ratio?

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ratio?

by jopup » Tue Jun 17, 2008 2:08 pm
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i dont understand the question

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by II » Tue Jun 17, 2008 2:35 pm
Lets break down the question:

"If n denotes a number to the left of 0 on the number line"
- This basically is another way of saying that n is a negative number. It is less than zero. So we can write n<0.

"...the square of n is less than 1/100..."
This is saying that n squared must be less than 1/100.
We can write this as: "n^2 < 1/100"

From this we can say that if n^2 < 1/100, then n < 1/10, or n < -(1/10)
Note: 1/10 * 1/10 = 1/100, and -(1/10) * -(1/10) = 1/100.

Since the question tells us that n is a negative number ... lets take the negative value -(1/10).
The reciprocal of -(1/10) is -(10/1). So n has to be less than -10.

Hope this makes it clear.

Another method to solve this question is to backsolve ... so take the figures from the answer choices and check to see if they satisfy the criteria specified in the questions.
So lets take a look at answer choice A. If n = -10 ... then the reciprocal of n = -1/10.
-(1/10)^2 = 1/100. So this tells us that n has to be less than -10 for n squared to be less than 1/100.

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by lunarpower » Wed Jun 18, 2008 1:32 am
one mistake in here that deserves attention:
II wrote:From this we can say that if n^2 < 1/100, then n < 1/10, or n < -(1/10)
whoa, no.

if you're given that n^2 is less than a given value, then n must be less than the positive square root (as you've indicated), but greater than the negative square root.

so, in this case, we have -1/10 < n < 1/10. if you write this out as two separate inequalities (as you did here), then those two inequalities should be n <1> [/b]-1/10.

because n has to be negative, then, n itself must be between -1/10 and 0. this means that the reciprocal of n is between negative infinity and -10.

you got lucky here, because there's only one answer choice that even has -10 as a boundary value. notice that you managed to solve the problem without even thinking about whether your reciprocal must be greater or less than -10; it was good enough just to look for the answer with -10 in it.
if there were confounding answer choices, such as 'between -10 and 0', you'd have to think about whether the reciprocal itself is less/greater than -10.
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by lunarpower » Wed Jun 18, 2008 1:34 am
lunarpower wrote:one mistake in here that deserves attention:
II wrote:From this we can say that if n^2 < 1/100, then n < 1/10, or n < -(1/10)
whoa, no.

if you're given that n^2 is less than a given value, then n must be less than the positive square root (as you've indicated), but greater than the negative square root.

so, in this case, we have -1/10 < n < 1/10. if you write this out as two separate inequalities (as you did here), then those two inequalities should be n < 1/10 and n > -1/10.

because n has to be negative, then, n itself must be between -1/10 and 0. this means that the reciprocal of n is between negative infinity and -10.

you got lucky here, because there's only one answer choice that even has -10 as a boundary value. notice that you managed to solve the problem without even thinking about whether your reciprocal must be greater or less than -10; it was good enough just to look for the answer with -10 in it.
if there were confounding answer choices, such as 'between -10 and 0', you'd have to think about whether the reciprocal itself is less/greater than -10.
Ron has been teaching various standardized tests for 20 years.

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by II » Wed Jun 18, 2008 2:44 am
Hi Lunarpower ... sorry ... dont think I understand your comments. Can you please elaborate.

n^2 < 1/100
n < 1/10
reciprocal of 1/10 is 10.
so n < 10
n is a negative number ... so we get the -10.
n < -10.

How would you solve this question ?

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by lunarpower » Wed Jun 18, 2008 1:04 pm
II wrote:Hi Lunarpower ... sorry ... dont think I understand your comments. Can you please elaborate.

n^2 < 1/100
n < 1/10
reciprocal of 1/10 is 10.
so n < 10
n is a negative number ... so we get the -10.
n < -10.

