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# In the xy-coordinate plane, line A is defined by the equation j*x-y=-7

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In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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Updated on: 01 Sep 2015, 08:30
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27% (02:42) correct 73% (02:25) wrong based on 386 sessions

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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)

Originally posted by Tornikea on 01 Sep 2015, 08:21.
Last edited by ENGRTOMBA2018 on 01 Sep 2015, 08:30, edited 1 time in total.
Formatted the question
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In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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Updated on: 03 Sep 2015, 04:23
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Tornikea wrote:
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)

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 x, we get -1.5x+7=jx+7 ----> x(j+1.5) =0 ---> either x =0 or j=-1.5 but as per (a) above, j $$\neq$$-1.5

The only case possible is for x=0 ---> y =7 to be the actual 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.

Originally posted by ENGRTOMBA2018 on 01 Sep 2015, 08:44.
Last edited by ENGRTOMBA2018 on 03 Sep 2015, 04:23, edited 1 time in total.
Edited the wording to make it clearer.
##### General Discussion
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Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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02 Sep 2015, 06:23
Engr2012 wrote:
Tornikea wrote:
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)

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.

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
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Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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02 Sep 2015, 06:36
1
Engr2012 wrote:
Tornikea wrote:
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)

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.

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.
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In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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Updated on: 06 Sep 2015, 20:50
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.
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"Only $149 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" Originally posted by MathRevolution on 03 Sep 2015, 04:47. Last edited by MathRevolution on 06 Sep 2015, 20:50, edited 1 time in total. CEO Joined: 20 Mar 2014 Posts: 2625 Concentration: Finance, Strategy Schools: Kellogg '18 (M) GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7 [#permalink] ### Show Tags 03 Sep 2015, 07:24 MathRevolution wrote: 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. Manager Joined: 24 Jan 2017 Posts: 142 GMAT 1: 640 Q50 V25 GMAT 2: 710 Q50 V35 GPA: 3.48 Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7 [#permalink] ### Show Tags 28 Jun 2017, 20:54 1 1 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":'( EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14188 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7 [#permalink] ### Show Tags 25 Jun 2018, 21:00 3 1 Hi All, We're told that In the XY-coordinate plane, line A is defined by the equation (J)(X) - Y = -7 and line B is defined by the equation 3X + 2Y = K and that the two lines are NOT parallel. We're asked at what point they intersect. to start, with graphing questions, it usually helps to convert the given equations into slope-intercept format (re: Y = MX + B). In this case, the lines would be: Y = (J)(X) + 7 Y = 3x/2 + K/2 We have to figure out at what point will X and Y be the same for BOTH equations. 1) Line A passes through the point (3,1) With the co-ordinate in Fact 1, we can plug in and get.... 1 = J(3) + 7 -6 = 3J -2 = J Now our two equations are: Y = -2X + 7 Y = 3X/2 + K/2 Unfortunately, we still have an unknown (K) and that keeps us from figuring out where the lines intersect. Fact 1 is INSUFFICIENT 2) Line B passes through the point (0,7) With the co-ordinate in Fact 2, we can plug in and get.... 7 = 0 + K/2 K = 14 Now, the two lines are; Y = 3X/2 + 7 Y = (J)(X) + 7 Notice how the y-intercepts are the same in both equations? We now KNOW where the two lines intersect: at (0,7) Fact 2 is SUFFICIENT Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save$75 + GMAT Club Tests Free
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In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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21 Jul 2018, 10:48
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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.
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In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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21 Jul 2018, 16:17
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.
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Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7  [#permalink]

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24 Aug 2018, 03:37
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
Re: In the xy-coordinate plane, line A is defined by the equation j*x-y=-7   [#permalink] 24 Aug 2018, 03:37
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