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by theCodeToGMAT » Wed Sep 18, 2013 12:46 am
let X be _ _ . _ _ _ (assume it to be till thousandth digit; Question Stem doesn't tell anything)


Statement 1: 10x is not an integer
Consider values:
0.304 --> 10 (0.304) --> 3.04
0.314 --> 10 (0.314) --> 3.14
Hence, Insufficient

Statement 2: 100x is an integer; so our assumption is wrong.. the number is in fact _ _ . _ _ (till hundredth digit)
Consider Values
.30 --> 100(0.30) --> 30
.31 --> 100(0.31) --> 31
Hence, Insufficient


Combining,
Conclusion--> Number is two places after decimal; Confirmed _ _ . _ _
Now, since 10x is not a decimal.. so the Hundred digit cannot be "0"
i.e. .30 is not applicable.. since 10(0.30) = 3
So, Hundredth can be 1 to 9

hence, [C]


Yaj wrote:Hi Guys,

Can anyone explain any way to approach this problem!

If x2 < x and x is written as a terminating decimal, does x have a nonzero hundredths digit?

(1) 10x is not an integer.

(2) 100x is an integer.

[spoiler]OA:C[/spoiler]

Thanks so much!
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by sanju09 » Wed Sep 18, 2013 1:43 am
Yaj wrote:Hi Guys,

Can anyone explain any way to approach this problem!

If x2 < x and x is written as a terminating decimal, does x have a nonzero hundredths digit?

(1) 10x is not an integer.

(2) 100x is an integer.

[spoiler]OA:C[/spoiler]

Thanks so much!
It's possible only when 0<x<1. If x may be written as a terminating decimal, then 0.25 is a good example for x, which does have a nonzero hundredths digit. Further, 0.2 and 0.205 are another good examples for x, which don't have a nonzero hundredths digit.

Our question reduces to, "Does x have a hundredths digit and it's not zero?"

I. 10x is an integer only when x is 0.1, 0.2, ... 0.9, and 10x is not an integer when x has two or more significant figures to the right of decimal point. According to this statement, x may assume 0.25, a nonzero hundredths digit, or x may assume 0.205, a zero in hundredths place. Insufficient

II. 100x is an integer only when x is of 0.2, 0.12 or 0.09 kind of forms, otherwise NOT. According to this statement, x may assume 0.12, a nonzero hundredths digit, or x may assume 0.2, a zero in hundredths place. Insufficient

When taken together, we find that x cannot be of the 0.2 or 0.205 types, but x can be of the 0.12 or 0.25 types; and [spoiler]YES! x does have a hundredths digit and it's not zero. Sufficient

Take C
[/spoiler]
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by GMATGuruNY » Wed Sep 18, 2013 1:42 pm
Yaj wrote:Hi Guys,

Can anyone explain any way to approach this problem!

If x2 < x and x is written as a terminating decimal, does x have a nonzero hundredths digit?

(1) 10x is not an integer.

(2) 100x is an integer.

[spoiler]OA:C[/spoiler]

Statement 1: 10x is not an integer
In other words, when the decimal point is moved ONE PLACE TO THE RIGHT, the result is a NON-INTEGER.
It's possible that x = .01, since moving the decimal point one place to the right yields non-integer 0.1.
In this case, the hundredths digit of x is nonzero.
It's possible that x = .001, since moving the decimal point one place to the right yields non-integer 0.01.
In this case, the hundredths digit of x is 0.
INSUFFICIENT.

Statement 2: 100x is an integer
In other words, when the decimal point is moved TWO PLACES TO THE RIGHT, the result is an INTEGER.
It's possible that x = .01, since moving the decimal point two places to the right yields 1.
In this case, x has a nonzero hundredths digit.
It's possible that x = .1, since moving the decimal point two places to the right yields 10.
In this case, x does not have a nonzero hundredths digit.
INSUFFICIENT.

Statements combined:
Since moving the decimal point one place to the right yields a NONINTEGER, while moving the decimal point two places to the right yields an INTEGER, x must be of the form 0.AB, where B is a NONZERO digit.
To illustrate:
0.01 --- one place to the right --> 0.1 --- two places to the right --> 1.
0.32 --- one place to the right --> 3.2 --- two places to the right --> 32.
0.15 --- one place to the right --> 1.5 --- two places to the right --> 15.
SUFFICIENT.

The correct answer is C.
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