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

It is currently 22 Oct 2019, 16:58

Close

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
Your Progress

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.

Close

Request Expert Reply

Confirm Cancel

The Discreet Charm of the DS

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58427
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 10 Jun 2017, 12:47
KARISHMA315 wrote:
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.



Hi Bunuel, a small doubt instead of \(2x+8\geq{0}\) --> \(x\geq{-4}\) this logic if we take |x+10| to be both positive and negative we get 2 vale by solving equation x=2 and x=-6. however by putting values back in eqtn we can see that only for x=2 equation is satisfying. Is this correct approach?


Yes, that's also a correct way of solving. Good thing you did is that you did not forget to test both values after you got them.
_________________
Intern
Intern
avatar
B
Joined: 16 Feb 2017
Posts: 2
GMAT 1: 730 Q49 V41
Reviews Badge
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 06 Sep 2017, 00:43
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Answer: A.


Hi Bunuel,

I understood that x*|y|=positive; but if y<=0 then |y|=-y and in this case x can be negative.
Am I missing something?

Thanks.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58427
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 06 Sep 2017, 00:49
1
Yashodhan123 wrote:
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Answer: A.


Hi Bunuel,

I understood that x*|y|=positive; but if y<=0 then |y|=-y and in this case x can be negative.
Am I missing something?

Thanks.


Yes.

An absolute value of a number cannot be negative: |a| is positive or 0, no matter whether a itself is negative or not.

If y <= 0, then |y| = -y, yes, but even in this case -y = -negative = positive.
_________________
Intern
Intern
avatar
B
Joined: 21 Mar 2012
Posts: 23
Location: India
GMAT 1: 710 Q49 V38
GMAT ToolKit User Reviews Badge
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 08 Jun 2018, 15:04
12. If 6a=3b=7c, what is the value of a+b+c?
(1) ac=6b
(2) 5b=8a+4c


Let 6a=3b=7c=K (where K is a constant)
Thus, a=K/6
b= K/3
c=K/7

Therefore, we need to find the value of K/6 + K/3 + K/7 = 9K/14. OR if we find the value of just K it will give us the answer.

Stmt 1: ac=6b
\(K/6*K/7 = 6*K/3;\)

\(K^2/42-2K=0\)

\(K(K/42-2)=0\)

So, K = 0 or K= 84. NOT Sufficient.

Stmt 2:
5b=8a+4c
5K/3=8K/6 +4K/7
40K/21-5K/3=0
5K/3=0
K=0 SUFFICIENT

Ans. B
Manager
Manager
User avatar
S
Joined: 11 Feb 2013
Posts: 147
Location: Bangladesh
GMAT 1: 490 Q44 V15
GMAT 2: 690 Q47 V38
GPA: 3.05
WE: Analyst (Commercial Banking)
GMAT ToolKit User Premium Member Reviews Badge
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 09 Jun 2018, 01:36
Bunuel wrote:
2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.


Dear sir,
Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58427
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 10 Jun 2018, 00:16
BelalHossain046 wrote:
Bunuel wrote:
2. Is xy<=1/2?

(1) x^2+y^2=1. Recall that \((x-y)^2\geq{0}\) (square of any number is more than or equal to zero) --> \(x^2-2xy+y^2\geq{0}\) --> since \(x^2+y^2=1\) then: \(1-2xy\geq{0}\) --> \(xy\leq{\frac{1}{2}}\). Sufficient.

(2) x^2-y^2=0 --> \(|x|=|y|\). Clearly insufficient.

Answer: A.


Dear sir,
Why are not you taking (x+y)^2 >=0 i.e. x^2+y^2+2xy>=0?


The point is that if you consider \((x+y)^2 \ge 0\), you'll get \(xy\geq{-\frac{1}{2}}\), which is not helpful at all. Actually since both \(xy\geq{-\frac{1}{2}}\) (from \((x+y)^2 \ge 0\)) and \(xy \le \frac{1}{2}\) (from \((x-y)^2 \ge 0\)) are true, then we get that \(-\frac{1}{2} \leq{xy} \leq{\frac{1}{2}}\). But only the approach given in the solution above gives you the answer we are looking for, while another one gives you an inequality which is not helpful.

Hope it's clear.
_________________
Manager
Manager
avatar
S
Joined: 24 Mar 2018
Posts: 246
CAT Tests
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 18 Jul 2018, 05:26
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Answer: A.


Shouldn't we consider zero as positive ?
2 is not sufficient as |y| can be zero so x can be negative or positive
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58427
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 18 Jul 2018, 05:29
1
teaserbae wrote:
Bunuel wrote:
11. If x and y are integers, is x a positive integer?

(1) x*|y| is a prime number --> since only positive numbers can be primes, then: x*|y|=positive --> x=positive. Sufficient

(2) x*|y| is non-negative integer. Notice that we are told that x*|y| is non-negative, not that it's positive, so x can be positive as well as zero. Not sufficient.

Answer: A.


Shouldn't we consider zero as positive ?
2 is not sufficient as |y| can be zero so x can be negative or positive


ZERO.

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.
_________________
Intern
Intern
avatar
B
Joined: 16 Feb 2018
Posts: 14
Location: India
Schools: DeGroote'21 (S)
GPA: 3.5
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 03 Oct 2018, 07:33
Bunuel wrote:
5. What is the value of integer x?

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.



