Last visit was: 24 Apr 2024, 18:38 It is currently 24 Apr 2024, 18:38

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
SORT BY:
Date
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 92900
Own Kudos [?]: 618819 [25]
Given Kudos: 81588
Send PM
Most Helpful Reply
Retired Moderator
Joined: 18 Sep 2014
Posts: 1015
Own Kudos [?]: 2755 [10]
Given Kudos: 79
Location: India
Send PM
General Discussion
avatar
Intern
Intern
Joined: 03 Mar 2016
Posts: 13
Own Kudos [?]: 50 [1]
Given Kudos: 5
Location: India
Schools: ISB '19
Send PM
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11169
Own Kudos [?]: 31890 [1]
Given Kudos: 290
Send PM
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
1
Kudos
Expert Reply
vardhanindaram wrote:
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E



Hi,
there is some info in statement 1, which can answer the Q..

Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

INFO from this:-


A)the intercept on x axis is (3,0) and y axis is (0,4)..
B) the slope is 4/3, which tells us that y increases at a higher rate than x..

Inference:-


1)Since the Line L is perpendicular to K, If its x-intercept is less than 3, the two lines will intersect above x-axis and if more than 3, it will intersect below X-axis..

lets see the statements


(1) x-intercept of Line L is less than that of Line K
see att fig, p+q will be always negative
a) in Quad III, both p and q will be -ive, so p+q will be NEGAIVE.

Suff..

(2) y-intercept of Line L is less than that of Line K
not suff
insuff..

ans A
avatar
Intern
Intern
Joined: 03 Mar 2016
Posts: 13
Own Kudos [?]: 50 [0]
Given Kudos: 5
Location: India
Schools: ISB '19
Send PM
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
chetan2u wrote:
vardhanindaram wrote:
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E



Hi,
there is some info in statement 1, which can answer the Q..

Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

INFO from this:-


A)the intercept on x axis is (3,0) and y axis is (0,4)..
B) the slope is 4/3, which tells us that y increases at a higher rate than x..

Inference:-


1)Since the Line L is perpendicular to K, If its x-intercept is less than 3, the two lines will intersect above x-axis and if more than 3, it will intersect below X-axis..

lets see the statements


(1) x-intercept of Line L is less than that of Line K
see att fig, p+q will be always negative
a) in Quad III, both p and q will be -ive, so p+q will be NEGAIVE.

Suff..

(2) y-intercept of Line L is less than that of Line K
not suff
insuff..

ans A


The line that you drew in the picture is wrong.It has to meet x axis at -3.
And I would like someone to clarify if intercept is said to be less than -3,does that mean intercept of line should be between 3 and -3 considering the definition of intercept to be distance between origin and point where line meets axis.This is if we ignore the sign as it's the convention to denote the direction.
RC & DI Moderator
Joined: 02 Aug 2009
Status:Math and DI Expert
Posts: 11169
Own Kudos [?]: 31890 [0]
Given Kudos: 290
Send PM
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
Expert Reply
vardhanindaram wrote:
chetan2u wrote:
vardhanindaram wrote:
Clue 1 and 2 only suggest that line L can meet K in either quadrant 1 or quadrant 3 as intercepts constraints apply in both the cases.Therefore meeting point p+q can be either positive or negative which can't be decided by given clues.
Hence Answer E



Hi,
there is some info in statement 1, which can answer the Q..

Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

INFO from this:-


A)the intercept on x axis is (3,0) and y axis is (0,4)..
B) the slope is 4/3, which tells us that y increases at a higher rate than x..

Inference:-


1)Since the Line L is perpendicular to K, If its x-intercept is less than 3, the two lines will intersect above x-axis and if more than 3, it will intersect below X-axis..

lets see the statements


(1) x-intercept of Line L is less than that of Line K
see att fig, p+q will be always negative
a) in Quad III, both p and q will be -ive, so p+q will be NEGAIVE.

Suff..


(2) y-intercept of Line L is less than that of Line K
not suff
insuff..

ans A


The line that you drew in the picture is wrong.It has to meet x axis at -3.
And I would like someone to clarify if intercept is said to be less than -3,does that mean intercept of line should be between 3 and -3 considering the definition of intercept to be distance between origin and point where line meets axis.This is if we ignore the sign as it's the convention to denote the direction.


