Question

Professor Burgus compared two fractions, but made mistakes in three of the comparisons. Choose the option that is correct.

• A. $\frac{3}{4}>\frac{6}{7}$
• B. $\frac{6}{5}<\frac{8}{9}$
• C. $\frac{5}{7}<\frac{7}{9}$
• D. $\frac{1}{9}<\frac{2}{21}$

Answer 1

The correct option is C $\frac{5}{7}<\frac{7}{9}$

Let's apply the cross multiplication method to the fractions.
If we solve each and every option, we will find that only option, $\frac{5}{7}<\frac{7}{9}$, is correct.
$\frac{5}{7}<\frac{7}{9}\to \frac{5\times 9}{7\times 9}<\frac{7\times 7}{9\times 7}\frac{45}{63}<\frac{49}{63}$
Since these are like fractions, we can compare the numerators to get the answer.
We know that 45 < 49.
Therefore, option c is correct.