Last visit was: 24 Apr 2026, 00:30 It is currently 24 Apr 2026, 00:30
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Tornikea
Joined: 19 May 2015
Last visit: 13 Oct 2020
Posts: 6
Own Kudos:
261
 [126]
Given Kudos: 4
Posts: 6
Kudos: 261
 [126]
6
Kudos
Add Kudos
119
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,777
Own Kudos:
13,047
 [21]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,777
Kudos: 13,047
 [21]
11
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
User avatar
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Own Kudos:
3,890
 [9]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
GMAT 1: 750 Q49 V44
Posts: 2,319
Kudos: 3,890
 [9]
8
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
General Discussion
avatar
anupamadw
Joined: 31 Jul 2014
Last visit: 29 Jun 2016
Posts: 104
Own Kudos:
Given Kudos: 373
GMAT 1: 630 Q48 V29
GMAT 1: 630 Q48 V29
Posts: 104
Kudos: 140
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Engr2012
Tornikea
In the XY-coordinate plane, line A is defined by the equation j*x-y=-7 and line B is defined by the equation 3*x+2*y=k

If line A and line B are not parallel, at what point do they intersect?

(1) Line A passes through the point (3,1)

(2) Line B passes through the point (0,7)

Please follow the posting guidelines.

Given: A: y = jx+7, B : y = -1.5x+k/2

As A is NOT parallel to B ---> j \(\neq\)-1.5 ....(a)

Per statement 1, A passes through (3,1), without complete equation of B, we will not be able to solve this question. Not sufficient.

Check: solving the 2 equations y = -2x+7 and y = -1.5x+k/2, we get x = 14-k and y = 2k-21. Thus without the value 'k' we dont know the exact point of intersection.

Per statement 2, B passes through (0,7), thus 7 = 0+k/2 ---> k = 14. Thus, equation of B is y = -1.5x+7.

Solving for A and B , we get j = -1.5 but as per (a) above, j can not be = -1.5 The only case possible is for (0,7) to be the actual point of interesection. Sufficient.

B is the correct answer.

Hope this helps.

Sorry, red part is not clear, you solved A and B , got j=-1.5 --> The only case possible is for (0,7) to be the actual point of interesection, can you please elaborate ..please
B: y = -1.5x+7
A: y = jx+7
User avatar
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Own Kudos:
3,890
 [1]
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
GMAT 1: 750 Q49 V44
Posts: 2,319
Kudos: 3,890
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
anupamadw
Engr2012
Tornikea
In the XY-coordinate plane, line A is defined by the equation j*x-y=-7 and line B is defined by the equation 3*x+2*y=k

If line A and line B are not parallel, at what point do they intersect?

(1) Line A passes through the point (3,1)

(2) Line B passes through the point (0,7)

Please follow the posting guidelines.

Given: A: y = jx+7, B : y = -1.5x+k/2

As A is NOT parallel to B ---> j \(\neq\)-1.5 ....(a)

Per statement 1, A passes through (3,1), without complete equation of B, we will not be able to solve this question. Not sufficient.

Check: solving the 2 equations y = -2x+7 and y = -1.5x+k/2, we get x = 14-k and y = 2k-21. Thus without the value 'k' we dont know the exact point of intersection.

Per statement 2, B passes through (0,7), thus 7 = 0+k/2 ---> k = 14. Thus, equation of B is y = -1.5x+7.

Solving for A and B , we get j = -1.5 but as per (a) above, j can not be = -1.5 The only case possible is for (0,7) to be the actual point of interesection. Sufficient.

B is the correct answer.

Hope this helps.

Sorry, red part is not clear, you solved A and B , got j=-1.5 --> The only case possible is for (0,7) to be the actual point of interesection, can you please elaborate ..please
B: y = -1.5x+7
A: y = jx+7

Sure, look below.

From statement 2, you get the equation of line B: y = -1.5x+7 and A: y= jx+7

Now,when you plot A and B such that j \(\neq\)-1.5, you will see that

y = -1.5x+7 and
y = jx+7 (with j = anything BUT -1.5)

will intersect at (0,7) ONLY. Try with j = 2 or 5 or 10 or -4.

