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If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
 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
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Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
10 Oct 2012, 13:39
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.
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Re: If x, y, and z lie between 0 and 1 on the number line, with [#permalink]
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
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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
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Re: Distinct Range [#permalink]
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: Distinct Range [#permalink]
29 Oct 2012, 03:22
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Re: Distinct Range
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29 Oct 2012, 03:22
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