Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

 It is currently 23 Jul 2019, 07:20 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M22-35

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 56366

### Show Tags

1
8 00:00

Difficulty:   95% (hard)

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

### HideShow timer Statistics 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) $$x + z \gt 14$$

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 56366

### Show Tags

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.

_________________
Manager  Joined: 14 Jul 2014
Posts: 91

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 56366

### Show Tags

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.
_________________
Manager  Joined: 14 Jul 2014
Posts: 91

### Show Tags

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  B
Joined: 05 Sep 2013
Posts: 3

### Show Tags

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 V
Joined: 02 Sep 2009
Posts: 56366

### Show Tags

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.
_________________
Intern  B
Joined: 14 Aug 2017
Posts: 6

### Show Tags

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
Manager  S
Joined: 28 Jun 2018
Posts: 134
Location: Bouvet Island
GMAT 1: 490 Q39 V18 GMAT 2: 640 Q47 V30 GMAT 3: 670 Q50 V31 GMAT 4: 700 Q49 V36 GPA: 4

### Show Tags

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.

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! Intern  B
Joined: 18 Dec 2012
Posts: 4

### Show Tags

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

# M22-35  