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# k is an integer for which

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k is an integer for which [#permalink]  24 Feb 2017, 02:43
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Question Stats:

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]  03 Mar 2017, 05:40
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Expert's post
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]  27 Nov 2017, 11:47
Expert's post
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]  30 Nov 2017, 13:53
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Expert's post
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

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Re: k is an integer for which [#permalink]  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]  07 Mar 2018, 15:48
Expert's post
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]  12 Mar 2018, 01:29
B
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Re: k is an integer for which [#permalink]  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]  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]  22 Jun 2019, 23:54
Expert's post
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]  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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