If 8 5/x 16, and 10 y 25, what is the smallest possible value

From the question, we have
(i) 8 ≤ \(\frac{5}{x}\) ≤ 16
(ii) 10 ≤ y ≤ 25

To find: Minimum possible value of \(\frac{x}{y}\)

In order to minimize a fraction, the numerator should be minimized and the denominator should be maximized.
i.e., \(\frac{x}{y}\) will be minimum when x is minimum and y is maximum.

Let us try and deduce the minimum value of x and the maximum value of y.

We have,

8 ≤ \(\frac{5}{x}\) ≤ 16
Here the fraction is greater than 8, which is greater than 0. If \(\frac{5}{x}\) is greater than 0, then x is clearly a positive value. (If x is negative, then the fraction will also be negative which objects the condition)

We can multiply the entire inequality by \(\frac{x}{5}\). Since \(\frac{x}{5}\) is positive, the inequality won't be altered.

8 * \(\frac{x}{5}\) ≤ \(\frac{5}{x}\) * \(\frac{x}{5}\) ≤ 16 * \(\frac{x}{5}\)

\(\frac{8x}{5}\) ≤ 1 ≤ \(\frac{16x}{5}\)

We need the minimum value of x, so taking the second half of the inequality (Since the first half will give us only the maximum limit for x)

1 ≤ \(\frac{16x}{5}\)

1 * \(\frac{5}{16}\) ≤ \(\frac{16x}{5}\) * \(\frac{5}{8}\) (Multiplying by \(\frac{5}{16}\))

\(\frac{5}{16}\) ≤ x.
We get, x should be greater than or equal to \(\frac{5}{16}\)

Thus minimum value x can take is \(\frac{5}{16}\)

From the question, we have 10 ≤ y ≤ 25
Thus maximum value y can take is 25.

The minimum possible value of \(\frac{x}{y}\) = \(\frac{5}{16}\)\(\frac{1}{25}\) (Substituting x, y)

= \(\frac{5}{16*25}\)

= \(\frac{1}{80}\)

Answer is Choice-C \(\frac{1}{80}\)