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non zero hundredths digit.

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non zero hundredths digit.

by j_shreyans » Sun Nov 02, 2014 3:01 am
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
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by GMATGuruNY » Sun Nov 02, 2014 3:45 am
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.
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by Mathsbuddy » Mon Nov 03, 2014 8:51 am
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
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by ceilidh.erickson » Wed Nov 05, 2014 8:06 am
Your setup of the problem and your breakdown of the statements was perfect here. You came up with examples that quickly proved that each statement was insufficient. But I think you weren't combining the statements correctly...

You eliminated each statement because you came up with a YES case and a NO case. Then (although you don't explain your reasoning, so forgive me if I'm putting thoughts in your head), you say that you don't understand why C works, but it doesn't seem like you've combined the statements. You might still have been looking at all 4 examples.

If 100x is an integer, that must mean that x can't have more than 2 digits after the decimal point. So the examples you used for statement (1), 0.318 and 0.308, are irrelevant. Looking at the two examples you used for statement (2), 0.38 and 0.30, do they also satisfy statement (1)?

x = 0.38 does, because 100x is an integer (38.0) but 10x is NOT and integer (3.8).
x = 0.30 does NOT, because 100x is an integer (30.0), but so is 10x (3.0).

So, if you example with a zero hundreds digits cannot work, then together they must be SUFFICIENT.
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by GMATinsight » Wed Nov 05, 2014 10:04 pm
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)
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by GMATinsight » Wed Nov 05, 2014 10:15 pm
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
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