Data Sufficiency

In the xy-plane, does the line with the equation y = 2x - 4
contain the point (a; b) ?
(1) (2a - b - 4)(a + 5b + 2) = 0
(2) (4a + 3b - 1)(2a - b - 4) = 0

"Taken together, the statements are su¢ cient. The only way both equations
are true is if 2a - b - 4 = 0, and if that is the case, we know that y = 2x - 4
contains the point (a; b). Choice (C) is correct."

Why can't a+5b+2=0 and 4a+3b-1=0? I don't understand why 2a-b-4 has to be zero.

This can occur when a=2/3b+1

When you say contradict, are you implying that there's no value of a and b that will make a+5b+2 and 4a+3b-1 BOTH zero? How can you tell? Did you think about this scenario when you did this problem? OR I'm I over thinking this.

Let's put it another way:

Is X= 0?

1) X*Y=0
2) Z*X=0

To me, maybe but not definite. What if Y=Z=0.


[quote="Rahul@gurome"][quote]In the xy-plane, does the line with the equation y = 2x - 4 contain the point (a; b) ?
(1) (2a - b - 4)(a + 5b + 2) = 0
(2) (4a + 3b - 1)(2a - b - 4) = 0 [/quote]

If the line y = 2x - 4 contains the point (a, b), then b = 2a - 4
Hence, (2a - b - 4) must be equal to zero.

[b]Statement 1:[/b] Either (2a - b - 4) = 0 or (a + 5b + 2) = 0

Not sufficient

[b]Statement 2:[/b] Either (4a + 3b - 1) = 0 or (2a - b - 4) = 0

Not sufficient

[b]1 & 2 Together:[/b] (2a - b - 4) = 0

Sufficient

[spoiler]The correct answer is C.[/spoiler]

Note : While you are considering both the statements together, then they must not contradict each other. From each of the statements we have two possible results. When you are considering both of them together, you have to take the common result as that satisfies both the statement.

Thus, while considering both of them together, you cannot assume (a + 5b + 2) = 0 as the values of of a and b satisfying this relation doesn't satisfy statement 2. For example, a = -2 and b = 0. Same with the other one.[/quote]

Agree with yellowho.

Rahul,

at the point (11/17 , -9/17)
both
a + 5b + 2 = 0
and
4a + 3b - 1 = 0

where as 2a - b - 4 = -37/17

This point satisfies both statement 1 and 2 and still does not satisfy the eq y = 2x - 4.

So here,even after combining (1) and (2) the eq (2a - b - 4) need not necessarily be Zero.

Please correct me if im wrong.

Thanks for the confirmation. YOu are still awesome btw!

[quote="Rahul@gurome"]Point taken guys.
Even after combining both the statements, (2a - b - 4) need not necessarily to be equal to zero.[/quote]

So the answer is E?

yellowho wrote:In the xy-plane, does the line with the equation y = 2x - 4
contain the point (a; b) ?
(1) (2a - b - 4)(a + 5b + 2) = 0
(2) (4a + 3b - 1)(2a - b - 4) = 0

"Taken together, the statements are su¢ cient. The only way both equations
are true is if 2a - b - 4 = 0, and if that is the case, we know that y = 2x - 4
contains the point (a; b). Choice (C) is correct."

Why can't a+5b+2=0 and 4a+3b-1=0? I don't understand why 2a-b-4 has to be zero.

This can occur when a=2/3b+1

If the OA is C, then the question is flawed. The correct answer is E.

The question above seems to be modeled on a question from GMAT Prep. If you don't want to see the GMAT Prep question to which I'm referring, please read no further. Here it is:

In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

1) (3r + 2 - s)(4r + 9 - s) = 0
2) (4r - 6 - s)(3r + 2 - s) = 0

The correct answer to the question from GMATPrep is C. Rewritten, the question is asking: Does s = 3r + 2?

The GMAT writers were careful. There is no combination of values for (r,s) that will satisfy both 4r + 9 - s = 0 and 4r - 6 - s = 0. Thus, the only way to satisfy both statements is if 3r + 2 - s = 0 -- which can be rewritten as s = 3r + 2 -- giving us sufficient information to answer the question.

Hi Mitch,

That' what I was going at. How were you able to check quickly that there were no r and s that satisfy both?

[quote="GMATGuruNY"][quote="yellowho"]In the xy-plane, does the line with the equation y = 2x - 4
contain the point (a; b) ?
(1) (2a - b - 4)(a + 5b + 2) = 0
(2) (4a + 3b - 1)(2a - b - 4) = 0

"Taken together, the statements are su¢ cient. The only way both equations
are true is if 2a - b - 4 = 0, and if that is the case, we know that y = 2x - 4
contains the point (a; b). Choice (C) is correct."

Why can't a+5b+2=0 and 4a+3b-1=0? I don't understand why 2a-b-4 has to be zero.

This can occur when a=2/3b+1[/quote]

If the OA is C, then the question is flawed. The correct answer is [spoiler]E[/spoiler].

The question above seems to be modeled on a question from GMAT Prep. If you don't want to see the GMAT Prep question to which I'm referring, please read no further. Here it is:

[quote]In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

1) (3r + 2 - s)(4r + 9 - s) = 0
2) (4r - 6 - s)(3r + 2 - s) = 0[/quote]

The correct answer to the question from GMATPrep is [spoiler]C[/spoiler]. Rewritten, the question is asking: Does s = 3r + 2?

The GMAT writers were careful. There is no combination of values for (r,s) that will satisfy both 4r + 9 - s = 0 and 4r - 6 - s = 0. Thus, the only way to satisfy both statements is if 3r + 2 - s = 0 -- which can be rewritten as s = 3r + 2 -- giving us sufficient information to answer the question.[/quote]

yellowho wrote:Hi Mitch,

That' what I was going at. How were you able to check quickly that there were no r and s that satisfy both?

