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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

============================================== Do not answer without sharing the reasoning behind ur choice ----------------------------------------------------------- Working on my weakness : GMAT Verbal ------------------------------------------------------------ Ask: Why, What, How, When, Where, Who ==============================================

Last edited by Bunuel on 28 Apr 2015, 04:31, edited 1 time in total.

(1) If a/b = c/d, a and b could be 100 and c and d could be 1 or a, b, c and d could be 1. NSF.

(2) If (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2), then a+b = c+d. In this case, if a = 9 and b = 1, and c =d=5, a+b = c+d is true. but they have different distance.NSF.

From 1 and 2, a must be equal to c and b must be equal to d. So Suff.

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

02 Jun 2009, 12:13

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

This singnifies that the sign combination on both sides of = is the same. i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well. However since a,b / c,d can take any value , INSUFFICIENT.

(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) Since we do not know anything about signs of a,b / c,d - INSUFFICIENT.

Using 1 and 2 together , using (1) the second statement drills down to a+b=c+d (the sign combination on both sides of = is the same.)

Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point. _________________

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

02 Jun 2009, 21:44

4

This post received KUDOS

goldeneagle94 wrote:

amolsk11 wrote:

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

(1) a/b = c/d

This singnifies that the sign combination on both sides of = is the same. i.e. if one of the numbers on LHS is -ve , one of the numbers on RHS has to -ve as well. However since a,b / c,d can take any value , INSUFFICIENT.

(2) (a^2)^(1/2) + (b^2)^(1/2) = (c^2)^(1/2) + (d^2)^(1/2) Since we do not know anything about signs of a,b / c,d - INSUFFICIENT.

Using 1 and 2 together , using (1) the second statement drills down to a+b=c+d (the sign combination on both sides of = is the same.)

Since a/b=c/d and a+b=c+d , (a, b) and (c, d) equidistant from the origin and infact represent the same point.

Is there a way we can derive, using these two equations, that the two points are equidistant ?

therefore points are equidistant from (0,0) or Origin. _________________

============================================== Do not answer without sharing the reasoning behind ur choice ----------------------------------------------------------- Working on my weakness : GMAT Verbal ------------------------------------------------------------ Ask: Why, What, How, When, Where, Who ==============================================

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Bunuel, had condition (2) simply said a+b=c+d (instead of the squares and square root) how would have the answer changed?

Also in the current question - is a^2+b^2=c^2+d^2? When I square both sides of (2), I get a^2+b^2+2sqrt(a^2b^2)=c^2+d^2+2sqrt(c^2d^2) so if ab=cd then this is satisfied. however (1) only gives me ad=bc, how do I infer ab=cd from that? I am following a different process, but I should end up with the same answer. Not sure where am I wrong? _________________

I have a question here bunuel. How did you get a = c * x and b= d*x??

Because from this it means that x = b/d = a/c

Can you please explain??

It's the same: \(\frac{b}{d} = \frac{a}{c}\) --> \(bc=ad\)--> \(\frac{c}{d}=\frac{a}{b}\).

Given: \(\frac{a}{b}=\frac{c}{d}=\frac{cx}{dx}\) (as the ratios are equal then there exist some \(x\) for which \(a=cx\) and \(b=dx\)). _________________

(1) knowing these proportions does not help me solve it, because for example if 3/1 = 9/3 , point (a,b) will be closer to the origin than point (b,c)

(2) this statement tells that |a|+|b|=|c|+|d| , which is still not sufficient because we lack information about the correlation between |a| and |b|,and |c|and|d|.

But if we combine the two statements together we will have this correlation from statement (1) and then both statements taken together will be sufficient.

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

a. \(a/b = c/d\)

b. \(\sqrt{a^2}+\sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)

Distance of \((x,y)\) from origin is \(\sqrt{x^2+y^2}\) So we need to answer \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\) ?

(1) a/b=c/d ... doesnt really help in proving or disproving. Insufficient

(2) This is equivalent to saying \(|a|+|b|=|c|+|d|\). Again insufficient to say anything about the statement we have.

(1+2) a/c=b/d=x say (needs, c,d to be non zero)

a = cx b = dx

|a|+|b|=|c|+|d| |cx| - |c| = |d| - |dx| |c|(|x|-1)=|d|(1-|x|) (|c|+|d|)(|x|-1)=0 Since c,d are non-zero means |x|=1 So either a=c & b=d OR a=-c or b=-d Either case a^2=c^2 and b^2=d^2 Hence \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\) Sufficient

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. C. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. D. EACH statement ALONE is sufficient. E. Statements (1) and (2) TOGETHER are NOT sufficient.

You can take values to solve this question quickly:

Statement 1: a/b = c/d (a,b) and (c,d) may be equidistant from the origin e.g. (1, 1) and (-1, -1) or they may not be e.g. (1, 1) and (2, 2). Not sufficient.

