February 20, 2019 February 20, 2019 08:00 PM EST 09:00 PM EST Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST February 21, 2019 February 21, 2019 10:00 PM PST 11:00 PM PST Kick off your 2019 GMAT prep with a free 7day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52971

Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
15 Mar 2015, 22:04
Question Stats:
79% (01:55) correct 21% (02:20) wrong based on 249 sessions
HideShow timer Statistics



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 13559
Location: United States (CA)

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
16 Mar 2015, 21:55
Hi All, Since it looks like this Ds question has gotten 'buried', I'm going to bump it back up with a 'nudge' about how to solve it: When dealing with graphing and lines, it helps to organize your data using slopeintercept format: Y = (M)(X) + B We're given the slope of a line and one of the points that it runs through. From this, you can calculate the value of B. With the equation of the full line, you should then be able to calculate the Xintercept.... 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
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/
*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****



Director
Joined: 23 Jan 2013
Posts: 556

Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
16 Mar 2015, 23:04
slope 5/3 and point (2;7), so equation is
7=(5/3)*2+b => b=11/3
to find X intercept, rewrite the equation
0=(5/3)*x+11/3 => x=11/5=2 1/5
B



Senior Manager
Joined: 28 Feb 2014
Posts: 294
Location: United States
Concentration: Strategy, General Management

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
17 Mar 2015, 08:31
Here is my algebraic approach: y = (5/3)x + b 7 = (5/3)(2) + b b = 11/3
0 = (5/3)x + 11/3 x = 11/5 or 2 1/5



Math Expert
Joined: 02 Sep 2009
Posts: 52971

Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
23 Mar 2015, 02:58
Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Think about this visually. A slope of – 5/3 means, among other things, right 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is much closer to the xintercept. Let’s think about this in the vicinity of the xintercept. Obviously, (1, 2) is at a height of 2 above the xaxis. Let b be the distance from (1, 0) to the xintercept. We know –h/b must equal the slope. h/b = 5/3 > 2/b = 5/3. 5b = 6 > b = 6/5 = 1 1/5. Now, we just have to add one to that to get the horizontal distance from the origin. xintercept = 2 1/5. Answer = (B) Attachment:
ppocg_img6.png [ 9.41 KiB  Viewed 5698 times ]
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 26 Dec 2011
Posts: 114

Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
25 May 2015, 04:05
Bunuel wrote: Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. Let’s think about this in the vicinity of the xintercept. Attachment: ppocg_img6.png Obviously, (1, 2) is at a height of 2 above the xaxis. Let b be the distance from (1, 0) to the xintercept. We know –h/b must equal the slope. h/b = 5/3 > 2/b = 5/3. 5b = 6 > b = 6/5 = 1 1/5. Now, we just have to add one to that to get the horizontal distance from the origin. xintercept = 2 1/5. Answer = (B) Hi Bunuel, Please make me understand the below point: A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. I believe, for slope  5/3, right 3 spaces and down 5 so that the line goes through (1,2) from (2,7) downward slope. Please explain if iam wrong. Thanks
_________________
Thanks, Kudos Please



Math Expert
Joined: 02 Sep 2009
Posts: 52971

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
25 May 2015, 07:54
balamoon wrote: Bunuel wrote: Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. MAGOOSH OFFICIAL SOLUTION:Think about this visually. A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. Let’s think about this in the vicinity of the xintercept. Obviously, (1, 2) is at a height of 2 above the xaxis. Let b be the distance from (1, 0) to the xintercept. We know –h/b must equal the slope. h/b = 5/3 > 2/b = 5/3. 5b = 6 > b = 6/5 = 1 1/5. Now, we just have to add one to that to get the horizontal distance from the origin. xintercept = 2 1/5. Answer = (B) Hi Bunuel, Please make me understand the below point: A slope of – 5/3 means, among other things, left 3 spaces, down 5. If the line goes through the point (–2, 7), then it must also go through (1, 2), which is must closer to the xintercept. I believe, for slope  5/3, right 3 spaces and down 5 so that the line goes through (1,2) from (2,7) downward slope. Please explain if iam wrong. Thanks Yes, that's true. Edited. Thank you.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 21 Feb 2012
Posts: 56

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
26 May 2015, 00:55
My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.
_________________
Regards J
Do consider a Kudos if you find the post useful



Intern
Joined: 23 Apr 2014
Posts: 21

Re: Line A has a slope of 5/3 through the point (–2, 7). What
[#permalink]
Show Tags
26 May 2015, 04:26
With the equation y=mx+c we get 7=10/3 + c So c = 11/3
To find the x intercept put y=0 in the equationy=mx+c
0= 5/3 x +11/3 Multiplying by 3 throughout 11 = 5x So x=11/5 Which is 2 1/5



Intern
Joined: 01 Apr 2014
Posts: 4

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
07 Apr 2016, 22:57
Jackal wrote: My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations. HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN



Math Expert
Joined: 02 Aug 2009
Posts: 7334

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
07 Apr 2016, 23:04
akshay4gmat wrote: HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN Hi here 11/5 is not 21/5 BUT \(2 \frac{1}{5}\)that is 2 and 1/5.. \(2 + \frac{1}{5} = \frac{(10+1)}{5} =\frac{11}{5}\)
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html 4) Base while finding % increase and % decrease : https://gmatclub.com/forum/percentageincreasedecreasewhatshouldbethedenominator287528.html
GMAT Expert



Manager
Joined: 21 Feb 2012
Posts: 56

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
08 Apr 2016, 06:46
akshay4gmat wrote: Jackal wrote: My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations. HOW YOU ARE EQUATING 11/5 TO 21/5 PLEASE EXPLAIN Hi Akshay Just as Chetan explained. \(= \frac{11}{5}\) \(= \frac{(10 + 1)}{5}\) \(= 2 \frac{1}{5}\) Apologies for not using Latex equivalent symbols
_________________
Regards J
Do consider a Kudos if you find the post useful



Director
Joined: 04 Jun 2016
Posts: 568

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
14 Jul 2016, 06:55
Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. Point intercept form a line is y=mx+b slope = m, y intercept =b (given that x=0) x intercept =b/m Here we know m=5/3 ; we need to find b and we are done lets use the equation by plugging values of x,y and m from the question stem 7=5/3*2 +b ==> 7=10/3+b b=11/3 x intercept = b/m = 11/3/5/3=11/5==> 11/5 = 2 \(\frac{1}{5}\) Answer is B
_________________
Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE : 17th SEPTEMBER 2016. .. 16 March 2017  I am back but for all purposes please consider me semiretired.



Intern
Joined: 29 Dec 2016
Posts: 12

Re: Line A has a slope of 5/3
[#permalink]
Show Tags
03 Jun 2017, 23:39
slope m= 5/3 and point (2;7), so equation is
y=mx+b
We need to find b 7=(5/3)*2+b 710/3=b
b=11/3 to find x intercept, rewrite the equation
0=(5/3)*x+11/3 => x=11/5
Answer B
Press +1 Kudos if you like my answer



Intern
Joined: 08 Sep 2016
Posts: 32

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
05 Nov 2017, 02:45
Hi @bunnel, Could please explain the how you arrived at x intercept. The equation of the line is 5x3y+11 = 0 So should the y intercept be a/b = 5/3? Thanks in advance! S Jackal wrote: My attemptI am very fond of coordinate geometry Using slope intercept form of equation y=mx+c In this case y=(5/3)x+c We know that the line passes through (2,7) So solving for c c=7(5*2/3)=11/3 So equation of line is y=(5/3)x+(11/3) Changing the equation of line to intercept form x/(11/5) + y/(11/3) = 1 So x intercept is 11/5=2 1/5 which is B Some theory to keep in mindAn equation of line can be represented in multiple forms 1. y=mx+c where m is the slope and c is a constant 2. x/a + y/b = 1 here a and b are the intercepts on x and y axis respectively 3. (yy1)/(y2y1) = (xx1)/(x2x1) equation of line passing through (x1,y1) and (x2,y2) I think basis these it becomes easier to solve and one must know how to change forms of these equations quickly. All these are linear equations.



Director
Joined: 02 Sep 2016
Posts: 670

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
10 Nov 2017, 02:46
Is this solution fine? y= mx+c y= 5/3*x +c Using formula to find slope: (7y)/(2x) = 5/3 3(7y)= 5(2x) 213y= 10+5x 5x+3y= 11 To find xintercept, plug y= 0 x= 11/5
_________________
Help me make my explanation better by providing a logical feedback.
If you liked the post, HIT KUDOS !!
Don't quit.............Do it.



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2827

Re: Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
14 Nov 2017, 06:10
Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5 The equation of Line A is y = (5/3)x + b. Let’s now plug in (2,7) and determine b: 7 = (5/3)(2) + b 7 = 10/3 + b Multiplying by 3, we have: 21 = 10 + 3b 11 = 3b 11/3 = b Thus, we have: y = (5/3)x + 11/3 We can substitute 0 for y to determine the xintercept, and we have: 0 = (5/3)x + 11/3 11/3 = 5x/3 11 = 5x x = 11/5 = 11/5 = 2 1/5 Answer: B
_________________
Jeffery Miller
Head of GMAT Instruction
GMAT Quant SelfStudy Course
500+ lessons 3000+ practice problems 800+ HD solutions



Senior SC Moderator
Joined: 22 May 2016
Posts: 2484

Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
Show Tags
15 Nov 2017, 12:39
Bunuel wrote: Line A has a slope of 5/3 and passes through the point (–2, 7). What is the xintercept of Line A?
A. 2 1/3 B. 2 1/5 C. 2 2/3 D. 2 2/5 E. 2 3/5
Kudos for a correct solution. With a slope and a point, and asked to find the xintercept of a line, I use the slope intercept form. (It's quick). 1) Start with just the slope to construct a slopeintercept equation for the line: \(y = mx + b\), m = slope, b = yintercept Slope = \(\frac{5}{3}\),so \(y = \frac{5}{3}x + b\) 2) Next, plug in the given point (2,7) to find \(b\) \(7 = \frac{5}{3}(2) + b\) \(7 = \frac{10}{3} + b\) \((7  \frac{10}{3}) = \frac{11}{3} = b\) 3) Plug \(b\) in. The line's equation is now \(y = \frac{5}{3}x + \frac{11}{3}\) 4) Set y equal to 0 to find xintercept \(0 = \frac{5}{3}x + \frac{11}{3}\) \(\frac{5}{3}x=\frac{11}{3}\) \(x=\frac{11}{3}*\frac{3}{5}=\frac{11}{5}\) \(=2\frac{1}{5}\) Answer B Shiv2016 , yes, it's correct. I had to grin, though (with relief!) because I have done what you did . . . There is no need to find the slope. It's given. shinrai , you wrote Quote: Could please explain the how you arrived at x intercept?
The equation of the line is 5x3y+11 = 0
So should the y intercept be a/b = 5/3? Mmm, no. The yintercept is 11/3. 5/3 is the slope. You were quoting Jackal, who wrote, "So equation of line is y=(5/3)x+(11/3)" Jackal got the xintercept by setting y equal to 0: 0 = (5/3)x+(11/3) \(\frac{5}{3}x = \frac{11}{3}\) \(x=\frac{11}{3} * \frac{3}{5}=\frac{11}{5}\) The line crosses the xaxis at \(\frac{11}{5}\) (and when it crosses the xaxis, y = 0) You used this equation for the line: 5x3y+11 = 0. (A more standard form is 5x + 3y = 11.) It doesn't matter which form of equation you use; to find the yintercept, set x = 0. (x IS zero when the line crosses the yaxis.) Your way: 5x  3y + 11 = 0 (5)(0) (3)(y) + 11 = 0 3y = 11 y = \(\frac{11}{3}\) Hope it helps.
_________________
To live is the rarest thing in the world. Most people just exist. —Oscar Wilde




Line A has a slope of 5/3 and passes through the point (–2, 7). What
[#permalink]
15 Nov 2017, 12:39






