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 Author: thevenus [ 10 Oct 2012, 13:26 ] Post subject: If x, y, and z lie between 0 and 1 on the number line, with Manhattan weekly challenge oct 1st week, 2012If x, y, and z lie between 0 and 1 on the number line, with no two variables equal, then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between(A) 0 and 1/3(B) 1/3 and 2/3(C) 2/3 and 1(D) 1 and 5/3(E) 5/3 and 7/3

 Author: Bunuel [ 10 Oct 2012, 13:39 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with If x, y, and z lie between 0 and 1 on the number line, with no two variables equal, then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between(A) 0 and 1/3(B) 1/3 and 2/3(C) 2/3 and 1(D) 1 and 5/3(E) 5/3 and 7/3We need to find the value of $$\frac{xyz}{xy+xz+yz}$$.Consider the reciprocal of this fraction: $$\frac{xy+xz+yz}{xyz}$$.Split it: $$\frac{xy+xz+yz}{xyz}=\frac{xy}{xyz}+\frac{xz}{xyz}+\frac{yz}{xyz}=\frac{1}{z}+\frac{1}{y}+\frac{1}{x}$$.Now, since all variables are between 0 and 1, then all reciprocals $$\frac{1}{z}$$, $$\frac{1}{y}$$ and $$\frac{1}{x}$$, are more than 1, thus $$\frac{1}{z}+\frac{1}{y}+\frac{1}{x}$$ is more than 3. Which means that our initial fraction is between 0 and 1/3.For example if $$\frac{xy+xz+yz}{xyz}=\frac{1}{z}+\frac{1}{y}+\frac{1}{x}$$ is 4 (so more than 3), then $$\frac{xyz}{xy+xz+yz}$$ is 1/4 which is between 0 and 1/3.Answer: A.Of course one can also assign some values to x, y, and z and directly calculate $$\frac{xyz}{xy+xz+yz}$$.Hope it's clear.

 Author: thevenus [ 10 Oct 2012, 14:37 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with I chose the smart numbers and went off as follows :-let x=1/2, y=1/3,z=1/4 ( all between 0 and 1)now , as per the question ; xyz/(xy+yz+zx) = (1/2)x(1/3)x(1/4)/{(1/2)x(1/3)+(1/3)x(1/4)+(1/4)x(1/2)}=1/9=0.11which is between 0 and 0.33 (i.e 1/3)so, Answer: A

 Author: Vips0000 [ 28 Oct 2012, 01:27 ] Post subject: Re: Distinct Range Pansi wrote:If a, b & c are distinct variables on the number line that are between 0&1. Then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between(A) 0 and 1/3(B) 1/3 and 2/3(C) 2/3 and 1(D) 1 and 5/3(E) 5/3 and 7/3Basically we want value of abc/(ab+bc+ca)Lets denote it as V for ease.$$V=abc/(ab+bc+ca)$$$$=> V = 1/((ab/abc)+(bc/abc)+(ca/abc))$$=>$$V = 1/((1/c)+(1/a)+(1/b))$$=> $$V = 1/N$$ , where N >3note, since a,b and c are each less than one, therefore 1/a, 1/b and 1/c each are more than 1. (eg. 0.5 is less than 1 so 1/0.5 =2 is greater than 1)Therefore the denominator of ((1/c)+(1/a)+(1/b)) is a number N greater than 3. If we are dividing 1 by 3 result is 1/3; if we divide 1 by a number greater than 3 , result would be less than 1/3=> V <1/3Hence ans A.Hope it helps

 Author: jlgdr [ 03 Jan 2014, 08:57 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with thevenus wrote:I chose the smart numbers and went off as follows :-let x=1/2, y=1/3,z=1/4 ( all between 0 and 1)now , as per the question ; xyz/(xy+yz+zx) = (1/2)x(1/3)x(1/4)/{(1/2)x(1/3)+(1/3)x(1/4)+(1/4)x(1/2)}=1/9=0.11which is between 0 and 0.33 (i.e 1/3)so, Answer: AIs that valid? They mentioned all variables different. I guess numerators as well as denominatorsI tried with 1/2 * 3/4 * 5/6 and 1/2 + 3/4 + 5/6Got 3/20, of course answer ACheers!J

 Author: Bunuel [ 03 Jan 2014, 09:11 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with jlgdr wrote:thevenus wrote:I chose the smart numbers and went off as follows :-let x=1/2, y=1/3,z=1/4 ( all between 0 and 1)now , as per the question ; xyz/(xy+yz+zx) = (1/2)x(1/3)x(1/4)/{(1/2)x(1/3)+(1/3)x(1/4)+(1/4)x(1/2)}=1/9=0.11which is between 0 and 0.33 (i.e 1/3)so, Answer: AIs that valid? They mentioned all variables different. I guess numerators as well as denominatorsI tried with 1/2 * 3/4 * 5/6 and 1/2 + 3/4 + 5/6Got 3/20, of course answer ACheers!J We are simply told that x, y, and z are distinct numbers, it's not necessary denominators and numerators to be distinct. For example x can be 1/2 and y can be 1/3.

