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# M22-35

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Math Expert
Joined: 02 Sep 2009
Posts: 49208

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16 Sep 2014, 01:17
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Difficulty:

95% (hard)

Question Stats:

46% (01:31) correct 54% (01:20) wrong based on 138 sessions

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If $$a=0.xyz$$, where $$x$$, $$y$$ and $$z$$ are digits from 0 to 9, inclusive, is $$a \gt \frac{2}{3}$$?

(1) $$x + y \gt 13$$

(2) $$x + z \gt 14$$

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Joined: 02 Sep 2009
Posts: 49208

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16 Sep 2014, 01:17
Official Solution:

If $$a=0.xyz$$, where $$x$$, $$y$$ and $$z$$ are digits from 0 to 9, inclusive, is $$a \gt \frac{2}{3}$$?

First of all $$\frac{2}{3}$$ is a recurring decimal 0.666...

(1) $$x+y \gt 13$$. The least value of x is $$5 (5+9=14 \gt 13)$$, so in this case $$x=0.59z \lt 0.66(6)$$ but $$x=7$$ and $$y=9$$ is also possible, and in this case $$x=0.79z \gt 0.66(6)$$. Not sufficient.

(2) $$x+z \gt 14$$. The least value of $$x$$ is $$6 (6+9=15 \gt 14)$$, but we don't know the value of $$y$$. Not sufficient.

(1)+(2) The least value of $$x$$ is 6 and in this case from (1) the least value of $$y$$ is $$8 (6+8=14 \gt 13)$$, hence the least value of $$a$$ is $$0.68z \gt 0.66(6)$$. Sufficient.

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Joined: 14 Jul 2014
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24 Mar 2015, 09:17
Bunuel wrote:
Official Solution:

First of all $$\frac{2}{3}$$ is a recurring decimal 0.666...

(1) $$x+y \gt 13$$. The least value of x is $$5 (5+9=14 \gt 13)$$, so in this case $$x=0.59z \lt 0.66(6)$$ but $$x=7$$ and $$y=9$$ is also possible, and in this case $$x=0.79z \gt 0.66(6)$$. Not sufficient.

(2) $$x+z \gt 14$$. The least value of $$x$$ is $$6 (6+9=15 \gt 14)$$, but we don't know the value of $$y$$. Not sufficient.

(1)+(2) The least value of $$x$$ is 6 and in this case from (1) the least value of $$y$$ is $$8 (6+8=14 \gt 13)$$, hence the least value of $$a$$ is $$0.68z \gt 0.66(6)$$. Sufficient.

Hi Bunuel

When combining, why cant I ADD the inequalities, which gives

2x + y + z > 27

least value can be 28

2 (6) + 7 + 9 = 28 ........ Here I get 0.679 ---> Yes
2 (5) + 9 + 9 = 28 ........ Here I get 0.599 ----> No

Hence (E) is the ans

Where am I going wrong?

Thanks
Math Expert
Joined: 02 Sep 2009
Posts: 49208

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24 Mar 2015, 09:46
1
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

First of all $$\frac{2}{3}$$ is a recurring decimal 0.666...

(1) $$x+y \gt 13$$. The least value of x is $$5 (5+9=14 \gt 13)$$, so in this case $$x=0.59z \lt 0.66(6)$$ but $$x=7$$ and $$y=9$$ is also possible, and in this case $$x=0.79z \gt 0.66(6)$$. Not sufficient.

(2) $$x+z \gt 14$$. The least value of $$x$$ is $$6 (6+9=15 \gt 14)$$, but we don't know the value of $$y$$. Not sufficient.

(1)+(2) The least value of $$x$$ is 6 and in this case from (1) the least value of $$y$$ is $$8 (6+8=14 \gt 13)$$, hence the least value of $$a$$ is $$0.68z \gt 0.66(6)$$. Sufficient.