Case 1: j = 2 ---> A: y = 2x+7 and B: y = -1.5x+7 ---> (0,7) is the ONLY point of intersection.

Case 2: j = 10 ---> A: y = 10x+7 and B: y = -1.5x+7 ---> (0,7) is the ONLY point of intersection.

Case 3: j = -10 ---> A: y = -10x+7 and B: y = -1.5x+7 ---> (0,7) is the ONLY point of intersection.

Alternately, you can see that once you get the equations of A and B as

B: y = -1.5x+7
A: y = jx+7,

X-coordinate of intersection ----> -1.5x+7=jx+7 ---> x = 0. Put this value of x back into any 1 of the 2 equations, you will get y = 7 . Finally, note that the point of intersection is a constant/unique value without 'j' or 'k'

Hence B is sufficient to arrive at a unique answer.

Hope this helps.
User avatar
MathRevolution
User avatar
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Last visit: 27 Sep 2022
Posts: 10,063
Own Kudos:
Given Kudos: 4
GMAT 1: 760 Q51 V42
GPA: 3.82
Expert
Expert reply
GMAT 1: 760 Q51 V42
Posts: 10,063
Kudos: 20,000
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


In the XY-coordinate plane, line A is defined by the equation j*x-y=-7 and line B is defined by the equation 3*x+2*y=k

If line A and line B are not parallel, at what point do they intersect?

(1) Line A passes through the point (3,1)

(2) Line B passes through the point (0,7)

Transforming the original condition and question, jx-y=-7, 3x+2y=k and we have 2 variable (j,k), 1 equation. Since we need to match the number of variables and equations, we need 1 more equation and sine we have 1 each in 1) and 2), D is likely the answer.

In case of 1), 3j-1=-7, j=-2 but we don't know what k is, thus we can't find the point of interaction
In case of 2), 3*0+2*7=k gives us k=14 and line B: 3x+2y=14, y=-1.5x+7. Since line A cross (0,7) in jx-y=-7, line A and line B are not parallel and they meet in (0,7). Thus the condition is sufficient.Therefore the answer is B.
User avatar
ENGRTOMBA2018
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Own Kudos:
Given Kudos: 816
Concentration: Finance, Strategy
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
GMAT 1: 750 Q49 V44
Posts: 2,319
Kudos: 3,890
Kudos
Add Kudos
Bookmarks
Bookmark this Post
MathRevolution
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and equations ensures a solution.


In the XY-coordinate plane, line A is defined by the equation j*x-y=-7 and line B is defined by the equation 3*x+2*y=k

If line A and line B are not parallel, at what point do they intersect?

(1) Line A passes through the point (3,1)

(2) Line B passes through the point (0,7)

Transforming the original condition and question, jx-y=-7, 3x+2y=k and we have 2 variable (j,k), 1 equation. Since we need to match the number of variables and equations, we need 1 more equation and sine we have 1 each in 1) and 2), D is likely the answer.

In case of 1), 3j-1=-7, j=-2 but we don't know what k is, thus we can't find the point of interaction
In case of 2), 3*0+2*7=k gives us k=14 and line B: 3x+2y=14, y=-1.5x+7. Since line A cross (0,7) in jx-y=-7, line A and line B are not parallel and they meet in (0,7). Thus the condition is sufficient.Therefore the answer is B.

IMO, your quote "Remember equal number of variables and equations ensures a solution." is a bit misleading as it should be stated as "Remember equal number of variables and DISTINCT equations ensures a solution.".

Case in point,

2a+3b =16
4a+6b=-3.5. Although you have 2 equations and 2 variables, you still can not find the solution as the 2 equations are necessarily the same.
User avatar
Lucy Phuong
Joined: 24 Jan 2017
Last visit: 12 Aug 2021
Posts: 111
Own Kudos:
351
 [2]
Given Kudos: 106
GMAT 1: 640 Q50 V25
GMAT 2: 710 Q50 V35
GPA: 3.48
Products:
GMAT 2: 710 Q50 V35
Posts: 111
Kudos: 351
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Good question. Just learn my own lesson that never ever overlook a single detail of information provided that is, in this case, (0,7)

General equation for any line on x-y coordinate: y=ax+b
--> Line A: y=j*x+7
Line B: y=-1.5x + 0.5k

(0,7) is definitely a turning point, which helps 0.5k=7 matched exactly with 7 in equation of line A. Remember (0,7) is the one and only point that makes statement (2) become sufficient. Otherwise, if that is another value, let's say (0,5), we cannot find intersect point with only statement (2), because j*x+7=-1.5x+5, then (j+1.5)x = -2. In this case, we cannot find out a consistent answer.

