Tough Nut !

[himu]

At the beginning of year 1, an investor puts p dollars into an investment whose value increases at a variable rate of Xn(X suffix n) per year, where n is an integer ranging from 1 to 3 indicating the year. If 85<(Xn)<110 for all n between 1 and 3, inclusive, then at the end of 3 years, the value of the investment must be between

$p and $2p

$2p and $5p

$5p and $10p

$10p and $25p


$25p and $75p

[mkdureja]

I have assumed the rate given in in percentage terms, not sure if question clarifies it.
Xn minimum = 85%
Xn maximum = 110%
For an investment of p, for 3 years, at 85% rate, compounded annually, final value will be
p(1.85)^3 = 6.33p
Similarly, Maximum will be p(2.10)^3 = 9.26p
so answer would be
between 5p and 10p.

[lunarpower]

Yeah, this problem should mention percentages. Perhaps the original poster just forgot the percent sign; if not, then the problem is flawed.

Calculating 1.85^3 or 2.1^3 manually is unappetizing, to say the least. (Remember that there's no calculator on this test.)

The answer ranges are mutually exclusive, folks! So all you have to do is pick the easiest possible number, and see where it ends up.

Let's just make the investment grow by 100% each year. That's within the range of 85%-110%, and, gosh darnit, it's way easier than any other number in that range.
If that happens, then ...
... after 1 year, it's worth 2p
... after 2 years, it's worth 4p
... after 3 years, it's worth 8p.
So pick the range that has 8p in it. And you're done.

[sana.noor]

is it a real gmat question?

[shailendra.sharma]

Yeah, this problem should mention percentages. Perhaps the original poster just forgot the percent sign; if not, then the problem is flawed.

Calculating 1.85^3 or 2.1^3 manually is unappetizing, to say the least. (Remember that there's no calculator on this test.)

The answer ranges are mutually exclusive, folks! So all you have to do is pick the easiest possible number, and see where it ends up.

Let's just make the investment grow by 100% each year. That's within the range of 85%-110%, and, gosh darnit, it's way easier than any other number in that range.
If that happens, then ...
... after 1 year, it's worth 2p
... after 2 years, it's worth 4p
... after 3 years, it's worth 8p.
So pick the range that has 8p in it. And you're done.

Your estimation technique is very good given the answer choices. I think I need to learn estimation in context of answer choices. Is it possible for you to point on more practice questions which can have can help me estimation techniques in context of answer choices ?

In this particular question,

I calculated 1.85 < 2, so cube(1.85) < 8 and since root(3) is 1.73, so cube(1.85) > 3 root(3) > 2

And similarly 2.10 > 2, so cube(2.10) > 8, but not as high as 25. So I could choose correct answer - but it definitely took more time.

[lunarpower]

Shailendra-
Your post included the word "estimation", but I don't view my technique here as an example of estimation -- since I have not, in fact, estimated anything. I've selected one exact allowed value for the percentage change, and then propagated that value through the problem.

In other words, this isn't estimation; rather, it's the "pick your own numbers" method (referred variously in our MGMAT materials as "VIC" and/or "smart numbers", depending on the context).
This method is an extremely helpful tool for solving multiple-choice items; honestly, it's one of the most helpful tools out there. If you apply it with flexibility and panache, you can get it to work on as many as 25-30% of all the multiple-choice problems (!!).

I wouldn't worry too much about looking for positive signs that you can use the method. Instead, because the method is so common, I'd just have it as an automatic backup plan, in case some other approach fails -- that way you don't have to think of it. It'll just happen, as "plan B".

If you have the 12th edition of the OG, here is just a small selection of problems on which you can use "pick your own numbers":
228
227
226
224
220
219
213
212
208
204
202
198
187
186
185
181
180
179
165
163
158

That's a huge selection of problems, so this method clearly has a lot going for it.

Note what all of these problems have in common:
* There is/are an undetermined quantity or quantities in the problem. (Not just an unknown -- but a quantity that's actually undetermined. I.e., even if you solve the problem, you still don't get a definite value for that quantity.)
* You can easily find a value to plug in for that quantity or quantities.

The same is true above; the unknown percentage here could be anything from 85-110, and, even if you solve the problem, it's still not a single number -- it's still anything from 85 to 110.

As for how to pick the numbers, that's mostly just a matter of "playing smart" -- just look around the problem, see what's in there, and make decisions according to your own sense of which numbers are "nice" and which aren't.
Here, for instance, the given range is 85% to 110%, so I think it's pretty obvious that 100% is the "nicest" percentage in that range -- especially given that you have to execute the change three times.

That decision is going to vary individually, of course -- some things are "easy" for some people but not for others. For instance, I have a student who knows all of the cubes 1^3 through 20^3 by heart, and also happens to know right away that 7/8 is 87.5%. So, for that student, it would be "easy" to pick 87.5%, realize that a 87.5% increase is equivalent to multiplying by 15/8, and then evaluate (p)(15/8)(15/8)(15/8) = 3375p/512, which is somewhere around 6p.
This would not, of course, be "easy" for 99.99% of people, but who cares what other people find "easy". Take what works for you, leave what doesn't.