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If x and y are positive integers and 1/x + 1/y = 1/9, what is

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If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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New post 19 Aug 2017, 07:44
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If x and y are positive integers and 1/x + 1/y = 1/9, what is the difference between the maximum and minimum value of x ?

a) 40
b) 60
c) 80
d) 100
e) 120
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If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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New post 19 Aug 2017, 09:01
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rakibgmat wrote:
If x and y are positive integers and 1/x + 1/y = 1/9, what is the difference between the maximum and minimum value of x ?

a) 40
b) 60
c) 80
d) 100
e) 120


Hi..

\(\frac{1}{x}+\frac{1}{y}=\frac{1}{9}........\frac{1}{x}=\frac{1}{9}-\frac{1}{y}=\frac{y-9}{9y}\)...
so \(x=\frac{9y}{y-9}\)...
when the denominator is the least the fraction is the max and so will x be max..
y is positive and x is positive so y-9 will be least as 1..... so y-9=1..y=10
substitute y as 10, then 11, you will realize that as you increase y, x decreases..


so x is MAX at y =10 .....\(x=\frac{9y}{y-9}=\frac{9*10}{10-9}=90\)
when will x be MIN, when y is max... x will be min when y is 90, and so x is 10 then..

difference = 90-10=80

C
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If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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New post 19 Aug 2017, 10:03
Thnx for ur great solution..How could u determine that x will be minimum when y is 90...why not the other integers? or is it generally opposite of 1st condition (when y is 10, x will be maximum)? Or is their any method to determine that?chetan2u
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If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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New post 19 Aug 2017, 11:42
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rakibgmat wrote:
If x and y are positive integers and 1/x + 1/y = 1/9, what is the difference between the maximum and minimum value of x ?

a) 40
b) 60
c) 80
d) 100
e) 120

If \(\frac{1}{x}\) + \(\frac{1}{y}\) = \(\frac{1}{9}\), and x and y are positive integers, to find maximum and minimum for x:

We need maximum and minimum difference between \(\frac{1}{x}\) and \(\frac{1}{y}\). So we need the largest \(\frac{1}{x}\) possible.*

The next largest fraction (with prompt's restrictions re integers) less than \(\frac{1}{9}\) is \(\frac{1}{10}\). So minimum possible value for x is 10.

If x = 10 and \(\frac{1}{x}\) = \(\frac{1}{10}\), find \(\frac{1}{y}\).

\(\frac{1}{9}\) - \(\frac{1}{10}\) = \(\frac{1}{90}\)

\(\frac{1}{90}\) = \(\frac{1}{y}\)

Because \(\frac{1}{10}\) is the largest possible addend under these conditions, the difference between \(\frac{1}{9}\) and \(\frac{1}{10}\) will be the smallest addend --

namely, \(\frac{1}{90}\)

But those addends' order doesn't matter.

\(\frac{1}{x}\) + \(\frac{1}{y}\) =
\(\frac{1}{y}\) + \(\frac{1}{x}\)

They sum to \(\frac{1}{9}\).

So just switch \(\frac{1}{x}\) and \(\frac{1}{y}\) from previous calculations.

Let \(\frac{1}{x}\) be \(\frac{1}{90}\).

\(\frac{1}{9}\) - \(\frac{1}{90}\) = \(\frac{1}{10}\) =\(\frac{1}{y}\)

\(\frac{1}{x}\) = \(\frac{1}{90}\): maximum value of x is 90.

Max\(_{x}\) - Min\(_{x}\) = (90 - 10) = 80

Answer C

*If A + B = 10, where A and B are positive integers, and we want the maximum and minimum values of A, we find the difference between the possible extremes: 10 - 9 = 1. So max for A = 9, min = 1. Same idea with fractions where the integer restriction actually allows you to find the extremes.
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Re: If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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New post 25 Aug 2017, 03:10
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This is how we can figure out the minimum value of x.
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Re: If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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