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Given x,y,z are distinct positive real numbers
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Updated on: 25 Sep 2016, 03:23
Updated on: 25 Sep 2016, 03:23
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Question Stats:
42% (01:06) correct
57% (01:51) wrong based on 19 sessions
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Given x,y,z are distinct positive real numbers
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
(x²(y+z)+y²(x+z)+z²(x+y))/(xyz) |
6 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Originally posted by Sonalika42 on 23 Sep 2016, 09:10.
Last edited by Sonalika42 on 25 Sep 2016, 03:23, edited 2 times in total.
Last edited by Sonalika42 on 25 Sep 2016, 03:23, edited 2 times in total.
Re: Given x,y,z are distinct positive real numbers
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23 Sep 2016, 20:40
23 Sep 2016, 20:40
D is the answer
Posted from my mobile device
Posted from my mobile device
Re: Given x,y,z are distinct positive real numbers
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24 Sep 2016, 06:22
24 Sep 2016, 06:22
Expert Reply
The best way to approach this problem is by substitution.
Quantity A is \(\frac{(x^2(y+z)+y^2(x+z)+z^2(x+y))}{(xyz)}\)
or \(x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{x}+\frac{1}{z})+z(\frac{1}{x}+\frac{1}{y})\)
or \(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}\)
rearrange
\(\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{z}\).
now Let \(\frac{x}{y}\) = n. Then the first two terms become \(n + \frac{1}{n}\). The minimum value of \(n + \frac{1}{n}\) is 2 when n =1. Or and n can be one if x=y.
Hence a case x=y=z= 5( or any other real positive number) is not possible. Hence Quantity A is always greater
for x=1 y=1 z=2
\(\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{2}{1}+\frac{2}{1}+\frac{1}{2}\)
\(1+1+0.5+2+2+0.5\) =\(7\). Now Quantity A is 7.
Hence A is the correct option.
Quantity A is \(\frac{(x^2(y+z)+y^2(x+z)+z^2(x+y))}{(xyz)}\)
or \(x(\frac{1}{y}+\frac{1}{z})+y(\frac{1}{x}+\frac{1}{z})+z(\frac{1}{x}+\frac{1}{y})\)
or \(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{z}{x}\)
rearrange
\(\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{z}{y}+\frac{y}{z}\).
now Let \(\frac{x}{y}\) = n. Then the first two terms become \(n + \frac{1}{n}\). The minimum value of \(n + \frac{1}{n}\) is 2 when n =1. Or and n can be one if x=y.
Hence a case x=y=z= 5( or any other real positive number) is not possible. Hence Quantity A is always greater
for x=1 y=1 z=2
\(\frac{1}{1}+\frac{1}{1}+\frac{1}{2}+\frac{2}{1}+\frac{2}{1}+\frac{1}{2}\)
\(1+1+0.5+2+2+0.5\) =\(7\). Now Quantity A is 7.
Hence A is the correct option.
