The product of two fractions is $\left(\frac{25}{26}\right)$ and their quotient is $\left(\frac{13}{32}\right)$, find the larger fraction?

The correct option is B: $\left(\frac{15}{13}\right)$

Let the two fractions be $\left(\frac{a}{b}\right)$ and $\left(\frac{c}{d}\right)$

$\left(\frac{a}{b}\right)\times \left(\frac{c}{d}\right)=\left(\frac{25}{26}\right)\dots \dots \dots \left(i\right)$ [Product $=\frac{25}{26}$ ]

$(\frac{a}{b}÷\frac{c}{d})=\left(\frac{13}{32}\right)$ [Quotient $=\frac{13}{32}$]

$(\frac{a}{b}\times \frac{d}{c})=\left(\frac{13}{32}\right)\dots \dots \dots \left(ii\right)$

Multiplying $eq.\left(i\right)$ and $\left(ii\right)$ we get

$\frac{a}{b}\times \frac{c}{d}\times \frac{a}{b}\times \frac{d}{c}=\left(\frac{25}{26}\right)\left(\frac{13}{32}\right)$

${\left(\frac{a}{b}\right)}^2=\left(\frac{25}{64}\right)$

$\left(\frac{a}{b}\right)=+\left(\frac{5}{6}\right)$ or $-\left(\frac{5}{6}\right)$

Since, $\left(\frac{a}{b}\right)$ is a fraction therefore, $\left(\frac{a}{b}\right)=\frac{5}{6}$

Substituting the above value in $\left(i\right)$ we get

$+\left(\frac{5}{6}\right)\times \left(\frac{c}{d}\right)=\left(\frac{25}{26}\right)$

$\left(\frac{c}{d}\right)=\left(\frac{25}{26}\right)\times \left(\frac{6}{5}\right)$

$\left(\frac{c}{d}\right)=+\left(\frac{15}{13}\right)$

Now, we have to compare $\left(\frac{5}{6}\right)$ and $\left(\frac{15}{13}\right)$ to find the bigger fraction.

$+\left(\frac{5\times 13}{6\times 13}\right)=\frac{65}{78}$

$+\left(\frac{15\times 6}{13\times 6}\right)=\frac{90}{78}$

$\frac{90}{78}>\frac{65}{78}$

so, $\left(\frac{15}{13}\right)>\left(\frac{5}{6}\right)$

So the larger of the two is $\left(\frac{15}{13}\right)$