Find all School-related info fast with the new School-Specific MBA Forum

It is currently 01 Jul 2016, 01:25
GMAT Club Tests

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

m23#17

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Intern
Intern
avatar
Joined: 19 Jun 2008
Posts: 20
Followers: 0

Kudos [?]: 20 [0], given: 0

m23#17 [#permalink]

Show Tags

New post 23 Dec 2008, 12:41
2
This post was
BOOKMARKED
How many different numbers \(y\) are there such that \(Ay + B = C\) ( \(A\) , \(B\) , and \(C\) are known) ?

1. \(C \gt B\)
2. \(A \gt 1\)

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

S1 is not sufficient. Consider \(A = 0\) (the answer is 0) and \(A = 1\) (the answer is 1).

S2 is sufficient. Because \(A \gt 1\) , \(A \gt 0\) and the only \(y\) that satisfies the condition is \(\frac{C - B}{A}\) .
The correct answer is B.


Is the explanation for statement 2 correct? Why couldn't y = 0 and some other number as well?
1 KUDOS received
Manager
Manager
avatar
Joined: 26 Nov 2009
Posts: 176
Followers: 3

Kudos [?]: 56 [1] , given: 5

GMAT ToolKit User
Re: m23#17 [#permalink]

Show Tags

New post 04 Jan 2010, 07:10
1
This post received
KUDOS
My choice B

I started with option B as it is easier to approach

ii) A> 1
if A > 1 then for y we can have only one possible value so option 2 is sufficient

i) C >B
if A is 0 then we can have 'n' number of different values for y so option 1 is insufficient
1 KUDOS received
Math Forum Moderator
avatar
Joined: 20 Dec 2010
Posts: 2021
Followers: 156

Kudos [?]: 1489 [1] , given: 376

Re: m23#17 [#permalink]

Show Tags

New post 31 May 2011, 10:36
1
This post received
KUDOS
forsaken11 wrote:
Geronimo wrote:
I feel lonely this time, because I went for D....

ay+b=c => ay=c-b
(1) c>b => c-b > 0, so ay>0 and hence a and y are different from 0
Sufficient

(2) a is different from 0, so sufficient



I also used the same approach. I am unable to find any flaws in it. Any idea, anyone?


In my opinion, the equation should not be considered as true.

Thus,
ay+b=c ---- Is not true. We need to make it true by providing appropriate value to "y".

a=1
b=2
c=5
c-b=3(+ve). Agree.

Now,
y=(c-b)/a= 3/1=3.
y=3;

But, if a=0,
y=(c-b)/a=3/0 -> undefined.

Thus, just by knowing that a != 0, we would know that "y" has one solution because it is a linear equation.
_________________

~fluke

GMAT Club Premium Membership - big benefits and savings

1 KUDOS received
Director
Director
User avatar
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)
Followers: 90

Kudos [?]: 788 [1] , given: 43

Quant - Linear equations with parameters [#permalink]

Show Tags

New post 01 Aug 2012, 09:16
1
This post received
KUDOS
Relevant to an old question (was M03-Q23):
m23-74149.html

Consider the the equation \(Ax=B\), where \(A\) and \(B\) are given real numbers, and \(x\) is to be found.
There are three possible scenarios:

1) \(A\neq0\). In this case, the given equation has a unique solution given by \(\frac{B}{A}\). Doesn't matter who is \(B\), it can be also 0, as 0 divided by a non-zero number is 0.
2) \(A = 0\) and \(B = 0\). In this case, the given equation has infinitely many solutions, as for any number \(x\), \(0 * x = 0\).
3) If \(A = 0\), but \(B\neq0\), then the given equation has no solution, because \(Ax = 0 * x = 0 \neq{B}\).

