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If x^2 < x and x is written as a terminating decimal,does x
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27 Jul 2014, 10:54
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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.
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If x^2 < x and x is written as a terminating decimal,does x
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27 Jul 2014, 13:20
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=0.0 1, then the answer is YES but if x=0.0 01, then the answer is NO. Not sufficient. (2) 100x is an integer. This means that x is 0.0 1, 0.02, ..., 0.1 0, ..., 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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Re: If x^2 < x and x is written as a terminating decimal,does x
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08 Oct 2014, 00:51
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.
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Re: If x^2 < x and x is written as a terminating decimal,does x
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Re: If x^2 < x and x is written as a terminating decimal,does x
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09 Oct 2014, 00:47
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 .
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If x^2 < x and x is written as a terminating decimal,does x
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09 Oct 2014, 01:34
shreyagmat wrote: 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 0.9 9, 0.4 5, or 0.0 1 zeo?
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Re: If x^2 < x and x is written as a terminating decimal,does x
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09 Oct 2014, 02:08
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.
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Re: If x^2 < x and x is written as a terminating decimal,does x
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Re: If x^2 < x and x is written as a terminating decimal,does x
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09 Oct 2014, 03:08
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 nonzero (.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?
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If x^2 < x and x is written as a terminating decimal,does x
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09 Oct 2014, 09:12
WoundedTiger 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. 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



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Re: If x^2 < x and x is written as a terminating decimal,does x
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15 Oct 2014, 08:43
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 !



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Re: If x^2 < x and x is written as a terminating decimal,does x
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15 Oct 2014, 08:58



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Re: If x^2 < x and x is written as a terminating decimal,does x
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28 Dec 2016, 02:47
WoundedTiger 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. 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



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