Well in the first place, I just thought we cannot work out intersect point without knowing proper equations of lines A and B. That's why I ended up with option (C) in only 20s. Oh goshhhh too fast too "dangerous":'(
User avatar
AdityaHongunti
Joined: 20 Sep 2016
Last visit: 31 Mar 2021
Posts: 533
Own Kudos:
1,084
 [1]
Given Kudos: 632
Location: India
Concentration: Strategy, Operations
GPA: 3.6
WE:Operations (Consumer Packaged Goods)
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Cool Cool Cool..

First lets absorb the info and infer as much as we can >
equation of line >> y=mx+b >> m=slope >> b = y intercept (0,b)
line A: jx-y=-7 >> y=jx+7 >> get y intercept>> put x=0 > y=7 (y intercept)> Line A intersects Y axis at (0,7).
Line B: 3x+2y=k >> y= -1.5x+k/2 >> K is the Y intercept of line B.

Line A NOT PARALLEL to line B >> slopes of parallel lines are equal>> given- lines not parallel. Therefore slopes are not equal >> slope of B =-1.5 >> slope of A CANNOT BE -1.5.

Question- Point of intersection of lines A & B.

St 1. line A passes through the point (3,1) >> we can get the slope of line A >> m=(y1-y2/x1-x2) >> but line B can pass through anywhere . INSUFFICIENT.

st2. Line B passes through point (0,7) >> but (0,7) is also a point on Line A as (0,7) is y intercept of line A >> Therefore point of intersection = (0,7) >> SUFFICIENT.
User avatar
Hero8888
Joined: 29 Dec 2017
Last visit: 14 Apr 2019
Posts: 299
Own Kudos:
Given Kudos: 273
Location: United States
Concentration: Marketing, Technology
GMAT 1: 630 Q44 V33
GMAT 2: 690 Q47 V37
GMAT 3: 710 Q50 V37
GPA: 3.25
WE:Marketing (Telecommunications)
GMAT 3: 710 Q50 V37
Posts: 299
Kudos: 348
Kudos
Add Kudos
Bookmarks
Bookmark this Post
To those who struggle with questions about xy-plane: 99% of such questions can be solved drawing lines on xy plane, without messing with formulae. Just start drawing, and the answer will jump on you.
User avatar
thangvietnam
Joined: 29 Jun 2017
Last visit: 04 Apr 2026
Posts: 743
Own Kudos:
Given Kudos: 2,198
Posts: 743
Kudos: 419
Kudos
Add Kudos
Bookmarks
Bookmark this Post
this is tricky. we do not need to have 2 identified lines. we need just that the two line intersect at a specific point. mayber one line is unidentifie or mayby both lines are unidentified but both of them need to intersect at one specific point for us to find out the point.

with this thinking in mind, we anticipate that we need to find the specific point, not find 2 identified lines.

i dont think that we can do so on test day
User avatar
CEdward
Joined: 11 Aug 2020
Last visit: 14 Apr 2022
Posts: 1,162
Own Kudos:
289
 [1]
Given Kudos: 332
Posts: 1,162
Kudos: 289
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
In the XY-coordinate plane, line A is defined by the equation j*x-y=-7 and line B is defined by the equation 3*x+2*y=k

A: y = jx + 7
B: y = -3/2x + k/2

If line A and line B are not parallel, at what point do they intersect?

(1) Line A passes through the point (3,1)
y = jx + 7
1 = 3j + 7
j = -2

y = -2x + 7

Insufficient

(2) Line B passes through the point (0,7)
y = -3/2x + k/2
7= -3/2(0) + k/2
k = 14

y = -3/2x + 7

Sufficient (A and B each have the same y-intercept: 0,7)

B
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,964
Own Kudos:
Posts: 38,964
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109802 posts
498 posts
212 posts