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A=0.abc, where a, b and c are distinct digits, is A>34?
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Updated on: 07 May 2019, 23:25
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A=0.abc, where a, b and c are distinct digits, is \(A>\frac{3}{4}\)? (1) a/b=2 and b/c=3/4 (2) 2(a+b)>10a+b
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Originally posted by kiran1213 on 24 Dec 2018, 11:09.
Last edited by chetan2u on 07 May 2019, 23:25, edited 1 time in total.
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A=0.abc, where a, b and c are distinct digits, is A>34?
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Updated on: 13 May 2019, 01:15
Statement 1:if \(a:b = 2:1\) & \(b:c = 3:4\), then \(a:b:c = 6:3:4\) since a,b&c are singledigit numbers, the only valid value for \(A = 0.634\) , which is \(<\frac{3}{4}\) > sufficientStatement 2:if \(2(a+b)>10a+b\), then it can be simplified to \(b>8a\). so if \(a=1\), then \(b>8\) (9,10,...) and if \(a=2\), then \(b>9\) (10,11,...) since a&b are singledigit numbers, the only valid value for \(b = 9\) and \(a = 1\) so \(A = 0.19c\), which is \(<\frac{3}{4}\) regardless of the value of \(c\) > sufficientD
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Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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27 Dec 2018, 04:27
I think the question is , "is A>3/4?". A cannot be greater than 2.



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Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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07 May 2019, 22:21
poor quality question.
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Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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07 May 2019, 23:30
A=0.abc, where a, b and c are distinct digits, is \(A>\frac{3}{4}\)? So, A>0.75, which means is a\(\geq{7}\) (1) a/b=2 and b/c=3/4... a=2b, and 4b=3c... 4b=3c tells us that c can be 4 or 8.. a) if c is 4 4b=3*c=3*4...b=3, and a=2b=2*3=6 A=0.634...Ans NO b) if c is 8 4b=3*c=3*8...b=6, and a=2b=2*6=12.. BUT a is a digit, so cannot be >9. So, only possibility is A=0.634...Ans NO Suff (2) 2(a+b)>10a+b 2a+2b>10a+b....b>8a.... Since b is <10, only possibility for a is 1.. Hence ans NO suff D NOTE.. You have different values of a and b from statement I and II. Will not be the case in actuals.
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Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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11 May 2019, 23:39
kiran1213 wrote: A=0.abc, where a, b and c are distinct digits, is \(A>\frac{3}{4}\)? (1) a/b=2 and b/c=3/4 (2) 2(a+b)>10a+b Can somebody help me understand the question? A=0.abc I read this as A is equal to 0 multiplied by a*b*c.



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Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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12 May 2019, 00:14
saikeish15 wrote: kiran1213 wrote: A=0.abc, where a, b and c are distinct digits, is \(A>\frac{3}{4}\)? (1) a/b=2 and b/c=3/4 (2) 2(a+b)>10a+b Can somebody help me understand the question? A=0.abc I read this as A is equal to 0 multiplied by a*b*c. Firstly, the question will mention the same way you have written (A*B*C) if the intention is to multiply the digits or it will be mentioned explicitly in the question stem itself ( eg multiplication of three numbers). Secondly, do you think that there would any point to ask such question because any number multiplied by 0 will always yield you zero, then how can it be greater than 0; So you could have answered this question even without the 2 statements given. Give a thought!
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Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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12 May 2019, 21:57
BarcaForLife wrote: saikeish15 wrote: kiran1213 wrote: A=0.abc, where a, b and c are distinct digits, is \(A>\frac{3}{4}\)? (1) a/b=2 and b/c=3/4 (2) 2(a+b)>10a+b Can somebody help me understand the question? A=0.abc I read this as A is equal to 0 multiplied by a*b*c. Firstly, the question will mention the same way you have written (A*B*C) if the intention is to multiply the digits or it will be mentioned explicitly in the question stem itself ( eg multiplication of three numbers). Secondly, do you think that there would any point to ask such question because any number multiplied by 0 will always yield you zero, then how can it be greater than 0; So you could have answered this question even without the 2 statements given. Give a thought! Ah. Sorry. I had a brain fade. Reread the question again today and realized the absurdity of my doubt!




Re: A=0.abc, where a, b and c are distinct digits, is A>34?
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12 May 2019, 21:57






