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m23#17

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m23#17 [#permalink] New post 23 Dec 2008, 11:41
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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

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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?
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Re: m23#17 [#permalink] New post 23 Dec 2008, 14:18
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.
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Re: m23#17 [#permalink] New post 24 Dec 2008, 01:38
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.
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Re: m23#17 [#permalink] New post 01 Jan 2010, 06:15
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.
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Re: m23#17 [#permalink] New post 04 Jan 2010, 06:10
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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
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Re: m23#17 [#permalink] New post 04 Jan 2010, 16:52
"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
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Re: m23#17 [#permalink] New post 19 Mar 2010, 07:16
Answer is B.
stmt1 gives indefinite values for a=0
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Re: m23#17 [#permalink] New post 07 Jan 2011, 13:14
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.
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Re: m23#17 [#permalink] New post 08 Jan 2011, 00:59
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
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Re: m23#17 [#permalink] New post 08 Jan 2011, 02:06
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
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Re: m23#17 [#permalink] New post 08 Jan 2011, 02:14
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 ?
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Re: m23#17 [#permalink] New post 31 May 2011, 09:05
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?
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Re: m23#17 [#permalink] New post 31 May 2011, 09:36
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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.
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Re: m23#17 [#permalink] New post 23 Jul 2011, 07:25
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.
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Re: m23#17 [#permalink] New post 11 Jan 2012, 05:45
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.
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Re: m23#17 [#permalink] New post 11 Jan 2012, 07:20
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....
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Re: m23#17 [#permalink] New post 18 Feb 2012, 02:30
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?.
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Re: m23#17 [#permalink] New post 30 Mar 2012, 01:29
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.
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Re: m23#17 [#permalink] New post 18 Apr 2012, 20:33
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.
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Re: m23#17 [#permalink] New post 01 Aug 2012, 06:24
Bunuel,
I am a bit confused about this question.

If Ay + B = C and A, B and C are known, then there should be only one value of y = (C-B)/A.

1) C>B This should be sufficient because A, B and C are known. Correct? If C = 5, B = 4, A = 2, I will get a unique value of 'y'.

2) A>1 => Again,...All three variables are known. Hence, Sufficient.

I am really not able to follow the OE about #1 - Consider A= 0 and A=1..Why? A,B and C are known...I am lost.

Can you please help me?

Thanks
Re: m23#17   [#permalink] 01 Aug 2012, 06:24
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