If x, y, and z lie between 0 and 1 on the number line, with

Question

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

Bunuel · Accepted Answer

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.

thevenus · Answer

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

Vips0000 · Answer

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

jlgdr · Answer

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

Bunuel · Answer

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.

cssk · Answer

Got it right. +A

I followed the same approach as did Bunuel.

jlgdr · Answer

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

Bunuel · Answer

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 .

jlgdr · Answer

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

b2bt · Answer

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.

KarishmaB · Answer

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.

Temurkhon · Answer

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

shashankism · Answer

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

AlphaBeta1988 · Answer

I just picked these sample numbers:

0.1
0.3
0.5

Adds up to 0.8

Then these sample numbers for 'sum of two of the variables

0.1
0.3

Adds up to 0.4

0.4 divided by 0.8 = 1/2

Arrived 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

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