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Bunuel
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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

In the xy-plane, does the line with the equation y=2x-4 contain the point (a,b)?

Line with equation \(y=2x-4\) contains the point \((a,b)\) means that when substituting \(a\) ans \(b\) in line equation: \(b=2a-4\) (or \(2a-b-4=0\)) holds true.

So, basically we are asked to determine whether \(2a-b-4=0\) is true.

(1) (2a-b-4)(a+5b+2)=0 --> either \(2a-b-4=0\) OR \(a+5b+2=0\) OR both. Not sufficient.

(2) (4a+3b-1)(2a-b-4)=0 --> either \(2a-b-4=0\) OR \(4a+3b-1=0\) OR both. Not sufficient.

(1)+(2) Now, since \(a+5b+2=0\) and \(4a+3b-1=0\) can simultaneously be true (for a=11/17 and b=-9/17), then we have that EITHER both \(a+5b+2=0\) and \(4a+3b-1=0\) are true OR \(2a-b-4=0\) is true. Not sufficient.
Answer: E (C is not correct).

Similar question to practice from GMAT Prep: in-the-xy-plane-does-the-line-with-equation-y-3x-100399.html

Hope it helps.


Hello Bunuel,

Can you plz explain the highlighed part once again ?

We have two equations,

We are asked whether a = 0.

1) ab = 0

2) ac = 0

From 1, the following situations are possible
a = 0 or
b = 0 or
Both a & b are 0

So a can be 0 or non zero. So insufficient.

From 2, the following situations are possible
a = 0 or
c = 0 or
Both a & c are 0

So a can be 0 or non zero. So insufficient.

1 & 2 together

a can be non zero while b and c could be 0. or
a can be 0 while b and c could be non zero. or
a,b & c all three can be 0.

So it is still not possible to determine whether a is zero or not.

Hence answer is E
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Got it buddy !!
Kudos for you :-D .. cheers
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1 & 2 together

a can be non zero while b and c could be 0. or
a can be 0 while b and c could be non zero. or
a,b & c all three can be 0.

So it is still not possible to determine whether a is zero or not.

Hence answer is E

Note: Look out for the case where b and c cannot be 0 at the same time. In that case, a must be 0.

e.g. the same question if slightly modified could result in (C)

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) (2a+10b-1)(2a-b-4)=0

From (1), (2a-b-4) = 0 OR (a+5b+2)=0 OR both are 0.
From (2), (2a-b-4)=0 OR (2a+10b-1)= 0 OR both are 0.

From no values of a and b, can (a+5b+2) and (2a+10b-1) be 0 simultaneously.
a + 5b + 2 = 0 implies a + 5b = -2
2a + 10b - 1 = 0 implies a + 5b = 1/2

a + 5b cannot take 2 different values at the same time. So (2a-b-4) must be 0.
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E is correct because there is an existence of a and b such that (a+5b+2) = 0 and (4a+3b-1) = 0. Also, the values of a and b in those 2 equations will not belong to a point on the line: y=2x-4
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Bunuel
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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

In the xy-plane, does the line with the equation y=2x-4 contain the point (a,b)?

Line with equation \(y=2x-4\) contains the point \((a,b)\) means that when substituting \(a\) ans \(b\) in line equation: \(b=2a-4\) (or \(2a-b-4=0\)) holds true.

So, basically we are asked to determine whether \(2a-b-4=0\) is true.

(1) (2a-b-4)(a+5b+2)=0 --> either \(2a-b-4=0\) OR \(a+5b+2=0\) OR both. Not sufficient.

(2) (4a+3b-1)(2a-b-4)=0 --> either \(2a-b-4=0\) OR \(4a+3b-1=0\) OR both. Not sufficient.

(1)+(2) Now, since \(a+5b+2=0\) and \(4a+3b-1=0\) can simultaneously be true (for a=11/17 and b=-9/17), then we have that EITHER both \(a+5b+2=0\) and \(4a+3b-1=0\) are true OR \(2a-b-4=0\) is true. Not sufficient.

Answer: E (C is not correct).

Similar question to practice from GMAT Prep: https://gmatclub.com/forum/in-the-xy-pla ... 00399.html

Hope it helps.

hello Bunuel,
superb explanation, just one question though, Are we interested in getting a only a unique solution for a and b, and only then we can say definitely if (a,b) lies on the line segment or not?
Thanx
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Bunuel
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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

In the xy-plane, does the line with the equation y=2x-4 contain the point (a,b)?

Line with equation \(y=2x-4\) contains the point \((a,b)\) means that when substituting \(a\) ans \(b\) in line equation: \(b=2a-4\) (or \(2a-b-4=0\)) holds true.

So, basically we are asked to determine whether \(2a-b-4=0\) is true.

(1) (2a-b-4)(a+5b+2)=0 --> either \(2a-b-4=0\) OR \(a+5b+2=0\) OR both. Not sufficient.

(2) (4a+3b-1)(2a-b-4)=0 --> either \(2a-b-4=0\) OR \(4a+3b-1=0\) OR both. Not sufficient.

(1)+(2) Now, since \(a+5b+2=0\) and \(4a+3b-1=0\) can simultaneously be true (for a=11/17 and b=-9/17), then we have that EITHER both \(a+5b+2=0\) and \(4a+3b-1=0\) are true OR \(2a-b-4=0\) is true. Not sufficient.

Answer: E (C is not correct).

