Question

Given that a, b, c, d, p and q are natural numbers such that a < b and d < c. It is also given that the following result holds true:

$\frac{a}{b}\times \frac{d}{c}=\frac{q}{p}$

Choose the correct option(s):

• A. $\frac{a}{b}>\frac{q}{p}and\frac{d}{c}>\frac{q}{p}$
• B. $\frac{a}{b}<\frac{q}{p}and\frac{d}{c}<\frac{q}{p}$
• C. $q>p$
• D. $q<p$

Answer 1

Answer: (A), (D)

The correct options are
A

$\frac{a}{b}>\frac{q}{p}and\frac{d}{c}>\frac{q}{p}$

D

$q<p$

Given that,
$\frac{a}{b}\times \frac{d}{c}=\frac{q}{p}$
here, $\frac{a}{b}>\frac{q}{p}and\frac{d}{c}>\frac{q}{p}andq<p$
For eg,
$\frac{a}{b}\times \frac{d}{c}=\frac{q}{p}$
Let a = 2, b= 3 and c = 6, d = 5,
$\frac{2}{3}\times \frac{5}{6}=\frac{10}{18}$
here, $\frac{2}{3}>\frac{10}{18}(as\frac{2\times 6}{3\times 6}=\frac{12}{18}>\frac{10}{18})$
and $\frac{5}{6}>\frac{10}{18}(as\frac{5\times 3}{6\times 3}=\frac{15}{18}>\frac{10}{18})$
Here, q = 10 and p = 18.
Hence, $q<p$ $(as10<18)$.