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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

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:

If x, y, and z lie between 0 and 1 on the number line, with [#permalink]

Show Tags

10 Oct 2012, 13:26

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

25% (medium)

Question Stats:

77% (02:43) correct
23% (02:15) wrong based on 122 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

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: \(\frac{xy+xz+yz}{xyz}\).

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 \(\frac{xyz}{xy+xz+yz}\).

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

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

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. _________________

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 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

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 ?

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. _________________

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. _________________

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

Time is a weird concept. It can stretch for seemingly forever (like when you are watching the “Time to destination” clock mid-flight) and it can compress and...