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

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

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10 Oct 2012, 13:26
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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
[Reveal] Spoiler: OA

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

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10 Oct 2012, 13:39
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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.

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

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24 Feb 2014, 22:26
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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.
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Director
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Kudos [?]: 290 [0], given: 75

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

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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)

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

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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)

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

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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)

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

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03 Jan 2014, 10:48
Got it right. +A

I followed the same approach as did Bunuel.
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Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]

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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)

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 ?

Thanks
Cheers!
J
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Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]

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

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

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

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

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