Similar question to practice from GMAT Prep: https://gmatclub.com/forum/in-the-xy-pla ... 00399.html

Hope it helps.

hello Bunuel,
superb explanation, just one question though, Are we interested in getting a only a unique solution for a and b, and only then we can say definitely if (a,b) lies on the line segment or not?
Thanx

I believe the question is already obvious. The question is not about the unique value of a or b, but about whether any point (a,b) with the value satisfies the statement 1 or 2 will belong to the line
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MacFauz
1 & 2 together

a can be non zero while b and c could be 0. or
a can be 0 while b and c could be non zero. or
a,b & c all three can be 0.

So it is still not possible to determine whether a is zero or not.

Hence answer is E

Note: Look out for the case where b and c cannot be 0 at the same time. In that case, a must be 0.

e.g. the same question if slightly modified could result in (C)

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) (2a+10b-1)(2a-b-4)=0

From (1), (2a-b-4) = 0 OR (a+5b+2)=0 OR both are 0.
From (2), (2a-b-4)=0 OR (2a+10b-1)= 0 OR both are 0.

From no values of a and b, can (a+5b+2) and (2a+10b-1) be 0 simultaneously.
a + 5b + 2 = 0 implies a + 5b = -2
2a + 10b - 1 = 0 implies a + 5b = 1/2

a + 5b cannot take 2 different values at the same time. So (2a-b-4) must be 0.



this question is not modified, so the answer is "C" ?
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MacFauz
1 & 2 together

a can be non zero while b and c could be 0. or
a can be 0 while b and c could be non zero. or
a,b & c all three can be 0.

So it is still not possible to determine whether a is zero or not.

Hence answer is E

Note: Look out for the case where b and c cannot be 0 at the same time. In that case, a must be 0.

e.g. the same question if slightly modified could result in (C)

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) (2a+10b-1)(2a-b-4)=0

From (1), (2a-b-4) = 0 OR (a+5b+2)=0 OR both are 0.
From (2), (2a-b-4)=0 OR (2a+10b-1)= 0 OR both are 0.

From no values of a and b, can (a+5b+2) and (2a+10b-1) be 0 simultaneously.
a + 5b + 2 = 0 implies a + 5b = -2
2a + 10b - 1 = 0 implies a + 5b = 1/2

a + 5b cannot take 2 different values at the same time. So (2a-b-4) must be 0.



this question is not modified, so the answer is "C" ?

the correct answer is E, and btw, what do you mean by "the question is not modified"?
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soodia
VeritasPrepKarishma
MacFauz
1 & 2 together

a can be non zero while b and c could be 0. or
a can be 0 while b and c could be non zero. or
a,b & c all three can be 0.

So it is still not possible to determine whether a is zero or not.

Hence answer is E

Note: Look out for the case where b and c cannot be 0 at the same time. In that case, a must be 0.

e.g. the same question if slightly modified could result in (C)

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) (2a+10b-1)(2a-b-4)=0

From (1), (2a-b-4) = 0 OR (a+5b+2)=0 OR both are 0.
From (2), (2a-b-4)=0 OR (2a+10b-1)= 0 OR both are 0.

From no values of a and b, can (a+5b+2) and (2a+10b-1) be 0 simultaneously.
a + 5b + 2 = 0 implies a + 5b = -2
2a + 10b - 1 = 0 implies a + 5b = 1/2

a + 5b cannot take 2 different values at the same time. So (2a-b-4) must be 0.



this question is not modified, so the answer is "C" ?

If the question is modified (as discussed above), the answer would be (C).
For the unmodified question, the answer would be (E).
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We are asked whether a = 0.

1) ab = 0

2) ac = 0

From 1, the following situations are possible
a = 0 or
b = 0 or
Both a & b are 0

So a can be 0 or non zero. So insufficient.

From 2, the following situations are possible
a = 0 or
c = 0 or
Both a & c are 0

So a can be 0 or non zero. So insufficient.

1 & 2 together

a can be non zero while b and c could be 0. or
a can be 0 while b and c could be non zero. or
a,b & c all three can be 0.

So it is still not possible to determine whether a is zero or not.
SO E

Posted from my mobile device
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Bunuel
In the xy-plane, does the line with the equation y=2x-4 contain the point (a,b)?

Line with equation \(y=2x-4\) contains the point \((a,b)\) means that when substituting \(a\) ans \(b\) in line equation: \(b=2a-4\) (or \(2a-b-4=0\)) holds true.

So, basically we are asked to determine whether \(2a-b-4=0\) is true.

(1) (2a-b-4)(a+5b+2)=0 --> either \(2a-b-4=0\) OR \(a+5b+2=0\) OR both. Not sufficient.

(2) (4a+3b-1)(2a-b-4)=0 --> either \(2a-b-4=0\) OR \(4a+3b-1=0\) OR both. Not sufficient.

(1)+(2) Now, since \(a+5b+2=0\) and \(4a+3b-1=0\) can simultaneously be true (for a=11/17 and b=-9/17), then we have that EITHER both \(a+5b+2=0\) and \(4a+3b-1=0\) are true OR \(2a-b-4=0\) is true. Not sufficient.

Answer: E (C is not correct).
Bunuel

IMO we should not conclude the answer as E at this stage but instead we should substitute the values of a=11/17 and b=-9/17 into 2a-b-4=0. If this holds true, then the answer would be C like in this ques
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