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by alltimeacheiver » Wed May 04, 2011 8:06 pm
ques If K is a integer less than 17 and k-1 is the square of an integer, what is value of k

1 k is an even integer Insufficeint

2 k+2 is the square of an ingteger
in second it can be two option if 2+2= 4 and 14+2= 16

So I marked e but the ans is but how. If something i missed
Source: — Data Sufficiency |

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by manpsingh87 » Wed May 04, 2011 8:40 pm
alltimeacheiver wrote:ques If K is a integer less than 17 and k-1 is the square of an integer, what is value of k

1 k is an even integer Insufficeint

2 k+2 is the square of an ingteger
in second it can be two option if 2+2= 4 and 14+2= 16

So I marked e but the ans is but how. If something i missed
as k is less than 17 and also k-1 is a square of an integer; therefore possible values of k would be; 1,2,5,10;

1) if k is an even integer than possible values of k would be 2,10 hence 1 is not sufficient to answer the question..!!!
2)k+2 is a square of an integer, now from the possible values of k as per the stem of the question(1,2,5,10) only 2 is the value which satisfies initial as well as the condition mentioned in statement..!!! (2-1=1 which is a perfect square and 2+2=4 which is a perfect square) therefore 2 alone is sufficient to answer the question..!!!

hence B
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by Anurag@Gurome » Wed May 04, 2011 8:43 pm
alltimeacheiver wrote:ques If K is a integer less than 17 and k-1 is the square of an integer, what is value of k

1 k is an even integer Insufficeint

2 k+2 is the square of an ingteger
in second it can be two option if 2+2= 4 and 14+2= 16

So I marked e but the ans is but how. If something i missed
k < 17 and k - 1 = (integer)²

(1) k is an even integer implies k can be 0, 2, 4, 6, 8, 10, 12, 14, 16.
Now k - 1 = (integer)² implies the possible values of k can be 2 (as 2 - 1 = 1 = 1²), 10 (10 - 1 = 9 = 3²).
We don't get a unique answer.
So, (1) is NOT SUFFICIENT.

(2) k + 2 is the square of an integer and k - 1 = (integer)² implies k can only take the value 2 (as 2 + 2 = 4 = 2² and 2 - 1 = 1 = 1²). No other value satisfies both the conditions.
So, k = 2
So, (2) is SUFFICIENT.

The correct answer is B.
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