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.

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 ==============================================

Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 12:53

16

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: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 20:44

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 ==============================================

Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 12:06

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 _________________

Mission: Be a force of good and make a positive difference to every life I touch!

Re: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 13:27

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]
21 Dec 2011, 09:34

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

(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: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 11:13

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: distance in rectangular coordinate system [#permalink]
02 Jun 2009, 17:23

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 ?

(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: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 11:45

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: distance in rectangular coordinate system [#permalink]
05 Feb 2011, 13:32

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: distance in rectangular coordinate system [#permalink]
19 Feb 2011, 18:12

Entwistle 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

I liked your explanation, of course Bunuel was as helpful as ever too.

Re: In the rectangular coordinate system, are the points (a, [#permalink]
27 Jul 2012, 20:49

Hi there, i was just wondering if the way i do it is correct!

Statement 1: insuff Statement 2 : |a|+|b|=|c|+|d|

Our goal is to prove that a^2 + b^2 = c^2 + d^2

(1+2)

Square both sides in stmt 2.

We have a^2 + b^2 + 2|a||b| = c^2 + d^2 + 2|c||d| ----------- * From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2|a||b| from LHS and 2|c||d| from equation *.

Re: In the rectangular coordinate system, are the points (a, [#permalink]
08 Aug 2012, 02:22

reagan wrote:

Hi there, i was just wondering if the way i do it is correct!

Statement 1: insuff Statement 2 : |a|+|b|=|c|+|d|

Our goal is to prove that a^2 + b^2 = c^2 + d^2

(1+2)

Square both sides in stmt 2.

We have a^2 + b^2 + 2|a||b| = c^2 + d^2 + 2|c||d| ----------- * From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2|a||b| from LHS and 2|c||d| from equation *.

Please tell me that i am correct! =)

Reagan

Absolutely, I found this approach much better. Infact I am wondering why we are targetting absolute values in the equation.

Re: In the rectangular coordinate system, are the points (a, [#permalink]
01 Nov 2013, 03:40

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________