Needgmat 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.
x rounded to the nearest ten will be greater than x if the units digit of x is 5 or greater.
x=5 rounded to the nearest ten --> 10, which is greater than x=5.
x=16 rounded to the nearest ten --> 20, which is greater than x=16.
x=157 rounded to the nearest ten --> 160, which is greater than x=157.
If the units digit of x is less than 5, then x rounded to the nearest ten will be less than or equal to x.
x=10 rounded to the nearest ten --> 10, which is equal to x=10.
x=24 rounded to the nearest ten --> 20, which is less than x=24.
x=101 rounded to the nearest ten --> 100, which is less than x=101.
Question stem, rephrased:
Is the units digit of x between 5 and 9, inclusive?
Statement 1: If x is divided by 10, the remainder is even.
It's possible that x=18, since 18/10 = 1 R8.
In this case, the units digit of x is between 5 and 9, inclusive.
It's possible that x=12, since 12/10 = 1 R2.
In this case, the units digit of x is NOT between 5 and 9, inclusive.
INSUFFICIENT.
Statement 2: If x is divided by 5, the remainder is odd.
It's possible that x=18, since 18/5 = 3 R3.
In this case, the units digit of x is between 5 and 9, inclusive.
It's possible that x=13, since 13/5 = 2 R3.
In this case, the units digit of x is NOT between 5 and 9, inclusive.
INSUFFICIENT.
Statements combined:
Statement 1 requires that the units digit of x be EVEN.
Between 10 and 20, inclusive, the following integers have an even units digit:
10, 12, 14,
16, 18.
Of these options, only the options in blue yield an odd remainder when divided by 5.
Implication:
To satisfy both statements, x must have a units digit of 6 or 8.
Thus, the units digit of x must be between 5 and 9, inclusive.
SUFFICIENT.
The correct answer is
C.
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