear alarm at Grizzly’s Peak ski resort sounds an average of once every thirty days, but the alarm is so sensitively calibrated that it sounds an average of ten false alarms for every undetected bear. Despite this, the alarm only sounds for three out of four bears that actually appear at the resort

There are then several questions based on this information. Let’s face it. It’s a lot of information to process and synthesize.

Let’s figure out what the pieces of data are:

Total Alarms

Total Bears

Good Alarms (where a bear is actually sighted)

False Alarms (where there is no bear)

Bears sighted (this should be the same as “good alarms”)

Sneaky bears (bears that don’t set off the alarm)

Days

Let’s make it into a chart:

Good Alarms | False Alarms | Total Alarms | Sneaky Bears | Total Bears | Days |

_{ } |
_{ } |
_{ } |
_{ } |
_{ } |
_{ } |

Ok – So now to fill in our chart.

The alarm has 10 false alarms for every undetected bear. So let’s put 10 under false alarms and 1 under sneaky bears.

The alarm sounds for 3 out of 4 bears that actually appear – so for every 4 bears, we have 3 good alarms and 1 sneaky bear.

Last, we have one alarm every 30 days. This is in a different scale than the other numbers – since we have 13 alarms total, and we have 1 alarm every thirty days, it takes us 390 days to get to 13 alarms. Let’s plug that in, too.

Good Alarms | False Alarms | Total Alarms | Sneaky Bears | Total Bears | Days |

3 | 10 | 13 | 1 | 4 | 390 |

Now, we can look at our questions:

87. If the alarm sounds, what are the odds a bear has actually been sighted?

So we have 3 alarms that signal bears out of 13 total alarms, which gives us (B) 3/13.

88. On any given day at the resort, what is the approximate probability that there is neither an alarm nor an undetected bear?

So out of our 390 days, we have 13 days with alarms and 1 more day that has a sneaky bear. This means that for 376 days (390 – 14) out of the 390, we have no alarms and no bears. This is answer (D).

89. If the alarm were to sound an average of 10 false alarms for every detected bear, the probability that a sounded alarm would indicate an actual bear would:

In this case, let’s adjust our chart:

Good Alarms | False Alarms | Total Alarms | Sneaky Bears | Total Bears | Days |

1 | 10 | 11 |

Now, our odds that an alarm signals a bear would be 1/11 which is 9%.

This is a decrease from our original numbers, which were 3/13 (see problem #87) or 23%, so our difference is 14%, or answer (C).

90. Approximately how many bears appear at the resort each year?

If we have 4 bears in 390 days, this is just slightly less than 1 bear every 100 days. Extrapolating this to a year, which is 365 days, we have approximately 3.6 bears in a year, which is closer to 4 than 3, so our **answer is (C).**

While this problem is overwhelming and daunting at first glance, a careful study of the information presented, and taking the time to organize it all into meaningful pieces makes the rest of the problem run really quite smoothly.

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*Valerie Browning has been teaching GMAT for Veritas Prep for 10 years. After graduating from the McCombs School of Business at UT Austin, Valerie is now based in Houston. Since graduating, she has been interviewing applicants to McCombs as an alumni volunteer.*