Hi Bunuel

When combining, why cant I ADD the inequalities, which gives

2x + y + z > 27

least value can be 28

2 (6) + 7 + 9 = 28 ........ Here I get 0.679 ---> Yes
2 (5) + 9 + 9 = 28 ........ Here I get 0.599 ----> No

Hence (E) is the ans

Where am I going wrong?

Thanks

You can add but numbers you choose should still fit the statements and your numbers do not.
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Joined: 14 Jul 2014
Posts: 93

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24 Mar 2015, 09:52
Bunuel wrote:
buddyisraelgmat wrote:
Bunuel wrote:
Official Solution:

First of all $$\frac{2}{3}$$ is a recurring decimal 0.666...

(1) $$x+y \gt 13$$. The least value of x is $$5 (5+9=14 \gt 13)$$, so in this case $$x=0.59z \lt 0.66(6)$$ but $$x=7$$ and $$y=9$$ is also possible, and in this case $$x=0.79z \gt 0.66(6)$$. Not sufficient.

(2) $$x+z \gt 14$$. The least value of $$x$$ is $$6 (6+9=15 \gt 14)$$, but we don't know the value of $$y$$. Not sufficient.

(1)+(2) The least value of $$x$$ is 6 and in this case from (1) the least value of $$y$$ is $$8 (6+8=14 \gt 13)$$, hence the least value of $$a$$ is $$0.68z \gt 0.66(6)$$. Sufficient.

Hi Bunuel

When combining, why cant I ADD the inequalities, which gives

2x + y + z > 27

least value can be 28

2 (6) + 7 + 9 = 28 ........ Here I get 0.679 ---> Yes
2 (5) + 9 + 9 = 28 ........ Here I get 0.599 ----> No

Hence (E) is the ans

Where am I going wrong?

Thanks

You can add but numbers you choose should still fit the statements and your numbers do not.

Oh I get it - wow this is really tricky !! Thanks a ton! +1 Kudos
Intern
Joined: 05 Sep 2013
Posts: 3

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13 Dec 2016, 08:52
I am not able to understand the exact question stem (If a=0.xyz)
Is this equation means a = 0*x*y*z ?
Math Expert
Joined: 02 Sep 2009
Posts: 49208

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13 Dec 2016, 10:19
VaibhavSeth wrote:
I am not able to understand the exact question stem (If a=0.xyz)
Is this equation means a = 0*x*y*z ?

0.xyz is a decimal where x is tenths, y is hundredths and z is thousandths digits.
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Intern
Joined: 14 Aug 2017
Posts: 4

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20 Jul 2018, 21:12
Bunuel wrote:
Official Solution:

If $$a=0.xyz$$, where $$x$$, $$y$$ and $$z$$ are digits from 0 to 9, inclusive, is $$a \gt \frac{2}{3}$$?

First of all $$\frac{2}{3}$$ is a recurring decimal 0.666...

(1) $$x+y \gt 13$$. The least value of x is $$5 (5+9=14 \gt 13)$$, so in this case $$x=0.59z \lt 0.66(6)$$ but $$x=7$$ and $$y=9$$ is also possible, and in this case $$x=0.79z \gt 0.66(6)$$. Not sufficient.

(2) $$x+z \gt 14$$. The least value of $$x$$ is $$6 (6+9=15 \gt 14)$$, but we don't know the value of $$y$$. Not sufficient.

(1)+(2) The least value of $$x$$ is 6 and in this case from (1) the least value of $$y$$ is $$8 (6+8=14 \gt 13)$$, hence the least value of $$a$$ is $$0.68z \gt 0.66(6)$$. Sufficient.

Hi Bunuel,

Here in S1 the value of x and y has been taken as 5,9 and 7,9.
and in S2 the value of x and z has been taken as 6,9.

Then how have we arrived at the conclusion that the lowest value of x is 6? I am not able to understand the logic, may you pl elaborate on it. Also why cant the value assumed in both the statements be reversed, after all there is nothing in the question stem that suggests that x is greater/smaller than y or z.

Regards
Re: M22-35 &nbs [#permalink] 20 Jul 2018, 21:12
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# M22-35

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