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

(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 ?

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

(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, 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, 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]
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: 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]
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

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

1

This post received KUDOS

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

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