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Re: ManhttanGMAT Practice CAT
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04 Jun 2012, 19:39
kunalbh19 wrote: 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)
Squaring both sides
\(a^2 + d^2  2\sqrt{a^2*d^2} = c^2 + b^2  2\sqrt{b^2*c^2}\)
from statement (1) we know that ad=bc > a^2*d^2 = b^2*c^2
Cancelling last term of both sides, we get a^2 + d^2 = c^2 + b^2 Thus, a^2  b^2 = c^2  d^2
Not able to figure out where I went wrong. Please suggest!
Thanks
There is nothing wrong with your approach. The relation you have got holds (a^2  b^2 = c^2  d^2). But if you start doing algebraic manipulations on the starting point without an eye on what you need to achieve at the end, you may not obtain the result you want. You can manipulate an expression in many ways to get seemingly different results. Notice that Bunuel got a^2 = c^2 and b^2 = d^2. He could have chosen to add them as a^2 + d^2 = c^2 + b^2 (or subtract them) which is same as your result. But he chose to add them as a^2 + b^2 = c^2 + d^2 to get the expression he desired. Using the same two expressions, Bunuel arrived at a^2 + b^2 = c^2 + d^2 and you arrived at a^2  b^2 = c^2  d^2, both of which are correct. But only one of them (a^2 + b^2 = c^2 + d^2) helps you answer the question directly since you know that the distance of a point (a, b) is given by square root of (a^2 + b^2). Try to find the relation between individual variables instead of working on the equations as a whole. Either use number plugging or Bunuel's algebraic approach.
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Re: In the rectangular coordinate system, are the points (a,
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27 Jul 2012, 21:49
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 + 2ab = c^2 + d^2 + 2cd  * From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2ab from LHS and 2cd from equation *.
Please tell me that i am correct! =)
Reagan



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Re: In the rectangular coordinate system, are the points (a,
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08 Aug 2012, 03:22
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 + 2ab = c^2 + d^2 + 2cd  * From one we know that a/b=c/d, therefore their LHS=RHS and therefore, this condition would allow us to cancel out 2ab from LHS and 2cd 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. (a+b)^2 = a^2+b^2+2ab [no absolute a, b needed] From 1) we know ab = cd 2) we know a + b = c + d. and hence (a+b)^2 = (c+d)^2 Combining 1) & 2) we can very well see that a^2+b^2 = c^2+d^2



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Re: In the rectangular coordinate system, are the points (a, b)
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15 Jan 2013, 08:45
1+2: 1) Ensures that the lines joining each of the two points to origin have same slope. 2) Ensures that absicca and ordinates correspondingly have equal magnitudes. {If one line has points (a,b) then the other line will have coordinates (ak,bk) but (2) ensures that k = 1.} Hence C. KUDOS IF YOU LIKE!
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Re: ManhttanGMAT Practice CAT
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22 Apr 2013, 12:55
Bunuel wrote: nsp007 wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
(2) \(\sqrt{a^2}+ \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)
Will post OA later. 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 nonzero \(x\). 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. Hi Bunnel i am not understanding how \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) > \(a+b=c+d\) have come?



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Re: ManhttanGMAT Practice CAT
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22 Apr 2013, 22:00
mun23 wrote: \(\sqrt{a^2}+\sqrt{b^2}=\sqrt{c^2} +\sqrt{d^2}\) > \(a+b=c+d\) have come? By convention, only positive roots are considered (at least in GMAT) \(\sqrt{4} = 2\) (and not 2) Similarly, \(\sqrt{a^2} = a\) i.e. only the positive value
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Re: coordinate geometry
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15 Aug 2013, 19:29
sudharsansuski wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin? (1) a/b = c/d (2) (a^2)^0.5 + (b^2)^0.5 = (c^2)^0.5 + (d^2)^0.5 I didnt understand the answer explanation given by MGMAT. Could someone please help. Origin on coordinate system is (0,0). The question is if distance between (0,0) to (a,b) is same as distance between (0,0) to (c,d) Case 1:  a/b = c/d let a=7, b= 14 then a/b = 1/2. So c/d fraction has to be 1/2. i.e. c can be 2 and d can be 4. or c can be 6 & d can be 12 or c can be 7 & d can be 14. So distance between origin can either be same to (c,d) or different from (a,b). Clearly insufficient Case 2: (a^2)^0.5 + (b^2)^0.5 = (c^2)^0.5 + (d^2)^0.5 translates to "a + b = c + d" let (a,b) = (4,4) and (c,d) = (4,4). Then it satisfies the condition a+b = c+d. Also distance from origin to (a,b) is same as (c,d). let (a,b) = (5,3) and (c,d) = (4,4). Then it satisfies the condition a+b = c+d. And distance from origin to (a,b) is not same as (c,d). Insufficient. Lets take 1 & 2 together we have a/b = c/d.......so a= bc/d Eq (1) we also have a + b = c + d substitute a=bc/d from Eq(1) bc/d + b = c + d b(c/d + 1) = c + d b(c + d) = d(c+d) so b = d similarly we get a=b. so taking 1 & 2 together, the distance between origin to both points are equal.
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Re: In the rectangular coordinate system, are the points (a, b)
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16 Aug 2013, 03:33
sudharsansuski wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin? (1) a/b = c/d (2) (a^2)^0.5 + (b^2)^0.5 = (c^2)^0.5 + (d^2)^0.5 I didnt understand the answer explanation given by MGMAT. Could someone please help. basically we need to prove \(a^2+b^2 = c^2+d^2\) stmnt 1: \(\frac{a}{b} = \frac{c}{d}\) 1/2 = 3/6 ==>using this we can prove NO 1/2 = 1/2 ==>using these numbers we can prove yes hence insufficient statement 2: \(\sqrt{(a^2)} + \sqrt{(b^2)} = \sqrt{(c^2)} + \sqrt{(d^2)}\)or\(a + b = c + d\) putting \(a=b=c=d=1\)..we will get YES. Putting \(a= 1, b=2, c= 0, d=3\)...we will get NO. HENCE INSUFFICIENT. combining. from 2nd statement: \(a + b = c + d\) rearrange: \(a  d = c  b\) and then square both sides. \(a^2 + d^2  2*a*d = c^2 + b^2  2*c*b.\) Since \(ad = bc\) as per statement 1 we can cancel few things: \(a^2 + d^2  2*a*d = c^2 + b^2  2*c*b.\) hence sufficient
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Re: In the rectangular coordinate system, are the points (a, b)
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16 Aug 2013, 09:13
nsp007 wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) \(\frac{a}{b}=\frac{c}{d}\)
(2) \(\sqrt{a^2}+ \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\) 1. adbc=0 means that the vector pointing to (c,d) is a scalar multiple of the vector pointing to (a,b). In other words they are collinear with the origin. If they are equidistant to the origin, the scalar will be 1 or 1. But given this information it could be any scalar. Not sufficient. 2. a+b=c+d. This means that (a,b) and (c,d) both lay on some square centered at the origin. But they could be anywhere on the square. Not sufficient. 1 and 2. (a,b) and (c,d) are collinear with the origin and both lay on a square centered at the origin. Therefore they must be the same distance from the origin. Sufficient. C