I hope the above can help when discussing any linear equation with one unknown and parameters (letters instead of numbers as coefficients).
_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Manager
Manager
avatar
Affiliations: Beta Gamma Sigma
Joined: 14 Aug 2008
Posts: 210
Schools: Harvard, Penn, Maryland
Followers: 5

Kudos [?]: 56 [0], given: 3

Re: m23#17 [#permalink]

Show Tags

New post 23 Dec 2008, 15:18
I would say A because

If A, B, and C are known, then as long as C>B, A and y could only be one number. the trick is that A,B, and C are all known, if A was unknown then the answer would be infinite. For Example, 4y + 2 = 3, A=4 y=1/4 B=2 C=3, A B and C are known and C>B, so y is an easy calculation.

unless I am misinterpreting it, A is already known, so its value being more than one shouldnt matter.
SVP
SVP
avatar
Joined: 17 Jun 2008
Posts: 1570
Followers: 11

Kudos [?]: 226 [0], given: 0

Re: m23#17 [#permalink]

Show Tags

New post 24 Dec 2008, 02:38
Stmt1 is not sufficient as y = (C-B)/A and unless we are sure A is non-zero we cannot say that y has a definite value.

From stmt2 only, A > 1 and that means, (C-B)/A will not be infinite. Hence, for any value of C and B, stmt2 is sufficient.
Manager
Manager
avatar
Joined: 20 Oct 2009
Posts: 113
Schools: MIT LGO (Admitted), Harvard (Admitted))
Followers: 8

Kudos [?]: 31 [0], given: 0

GMAT ToolKit User
Re: m23#17 [#permalink]

Show Tags

New post 01 Jan 2010, 07:15
y=(C-B)/A

1) is not sufficient because if A=0, no y value; otherwise, y is (C-B)/A
2) A>1 makes sure case 1 is excluded.
_________________

Dream the impossible and do the incredible.

Live. Love. Laugh.

Manager
Manager
avatar
Joined: 23 Nov 2009
Posts: 53
Followers: 1

Kudos [?]: 11 [0], given: 1

Re: m23#17 [#permalink]

Show Tags

New post 04 Jan 2010, 17:52
"How many different numbers y" requires a definite answer in GMAT. It should be for instance 2,3 or any other number.
It cannot be "can be determined" vs "cannot be determined"

All the explanations above make sense, but this question seems to me as non-standard as far as GMAT is concerned
_________________

A kudos would greatly help :)

Tuhin

Senior Manager
Senior Manager
User avatar
Joined: 13 Dec 2009
Posts: 263
Followers: 10

Kudos [?]: 165 [0], given: 13

Reviews Badge
Re: m23#17 [#permalink]

Show Tags

New post 19 Mar 2010, 08:16
Answer is B.
stmt1 gives indefinite values for a=0
_________________

My debrief: done-and-dusted-730-q49-v40

Senior Manager
Senior Manager
User avatar
Joined: 01 Nov 2010
Posts: 295
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Followers: 10

Kudos [?]: 72 [0], given: 44

Re: m23#17 [#permalink]

Show Tags

New post 07 Jan 2011, 14:14
its already given that A,B,C are known.

this equation is defined everywhere except, A=0.

from
statement 1 ; C>B , it doesn't make any sense , as it doesn't affect the value of y.

statement 2; A>0, this condition should exist since A can't be zero.

so, we can answer this question using statement 2.
the Ans is B.

but this question asks about the different values of y.
and, by using both these statement together or alone we cant say how many values does y have.
hence, if we look from this perspective, the answer will be E.

plz tell the OA and source of question.
_________________

kudos me if you like my post.

Attitude determine everything.
all the best and God bless you.

