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M22-35

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M22-35  [#permalink]

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New post 16 Sep 2014, 01:17
1
8
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

47% (02:09) correct 53% (01:59) wrong based on 122 sessions

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Math Expert
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M22-35  [#permalink]

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New post 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.


Answer: C
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New post 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.


Answer: C



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
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M22-35  [#permalink]

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New post 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.


Answer: C



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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Re: M22-35  [#permalink]

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New post 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.


Answer: C



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
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Re: M22-35  [#permalink]

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New post 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 ?
Please explain?
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New post 13 Dec 2016, 10:19
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Re: M22-35  [#permalink]

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New post 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.


Answer: C


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
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M22-35  [#permalink]

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New post 25 Nov 2018, 07:22
manishred07 wrote:
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.


Answer: C


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


Hi manishred07

Bunuel has arrived at the conclusion that minimum value of \(x=6\) because of statement 2.
If x has to be minimum 6.
If \(x<=5\)then statement 2 will not be satisfied.
(2) \(x+z \gt 14\)
Here if \(x<=5\) then we have to take \(Z>9\). But observe that we are given in the question that \(z\) value is between 0 to 9 inclusive only.

We are assuming values based on the statement 1 and 2. The values assumed must satisfy the statements. Also, please elaborate what u mean by typing out ur understanding of the question so that anyone can help u better!

Hope this helps! :)
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Re: M22-35  [#permalink]

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New post 19 Jul 2019, 01:01
Can somebody pls elaborate the combining of 1) & 2) , how come the value of X=6 & y=8 ??
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Re: M22-35   [#permalink] 19 Jul 2019, 01:01
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