positive integer x

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positive integer x

by j_shreyans » Wed Nov 05, 2014 9:43 am
If the positive integer x is rounded to the nearest ten, will the result be greater than x ?

(1) If x is divided by 10, the remainder is even.

(2) If x is divided by 5, the remainder is odd.

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by GMATGuruNY » Wed Nov 05, 2014 10:38 am
j_shreyans wrote:If the positive integer x is rounded to the nearest ten, will the result be greater than x ?

(1) If x is divided by 10, the remainder is even.

(2) If x is divided by 5, the remainder is odd.
Test a set of 10 consecutive integers:
10, 11, 12, 13, 14, 15, 16, 17, 18, 19.
The five red integers rounded to the nearest ten = 10.
The five green integers rounded to the nearest ten = 20.
Thus, if x is equal to any of the green integers, the result of rounding to the nearest ten -- 20 -- will be greater than x.

Question stem, rephrased:
Is x equal to a green integer?

Statement 1: If x is divided by 10, the remainder is even.
It's possible that x = 10, 12, 14, 16, 18.
INSUFFICIENT.

Statement 2: If x is divided by 5, the remainder is odd.
It's possible that x = 11, 13, 16, 18.
INSUFFICIENT.

Statements combined:
Statement 1: x = 10, 12, 14, 16, 18.
Statement 2: x = 11, 13, 16, 18.
Only 16 and 18 satisfy both statements.
Thus, x must be equal to a green integer.
SUFFICIENT.

The correct answer is C.

The reasoning used above will hold true for ANY set of 10 consecutive integers (20...29, 30...29, etc.).
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by Mathsbuddy » Wed Nov 12, 2014 7:06 am
Let V = rounded value
Let L = last digit of x

If L = 0,1,2,3,4 then V < x
If L = 5,6,7,8,9 then x < V

Using Statement 1:
Dividing by 10 to get an even remainder means that L = 2,4,6,8
So INSUFFICIENT

Using Statement 2:
Dividing by 5 to get an odd number means that L = 1,3,6,8
So INSUFFICIENT

Combining statements:
Only L = 6,8 matches both
Therefore x<V
SUFFICIENT