GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 10 Dec 2018, 12:42

### 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

## Events & Promotions

###### Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### Free lesson on number properties

December 10, 2018

December 10, 2018

10:00 PM PST

11:00 PM PST

Practice the one most important Quant section - Integer properties, and rapidly improve your skills.
• ### Free GMAT Prep Hour

December 11, 2018

December 11, 2018

09:00 PM EST

10:00 PM EST

Strategies and techniques for approaching featured GMAT topics. December 11 at 9 PM EST.

# M22-35

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51072

### Show Tags

16 Sep 2014, 00:17
1
8
00:00

Difficulty:

95% (hard)

Question Stats:

46% (01:23) correct 54% (01:22) wrong based on 158 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
Joined: 02 Sep 2009
Posts: 51072

### Show Tags

16 Sep 2014, 00:17
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.

_________________
Manager
Joined: 14 Jul 2014
Posts: 93

### Show Tags

24 Mar 2015, 08:17
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: 51072

### Show Tags

24 Mar 2015, 08:46
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.
_________________
Manager
Joined: 14 Jul 2014
Posts: 93

### Show Tags

24 Mar 2015, 08:52
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

### Show Tags

13 Dec 2016, 07:52
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: 51072

### Show Tags

13 Dec 2016, 09:19
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.
_________________
Intern
Joined: 14 Aug 2017
Posts: 4

### Show Tags

20 Jul 2018, 20:12
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
Manager
Joined: 28 Jun 2018
Posts: 66
GMAT 1: 490 Q39 V18
GMAT 2: 640 Q47 V30
GMAT 3: 670 Q50 V31
GMAT 4: 700 Q49 V36
GPA: 4

### Show Tags

25 Nov 2018, 06:22
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.

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!
M22-35 &nbs [#permalink] 25 Nov 2018, 06:22
Display posts from previous: Sort by