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Is f(n)>f(n−1) ?

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Is f(n)>f(n−1) ?  [#permalink]

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New post 04 Feb 2016, 17:50
1
7
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

56% (00:41) correct 44% (00:32) wrong based on 316 sessions

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Is f(n)>f(n−1)?

(1) n=8

(2) f(n) = n-1
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Re: Is f(n)>f(n−1) ?  [#permalink]

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New post 04 Feb 2016, 20:44
zxcvbnmas wrote:
Is f(n)>f(n−1)?

(1) n=8

(2) f(n) = n-1


Statement 1: It tells us nothing about the function. Insufficient.
Statement 2:
f(n) = n - 1
f(n-1) = n - 2

We have two numbers that can be compared and the relation can be found out.
SUFFICIENT
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Re: Is f(n)>f(n−1) ?  [#permalink]

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New post 05 Feb 2016, 07:05
zxcvbnmas wrote:
Is f(n)>f(n−1)?

(1) n=8

(2) f(n) = n-1


Question : Is f(n)>f(n−1)?

We need the definition of the function f(n) to answer the question

Statement 1: n=8
Since we have no definition of function to falculate f(n) hence
NOT SUFFICIENT

Statement 2: f(n) = n-1
i.e. f(n-1) = (n-1)-1 = n-2
and (n-1) will always be less than (n-2), Hence,
SUFFICIENT

Answer: Option B
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Re: Is f(n)>f(n−1) ?  [#permalink]

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New post 07 Feb 2016, 14:25
1
1. No information about the function itself. f(n) could be more or less than f(n-1). NOT SUFFICIENT
2. Gives us the general function. From this we have sufficient information to test the inequality. SUFFICIENT

ANSWER D



zxcvbnmas wrote:
Is f(n)>f(n−1)?

(1) n=8

(2) f(n) = n-1

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Re: Is f(n)>f(n−1) ?  [#permalink]

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New post 26 Aug 2016, 12:05
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zxcvbnmas wrote:
Is f(n) > f(n−1)?

(1) n = 8

(2) f(n) = n-1


Target question: Is f(n) > f(n−1)?

Statement 1: n = 8
Plug n = 8 into target question to get: Is f(8) > f(8−1)?
In other words, Is f(8) > f(7)?
We cannot answer the target question with certainty, because we don't know anything about the function f.
So, statement 1 is NOT SUFFICIENT

Statement 1: f(n) = n-1
At first glance, we might assume that statement 2 is not sufficient, since we don't know the value of n.
However, now that we know how the function f behaves, we can, indeed answer the target question.
If f(n) = n-1, then...
f(n) = n - 1 and f(n - 1) = (n - 1) - 1 = n - 2
So, the original target question Is f(n) > f(n−1)? becomes Is n - 1 > n − 2?
Sure, n - 1 is definitely greater than n - 2
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer =

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Re: Is f(n)>f(n−1) ?  [#permalink]

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New post 16 Apr 2018, 07:45
zxcvbnmas wrote:
Is f(n)>f(n−1)?

(1) n=8

(2) f(n) = n-1


Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 1 variable (n) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.

Condition 1) : n = 8
Since f(n) is not determined, condition 1) is not sufficient.

Condition 2) : f(n) = n - 1
Since f(n) = n - 1, we have f(n-1) = (n-1)-1 = n-2.
Since n - 1 > n - 2, f(n) > f(n-1).
Thus, the condition 2) is sufficient.

Therefore, B is the answer.

If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: Is f(n)>f(n−1) ? &nbs [#permalink] 16 Apr 2018, 07:45
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