 Author: cssk [ 03 Jan 2014, 10:48 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with Got it right. +AI followed the same approach as did Bunuel.

 Author: jlgdr [ 18 Jan 2014, 15:28 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with Bunuel wrote:jlgdr wrote:thevenus wrote:I chose the smart numbers and went off as follows :-let x=1/2, y=1/3,z=1/4 ( all between 0 and 1)now , as per the question ; xyz/(xy+yz+zx) = (1/2)x(1/3)x(1/4)/{(1/2)x(1/3)+(1/3)x(1/4)+(1/4)x(1/2)}=1/9=0.11which is between 0 and 0.33 (i.e 1/3)so, Answer: AIs that valid? They mentioned all variables different. I guess numerators as well as denominatorsI tried with 1/2 * 3/4 * 5/6 and 1/2 + 3/4 + 5/6Got 3/20, of course answer ACheers!J We are simply told that x, y, and z are distinct numbers, it's not necessary denominators and numerators to be distinct. For example x can be 1/2 and y can be 1/3.Well OK, but in the problem they mentioned that it is the sum of only two of the variables not the three of them. Am I understanding well ?ThanksCheers!J

 Author: Bunuel [ 19 Jan 2014, 09:04 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with jlgdr wrote:Well OK, but in the problem they mentioned that it is the sum of only two of the variables not the three of them. Am I understanding well ?ThanksCheers!J Not sure I understand your question...We are asked to find the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables.

 Author: jlgdr [ 24 Feb 2014, 12:12 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with Bunuel wrote:jlgdr wrote:Well OK, but in the problem they mentioned that it is the sum of only two of the variables not the three of them. Am I understanding well ?ThanksCheers!J Not sure I understand your question...We are asked to find the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables.Oh ok so then I guess I can choose 1/2, 1/3 and 1/4 then product will be 1/24Sum will be 9/24And therefore we will have 1/24 divided be 9/24 is equal to 1/9Thus 1/0 belongs to the first range. Answer is AHope its clearCheersJ

 Author: b2bt [ 24 Feb 2014, 20:51 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with Why are we deal with fractions here?Just take the numbers are --> 0.2, 0.3, 0.4You can convert them later into fractions very conveniently. If you are good with fraction conversion you might not even need to do that.

 Author: KarishmaB [ 24 Feb 2014, 22:26 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with thevenus wrote:Manhattan weekly challenge oct 1st week, 2012If x, y, and z lie between 0 and 1 on the number line, with no two variables equal, then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between(A) 0 and 1/3(B) 1/3 and 2/3(C) 2/3 and 1(D) 1 and 5/3(E) 5/3 and 7/3The question doesn't clarify whether 'between 0 and 1' is inclusive or not so I would assume that it doesn't matter whether you include them. If you do include them, you can take the numbers as 0, 1/2 and 1 and straight away get the answer as 0 since the product of the three numbers will be 0. This automatically gives us the range (A).On the other hand, if you want to be extra careful, you can assume the numbers to be 0.00000000000000001, 1/2, .99999999999 i.e. very close to 0, 1/2 and very close to 1. The product of these three will also be very close to 0 and when you divide it by approximately 1/2, you will still get something very close to 0. Hence the range will be (A) only.

 Author: Temurkhon [ 01 Oct 2014, 04:54 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with I got 0.5, 0.1, 0.2product=0.01sum of possible pairs=0.05+0.02+0.1=0.170.01/0.17=1/17A

 Author: shashankism [ 06 Apr 2018, 02:29 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with thevenus wrote:Manhattan weekly challenge oct 1st week, 2012If x, y, and z lie between 0 and 1 on the number line, with no two variables equal, then the product of all three variables divided by the sum of all the distinct products of exactly two of the three variables is between(A) 0 and 1/3(B) 1/3 and 2/3(C) 2/3 and 1(D) 1 and 5/3(E) 5/3 and 7/3xyz/(xy+yz+zx)= 1/(1/x+1/y+1/z)0

 Author: AlphaBeta1988 [ 28 Aug 2019, 16:09 ] Post subject: If x, y, and z lie between 0 and 1 on the number line, with I just picked these sample numbers:0.10.30.5Adds up to 0.8Then these sample numbers for 'sum of two of the variables0.10.3Adds up to 0.40.4 divided by 0.8 = 1/2Arrived at A like everyone else.Is there anything wrong with this approach? If these problems can reliably be done this way then I'd rather stick with it than use an algebraic approach

 Author: bumpbot [ 21 Jan 2023, 05:46 ] Post subject: Re: If x, y, and z lie between 0 and 1 on the number line, with 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.

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