GMATGuruNY wrote:

yellowho wrote:In the xy-plane, does the line with the equation y = 2x - 4
contain the point (a; b) ?
(1) (2a - b - 4)(a + 5b + 2) = 0
(2) (4a + 3b - 1)(2a - b - 4) = 0

"Taken together, the statements are su¢ cient. The only way both equations
are true is if 2a - b - 4 = 0, and if that is the case, we know that y = 2x - 4
contains the point (a; b). Choice (C) is correct."

Why can't a+5b+2=0 and 4a+3b-1=0? I don't understand why 2a-b-4 has to be zero.

This can occur when a=2/3b+1

If the OA is C, then the question is flawed. The correct answer is E.

The question above seems to be modeled on a question from GMAT Prep. If you don't want to see the GMAT Prep question to which I'm referring, please read no further. Here it is:

In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

1) (3r + 2 - s)(4r + 9 - s) = 0
2) (4r - 6 - s)(3r + 2 - s) = 0

The correct answer to the question from GMATPrep is C. Rewritten, the question is asking: Does s = 3r + 2?

The GMAT writers were careful. There is no combination of values for (r,s) that will satisfy both 4r + 9 - s = 0 and 4r - 6 - s = 0. Thus, the only way to satisfy both statements is if 3r + 2 - s = 0 -- which can be rewritten as s = 3r + 2 -- giving us sufficient information to answer the question.

If the 2 other factors are set equal to 0, we get:
4r + 9 - s = 0, which can be rewritten as 4r - s = -9.
4r - 6 - s = 0, which can be rewritten as 4r - s = 6.

It is not possible that 4r - s equals both -9 and 6. If we subtract the second equation from the first, we get 0 = -15.

Thus, to satisfy both statements, the other factor in each statement must equal zero: 3r + 2 - s = 0.

Does this help?

Got it. Its made easy because the variables have the same coefficients. If they have different coefficient, then theres some possibilities. Would you even have to check? (You have 1 and 5 in one set and 4 and 3 in one set. Theres no way to get them to equal)

[quote="GMATGuruNY"][quote="yellowho"]Hi Mitch,

That' what I was going at. How were you able to check quickly that there were no r and s that satisfy both?

[quote="GMATGuruNY"][quote="yellowho"]In the xy-plane, does the line with the equation y = 2x - 4
contain the point (a; b) ?
(1) (2a - b - 4)(a + 5b + 2) = 0
(2) (4a + 3b - 1)(2a - b - 4) = 0

"Taken together, the statements are su¢ cient. The only way both equations
are true is if 2a - b - 4 = 0, and if that is the case, we know that y = 2x - 4
contains the point (a; b). Choice (C) is correct."

Why can't a+5b+2=0 and 4a+3b-1=0? I don't understand why 2a-b-4 has to be zero.

This can occur when a=2/3b+1[/quote]

If the OA is C, then the question is flawed. The correct answer is [spoiler]E[/spoiler].

The question above seems to be modeled on a question from GMAT Prep. If you don't want to see the GMAT Prep question to which I'm referring, please read no further. Here it is:

[quote]In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

1) (3r + 2 - s)(4r + 9 - s) = 0
2) (4r - 6 - s)(3r + 2 - s) = 0[/quote]

The correct answer to the question from GMATPrep is [spoiler]C[/spoiler]. Rewritten, the question is asking: Does s = 3r + 2?

The GMAT writers were careful. There is no combination of values for (r,s) that will satisfy both 4r + 9 - s = 0 and 4r - 6 - s = 0. Thus, the only way to satisfy both statements is if 3r + 2 - s = 0 -- which can be rewritten as s = 3r + 2 -- giving us sufficient information to answer the question.[/quote][/quote]

If the 2 other factors are set equal to 0, we get:
4r + 9 - s = 0, which can be rewritten as 4r - s = -9.
4r - 6 - s = 0, which can be rewritten as 4r - s = 6.

It is not possible that 4r - s equals both -9 and 6. If we subtract the second equation from the first, we get 0 = -15.

Thus, to satisfy both statements, the other factor in each statement must equal zero: 3r + 2 - s = 0.

Does this help?[/quote]

jaxis wrote:Agree with yellowho.

Rahul,

at the point (11/17 , -9/17)
both
a + 5b + 2 = 0
and
4a + 3b - 1 = 0

where as 2a - b - 4 = -37/17

This point satisfies both statement 1 and 2 and still does not satisfy the eq y = 2x - 4.

So here,even after combining (1) and (2) the eq (2a - b - 4) need not necessarily be Zero.

Please correct me if im wrong.

How did jaxis calculate points (11/17 , -9/17) . Can someone please explain ?

For this question, you can get 3a-3=2b, so would the answer be E for this question?

I understand Mitch's question when the answer can only be 0, making C the answer.

Please provide insights. Thank you.

jainrahul1985 wrote:

jaxis wrote:Agree with yellowho.

Rahul,

at the point (11/17 , -9/17)
both
a + 5b + 2 = 0
and
4a + 3b - 1 = 0

where as 2a - b - 4 = -37/17

This point satisfies both statement 1 and 2 and still does not satisfy the eq y = 2x - 4.

So here,even after combining (1) and (2) the eq (2a - b - 4) need not necessarily be Zero.

Please correct me if im wrong.

How did jaxis calculate points (11/17 , -9/17) . Can someone please explain ?

Rahul,

there are 2 equations and 2 variables.
a + 5b + 2 = 0
and
4a + 3b - 1 = 0

Solve to get the point (11/17 , -9/17).