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C
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Difficulty:
95%
(hard)
Question Stats:
31% (02:38) correct
69%
(02:32)
wrong
based on 333
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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. Approximately how many bears appear at the resort each year?
A) 1
B) 3
C) 4
D) 10
E) 13
A) 1
B) 3
C) 4
D) 10
E) 13
There are actually two totally different ways to solve this problem: (1) Using a combination of rates and ratios to arrive at the answer, (2) Using a specialized table I call in my classes a "Matrix Box". Here is the full "GMAT Jujitsu" for this question, using both techniques to arrive at the same answer. For those of you preparing for the GMAT, pick which ever technique resonates with you the most.
First, note that the target of this question is the ratio of “Bears per year.” Thus, we know we need to combine ratios so that "Total bears" are on top and "year" is in the denominator. The problem gives us several ratios. First, from the phrase “the alarm…sounds an average of once very thirty days”, we get the ratio:
\(\frac{\text{1 Alarm}}{\text{30 Days}}\)
Multiplying this by (\(\frac{12}{12}\)) gives us: \(\frac{\text{12 Alarms}}{\text{360 Days}}\)
This is REALLY close to 12 alarms in a year (365 days), so in the spirit of approximating we are half-way there.
Now, we need to find out how many bears (both detected and undetected) happen over the course of a year. Then, we could transform alarms into bears. The problem tells us that there are ten false alarms for every undetected bear. Thus,
\(\frac{\text{False Alarms}}{\text{Undetected Bears}}=\frac{10}{1}\)
The problem also tells us that the alarm sounds for three out of four bears that appear. (One of those bears remains undetected.) Thus,
\(\frac{\text{Working Alarms}}{\text{Undetected Bears}}=\frac{3}{1}\)
With these two ratios, it is very easy to see that the ratio between false alarms and working alarms is \(10:3\), and that the total amount of alarms would be\(10x + 3x = 13x\). This means that:
\(\frac{\text{Working Alarms}}{\text{Total Alarms}}=\frac{3}{13}\)
Finally, the problem tells us that “the alarm only sounds for three out of four bears that actually appear", thus:
\(\frac{\text{Total Bears}}{\text{Working Alarms}}=\frac{4}{3}\)
Now, we just need to multiply the rates together, paying attention to the units to cancel them out as we go.
\(\frac{\text{12 Alarms}}{\text{1 Year}}*\frac{\text{3 Working Alarms}}{\text{13 Alarms}}*\frac{\text{4 Total Bears}}{\text{3 Working Alarms}}\)
This simplifies down to \(\frac{12}{13}*\frac{\text{4 Bears}}{\text{Year}}\)
The word “approximately” in the original question cues us in that we won’t need to necessarily do the math here. It is easy to see that it is closest to 4 bears per year. The answer is C.
We have two mutually exclusive groups (detected vs. undetected bears and alarms vs. no alarms.) If you can organize your thoughts around these mutually-exclusive groups by creating a specialized table called a "Matrix Box", the solution to this problem naturally unfolds.
Here is what the blank table might look like:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{____} & \text{____} & \text{____} \\
\text{No Alarms} & \text{____} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{____}
\end{matrix}
\]
Matrix boxes are simplest if we can use concrete numbers. The problem (or the opportunity) here is that we are only given ratios. This means we can invent our own total amount, using leverage inside the problem to guess what number would work the best. The problem states that an alarm sounds "an average of once every thirty days", so our total should be at least a multiple of 30 to avoid messy fractions. But we also want to have it be a multiple of the “total alarms” in order to make the math easy. It looks like the ratios involving total alarms have false alarms in ratios of 10 and real alarms in ratios of 3. So, total alarms should be at least in ratios of 13. Thus, we want a number that is a multiple of \(13\) and a multiple of \(30\). Thus, if we were to pick \(13*30=390\) as our total value, we wouldn't have to deal with ugly math.
