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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