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DS Mathematics OG #166

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DS Mathematics OG #166

by barrelbowl » Sat Dec 03, 2011 12:33 pm
Hey all,

This question pertains to #166. The question states:

If n is a positive integer is (1/10)^n < 0.01?

(1) n>2
(2) (1/10)^n-1 < 0.1


When I was reviewing this particular question I was curious as to what was wrong with the following mathematic manipulation. I picked D, but my math for (2) came out wrong.

Originally I rephrased the question as is (1/10)^n < (1/10)^2 or is n < 2?

(1) Was sufficient, because it clearly states that n is > 2.
(2) I manipulated this equation to (1/10)^n/(1/10)^1 < (1/10).
I multiplied the bottom over to the right hand side of the inequality to yield, (1/10)^n < (1/10)*(1/10).
Further reduction yielded (1/10)^n < (1/10)^2 or n < 2.

Because (1) and (2) are not allowed to contradict each other, I'm wondering what was wrong with my manipulation in (2)?

Thanks in advance!
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by pemdas » Sat Dec 03, 2011 1:09 pm
from this point (quoted) and over you had to reverse the sign
Further reduction yielded (1/10)^n < (1/10)^2 or n < 2.
you committed the typical inequality error -> taking reciprocals reverses the sign
1/10^n and 1/100 should be compared in reciprocal format, that's why 10^n>10^2 and n>2, Sufficient

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by tpr-becky » Sun Dec 04, 2011 9:46 am
Think about the numbers involved (1/10)^n <(1/10)^2

if n = 1 then 1/10 is not less than 1/100 but if n = 3 then 1/1000 is less than 1/100. This shows that n must be greater than 2, not less than. Thus, when dealing with fractions exponents will make the number smaller, not larger and you need to reverse the sign.
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by barrelbowl » Tue Dec 06, 2011 7:20 pm
Thanks guys, very helpful!
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by Ian Stewart » Tue Dec 06, 2011 10:34 pm
pemdas wrote:
you committed the typical inequality error -> taking reciprocals reverses the sign
I don't recommend learning rules for inequalities and reciprocals, because they aren't entirely straightforward. You can always just multiply and divide on both sides of your inequality to get the same result, so you don't need to learn any rules for reciprocation. It is true that you need to reverse your inequality if you take reciprocals on both sides *if* all the quantities involved are positive. For example, if we have this inequality, which is clearly true:

1/3 < 1/2

then when we take reciprocals on both sides, we need to reverse the inequality to get something which is true:

3 > 2.

That is *not* the case if you have a negative quantity in one of your fractions. For example, this inequality is clearly true:

-1/2 < 1/3

but if we take reciprocals on both sides, we do *not* want to reverse the inequality:

-2 < 3

Using rules for inequalities and reciprocals is especially dangerous when you have unknowns. For example, if you have the inequality:

2/x < 3/y

you certainly cannot take reciprocals on both sides and reverse the inequality *unless* you know that x and y are both positive (or both negative). In such situations, I find it safer not to use reciprocal rules at all. Instead I will try to multiply on both sides by x and by y, which will remind me that I need to know the signs of x and y to proceed.

barrelbowl wrote:
Originally I rephrased the question as is (1/10)^n < (1/10)^2 or is n < 2?
What you've done here would be correct if your base was a positive integer. You can get a positive integer base here, since 1/10 is just equal to 10^(-1). So you can rewrite the question as follows:

(1/10)^n < (1/10)^2
10^(-n) < 10^(-2)
-n < -2
n > 2

(in the last step, we reverse the inequality since we multiplied by -1 on both sides)
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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by barrelbowl » Sat Dec 10, 2011 12:16 pm
Thanks Ian. I think this was a case of 'equation mentality'.

Thinking about a problem that you're solving, can often be easier than trying to do things algebraically.
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