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M27-17

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M27-17  [#permalink]

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New post 16 Sep 2014, 01:27
1
10
00:00
A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

46% (02:09) correct 54% (01:45) wrong based on 188 sessions

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Re M27-17  [#permalink]

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New post 16 Sep 2014, 01:27
1
3
Official Solution:


First of all, \(\frac{7}{9}\) is a recurring decimal \(=0.77(7)\).

(1) \(a+b \gt 14\). The least value of \(a\) is 6 (\(6+9=15 \gt 14\)), so in this case, \(x=0.69d \lt 0.77(7)\), but \(a=7\) and \(b=9\) is also possible, and in this case \(x=0.79d \gt 0.77(7)\). Not sufficient.

(2) \(a-c \gt 6\). The least value of \(a\) is 7 (\(7-0=7 \gt 6\)), but we don't know the value of \(b\). Not sufficient.

(1)+(2) The least value of \(a\) is 7 and in this case from (1) the least value of \(b\) is 8 (\(7+8=15 \gt 14\)), hence the least value of \(x=0.78d \gt 0.77(7)\). Sufficient.


Answer: C
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Re: M27-17  [#permalink]

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New post 03 Oct 2017, 01:27
Hi Bunuel

Thank you for the explanation.

I am not able to follow when we combine the two statements.

For S1 a+b>14 , Can we not have 9+6=15 and in this case the least value of b will be 6 and we get 0.76 which is less than 0.77 and answer is E.

Please help me understand what am I doing wrong?

Appreciate your help. :-)
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Re: M27-17  [#permalink]

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New post 03 Oct 2017, 01:36
KB04 wrote:
Hi Bunuel

Thank you for the explanation.

I am not able to follow when we combine the two statements.

For S1 a+b>14 , Can we not have 9+6=15 and in this case the least value of b will be 6 and we get 0.76 which is less than 0.77 and answer is E.

Please help me understand what am I doing wrong?

Appreciate your help. :-)


When we combine the statements we get that the least value of a is 7, so it can be 7, 8 or 9. If a = 9, as it is in your example, then x = 0.9... not 0.76.., so x > 0.777...
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M27-17  [#permalink]

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New post 12 Jul 2018, 18:17
Here is my approach

Question is x > 7/9 which is 0.777..

(1). Given a+b>14

Only possible values are (Please note the max value is 9 for any digit)
9+6=15>14 --> 0.96.. (Greater than 0.77.., Yes)
8+7=15>14 --> 0.87..(greater than 0.77.., yes)
7+8=15>14 --> 0.78..(greater than 0.77.., Yes)
6+9=15>14 --> 0.69..(Lesser than 0.77.., NO)
Thus Statement 1 is not sufficient.

(2). Given a-b >6
similar way
9-2=7>6 --> 0.9b2d (greater than 0.7...,Yes)
8-1=7>6 --> 0.8b1d (greater than 0.7...,Yes)
7-0=7>6 --> 0.7b0d (lesser than 0.7777, NO)
Thus statement 2 is not sufficient.

(1) and (2) >> Given a+b>14 and a-c>6
Overlapping both, we can take a=6, as it will not satisfy (1) and (2) together.

a=7 and then b must be 8 from above.

so x=0.78cd >0.7777 Thus (1) and (2) are sufficient.

Answer is (C)
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Re: M27-17  [#permalink]

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New post 12 Jul 2018, 19:51
Is x>0.77777777 ?

(1) a=7; b=8
x=0.78 yes
a=6; b=9
x=0.69 no
Insufficient

(2) a>c+6
c=0; a=7
x=0.700d no
x=0.780d yes
Insufficient

(1) and (2) 7 is min value of a
x=0.7b0d
then b>=8
yes

Answer C
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Re: M27-17  [#permalink]

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New post 01 Oct 2018, 09:06
Very intersting... thank you Bunuel
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Re: M27-17   [#permalink] 01 Oct 2018, 09:06
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