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k is an integer for which

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k is an integer for which [#permalink] New post 24 Feb 2017, 02:43
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58% (00:49) correct 41% (00:55) wrong based on 36 sessions



k is an integer for which \(\frac{1}{2^{(1-k)}} < \frac{1}{8}\)

Quantity A
Quantity B
\(k\)
\(-2\)


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: k is an integer for which [#permalink] New post 03 Mar 2017, 05:40
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Explanation

We have the expression \(\frac{1}{2^{(1 - k)}}\) and the fraction \(\frac{1}{8}\).

Rewriting \(\frac{1}{8}\) as \(\frac{1}{2^3}\).

So if 1 -k > 3 (for example 1- k = 4) then the inequality \(\frac{1}{2^{(1 - k)}} < \frac{1}{8}\).

So \(1 -k > 3\) or \(-k > 2\) or \(k <-2\).

Thus Quantity B is greater.
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Re: k is an integer [#permalink] New post 27 Nov 2017, 11:47
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Explanation

the most common strategy here is testing number. Picking a number and substitute to k and see what happens. This is true but is time-consuming.

The best way to attack the question is to manipulate the stem

\(2^{1-k} > 8\)

\(2^{1-k} > 2^3\)

\(1-k > 3\)

\(-k > 2\)

\(k < -2\) which means that B is greater. Hence, B is the right answer
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Re: k is an integer [#permalink] New post 30 Nov 2017, 13:53
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Carcass wrote:
K is an integer for which

\(\frac{1}{2^{1-k}} < \frac{1}{8}\)

Quantity A
Quantity B
k
-2


A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.


Given: 1/[2^(1-k)] < 1/8

Since 2^(1-k) is POSITIVE for all values of k, we can safely take 1/[2^(1-k)] < 1/8 and multiply both sides by [2^(1-k)]
When we do this, we get: 1 < [2^(1-k)]/8
Rewrite 8 as follows: 1 < [2^(1-k)]/[2^3]
Since we have the same base, we can apply the quotient law to get: 1 < 2^(1 - k - 3)
In other words, 2^0 < 2^(1 - k - 3)
From this, we can conclude that 0 < 1 - k - 3
Simplify: 0 < -2 - k
Add k to both sides to get: k < -2

We get:
Quantity A: k (which is some number that's LESS THAN -2)
Quantity B: -2

Answer: B

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Re: k is an integer for which [#permalink] New post 07 Mar 2018, 08:42
I think the explanation and the problem post have a different fraction here
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Re: k is an integer for which [#permalink] New post 07 Mar 2018, 15:48
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chgre wrote:
I think the explanation and the problem post have a different fraction here

It was a formatting error. Its fixed now.
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Re: k is an integer for which [#permalink] New post 12 Mar 2018, 01:29
B
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Re: k is an integer for which [#permalink] New post 13 Mar 2018, 16:33
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correct: B
1 / 2^K-1 < 1/8
8 < 2 ^ 1-K
2^3 < 2 ^ 1-K
3 < 1-K
K < -2
so B is bigger

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Re: k is an integer for which [#permalink] New post 22 Jun 2019, 23:46
I m missing something here the fraction 1/2^(1-k)< 1/2^3 why r we fliping the sign when we ignore the numerator 1?

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Re: k is an integer for which [#permalink] New post 22 Jun 2019, 23:54
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Rule of exponents

\(\frac{1}{2^{(1-k)}} < \frac{1}{8}\)

Now, you can cross multiply

\(8 < 2^{1-k}\)

\(2^3 < 2^{1-k}\)

which means that the exponents look as the following

\(3 < 1-k\)

\(3-1 < -k\)

\(2 < -k\)

\(-2 > k\)

OR

\(k < -2\)

\(B > A\)
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Re: k is an integer for which [#permalink] New post 23 Jun 2019, 01:48
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thank you so much..that helped.

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Re: k is an integer for which   [#permalink] 23 Jun 2019, 01:48
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