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10 Oct 2012, 13:26
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A
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Difficulty:
35%
(medium)
Question Stats:
73% (02:22) correct
27%
(02:27)
wrong
based on 753
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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
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
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.
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.
(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.
General Discussion
10 Oct 2012, 14:37
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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
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
