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Bunuel
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If k is an integer, is 5^(-k) < 5^(1-2k)? 17 Sep 2015, 00:46
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A
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25% (medium)

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75% (01:57) correct 25% (02:09) wrong based on 340 sessions

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If k is an integer, is \(5^{(-k)} < 5^{(1-2k)}\)?

(1) 2 < 1 - k

(2) 2k < 3

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A
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If k is an integer, is 5^(-k) < 5^(1-2k)? 17 Sep 2015, 08:07
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If k is an integer, is \(5^{(-k)} < 5^{(1-2k)}\)?

(1) 2 < 1 - k

K is negative so the left hand side of the equation in the question will always be smaller than the right hand side
Sufficient

(2) 2k < 3
If k =1, they are equal. However if k is negative the left hand side is smaller
Insufficient

Answer: A
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If k is an integer, is 5^(-k) < 5^(1-2k)? 17 Sep 2015, 08:27
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Answer is A
the statement can be modified to 5^(k) < 5

1: k<-1
which satisfies the question

2: k<3/2
k could be 1 or any other value less than 1.5 which may or may not satisfy the question.

Hence A.
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If k is an integer, is 5^(-k) < 5^(1-2k)? 19 Sep 2015, 12:43
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If k is an integer, is \(5^{(-k)} < 5^{(1-2k)}\)?

(1) 2 < 1 - k

(2) 2k < 3

Kudos for a correct solution.

Is \(5^{(-k)} < 5^{(1-2k)}\) ?
or \(5^{(-k)} < \frac{5^1}{5^{2k}}\)
or Is \(5^k < 5\) ?

Statement 1:
2<1-k
or k<-1
as k<-1, \(5^k < 5\)
SUFFICIENT

Statement 2:
2k<3
or k<3/2
In this case, \(5^k\) may or may not be less than 5.
INSUFFICIENT

Answer:- A
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If k is an integer, is 5^(-k) < 5^(1-2k)? 20 Sep 2015, 20:50
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Simplifying what is to be found:

\(5^{-k} < 5^{1-2k}\)
\(=> \frac{1}{5^k} < \frac{5}{5^{2k}}\)

Statement 1:

2 < (1-k)
=> k < -1
Since we know that k is an integer, let us try the maximum possible value of k, which is -2

\(\frac{1}{5^{-2}} = 25\)
\(\frac{5}{5^{-4}} = 5^4 = 25^2 = 625\)

We see that for any negative value of k with k < -1, LHS will always be lesser than RHS. Hence, Sufficient. So, we now have A/D

Statement 2:
2k < 3
k < 1.5

Maximum possible value of k = 1
For k=1, LHS = RHS (ie., LHS is not < RHS and the answer to the question is No)
For k=0,
\(\frac{1}{5^0} = 1\)
\(\frac{5}{5^0} = 5\)
We see that LHS < RHS and the answer to the question is Yes. Hence, not sufficient

Leaving us with A.
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If k is an integer, is 5^(-k) < 5^(1-2k)? 22 Sep 2015, 14:52
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Bunuel
If k is an integer, is \(5^{(-k)} < 5^{(1-2k)}\)?

(1) 2 < 1 - k

(2) 2k < 3

Kudos for a correct solution.

\(5^{(-k)} < 5^{(1-2k)}\) will be true when the exponent of the first function is less than the exponent of the second function.
The question can be rewritten as Is -k<1-2k? \(==>\) Is k<1?

(1) 2< 1-k \(==>\) k<-1 Since k<-1 satisfies k<1, this statement is always sufficient
(2) 2k< 3 \(==>\) k<1.5 k may or may not be less than 1. This statement is insufficient

Correct answer is A
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If k is an integer, is 5^(-k) < 5^(1-2k)? 12 May 2017, 06:19
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But,

0 is also an integer.

5(−k)<5(1−2k).

Let's fill in 0. --> 5(-0)<5*5^(-0)
You get: 1<5*1. And that's correct.

Statement 1 ánd 2 allow us to use 0. So, neither is sufficient, even when they're combined, and the answer should be E.
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If k is an integer, is 5^(-k) < 5^(1-2k)? 12 May 2017, 07:06
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MaximD
But,

0 is also an integer.

5(−k)<5(1−2k).

Let's fill in 0. --> 5(-0)<5*5^(-0)
You get: 1<5*1. And that's correct.

Statement 1 ánd 2 allow us to use 0. So, neither is sufficient, even when they're combined, and the answer should be E.

That's not correct. For (1) k cannot be 0 because 2 < 1 - k means that k < -1.

If k is an integer, is \(5^{(-k)} < 5^{(1-2k)}\)?

Is \(5^{(-k)} < 5^{(1-2k)}\)?

Is \(\frac{5^{(-k)}}{5^{(1-2k)}}<1\)?

Is \(5^{(k-1)}<1\)?

Is k - 1 < 0?

Is k < 1?

(1) 2 < 1 - k --> k < -1. Sufficient.

(2) 2k < 3 --> k < 1.5. Not sufficient.

Answer: A.
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