Joined: 31 Dec 1969
Location: India
Concentration: Marketing, General Management
GMAT 1: 710 Q49 V0
GMAT 2: 700 Q V
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 7: Q42 V44
GMAT 8: Q42 V44
GMAT 9: 740 Q49 V42
GMAT 10: 740 Q V
GMAT 11: 500 Q47 V33
GPA: 3.3
WE: Sales (Investment Banking)
Followers: 0

Kudos [?]: 151 [0], given: 97736

Re: m23#17 [#permalink]

Show Tags

New post 08 Jan 2011, 01:59
I feel lonely this time, because I went for D....

ay+b=c => ay=c-b
(1) c>b => c-b > 0, so ay>0 and hence a and y are different from 0
Sufficient

(2) a is different from 0, so sufficient
Intern
Intern
avatar
Joined: 16 Dec 2010
Posts: 7
Followers: 0

Kudos [?]: 1 [0], given: 1

Re: m23#17 [#permalink]

Show Tags

New post 08 Jan 2011, 03:06
as both are not able to give a definite answers , like the question is how many different numbers Y are there...

this cannot be answered using any of the statements taken together and alone too..

Ans - E
Joined: 31 Dec 1969
Location: India
Concentration: Marketing, General Management
GMAT 1: 710 Q49 V0
GMAT 2: 700 Q V
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 7: Q42 V44
GMAT 8: Q42 V44
GMAT 9: 740 Q49 V42
GMAT 10: 740 Q V
GMAT 11: 500 Q47 V33
GPA: 3.3
WE: Sales (Investment Banking)
Followers: 0

Kudos [?]: 151 [0], given: 97736

Re: m23#17 [#permalink]

Show Tags

New post 08 Jan 2011, 03:14
zahuruddin wrote:
as both are not able to give a definite answers , like the question is how many different numbers Y are there...

this cannot be answered using any of the statements taken together and alone too..

Ans - E

Only one: y=(c-b)/a if a is different from 0. Or am I missing something ?
Intern
Intern
avatar
Joined: 10 Feb 2011
Posts: 49
Followers: 1

Kudos [?]: 15 [0], given: 18

Re: m23#17 [#permalink]

Show Tags

New post 31 May 2011, 10:05
Geronimo wrote:
I feel lonely this time, because I went for D....

ay+b=c => ay=c-b
(1) c>b => c-b > 0, so ay>0 and hence a and y are different from 0
Sufficient

(2) a is different from 0, so sufficient



I also used the same approach. I am unable to find any flaws in it. Any idea, anyone?
Director
Director
avatar
Joined: 01 Feb 2011
Posts: 757
Followers: 14

Kudos [?]: 95 [0], given: 42

Re: m23#17 [#permalink]

Show Tags

New post 23 Jul 2011, 08:25
My answer is D.

Y = (C-B)/A

Given A,B and C are all known => they are constants.

for any constant combination , we can only have one value for y. But we need to make sure that A is not equal to 0 (i.e undefined )

1. C>B

=> C-B>0
=> AY>0

=> A cannot be 0. (A can be positive or negative)

=> y = (C-B)/A will give one valid solution

2. A >1

=> A cannot be 0. (A can be positive or negative)

=> y = (C-B)/A will give one valid solution

So Answer is D.


fluke wrote:
forsaken11 wrote:
Geronimo wrote:
I feel lonely this time, because I went for D....

ay+b=c => ay=c-b
(1) c>b => c-b > 0, so ay>0 and hence a and y are different from 0
Sufficient

(2) a is different from 0, so sufficient



I also used the same approach. I am unable to find any flaws in it. Any idea, anyone?


In my opinion, the equation should not be considered as true.

Thus,
ay+b=c ---- Is not true. We need to make it true by providing appropriate value to "y".

a=1
b=2
c=5
c-b=3(+ve). Agree.

Now,
y=(c-b)/a= 3/1=3.
y=3;

But, if a=0,
y=(c-b)/a=3/0 -> undefined.

