Question

$\frac{5}{9}<\frac{p}{q}<\frac{4}{7}$
Find $q-p$ such that $q$ takes the least value.

Answer 1

Probability

Given = $\frac{5}{9}<\frac{P}{q}<\frac{4}{7}$

using componendo Divider do Rule :

$\frac{5+9}{5-9}<\frac{p+q}{p-q}<\frac{4+7}{4-7}$

$\frac{-14}{4}<\frac{p+q}{p-q}<\frac{-11}{3}$

$\frac{-7}{2}<\frac{p+q}{p-q}<\frac{-11}{3}$

$\frac{7}{2}<\frac{p+q}{q-p}<\frac{11}{3}$

$\frac{2}{7}>\frac{p+q-p}{p+q}>\frac{3}{11}$____(1)

Since we know, $p+q=1$

given, q takes least value

$q=0$

$p=1$

putting in (1)

$\frac{2}{7}>q-p>\frac{3}{11}$