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If A=0.abc, where a, b, and c are digits of A, is A greater

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If A=0.abc, where a, b, and c are digits of A, is A greater [#permalink]

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If A=0.abc, where a, b, and c are digits of A, is A greater than 2/3?

(1) a+b>14
(2) b+c>15
[Reveal] Spoiler: OA

Last edited by Bunuel on 03 Sep 2012, 03:12, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Value after Decimal point. [#permalink]

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If A=0.abc, where a, b, and c are digits of A, is A greater than 2/3?

Is \(A>\frac{2}{3}\)? --> Is \(A>0.666...\)?

(1) a+b>14 --> since a and b are single digits, then the least value of a is 6 and in this case b must be 9 (a+b=6+9=15>14), thus the least value of A is 0.69, so more than 0.666... Sufficient.

(2) b+c>15. Clearly insufficient, since we know nothing about a.

Answer: A.

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Hope it helps.
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Re: If A=0.abc, where a, b, and c are digits of A, is A greater [#permalink]

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Re: If A=0.abc, where a, b, and c are digits of A, is A greater [#permalink]

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Question: A greater than 2/3? or A greater than 0.6666?
This is again a YES or NO, question.
Getting either will be fine for the solution.

(1) a+b>14 or a+b>=15
The possible ways for a and b how we can get this is :

a b
6 9
7 8
8 7
9 6

these are the only possible combinations as anything less than 5 will make the other 10, and well anything more then 9 will result in a 10.

anyways, from the above possible values we can see that each is >=0.6666 =>Sufficient

(2) b+c>15

Here, the main digit that we are concerned about is a as that decides if the no is >=0.666 or not
even if b=c=9, but a=1, it will be less then 0.6666
or if a=b=c=9 the in that case it will be greater.

Anyway we need some information of A => Not Sufficient.

Ans : A
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Re: If A=0.abc, where a, b, and c are digits of A, is A greater [#permalink]

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Re: If A=0.abc, where a, b, and c are digits of A, is A greater [#permalink]

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New post 27 Jul 2016, 04:20
shrey2287 wrote:
If A=0.abc, where a, b, and c are digits of A, is A greater than 2/3?

(1) a+b>14
(2) b+c>15



(1) a+b>14

let a+b=15
also
10+5
9+6
8+7
6+9
lowest possible value for a = 6
therefore, let A =0.690=690/1000 > 2/3
since lowest possible is > 2/3
Sufficient

(2) b+c>15
here a can take 1 or a 9
A=0.199 < 2/3
but
A=0.999 > 2/3

insufficient

A
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Re: If A=0.abc, where a, b, and c are digits of A, is A greater [#permalink]

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New post 27 Jul 2016, 17:56
shrey2287 wrote:
If A=0.abc, where a, b, and c are digits of A, is A greater than 2/3?

(1) a+b>14
(2) b+c>15


Question- A>2/3 or .667

1) a+b>14
The minimum possible value for a+b can be 15

lets see the minimum possible value of ab. keeping maximum value of b will give minimum value of a

b = 9, hence a=6.

these values make the expression as .69c, which is >.667

Sufficient.

(2) b+c>15
b+c= 16 (suppose)

as in option 1, let's take maximum value for c i.e 9 and minimum value for b i.e 7.

But a can be anything from 1 to 9, making .abc > or <2/3
Not Sufficient.

A is the answer
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Re: If A=0.abc, where a, b, and c are digits of A, is A greater   [#permalink] 27 Jul 2016, 17:56
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