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

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Intern
Joined: 12 May 2014
Posts: 2
If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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19 Aug 2017, 07:44
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75% (hard)

Question Stats:

58% (02:47) correct 42% (02:32) wrong based on 78 sessions

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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
Math Expert
Joined: 02 Aug 2009
Posts: 7955
If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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19 Aug 2017, 09:01
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5
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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##### General Discussion
Intern
Joined: 12 May 2014
Posts: 2
If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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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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Joined: 22 May 2016
Posts: 3536
If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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19 Aug 2017, 11:42
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1
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

*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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Intern
Joined: 14 Dec 2016
Posts: 1
Re: If x and y are positive integers and 1/x + 1/y = 1/9, what is  [#permalink]

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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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20 Feb 2019, 15:01
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Re: If x and y are positive integers and 1/x + 1/y = 1/9, what is   [#permalink] 20 Feb 2019, 15:01
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