Author 
Message 
TAGS:

Hide Tags

Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 505
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)

If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
10 Oct 2012, 13:26
4
This post was BOOKMARKED
Question Stats:
79% (01:41) correct 21% (01:47) wrong based on 178 sessions
HideShow timer Statistics
Manhattan weekly challenge oct 1st week, 2012 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 7/3
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
" Make more efforts " Press Kudos if you liked my post



Math Expert
Joined: 02 Sep 2009
Posts: 44600

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
10 Oct 2012, 13:39
2
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
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 7/3 We 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 505
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
10 Oct 2012, 14:37
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=0.11 which is between 0 and 0.33 (i.e 1/3) so, Answer: A
_________________
" Make more efforts " Press Kudos if you liked my post



Director
Status: Done with formalities.. and back..
Joined: 15 Sep 2012
Posts: 617
Location: India
Concentration: Strategy, General Management
Schools: Olin  Wash U  Class of 2015
WE: Information Technology (Computer Software)

Re: Distinct Range [#permalink]
Show Tags
28 Oct 2012, 01:27
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 7/3 Basically 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 >3 note, 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 <1/3 Hence ans A. Hope it helps
_________________
Lets Kudos!!! Black Friday Debrief



Current Student
Joined: 06 Sep 2013
Posts: 1919
Concentration: Finance

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
03 Jan 2014, 08:57
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=0.11
which is between 0 and 0.33 (i.e 1/3)
so, Answer: A Is that valid? They mentioned all variables different. I guess numerators as well as denominators I tried with 1/2 * 3/4 * 5/6 and 1/2 + 3/4 + 5/6 Got 3/20, of course answer A Cheers! J



Math Expert
Joined: 02 Sep 2009
Posts: 44600

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
03 Jan 2014, 09:11
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=0.11
which is between 0 and 0.33 (i.e 1/3)
so, Answer: A Is 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 + 5/6 Got 3/20, of course answer A Cheers! 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 09 Apr 2013
Posts: 137
Location: India
WE: Supply Chain Management (Consulting)

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
03 Jan 2014, 10:48
Got it right. +A I followed the same approach as did Bunuel.
_________________
+1 KUDOS is the best way to say thanks
"Pay attention to every detail"



Current Student
Joined: 06 Sep 2013
Posts: 1919
Concentration: Finance

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
18 Jan 2014, 15:28
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=0.11
which is between 0 and 0.33 (i.e 1/3)
so, Answer: A Is 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 + 5/6 Got 3/20, of course answer A Cheers! 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 ? Thanks Cheers! J



Math Expert
Joined: 02 Sep 2009
Posts: 44600

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
19 Jan 2014, 09:04



Current Student
Joined: 06 Sep 2013
Posts: 1919
Concentration: Finance

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
24 Feb 2014, 12:12
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 ? Thanks Cheers! 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 1/24 Sum will be 9/24 And therefore we will have 1/24 divided be 9/24 is equal to 1/9 Thus 1/0 belongs to the first range. Answer is A Hope its clear Cheers J



Current Student
Joined: 25 Sep 2012
Posts: 276
Location: India
Concentration: Strategy, Marketing
GMAT 1: 660 Q49 V31 GMAT 2: 680 Q48 V34

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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8028
Location: Pune, India

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
24 Feb 2014, 22:26
thevenus wrote: Manhattan weekly challenge oct 1st week, 2012
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 7/3 The 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.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Director
Joined: 23 Jan 2013
Posts: 591

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



Director
Joined: 13 Mar 2017
Posts: 591
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)

Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
Show Tags
06 Apr 2018, 02:29
thevenus wrote: Manhattan weekly challenge oct 1st week, 2012
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 7/3 xyz/(xy+yz+zx) = 1/(1/x+1/y+1/z) 0<x<1... 1<1/x<inf. 0<y<1... 1<1/y<inf. 0<z<1... 1<1/z<inf. 3<1/x+1/y+1/z<inf. So, 0<1/(1/x+1/y+1/z)<1/3 Answer A
_________________
CAT 99th percentiler : VA 97.27  DILR 96.84  QA 98.04  OA 98.95 UPSC Aspirants : Get my app UPSC Important News Reader from Play store.
MBA Social Network : WebMaggu
Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".




Re: If x, y, and z lie between 0 and 1 on the number line, with
[#permalink]
06 Apr 2018, 02:29






