The number of antelope in a certain herd increases every year at a constant rate. If there are 500 antelope in the herd today, how many years will it take for the number of antelope to double?

(1) Ten years from now, there will be more than ten times the current number of antelope in the herd.

(2) If the herd were to grow in number at twice its current rate, there would be 980 antelope in the group in two years.


Hi - I hope I am posting this in the correct format. The problem I have with the above question is in the algebra of statement number (2).

The solution is:

According to the statement, 500y2 = 980
y2 = 980/500
y2 = 49/25
y = 7/5 OR 1.4 (y can’t be negative because we know the herd is growing)
This means that the hypothetical double rate from the statement represents an annual growth rate of 40%.
The actual growth rate is therefore 20%, so x = 1.2.

Unfortunately, I tried to use the equation: 500(2)x^2=980. When I work out my equation x=.1
I don't understand why my equation for statement (2) doesn't equal the MGMAT equation. The equation for the original question is 500x^n > 1000. If the herd grows at twice the current rate isn't that 2*x^n? I guess I am assuming that x^n is the rate. Is it not the rate?

I understand that I don't need to figure out the exact answer for a data sufficiency problem, but I would like to understand this concept.

Thank you.

hi - sorry this took so long to get to.

when you use MULTIPLIERS TO REPRESENT PERCENTAGE INCREASES/DECREASES, you should know that the multiplier is 1 plus the decimal that represents the percentage in question. for instance, to increase a quantity by 20%, you'd multiply by 1 + (0.2), which is 1.2, and to decrease a quantity by 20%, you'd multiply by 1 + (-0.2), which is 0.8.

the thing to notice here is that, because of the "1" added to the decimals in question, the multiplier corresponding to double the rate of increase is not just twice the multiplier corresponding to the original increase. for instance, the multiplier corresponding to a 30% increase is 1.3; the multiplier corresponding to a 60% increase, which is twice the original increase, is 1.6, which is not twice as much as 1.3. in fact, it's not an obvious multiple of 1.3 at all.
this is why you can't use (2x) to represent the rate multiplier, as you suggest; the multiplier isn't doubled.

there are 2 ways around this issue:
(1) just use "x" (or "y", as done in the mgmat solution, or whatever other letter) to represent the doubled rate multiplier, and then just cut the corresponding rate in half (this is what is done in the solution); or
(2) use (1 + x) to represent the original rate multiplier, and then use (1 + 2x) to represent the new multiplier. this will make the algebra a lil more awkward, but it's probably less likely to produce confusion as to what the "x" represents.

I used 500 * (2y)^2 = 980 hence y = .7. Ron can you confirm this is correct. After two yers we will have 500 * .49 antelopes

Also ans is B. PLs confirm

no, this is wrong.
...and the post directly above yours provides a detailed explanation of why it's wrong.
please read the threads before you post on them!

moreover, it should be obvious that this is wrong, because 500 * 0.49 is less than half of 500. since the size of the antelope herd is increasing, this result is absurd.



yes.

--

actually, the easiest way by far to determine that (2) is sufficient is to do the following reasoning:

* given "twice the growth rate", we have a starting number, AND an ending number, AND the amount of time it will take to get from that starting number to that ending number.

* therefore, since this is a complete set of information, we can determine "twice the growth rate".

* therefore, we can find the growth rate itself.

* once we find the growth rate, we'll have the answer to the problem.

sufficient!

again, since this is data sufficiency, there is no reason whatsoever to actually calculate any of these numbers.

Can we can also use the exponential growth formula here to determine which statements are sufficient? Especially statement #2?

Y(t) = Y(t=0) * (K)^t
where k is the constant multipler.

Please confirm ManhattanGMAT staff. Thanks

Hi Sudaif,
Yes, that's essentially what we want to do with this problem. Of course you're going to need to add a couple of steps..

_________________
Tim Sanders
Manhattan GMAT Instructor

Tim,
Why cant you use the linear growth model (y=mx+b) where m= constant growth rate
b = 500 (given from problem)
x = number of years
Therefore: y = rt + 500
and solve for t when y = 1000 (double the initial population)...???
How do you know to use the exponential growth rate formula instead of a linear growth rate formula to solve for t?

I still got the right answer but had I been asked to actually solve for the value t, I wouldn't have. Using the linear growth rate equation (above) I came up with t = 175/16. This can't be right given that you have already said that you can use the exponential growth rate formula.

Thank you!

It's because there's no such thing as a linear growth "rate". Linear growth means increase by the same NUMBER each year rather than the same RATE. When you see something growing at the same RATE, you are multiplying by a constant amount, hence the use of the exponential growth formula..

_________________
Tim Sanders
Manhattan GMAT Instructor