Thus, just by knowing that a != 0, we would know that "y" has one solution because it is a linear equation.
Senior Manager
Senior Manager
avatar
Joined: 08 Jun 2010
Posts: 397
Location: United States
Concentration: General Management, Finance
GMAT 1: 680 Q50 V32
Followers: 3

Kudos [?]: 77 [0], given: 13

Re: m23#17 [#permalink]

Show Tags

New post 11 Jan 2012, 06:45
If A, B and C are known the only case where y will have an unknown number of values is when A = 0 because there will be infinite possibilities. So, B correctly pinpoints that aspect.
Director
Director
avatar
Joined: 28 Jul 2011
Posts: 563
Location: United States
Concentration: International Business, General Management
GPA: 3.86
WE: Accounting (Commercial Banking)
Followers: 2

Kudos [?]: 154 [0], given: 16

Re: m23#17 [#permalink]

Show Tags

New post 11 Jan 2012, 08:20
I agree that B give answer, but question is asking for a specific number of Y values right? with Statement 2 we can say that we can find the value of y. but here we get multiples with different values of A B and C right????

Please correct me if if am wrong....
_________________

+1 Kudos If found helpful..

Manager
Manager
avatar
Joined: 10 Jan 2010
Posts: 117
GPA: 4
WE: Programming (Computer Software)
Followers: 0

Kudos [?]: 99 [0], given: 33

Re: m23#17 [#permalink]

Show Tags

New post 18 Feb 2012, 03:30
Since the questions talks about How many different values of y, shouldn't we look for a specific value for y?.

How to choose between "B" and "E" in this case?.
_________________

-If you like my post, consider giving KUDOS

Manager
Manager
avatar
Joined: 10 Jan 2011
Posts: 244
Location: India
GMAT Date: 07-16-2012
GPA: 3.4
WE: Consulting (Consulting)
Followers: 0

Kudos [?]: 41 [0], given: 25

Reviews Badge
Re: m23#17 [#permalink]

Show Tags

New post 30 Mar 2012, 02:29
snowy2009 wrote:
How many different numbers \(y\) are there such that \(Ay + B = C\) ( \(A\) , \(B\) , and \(C\) are known) ?

1. \(C \gt B\)
2. \(A \gt 1\)

[Reveal] Spoiler: OA
B

Source: GMAT Club Tests - hardest GMAT questions

S1 is not sufficient. Consider \(A = 0\) (the answer is 0) and \(A = 1\) (the answer is 1).

S2 is sufficient. Because \(A \gt 1\) , \(A \gt 0\) and the only \(y\) that satisfies the condition is \(\frac{C - B}{A}\) .
The correct answer is B.


Is the explanation for statement 2 correct? Why couldn't y = 0 and some other number as well?



IMO both statement 1 and 2 are insufficient becasue question ask about how many diffrent number. using both statement 1 and 2 still we can get infinite numbers of solution, hence answer should be E

GMAT experts please clarify.
_________________

-------Analyze why option A in SC wrong-------

Manager
Manager
avatar
Joined: 25 Feb 2012
Posts: 63
Followers: 1

Kudos [?]: 17 [0], given: 8

Re: m23#17 [#permalink]

Show Tags

New post 18 Apr 2012, 21:33
snowy2009 wrote:
How many different numbers \(y\) are there such that \(Ay + B = C\) ( \(A\) , \(B\) , and \(C\) are known) ?

1. \(C \gt B\)
2. \(A \gt 1\)

[Reveal] Spoiler: OA
B



S1 is not sufficient. Consider \(A = 0\) (the answer is 0) and \(A = 1\) (the answer is 1).

S2 is sufficient. Because \(A \gt 1\) , \(A \gt 0\) and the only \(y\) that satisfies the condition is \(\frac{C - B}{A}\) .
The correct answer is B.


Is the explanation for statement 2 correct? Why couldn't y = 0 and some other number as well?


S1 says C>B, and the question says Ay + B = C, doesn't this mean A cannot be equal to 0, because if it was, then B = C (from Ay + B = C) and thus contradicting Statement 1.
Re: m23#17   [#permalink] 18 Apr 2012, 21:33

Go to page    1   2    Next  [ 22 posts ] 

Display posts from previous: Sort by

m23#17

  post reply Question banks Downloads My Bookmarks Reviews Important topics  

Moderator: Bunuel



GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.