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Clarification

by knight247 » Sat Jul 23, 2011 12:10 am
If k and n are integers, is n divisible by 7?
(1) n/2-3=k
(2) 2k-1 is divisible by 7

The OA is C. But I disagree with it. It should be A according to me

Here's why
(1) n/2-3=k

Rewritten as (n-6)/2=k
n-6=2k
n=2k+6

Now 2k and 6 are both even numbers. Even+Even=Even. Hence, n is even. If n is even then it can't be divisible by 7. Am I making sense?

(2)
2k-1 is divisible by 7
Gives no hints on the divisibility of n hence insufficient.

Answer is A

Thoughts, Comments? lol
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Source: — Data Sufficiency |
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by cans » Sat Jul 23, 2011 12:27 am
(1) n/2-3=k

Rewritten as (n-6)/2=k
n-6=2k
n=2k+6

Now 2k and 6 are both even numbers. Even+Even=Even. Hence, n is even. If n is even then it can't be divisible by 7.
14 is even and also divisible by 7.
So answer is C
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by knight247 » Sat Jul 23, 2011 12:45 am
Whooops...My Bad..lol
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by Ozlemg » Mon Jul 25, 2011 2:18 am
knight247 wrote:If k and n are integers, is n divisible by 7?
(1) n/2-3=k
(2) 2k-1 is divisible by 7

The OA is C. But I disagree with it. It should be A according to me

Here's why
(1) n/2-3=k

Rewritten as (n-6)/2=k
n-6=2k
n=2k+6

Now 2k and 6 are both even numbers. Even+Even=Even. Hence, n is even. If n is even then it can't be divisible by 7. Am I making sense?

(2)
2k-1 is divisible by 7
Gives no hints on the divisibility of n hence insufficient.

Answer is A

Thoughts, Comments? lol
Here is my way :

plug K to statement 2 and the number will still be divisible by 7.

so, 2*(n/2-3)-1 = n-7 is divisible by 7. (for ex : 14-7, 28-7 etc.)

Hence, C
The more you suffer before the test, the less you will do so in the test! :)
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