Chapter 4, Problem 24MCQ

To determine

To determine which fraction is greater using LCD $\frac{5}{12},\frac{9}{20}$

Answer to Problem 24MCQ

  $\frac{5}{12}<\frac{9}{20}$

Explanation of Solution

Given information:

Expression $\frac{5}{12},\frac{9}{20}$

Formula Used: Least Common Denominator

Calculation:

The first step here is to identify the Least Common Denominator for both fractions. So, break the denominators into its multiples-

  $\frac{5}{12}=\frac{5}{3\times 2\times 2}$

For the other fraction, it is-

  $\frac{9}{20}=\frac{9}{5\times 2\times 2}$

In the above fractions, it is clear that the LCD is $2\times 2$ which is 4.

In the second step, extract the LCD, and rewrite the fractions as-

  $\frac{5}{12}=\frac{1}{4}\left(\frac{5}{3}\right)$

And the other one as-

  $\frac{9}{20}=\frac{1}{4}\left(\frac{9}{5}\right)$

Now our fractions are reduced to

  $\left(\frac{5}{3}\right)$ and $\left(\frac{9}{5}\right)$

now make the denominators of both fractions same by multiplying each fractions denominator, with the numerator & denominator of the other fraction.

So for first fraction-

  $\left(\frac{5\times 5}{3\times 5}\right)=\frac{25}{15}$

And for the second fraction

  $\left(\frac{9\times 3}{5\times 3}\right)=\frac{27}{15}$

Now, both the fractions have same denominator 12.

  $\frac{25}{15}$ , $\frac{27}{15}$

Since 25 is less than 27, so the first fraction is less than the second one.

  $\frac{5}{12}<\frac{9}{20}$

Conclusion-

  $\frac{5}{12}<\frac{9}{20}$