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GMAT Prep Prac 1 - Q19

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GMAT Prep Prac 1 - Q19

by mayuran23 » Tue Apr 14, 2009 3:32 pm
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

Since, when plugging (r,s) into y = 3x + 2 gives you s = 3r + 2, i'm assuming that this makes one of the terms in stmt 1 and 2 = 0 (3r + 2 - s = 0, since 3r + 2 = s).

Any suggestions please?
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by gmat740 » Tue Apr 14, 2009 7:36 pm
OA is D


I. (3r + 2 - s)(4r + 9 - s) = 0
so
(3r + 2 - s) = 0
(4r + 9 - s) = 0

solve both these eqn's
s= -19
r = -7

now our equation is

y = 3x +2

so for a point to lie on the lie, it must satisfy the equation of line

s = 3r + 2

Put the values of r and s

-19 = 3(-7)+2
-19= -19

Hence (r,s) lie on the line.


II. (4r - 6 - s)(3r + 2 - s) = 0

(4r - 6 - s) = 0

(3r + 2 - s) = 0

proceed just like the steps above and you will get two different values of r and s

put those values of r and s back into the eqn of the line and check whether they satisfy or not


So both statements are suff

So answer D


Hope this Helps

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by nasa » Tue Apr 14, 2009 8:27 pm
IMO C
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 => 3r+2 = s or 4r+9 = s
i.e. (1) is true either if (r,s) falls on line 3x+2=y
or if (r,s) falls on line 4r+9=y
(2) (4r - 6 - s)(3r + 2 - s) = 0 => 4r-6 = s or 3r+2 = s

Now we have 3 different lines
3x+2=y, 4x+9=y, 4x-6=y
and 1 and 2 becomes true only if (r,s) is contained in line 3x+2=y
So, C
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by shkusira » Tue Apr 14, 2009 11:24 pm
I agree with Nasa. Got C. what is the OA?
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by ketkoag » Wed Apr 15, 2009 4:14 am
IMO D, now lets rewrite the equation..
i.e. 3x + 2-y=0.
By statement 1, (3r + 2 - s)(4r + 9 - s) = 0
i.e. either (3r + 2 - s)=0 or (4r + 9 - s) = 0 or both are 0.
by this we can conclude that the equation y = 3x + 2 contains the point (r,s).

Similarly,by statement 2, (4r - 6 - s)(3r + 2 - s) = 0
again (3r + 2 - s)=0 i.e. equation y = 3x + 2 contains the point (r,s).

Hence D.
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by bluementor » Wed Apr 15, 2009 4:19 am
If point (r,s) is contained within y=3x + 2, then the following must be true:

y =3x + 2
s = 3r + 2 (by susbstituting values for x and y with r and s, respectively)
3r + 2 -s = 0

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

It could be that (3r + 2 - s) or (4r + 9 - s) or both are equal to zero. insufficient to conclude that (3r + 2 - s) is definitely equal to zero.

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

It could be that (3r + 2 - s) or (4r - 6 - s) or both are equal to zero. insufficient to conclude that (3r + 2 - s) is definitely equal to zero.

Both statements together:

Either

(3r + 2 - s)

OR

Both (4r - 6 - s) AND (4r + 9 - s)

must equal to zero.


If (4r - 6 - s) and (4r + 9 - s) are equal to zero, then:

4r - s = 6
4r - s = -9, both of these contradict each other.

Hence, (4r - 6 - s) and (4r + 9 - s) are NOT equal to zero, and (3r + 2 - s) must be equal to zero. So line y= 3x+2 must contain point (r,s) Sufficient.

Choose C.

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by pakaskwa » Wed Apr 15, 2009 2:29 pm
The OA is C. I just did the question yesterday on GMATPrep. I chose D, and I was wrong.
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by sanju09 » Sat Apr 18, 2009 2:54 am
Great explanation -BoMb-
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by sunnyd283 » Sun May 24, 2009 2:00 pm
'Gmat740', you really should make sure you don't put wrong official answers onto your posts. I use this forum for my own revision and it's annoying when people state wrong official answers as a matter of fact rather than their own misinformed opinion. Please take care not to do this again.

The official answer is C. You need both statements to see that r=(s-2)/3.
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by gmat740 » Sun May 24, 2009 6:41 pm
Gmat740', you really should make sure you don't put wrong official answers onto your posts. I use this forum for my own revision and it's annoying when people state wrong official answers as a matter of fact rather than their own misinformed opinion. Please take care not to do this again.

The official answer is C. You need both statements to see that r=(s-2)/3.
Well to my best knowledge at that time, the OA was D as I came across the same question in some of the book( I don't remember since it has been a long time!)

I did not put intentionally the wrong answer. It was after my post I came across other answers and realize that MY OA is wrong.