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Re: In the rectangular coordinate system, are the points (a,
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02 Jun 2014, 06:06
(Don't read my "solution" I am studying and got this one wrong. ) The first statement is telling me that the ratio between the x and y value are the same for both points. This could be (1,3) or (10,30) where the second one is obviously further away from the origin. Hence insufficient. Statement (2) I cannot fully understand intuitively. So I'll plug in some numbers. 0,1 =1 1/4,1/4 = 1/2+1/2=1 distance from origin in the second case = (1/2)^2+(1/2)^2=(2/4) square root of (1/2) =/= 1. So it differs. However, by using the same numbers I could make it work. So I can rule out both and go with E. Wrong answer above, will have a look at the solutions provided by other users



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Re: In the rectangular coordinate system, are the points (a, b)
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31 Jul 2014, 05:33
This was my view:
You see immediately that both statements alone are not sufficient. So start evaluating the combination of both:
Statement A says that that the points are on the same line through the origin. Statement B says that the different "possible" points all have an equal distance to the xaxis, but also to the yaxis. Therefore, combining both options say that either (a,b) = (c,d) or (a,b) = (c, d). So equidistant from origin.



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Re: In the rectangular coordinate system, are the points (a,
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11 Aug 2014, 00:37
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 nonzero \(x\). 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. Bunuel, how did you quickly determine that (2) was Not Sufficient?



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Re: In the rectangular coordinate system, are the points (a,
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12 Aug 2014, 08:09
TooLong150 wrote: 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 nonzero \(x\). 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. Bunuel, how did you quickly determine that (2) was Not Sufficient? You can do this with number plugging. The question asks whether \(a^2+b^2=c^2+d^2\) and (2) says that \(a+b=c+d\). If a = b = c = d = 0, then the answer would be YES but if a = 0, b = 2, c = 1 and d = 1, then the answer would be NO.
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Re: In the rectangular coordinate system, are the points (a, b)
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06 Jan 2015, 23:12
Bunuel Sir,
I Didn't understand this:
b/a=a/c>bc=ad>c/a=a/b given a/b=c/d=cx/dx
(as the ratios are equal then there exist some for which d=cx and b=dx)



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Re: In the rectangular coordinate system, are the points (a, b)
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Re: In the rectangular coordinate system, are the points (a,
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28 Apr 2015, 02:42
mbaMission wrote: In the rectangular coordinate system, are the points \((a, b)\) and \((c, d)\) equidistant from the origin? (1) \(\frac{a}{b} =\frac{c}{d}\) (2) \(\sqrt{a^2} + \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\) this question is pretty weid. is this question from gmatprep?
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Re: In the rectangular coordinate system, are the points (a,
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Re: In the rectangular coordinate system, are the points (a,
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15 May 2015, 01:56
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 nonzero \(x\). 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. the key point here is to know that a=xc this is simple but tricky wonderful explanation of this problem . thank you , Bunnu
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Re: In the rectangular coordinate system, are the points (a,
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26 May 2018, 01:40
Bunuel wrote: nsp007 wrote: In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?
(1) a/b = c/d
(2) \(\sqrt{a^2}+ \sqrt{b^2} = \sqrt{c^2} + \sqrt{d^2}\)
Will post OA later. 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 nonzero \(x\). 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. Hi y'all, I'm a newbie here. can you please review my method? I was disagreeing that only (2) is not sufficient. I use Trigonometry to solve (2). The (x,y) must make a right triangle with xaxis. Thus, the distance between the dot and the origin must be the hypotenuse. And (2) seems to be the equation for the hypotenuse. If the 2 triangle are equal, then the distance must be equal. Or am I missing something...?



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In the rectangular coordinate system, are the points (a,
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22 Aug 2018, 09:11
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