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# m23#17

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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

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 $$\frac{C - B}{A}$$ .

Is the explanation for statement 2 correct? Why couldn't y = 0 and some other number as well?
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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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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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Quant - Linear equations with parameters [#permalink]

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01 Aug 2012, 08:16
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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).
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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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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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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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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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19 Mar 2010, 07:16
stmt1 gives indefinite values for a=0
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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.

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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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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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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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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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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23 Jul 2011, 07:25

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

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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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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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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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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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 $$\frac{C - B}{A}$$ .

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

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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 $$\frac{C - B}{A}$$ .

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, 20:33

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