Statement 2: \(\sqrt{a^2}+\sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\) That is, |a|+|b|=|c|+|d| (a,b) and (c,d) may be equidistant from the origin e.g. (1, 3) and (-1, -3) or they may not be e.g. (1, 3) and (2, 2)

Using both together, |a|+|b|=|c|+|d| and a/b = c/d. This means that if a/c = 1/2, c/d cannot be 2/4 or -3/-6 etc. c/d has to be either 1/2 or (-1)/(-2). Similarly, if a/b = (-1)/2, c/d = (-1)/2 or 1/(-2)) Any pair of such points will be equidistant. Answer (C). _________________

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

05 Feb 2011, 12:45

Official explanation says: "If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

05 Feb 2011, 13:06

3

This post received KUDOS

1

This post was BOOKMARKED

tinki wrote:

Official explanation says: "If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses

a/b=c/d <---- Equation 1

|a|+|b|=|c|+|d|

Add 1 to both sides of Equation 1

(a+b)/b=(c+d)/d

Now a+b=c+d => 1/b=1/d

Thus, b=d Similarly, b/a=d/c Adding 1 to both sides of above equation: (b+a)/a=(d+c)/c a+b=d+c => 1/a=1/c Therefore a=c _________________

"Wherever you go, go with all your heart" - Confucius

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

05 Feb 2011, 13:53

17

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

tinki wrote:

Official explanation says: "If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

05 Feb 2011, 14:27

1

This post received KUDOS

Bunuel wrote:

tinki wrote:

Official explanation says: "If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Answer: C.

your explanation is great as always ! thaaanks + Kudo

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

05 Feb 2011, 14:32

Expert's post

AmrithS wrote:

tinki wrote:

Official explanation says: "If we know the proportion of a to b is the same as c to d and that |a| + |b| = |c| + |d|, then it must be the case that |a| = |c| and |b| = |d| ?

could someone elaborate how we are supposed to know |a| = |c| and |b| = |d|? a bit vague statement for me

thanks for responses

a/b=c/d <---- Equation 1

|a|+|b|=|c|+|d|

Add 1 to both sides of Equation 1

(a+b)/b=(c+d)/d

Now a+b=c+d => 1/b=1/d

Thus, b=d Similarly, b/a=d/c Adding 1 to both sides of above equation: (b+a)/a=(d+c)/c a+b=d+c => 1/a=1/c Therefore a=c

No that's not correct.

|a|+|b|=|c|+|d| doesn't mean that a+b=c+d (consider a=b=1 and c=d=-1). So from |a|+|b|=|c|+|d| and a/b=c/d we can not derive that a=c and b=d. What we can derive is that |a|=|c| and |b|=|d|. Refer to my post above for complete solution. _________________

Re: In the rectangular coordinate system, are the points (a, [#permalink]

Show Tags

21 Dec 2011, 10:34

1

This post received KUDOS

Here's a simpler solution......

We need to prove a^2 + d^2 = c^2 + b^2

Simpler way is to rearrange statement 2 viz. |a| + |b| = |c| + |d| as 1. |a| - |d| = |c| - |b| and then square both sides. We get - 2. a^2 + d^2 - 2*|a|*|d| = c^2 + b^2 - 2*|c|*|b|.

Since ad = bc as per statement 1,

3. |ad| = |bc| => |a|.|d| = |b|.|c| (Rule -> Abs of Product = Product of Abs)

So we can cancel the third term out from both sides of equation 2. to get the desired equation

In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?

Distance between the point A (x,y) and the origin can be found by the formula: \(D=\sqrt{x^2+y^2}\).

So we are asked whether \(\sqrt{a^2+b^2}=\sqrt{c^2+d^2}\)? Or whether \(a^2+b^2=c^2+d^2\)?

(1) \(\frac{a}{b}=\frac{c}{d}\) --> \(a=cx\) and \(b=dx\), for some non-zero \(x\). Not sufficient.

(2) \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) --> \(|a|+|b|=|c|+|d|\). Not sufficient.

(1)+(2) From (1) \(a=cx\) and \(b=dx\), substitute this in (2): \(|cx|+|dx|=|c|+|d|\) --> \(|x|(|c|+|d|)=|c|+|d|\) --> \(|x|=1\) (another solution \(|c|+|d|=0\) is not possible as \(d\) in (1) given in denominator and can not be zero, so \(d\neq{0}\) --> \(|c|+|d|>0\)) --> now, as \(|x|=1\) and \(a=cx\) and \(b=dx\), then \(|a|=|c|\) and \(|b|=|d|\) --> square this equations: \(a^2=c^2\) and \(b^2=d^2\) --> add them: \(a^2+b^2=c^2+d^2\). Sufficient.

Answer: C.

I am getting a different final result for answer (C). Here is my approach: \(\sqrt{a^2} - \sqrt{d^2} = \sqrt{c^2} - \sqrt{b^2}\) ----from statement (2)

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

A few weeks ago, the following tweet popped up in my timeline. thanks @Uber_Mumbai for showing me what #daylightrobbery means!I know I have a choice not to use it...

“This elective will be most relevant to learn innovative methodologies in digital marketing in a place which is the origin for major marketing companies.” This was the crux...