Let's start there:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{____} & \text{____} & \text{____} \\
\text{No Alarms} & \text{____} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{390}
\end{matrix}
\]
The problem tells us that we see 1 alarm happening about every \(30\) days. Thus over \(390\) days we get \(13\) alarms. (We can plug this in to the “Total alarms” slot.) The next two ratios tell us that we get \(10\) false alarms for every undetected bear (or failed alarm) and that \(3\) bears are detected for every one bear that is not detected (again, a failed alarm). Notice the “1 undetected bear” is common to both. Plugging all of these values into our Matrix box gives us:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{390}
\end{matrix}
\]
It is now very easy to fill in the rest of the box:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{376} & \text{377} \\
\text{Totals} & \text{4} & \text{386} & \text{390}
\end{matrix}
\]
It is now just a simple change of rates. If there are \(4\) bears in \(390\) days, then you can convert the days to years:
\(\frac{\text{4 Bears}}{\text{390 Days}}*\frac{\text{365 Days}}{\text{1 Year}}=\frac{4\text{ Bears}}{\text{Year}}*\frac{365}{390}\)
Now, don't fall for the trap and actually try to do this math. Remember: you don't have a calculator on the Quant portion of the GMAT. The GMAT often baits inexperienced test-takers into trying to solve very messy math, when looking down at the answer choices would be sufficient to answer the question. We can leverage the answer choices against each other and look for the approximate value.
It is also to see from this expression that we are still closer to 4 than we are to 3. (The actual answer is around 3.743.)
No matter how you look at it, the answer is still C.
By the way, for those of you that need extra practice, see these related questions:
https://gmatclub.com/forum/the-bear-ala ... l#p1386648
https://gmatclub.com/forum/the-bear-ala ... l#p2032644
https://gmatclub.com/forum/the-bear-ala ... l#p1536872
Technique #1: Using Rates and Ratios
First, note that the target of this question is the ratio of “Bears per year.” Thus, we know we need to combine ratios so that "Total bears" are on top and "year" is in the denominator. The problem gives us several ratios. First, from the phrase “the alarm…sounds an average of once very thirty days”, we get the ratio:
\(\frac{\text{1 Alarm}}{\text{30 Days}}\)
Multiplying this by (\(\frac{12}{12}\)) gives us: \(\frac{\text{12 Alarms}}{\text{360 Days}}\)
This is REALLY close to 12 alarms in a year (365 days), so in the spirit of approximating we are half-way there.
Now, we need to find out how many bears (both detected and undetected) happen over the course of a year. Then, we could transform alarms into bears. The problem tells us that there are ten false alarms for every undetected bear. Thus,
\(\frac{\text{False Alarms}}{\text{Undetected Bears}}=\frac{10}{1}\)
The problem also tells us that the alarm sounds for three out of four bears that appear. (One of those bears remains undetected.) Thus,
\(\frac{\text{Working Alarms}}{\text{Undetected Bears}}=\frac{3}{1}\)
With these two ratios, it is very easy to see that the ratio between false alarms and working alarms is \(10:3\), and that the total amount of alarms would be\(10x + 3x = 13x\). This means that:
\(\frac{\text{Working Alarms}}{\text{Total Alarms}}=\frac{3}{13}\)
Finally, the problem tells us that “the alarm only sounds for three out of four bears that actually appear", thus:
\(\frac{\text{Total Bears}}{\text{Working Alarms}}=\frac{4}{3}\)
Now, we just need to multiply the rates together, paying attention to the units to cancel them out as we go.
\(\frac{\text{12 Alarms}}{\text{1 Year}}*\frac{\text{3 Working Alarms}}{\text{13 Alarms}}*\frac{\text{4 Total Bears}}{\text{3 Working Alarms}}\)
This simplifies down to \(\frac{12}{13}*\frac{\text{4 Bears}}{\text{Year}}\)
The word “approximately” in the original question cues us in that we won’t need to necessarily do the math here. It is easy to see that it is closest to 4 bears per year. The answer is C.
Technique #2: Matrix Boxes
We have two mutually exclusive groups (detected vs. undetected bears and alarms vs. no alarms.) If you can organize your thoughts around these mutually-exclusive groups by creating a specialized table called a "Matrix Box", the solution to this problem naturally unfolds.
Here is what the blank table might look like:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{____} & \text{____} & \text{____} \\
\text{No Alarms} & \text{____} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{____}
\end{matrix}
\]
Matrix boxes are simplest if we can use concrete numbers. The problem (or the opportunity) here is that we are only given ratios. This means we can invent our own total amount, using leverage inside the problem to guess what number would work the best. The problem states that an alarm sounds "an average of once every thirty days", so our total should be at least a multiple of 30 to avoid messy fractions. But we also want to have it be a multiple of the “total alarms” in order to make the math easy. It looks like the ratios involving total alarms have false alarms in ratios of 10 and real alarms in ratios of 3. So, total alarms should be at least in ratios of 13. Thus, we want a number that is a multiple of \(13\) and a multiple of \(30\). Thus, if we were to pick \(13*30=390\) as our total value, we wouldn't have to deal with ugly math.
Let's start there:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{____} & \text{____} & \text{____} \\
\text{No Alarms} & \text{____} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{390}
\end{matrix}
\]
The problem tells us that we see 1 alarm happening about every \(30\) days. Thus over \(390\) days we get \(13\) alarms. (We can plug this in to the “Total alarms” slot.) The next two ratios tell us that we get \(10\) false alarms for every undetected bear (or failed alarm) and that \(3\) bears are detected for every one bear that is not detected (again, a failed alarm). Notice the “1 undetected bear” is common to both. Plugging all of these values into our Matrix box gives us:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{____} & \text{____} \\
\text{Totals} & \text{____} & \text{____} & \text{390}
\end{matrix}
\]
It is now very easy to fill in the rest of the box:
\[
\begin{matrix}
& \text{Bears} & \text{No Bears} & \text{Totals} \\
\text{Alarms} & \text{3} & \text{10} & \text{13} \\
\text{No Alarms} & \text{1} & \text{376} & \text{377} \\
\text{Totals} & \text{4} & \text{386} & \text{390}
\end{matrix}
\]
It is now just a simple change of rates. If there are \(4\) bears in \(390\) days, then you can convert the days to years:
\(\frac{\text{4 Bears}}{\text{390 Days}}*\frac{\text{365 Days}}{\text{1 Year}}=\frac{4\text{ Bears}}{\text{Year}}*\frac{365}{390}\)
Now, don't fall for the trap and actually try to do this math. Remember: you don't have a calculator on the Quant portion of the GMAT. The GMAT often baits inexperienced test-takers into trying to solve very messy math, when looking down at the answer choices would be sufficient to answer the question. We can leverage the answer choices against each other and look for the approximate value.
It is also to see from this expression that we are still closer to 4 than we are to 3. (The actual answer is around 3.743.)
No matter how you look at it, the answer is still C.
By the way, for those of you that need extra practice, see these related questions:
https://gmatclub.com/forum/the-bear-ala ... l#p1386648
https://gmatclub.com/forum/the-bear-ala ... l#p2032644
https://gmatclub.com/forum/the-bear-ala ... l#p1536872
General Discussion
30 Dec 2022, 21:09
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Bookmarks
[quote="AaronPond"]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. Approximately how many bears appear at the resort each year?
A) 1
B) 3
C) 4
D) 10
E) 13
[1 alarm per month avg, so 12 alarms a year
it shows that 12 bears a year
but due to its sensibility - let's assume an extra 10 alarms even it did not detect a bear so total of 22 alarms a year
as the alarm sounds for 3 out of 4 bears
so we have detected a total of 12 bears without device problem a year
now it detects only 3/4 of 12 a year =4
*if the qstn asks to find alarms- 12+8(no.of undetected bears)*10(false alarm per undetected bear) = 92]
A) 1
B) 3
C) 4
D) 10
E) 13
[1 alarm per month avg, so 12 alarms a year
it shows that 12 bears a year
but due to its sensibility - let's assume an extra 10 alarms even it did not detect a bear so total of 22 alarms a year
as the alarm sounds for 3 out of 4 bears
so we have detected a total of 12 bears without device problem a year
now it detects only 3/4 of 12 a year =4
*if the qstn asks to find alarms- 12+8(no.of undetected bears)*10(false alarm per undetected bear) = 92]
