Chapter 4, Problem 16CT

To determine

To determine which fraction is greater using LCD $\frac{29}{50},\frac{61}{100}$

Answer to Problem 16CT

  $\frac{29}{50}<\frac{61}{100}$

Explanation of Solution

Given information:

Expression $\frac{29}{50},\frac{61}{100}$

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-

  $\frac{29}{50}=\frac{29}{5\times 5\times 2}$

For the other fraction, it is-

  $\frac{61}{100}=\frac{61}{5\times 5\times 2\times 2}$

In the above fractions, it is clear that the LCD is 50.

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

  $\frac{29}{50}=\frac{1}{50}\left(\frac{29}{1}\right)$

And the other one as-

  $\frac{61}{100}=\frac{1}{50}\left(\frac{61}{2}\right)$

Now our fractions are reduced to

  $\left(\frac{29}{1}\right)$ and $\left(\frac{61}{2}\right)$

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-

  $\left(\frac{29}{1}\right)\times \left(\frac{2}{2}\right)=\frac{58}{2}$

And for the second fraction

  $\left(\frac{61}{2}\right)\times \left(\frac{1}{1}\right)=\frac{61}{2}$

Now, both the fractions have same denominator 63.

  $\frac{58}{2}$ , $\frac{61}{2}$

Since 58 is less than 61, so the first fraction is less than the second one.

  $\frac{29}{50}<\frac{61}{100}$

Conclusion-

  $\frac{29}{50}<\frac{61}{100}$