Hi, until now whenever I saw modulus, I was told to begin by removing the mod and considering both signs.
Therefore, I approached it as
x+10=2x+8 and -(x+10)=2x+8. Just wanted to understand when to use this method and when to use the take it as an absolute value method?
Thanks!
Intern
Intern
avatar
B
Joined: 09 Nov 2018
Posts: 3
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 10 Apr 2019, 01:20
With respect to question no 8 above, (If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?), why would the answer not just be A? Since we can determine the least possible value of both a and b, and thereby deduce that x would be greater than 0.77.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58427
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 10 Apr 2019, 01:28
tejasvis7 wrote:
With respect to question no 8 above, (If x=0.abcd, where a, b, c and d are digits from 0 to 9, inclusive, is x>7/9?), why would the answer not just be A? Since we can determine the least possible value of both a and b, and thereby deduce that x would be greater than 0.77.


Solution is here: https://gmatclub.com/forum/the-discreet ... l#p1039662

Here is what it says about (1): a+b>14 --> the least value of a is 6 (6+9=15>14), so in this case x=0.69d<0.77(7) but a=7 and b=9 is also possible, and in this case x=0.79d>0.77(7). Not sufficient.
_________________
Manager
Manager
avatar
B
Joined: 21 Nov 2018
Posts: 74
Location: India
GMAT 1: 680 Q48 V35
GMAT 2: 640 Q48 V29
CAT Tests
The Discreet Charm of the DS  [#permalink]

Show Tags

New post 13 Apr 2019, 06:41
1
Bunuel wrote:
12. If 6a=3b=7c, what is the value of a+b+c?

Given: \(6a=3b=7c\) --> least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) --> \(a=7x\), \(b=14x\) and \(c=6x\).

(1) ac=6b --> \(7x*6x=6*14x\) --> \(x^2=2x\) --> \(x=0\) or \(x=2\). Not sufficient.

(2) 5b=8a+4c --> \(5*14x=8*7x+4*14x\) --> \(70x=80x\) --> \(10x=0\) --> \(x=0\) --> \(a=b=c=0\) --> \(a+b+c=0\). Sufficient.

Answer: B.


Hi Bunuel
Could you please change 4*14x (as highlighted) to 4*6x because c=6x. (small typo)
_________________
Beautiful is the one who continues to try despite failure.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58427
Re: The Discreet Charm of the DS  [#permalink]

Show Tags

New post 13 Apr 2019, 10:02
dhritidutta wrote:
Bunuel wrote:
12. If 6a=3b=7c, what is the value of a+b+c?

Given: \(6a=3b=7c\) --> least common multiple of 6, 3, and 7 is 42 hence we ca write: \(6a=3b=7c=42x\), for some number \(x\) --> \(a=7x\), \(b=14x\) and \(c=6x\).

(1) ac=6b --> \(7x*6x=6*14x\) --> \(x^2=2x\) --> \(x=0\) or \(x=2\). Not sufficient.

(2) 5b=8a+4c --> \(5*14x=8*7x+4*14x\) --> \(70x=80x\) --> \(10x=0\) --> \(x=0\) --> \(a=b=c=0\) --> \(a+b+c=0\). Sufficient.

Answer: B.


Hi Bunuel
Could you please change 4*14x (as highlighted) to 4*6x because c=6x. (small typo)

_______________________
Edited. Thank you.
_________________
Intern
Intern
avatar
B
Joined: 10 Jun 2018
Posts: 13
The Discreet Charm of the DS  [#permalink]

Show Tags

New post 01 May 2019, 09:02
Bunuel wrote:
piyushksharma wrote:
Bunuel wrote:
5. What is the value of integer x?

(1) 2x^2+9<9x --> factor qudratics: \((x-\frac{3}{2})(x-3)<0\) --> roots are \(\frac{3}{2}\) and 3 --> "<" sign indicates that the solution lies between the roots: \(1.5<x<3\) --> since there only integer in this range is 2 then \(x=2\). Sufficient.

(2) |x+10|=2x+8 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: \(2x+8\geq{0}\) --> \(x\geq{-4}\), for this range \(x+10\) is positive hence \(|x+10|=x+10\) --> \(x+10=2x+8\) --> \(x=2\). Sufficient.

Answer: D.

Hope it helps.


Hi bunuel,
Isn't |x+10|=2x+8 be written as
Either x+10=2x+8 or x+10=-(2x+8) ? and then this should be solved?
Please help on this one.


We goth that x is more than or equal to 4. Now, for this range x+10>0 so |x+10| expands only as x+10 (|x+10|=x+10).


Hi Bunuel,
A doubt on the above note.
You have mentioned that absolute values are non-negative and hence you have considered 2x+8>=0
But in some questions we solve by taking 2 cases when modulus is present in the question. Using this case-based approach, in the above given equation, case 1 would be x+10=2x+8 and case 2 would be x-10=-(2x+8)
So, following are my questions:
1. Why have we ignored 2x+8<0 in your solution?
2. When do we use the 2 case approach and when do we use the 'absolute values are non-negative approach?

This is a bit confusing. Please assist on this.
GMAT Club Bot
The Discreet Charm of the DS   [#permalink] 01 May 2019, 09:02

Go to page   Previous    1   2   3   4   5   6   [ 114 posts ] 

Display posts from previous: Sort by

The Discreet Charm of the DS

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne