If k is an integer, is 5^(-k)

Question

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

(1) 2 < 1 - k

(2) 2k < 3

Kudos for a correct  solution .

peachfuzz · Answer

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

longfellow · Answer

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.

kunal555 · Answer

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

rajarams · Answer

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.

bhaskar438 · Answer

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

MaximD · Answer

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.

Bunuel · Answer

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.

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

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