How would you solve this question ?
one thing that appears to be going wrong here is that you're taking reciprocals, but NOT flipping the inequality sign around. that's bad.
for instance, if x < 1/2, then 1/x > 2. you can think about this in one of two ways: either
(1) in the abstract ("if i take the reciprocal of a smaller number, i get a bigger number") or
(2) by plugging in numbers ("ok, 1/4 is less than 1/2; if i take the reciprocal, i get 4, which is bigger than 2")

so, in this particular problem, if you have n < 1/10, then the reciprocal of n must be GREATER than 10.

if you have n > -1/10 (which is the other side of the inequality here, since the square gives -1/10 < n < 1/10), then the reciprocal likewise makes the sign flip, so you have that n is LESS than -10, which is the answer to this particular problem.
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by II » Fri Jun 27, 2008 11:47 am
Got it ! Thanks Ron !

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by mbaapplicant2008 » Fri Oct 31, 2008 12:07 am
lunarpower's explanation is great

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by alexdallas » Mon Jul 06, 2009 6:15 am
tks Lunar

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by ghacker » Mon Jul 06, 2009 7:14 am
n denotes a number to the left of zero = the number is negative

the square of n is less than (1/100) = n^2 < 1/100 ( But n^2 is positive)

n^2 < 1/100 ------------> 100 < 1/n^2 , so n must be fraction

n is a negative fraction , n = -( a/b)

Since 1/n^2 >100 , /b / must be greater 10

hence is we take the reciprocal of n then the reciprocal /n/ must be greater than 10

or res /n/ ={ 11,12,................} , but n is negative so the values should be negative

res n = { -11 ,-12......................)

Hence the values should be less than -10

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by Ian Stewart » Mon Jul 06, 2009 9:43 am
lunarpower wrote: one thing that appears to be going wrong here is that you're taking reciprocals, but NOT flipping the inequality sign around. that's bad.
for instance, if x < 1/2, then 1/x > 2.
That isn't always 'bad', because the above isn't always true - it's only true if x is positive. If x is negative, then x < 1/2 and 1/x is also less than 2.

Rather than learn rules for taking reciprocals in inequalities, you can just rewrite the inequality by multiplying or dividing on both sides:

x < 1/2
2x < 1
2 < 1/x if x is positive
2 > 1/x if x is negative (we need to reverse the inequality)

Of course, if x is negative, we know anyway that 1/x < 0, so we don't get any new information from the above in that case.
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by missrochelle » Sun Aug 29, 2010 3:47 pm
Can someone just pls explain simply...how we actually "take" the reciprocal?


if we have:

-n > -1/10

and want to TAKE THE RECIPROCAL...what do we do to get to n<10? multiply by ...."negative...???"

thanks!
Last edited by missrochelle on Mon Aug 30, 2010 10:45 am, edited 1 time in total.

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by euro » Fri Oct 01, 2010 12:35 am
missrochelle wrote:Can someone just pls explain simply...how we actually "take" the reciprocal?


if we have:

-n > -1/10

and want to TAKE THE RECIPROCAL...what do we do to get to n<10? multiply by ...."negative...???"

thanks!
1 divided by a number gives the reciprocal of the number.

The product of a number and its reciprocal equals 1.

Reciprocal of n is 1/n.

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by GMATGuruNY » Fri Oct 01, 2010 2:54 am
If n denotes a number to the left of 0 on the number line such that the square of n is less than 1/100, then the reciprocal of n must be

A. less than -10
B. between -1 and -1/10
C. between -1/10 and 0
D. between 0 and -1/10
E. greater than 10


I would just plug in a value for n that satisfies the conditions of the problem.

Let n = -1/20.
This works because (-1/20)^2 = 1/400, and 1/400 < 1/100.
Reciprocal of -1/20 is -20.

Only answer choice A works:
-20 < -10.
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by katknms9 » Sat May 07, 2011 3:31 pm
Hopefully not beating a dead horse with this clarification, but if after simplifying the initial equation to get:

-1/10 < n

How is taking the reciprocal of n (1/n) not equal to -10 < n?

My logic is:
-1/10 < n so...
1/(-1/10) < 1/n
-10 < 1/n

I'm obviously missing something, so if someone could help me out that would be great. I understand the logic of plugging values to see why "A" makes sense, just wanted to try and see why the above rationale doesn't make sense.