Chapter 4, Problem 15CT

To determine

To determine which fraction is greater using LCD $\frac{3}{35},\frac{7}{45}$

Answer to Problem 15CT

  $\frac{3}{35}<\frac{7}{45}$

Explanation of Solution

Given information:

Expression $\frac{3}{35},\frac{7}{45}$

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-

  $\frac{3}{35}=\frac{3}{7\times 5}$

For the other fraction, it is-

  $\frac{7}{45}=\frac{7}{5\times 3\times 3}$

In the above fractions, it is clear that the LCD is 5.

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

  $\frac{3}{35}=\frac{1}{5}\left(\frac{3}{7}\right)$

And the other one as-

  $\frac{7}{45}=\frac{1}{5}\left(\frac{7}{9}\right)$

Now our fractions are reduced to

  $\left(\frac{3}{7}\right)$ and $\left(\frac{7}{9}\right)$

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-

  $\left(\frac{3}{7}\right)\times \left(\frac{9}{9}\right)=\frac{27}{63}$

And for the second fraction

  $\left(\frac{7}{9}\right)\times \left(\frac{7}{7}\right)=\frac{49}{63}$

Now, both the fractions have same denominator 63.

  $\frac{27}{63}$ , $\frac{49}{63}$

Since 27 is less than 49, so the first fraction is less than the second one.

  $\frac{3}{35}<\frac{7}{45}$

Conclusion-

  $\frac{3}{35}<\frac{7}{45}$