Problem Solving

If x^2 < 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.

Guys ,

it is given that x^2 < x means 0 < x < 1

and also x is terminating decimal.

Statement 1 - 10x is not an integer.

Case 1 x = 0.318 which is terminating decimal, here hundredths digit is non zero

10x = 3.18 is not integer and satisfies .

YES CASE

case 2 - x = 0.308 which is terminating decimal

10x = 3.08 is not integer and satisfies

but 0.308 here hundredths is zero so NO CASE.


Statement 2 - 100x is an integer

Case 1 - x = 0.38

100x = 38 which is integer satisfies

here hundredths is 8 which is non zero

case YES

case 2 - x = 0.30

100x = 30 which is integer satisfies.

here hundredths is zero so CASE NO.

From this i can easily eliminate option A,B,and D

I am not able to understand why optionC

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.

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.

If the hundreds digit of x is 0, then both statements cannot be satisfied.
To illustrate:
0.001 --- one place to the right --> 0.01 --- two places to the right --> 0.1.
Since moving two places to the right yields a NON-INTEGER, statement 2 is not satisfied.

Thus, to satisfy both statements, x must have a nonzero hundredths digit.
SUFFICIENT.

The correct answer is C.

x² < x means 0 < x < 1
With Statement (1) then 0 < 10x < 10 is INSUFFICIENT (as x could be for example 0.01 or 0.001)
With Statement (2) then 0 < 100x < 100 is INSUFFICIENT (as x could be for example 0.01 or 0.10)
Statement (1) tells us that x must have no less than 2 digits
Statement (2) tells that us x must have no more than 2 digits
Therefore, combined statements tell us that x must have 2 digits (non-zero for the second one) = SUFFICIENT

j_shreyans wrote:If x^2 < 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.
optionC

Please post the question in the right forum (Data Sufficiency for this Question)

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.

You may think a little for majority of Inequality question rather than jumping on the maths for solving
e.g.

x² < x is possible only if
1) x is positive and because x² is always positive which is less than x and
2) x is between o and 1 because higher powers always result in smaller number in the range
therefore, x² < x ==> 0<x<1 i.e. a number of type x = 0.abcde...Terminating

Question : if x = 0.abcde then Is b=0?

Statement 1) 10x is not an Integer
i.e. 10x = a.bcde but b may(a.0c) or may not (a.2c) be zero so not sufficient

Statement 2) 100x is an Integer
i.e. 100x = ab but b may(a0) or may not (a2) be zero so not sufficient

Combining the two statements

Now we know that number is terminating without c,d,e and is of the type 0.ab therefore b can't be zero
SUFFICIENT

Answer: Option C