If x^2 < x and x is written as a terminating decimal,does x

If \(x^{2} < x\) and x is written as a terminating decimal, does x have a nonzero hundredths digit?

\(x^{2} < x\) means that \(0<x<1\).

(1) 10x is not an integer. This only means that x is NOT 0.1, 0.2, ...., 0.9. Now, if x=0.01, then the answer is YES but if x=0.001, then the answer is NO. Not sufficient.

(2) 100x is an integer. This means that x is 0.01, 0.02, ..., 0.10, ..., 0.99. The hundredths digit could be 0 as well as nonzero. Not sufficient.

(1)+(2) From (1) 0.10, 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80 and 0.90 are excluded, thus from (2) only those decimals are left which have a nonzero hundredths digit: 0.01, 0.02, ..., 0.09, 0.11, ..., 0.19, ..., 0.21, ..., 0.99. Sufficient.

Answer: C.

Hope it's clear.
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Here's how I did it:

I represented x as x = .abc (I didn't go further because the question didn't ask for more than tens/hundredths digits).

1) 10x is NOT an integer. This means that a.bc is NOT an integer. Therefore, b and c cannot BOTH be zero; otherwise 10x would be an integer. Not sufficient.

2) 100x IS an integer. This means that ab.c is an integer, and therefore that c = 0. This just gives us x = .ab0 - no information about b. Not sufficient.

Together though, we know that c = 0. However, we also know that b and c cannot BOTH be zero. Therefore, b has to be nonzero.

Answer: C

Hi. Suppose i choose two numbers 0.23 and 0.03 for both of them x^2<x

Both number satisfy both the statements:

x=0.23 10x=2.3, not an integer
x=0.23 100X=23 is an integer

x=0.03 10x=0.3 is not an integer
x=0.03 100x=3 is an integer

in case 1 100 digit is 2 and in case 2 100 digit is 0.

Please help me by explaining where is the mistake.

In both cases the hundredths digit is 3: 0.23 and 0.03.


1234.567

1 - THOUSANDS
2 - HUNDREDS
3 - TENS
4 - UNITS
. - decimal point
5 - TENTHS
6 - HUNDREDTHS
7 - THOUSANDTHS

Hope it helps.
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Hi Bunuel, could u elaborate the explanation for this one? As soon as I saw that "100x is an integer" , figures like .99, .45, .01 come to mind and thus 100x always has a zero hundredth digit.

example: if x=.abcd
100x is an integer then 100x= ab.cd
for ab.cd to be an integer, cd has to be zero. therefore, hundredth place of x is a zero integer.

Please explain .

How is hundredths digit of 0.99, 0.45, or 0.01 zeo?
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say x is .45
then statement says that 100x is an integer ( of the form xx.0)
if thats true then x cant be .455 or .555, it has to be something like .45 or .10 ( since .45*100 = 45)

I dont even know where im wrong.

x CANNOT be 0.1 because (1) say that 10x is NOT an integer and if x=0.1, then 10x=1=integer. Anyway, what's your question?
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I just realized the mistake i was making.
i was counting a unit digit after decimal point and thought that B is sufficient. But now i realized its not the case.

Here is what i understand now:
.abc . question : is b zero or nonzero?

statement 1: a.b is not integer.
x could be :: .001 or .01
b can be zero or non zero if c is non-zero (.011 , .001) ; but if c is zero then b has to be non zero in order for 10x to not be an integer
2 different answers
Insuff
statement 2: c is zero
but no info about b


(1+2)
when c is zero , b is non zero


Is my logic correct this time?

Bunuel plz explain where am i going wrong, if I select two numbers i.e 0.105 and 0.125.
Both when multiplied by 10 become 10.5 and 12.5 (non - integers)
both when multiplied by 100 become integers 105 and 125.
But in one the hundreth digit and 0 and the other one its 2..What am i missing??? Thanks in advance !

Check the red parts.

10x = 1.05 and 1.25, receptively.
100x = 10.5 and 12.5, receptively.
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x is a +ve fraction , in decimal form ... 0.abcd... , is b not 0

from 1

a.bcd... is not integer ... b or c or d could be any digit , i.e b = 0 but c is not or b a no zero digit and c,d = 0 ... insuff

from 2

ab.cd.. = ab thus c,d...etc = 0 ... no idea about b itself .. insuff

both

the decimal representation of x is in the form 0.ab and a.b is not integer thus b aint zero ... suff

C

Don't you think you've made it quite complicated?

Given that x is a terminating decimal, the only options we have are

1/2

1/4

1/5

1/8

Now, from (1) we only have

1/4

1/8

As 10x is NOT and INTEGER

From (2) we have multiple options, including

1/4

and etc.

But Combining (1) and (2)
We have

1/4

And both of these numbers have a non-zero hundredth digit.

Great question. I narrowed down to using x with values only inclusive of 2 and 5, considering those are terminating decimals as required by X

From x^2<x it should be obvious x is between 0 and 1

Statement 1: 10x is not an integer. Cases used, x=1/20 (0.05), 1/200 (0.005) Hence Insufficient.
Statement 2: 100x is an integer. Cases used, 1/2,1/5, 1/20,1/50. 1/2 will give 0.50 and 1/20 will give 0.05. Hence Insufficient

Collab 1 and 2, Cases with 1/20 survive. Definite Answer. C

D4kshGargas

There are many more options for a terminating decimal:
1/20
1/25
1/40
1/50
1/80
1/100
1/200
...and so on

But anyway, it doesn't change the validity of your explanation.
Just saying...

Just thought of another approach:

If x^2<x, that means x must lie between 0 and 1.
We know that x is a terminating decimal.
Some examples we can use for trial and error: 0.2, 0.75, 0.625, 0.101

What the question really means: "Is the second digit to the right of the decimal point NOT equal to zero?"

Since this is a yes-no question, we can also ask, "Is the second digit to the right of the decimal point EQUAL to zero?"
(We'll get yes instead of no, but it won't affect our answer).

Statement 1 tells us that 10 x is NOT an integer.
All it means is that x MUST have more than one digit to the right of the decimal point.
If it has two digits to the right of the decimal point, then we get a YES answer (e.g. x = 0.11)
If it has three digits to the right of the decimal point, then we CAN get a NO answer (e.g. x = 0.101)
Not sufficient.

Statement 2 tells us that 100 x is an integer.
All it means is that x has either one OR two digits to the right of the decimal point.
It cannot have 3 or more digits to the right of the decimal point.
If it has one digit to the right of the decimal point, then we get a NO answer (e.g. x = 0.1)
If it has two digits to the right of the decimal point, then we get a YES answer (e.g. x = 0.11)
Not sufficient.

Combined: For both statements to be satisfied:
x CANNOT have one digit to the right of the decimal point (this would violate statement 1)
x CANNOT have 3 or more digits to the right of the decimal point (this would violate statement 2)
Therefore, x MUST have exactly two digits to the right of the decimal point...
...and the second digit to the right of the decimal place cannot be zero
i.e., x MUST have a nonzero hundredths digit.

Always YES.
Sufficient.
Answer choice: C