We are here to learn from mistakes and I learned from mine. Until and unless we don't try, how are we going to learn??

I should have used IMO instead of OA, I will take care about that.

Cheers
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by aj5105 » Wed May 27, 2009 3:32 am
BOMBAT!!
bluementor wrote:If point (r,s) is contained within y=3x + 2, then the following must be true:

y =3x + 2
s = 3r + 2 (by susbstituting values for x and y with r and s, respectively)
3r + 2 -s = 0

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

It could be that (3r + 2 - s) or (4r + 9 - s) or both are equal to zero. insufficient to conclude that (3r + 2 - s) is definitely equal to zero.

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

It could be that (3r + 2 - s) or (4r - 6 - s) or both are equal to zero. insufficient to conclude that (3r + 2 - s) is definitely equal to zero.

Both statements together:

Either

(3r + 2 - s)

OR

Both (4r - 6 - s) AND (4r + 9 - s)

must equal to zero.


If (4r - 6 - s) and (4r + 9 - s) are equal to zero, then:

4r - s = 6
4r - s = -9, both of these contradict each other.

Hence, (4r - 6 - s) and (4r + 9 - s) are NOT equal to zero, and (3r + 2 - s) must be equal to zero. So line y= 3x+2 must contain point (r,s) Sufficient.

Choose C.

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by Brent@GMATPrepNow » Thu May 12, 2011 7:25 am
mayuran23 wrote: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

Since, when plugging (r,s) into y = 3x + 2 gives you s = 3r + 2, i'm assuming that this makes one of the terms in stmt 1 and 2 = 0 (3r + 2 - s = 0, since 3r + 2 = s).

Any suggestions please?
If (r,s) is on the line defined by the equation y=3x+2, then (r,s) must satisfy the equation y=3x+2. In other words, it must be true that s=3r+2
For example: We know that the point (5, 17) is on the line y=3x+2, because when we plug x=5 and y=17 into the equation, we get 17 = 3(5)+2 and the equation holds true.

So, we can reword the target question to be "Does s = 3r + 2?"

1. (3r+2-s)(4r+9-s) = 0
From this, we know that either (3r+2-s) = 0 or (4r+9-s) = 0
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
If (4r+9-s) = 0 then s = 4r+9, in which case the answer to our new target question is no
Since we get two different answers to the target question, statement 1 is NOT SUFFICIENT

2. (4r-6-s)(3r+2-s) = 0
From this, we know that either (4r-6-s) = 0 or (3r+2-s) = 0
If (4r-6-s)) = 0 then s = 4r-6, in which case the answer to our new target question is no
If (3r+2-s) = 0 then s = 3r+2, in which case the answer to our new target question is yes
Since we get two different answers to the target question, statement 2 is NOT SUFFICIENT

Statements 1&2 combined: Since (3r+2-s) is the only expression common to both statements, it must be true that 3r+2-s = 0, in which case y MUST equal 3r+2
As such the answer is C
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by smackmartine » Thu May 12, 2011 6:05 pm
I found that there is a variation in these type of questions. If the same question comes as problem solving , it would be :

In the xy-plane, if the line with equation y = 3x + 2 contain the point (r,s), which of the following must be true?

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

A) 1 only
B) 2 only
C) 3 only
D) ALL
E) None

In this case [spoiler]D) ALL[/spoiler] would be correct answer because we can neglect the other solutions (4r + 9 - s)=0 and (4r - 6 - s)=0 in statement 1 and 2 respectively.

Experts please comment, if there are any discrepancies.
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by artistocrat » Fri May 13, 2011 4:02 pm
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

(1)either 3r+2=s (contains r,s) and/or 4r+9=s
INSUFF

(2)either 3r+2=s (contains r,s) and/or 4r-6=s
INSUFF

(1)&(2) 3r+2=s (contains r,s), because you can't have both 4r+9=s and 4r-6=s
SUFF

Answer: C
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by Brent@GMATPrepNow » Sat May 14, 2011 6:34 am
smackmartine wrote:I found that there is a variation in these type of questions. If the same question comes as problem solving , it would be :

In the xy-plane, if the line with equation y = 3x + 2 contain the point (r,s), which of the following must be true?

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

A) 1 only
B) 2 only
C) 3 only
D) ALL
E) None

In this case [spoiler]D) ALL[/spoiler] would be correct answer because we can neglect the other solutions (4r + 9 - s)=0 and (4r - 6 - s)=0 in statement 1 and 2 respectively.

Experts please comment, if there are any discrepancies.
You're 100% correct, smackmartine,

If we know that the line with equation y = 3x + 2 contains the point (r,s), then is must be true that s=3r+2
If s=3r+2 when we can also conclude that 3r + 2 - s = 0

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

So, the answer is D
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