DS #500 Test 19 q - 18 - The Beat The GMAT Forum - Expert GMAT Help & MBA Admissions Advice

DS #500 Test 19 q - 18

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f2001290: Master | Next Rank: 500 Posts; Posts: 400; Joined: Sat Mar 10, 2007 4:04 amFollowed by: 1 member; Thanked: 1 times

by f2001290 » Mon May 14, 2007 3:32 am

18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 – r and v = 1 – s

Please Explain

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Source: Beat The GMAT — Data Sufficiency | Mon May 14, 2007 3:32 am

gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member; Location: India; Thanked: 51 times

Re: DS #500 Test 19 q - 18

by gabriel » Mon May 14, 2007 10:59 am

f2001290 wrote:18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 – r and v = 1 – s

Please Explain

… if we join the origin to the points (r,s) and (u,v) .. then we have two triangles both of them wuld be right angle triangles … and the hypotenuse of the triangles wuld be the distance from the center … and the sides for the 2 triangle wuld be r, s for the first one and u,v for the second one ..

so what the q basically asks is if r^2+s^2 = u^2+v^2 …

.. the first statement says that r+s=1 .. which does not really help .. so insufficient ..

… from the second statement we have
u^2= 1+r^2-2r
v^2=1+s^2-2s …
adding both of them we have u^2+v^2=2+r^2+s^2-2(r+s) .. so we have
(u^2+v^2) - (r^2+s^2) = 2 – 2(r+s) now remember if the 2 points were equidistant then (u^2+v^2) – (r^2+s^2) wuld be equal to 0 .. but for it to be 0 (r+s) has to be equal to 1 but we don’t have anything in the second statement that says so .. so second statement is also insufficient

now combine both the statement and from the first statement we have r+s=1 … and by substituting this in the eqn (u^2+v^2)-(r^2+s^2)=2-2(r+s) .. we get (u^2+v^2)-(r^2+s^2)=0 .. and that means the two points are equidistant from the origin .. so the answer is C .. post if any doubts ..

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mendiratta: Junior | Next Rank: 30 Posts; Posts: 26; Joined: Tue May 08, 2007 8:22 pm; Thanked: 1 times

Re: DS #500 Test 19 q - 18

by mendiratta » Mon May 14, 2007 8:33 pm

gabriel wrote:
f2001290 wrote:18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 – r and v = 1 – s

Please Explain
… if we join the origin to the points (r,s) and (u,v) .. then we have two triangles both of them wuld be right angle triangles … and the hypotenuse of the triangles wuld be the distance from the center … and the sides for the 2 triangle wuld be r, s for the first one and u,v for the second one ..

so what the q basically asks is if r^2+s^2 = u^2+v^2 …

.. the first statement says that r+s=1 .. which does not really help .. so insufficient ..

… from the second statement we have
u^2= 1+r^2-2r
v^2=1+s^2-2s …
adding both of them we have u^2+v^2=2+r^2+s^2-2(r+s) .. so we have
(u^2+v^2) - (r^2+s^2) = 2 – 2(r+s) now remember if the 2 points were equidistant then (u^2+v^2) – (r^2+s^2) wuld be equal to 0 .. but for it to be 0 (r+s) has to be equal to 1 but we don’t have anything in the second statement that says so .. so second statement is also insufficient

now combine both the statement and from the first statement we have r+s=1 … and by substituting this in the eqn (u^2+v^2)-(r^2+s^2)=2-2(r+s) .. we get (u^2+v^2)-(r^2+s^2)=0 .. and that means the two points are equidistant from the origin .. so the answer is C .. post if any doubts ..

Agree with C. - Excellent post gabriel.

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abkhan: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Thu May 03, 2007 2:20 am; Location: India

by abkhan » Tue May 15, 2007 9:57 am

an excellent explanation. but sometime's its faster to try and reason it out with a random set of values and see if the conditions are satisified for example let (r,s)=(4,3) and (u,v)=(3,4) C gets satisfied. But dont depend on only one value take more sets

I am ..therefore I am..

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gabriel: Legendary Member; Posts: 986; Joined: Wed Dec 20, 2006 11:07 amFollowed by: 1 member; Location: India; Thanked: 51 times

by gabriel » Tue May 22, 2007 7:40 am

abkhan wrote:an excellent explanation. but sometime's its faster to try and reason it out with a random set of values and see if the conditions are satisified for example let (r,s)=(4,3) and (u,v)=(3,4) C gets satisfied. But dont depend on only one value take more sets

... hmmm .. i dont know abt that .. first of the vlues u have chosen do not match the condiiton given .. second of ... somehow i am not in favour of using ' random values" for a DS .. especially tricky ones ... nywayz watever works for u ..

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Cybermusings: Legendary Member; Posts: 559; Joined: Tue Mar 27, 2007 1:29 amFollowed by: 2 members; Thanked: 5 times

by Cybermusings » Tue May 22, 2007 10:59 pm

thanks a lot gabriel...great explanation

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gmatmachoman: Legendary Member; Posts: 2326; Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members; Thanked: 173 times; GMAT Score:710

by gmatmachoman » Sat Apr 03, 2010 11:01 pm

gabriel wrote:
f2001290 wrote:18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 ï¿½ r and v = 1 ï¿½ s

Please Explain
ï¿½ if we join the origin to the points (r,s) and (u,v) .. then we have two triangles both of them wuld be right angle triangles ï¿½ and the hypotenuse of the triangles wuld be the distance from the center ï¿½ and the sides for the 2 triangle wuld be r, s for the first one and u,v for the second one ..

.

Can anyone explain how 2 triangles are formed??

R&D on GMAT algorithm:

https://www.beatthegmat.com/r-d-on-gmat- ... tml#305739

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eaakbari: Master | Next Rank: 500 Posts; Posts: 435; Joined: Mon Mar 15, 2010 6:15 amFollowed by: 1 member; Thanked: 32 times

by eaakbari » Sat Apr 03, 2010 11:10 pm

Instead of using the two triangle thing which is a little confusing

Think about the distance formula. If 2 points are equidistant , obviously the distance is the same

D = sqrt ( (x-x1)^2 - (y-y1)^2 )

When one point is origin, hence
d = sqrt (x^2 - y^2)

And which gives
r^2 + s^2 = u^2 + v^2

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Sun Apr 04, 2010 12:50 am

18. In the rectangular coordinate system, are the points (r, s) and (u, v) equidistant from the origin ?
(1) r + s = 1
(2) u = 1 - r and v = 1 - s

If I such substitute (1) is each of the 2 eq. of (2)
u = 1-r but from (1) r+s = 1 so
u = r+s-r = s
v = 1-s = r+s-s = r
so (r,s) and (u,v) are same points
somethings funny

The points are actually on a circle with origin (0,0)

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