Hi,
Thanks for pointing out..
Firstly less than -3 should mean <-3 in literal sense..
now the point of intersection would be in III quad, where both x and y are -ive, so the SUM will always be -ive..
Director
Director
Joined: 05 Mar 2015
Posts: 852
Own Kudos [?]: 860 [0]
Given Kudos: 45
Send PM
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
Bunuel wrote:
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

(1) x-intercept of Line L is less than that of Line K
(2) y-intercept of Line L is less than that of Line K


Please refer fig. with above solution for better understanding
Attachments

coordinate (1).png
coordinate (1).png [ 12.68 KiB | Viewed 7963 times ]


Originally posted by rohit8865 on 10 Oct 2016, 10:21.
Last edited by rohit8865 on 23 Oct 2016, 08:09, edited 1 time in total.
Intern
Intern
Joined: 13 Apr 2015
Posts: 27
Own Kudos [?]: 15 [0]
Given Kudos: 2
Send PM
Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
Nevernevergiveup wrote:
Quote:
Line L is perpendicular to line K whose equation is 3y = 4x + 12; Lines L and K intersect at (p, q). Is p + q > 0?

(1) x-intercept of Line L is less than that of Line K
(2) y-intercept of Line L is less than that of Line K


Line K

Equation \(3y=4x+12\)

Point \((p,q)\) is on line L so \(3q=4p+12\)

X intercept is \(-3\)
Y intercept is \(4\)

Line L

Equation \(y-q=-\frac{3}{4}(x-p)\)
\(3x+4y=3p+4q\)

X intercept is \(\frac{3p+4q}{3}\)
Y intercept is \(\frac{3p+4q}{4}\)

Statement 1: x-intercept of Line L is less than that of Line K

\(\frac{3p+4q}{3}<-3\)

\(3p+4q<-9\)

Sum of p and q i.e., p+q is negative in all cases so sufficient.

Statement 2: y-intercept of Line L is less than that of Line K

\(\frac{3p+4q}{4}<4\)

\(3p+4q<16\)

Sum of p and q i.e., p+q can be both positive and negative hence statement 2 is insufficient.

A for me.



can you explain how did you deduce 3p+4q<-9[/m]

Sum of p and q i.e., p+q is negative in all cases so sufficient.
and
3p+4q<16[/m]

Sum of p and q i.e., p+q can be both positive and negative hence statement 2 is insufficient.
Director
Director
Joined: 16 Jun 2021
Posts: 994
Own Kudos [?]: 183 [0]
Given Kudos: 309
Send PM
Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
First of let us pen down the equation of the lines
k-3y = 4x + 12(given)
=>y=-3/4x+c since it's perpendicular to k
now we have to figure out whether p+q point of intersection of k and l that is p+q>0

state 1 provides :
c*4/3<-3 that is let us assume c=-10/4 <-9/4
gives us p+q<0 when substitued in L
sufficient

state 2 provides :
in a similar method of substitution provides c<-3
however when finding out for p and q provides both p+q>0
and p+q<0
hence Insuff

IMO A
Intern
Intern
Joined: 19 Sep 2022
Posts: 5
Own Kudos [?]: 1 [0]
Given Kudos: 29
Send PM
Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
Hi, please tell me what assumption that I am making is wrong here. I don't seem to need any of the statements to find a value for p and q. :-?

Can we not form the following equations based on the info given?
Info given in the question stem- Lines L and K intersect at point (p,q), so this pair should follow the equations of both Line K and line L
Line K equation is : y=4/3x +12
Line L equation b = -3/4a + k, where k is a constant
Now, since (p,q) satisfies both the equations,
q = 4/3p +12 = -3/4p + k
Using these three equations, we can find out p and k and q. We don't seem to need the Statements. Pls tell me where I am wrong. I am making an unfair assumption somewhere.

(how i did it was, using the last two terms of the equation in bold, i calculated the ratio of p and k. Then I put the value of k in terms of p in the equation of first and last terms and found out the ratio of p and q and finally, put this ratio in the first two terms and found out p =1.)


Thanks
GMAT Club Bot
Re: Line L is perpendicular to line K whose equation is 3y = 4x + 12; Line [#permalink]
Moderator:
Math Expert
92900 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne