Chapter 4.4, Problem 41E

To determine

To find the LCD

Answer to Problem 41E

  $\frac{11}{15}$ is greater than $\frac{7}{10}$ .

Explanation of Solution

Given information: $\frac{7}{10},\frac{11}{15}$

Formula Used: Factorisation

Calculation:

Convert these fractions to equivalent fractions with common denominator in order to compare them more easily.

LCM of 10 and 15 is 30

So ,denominator of both fractions 30.

  $\frac{7\times 3}{10\times 3}=\frac{15}{30}$

  $\frac{11\times 2}{15\times 2}=\frac{22}{30}$

  $\frac{22}{30}>\frac{15}{30}$

So, $\frac{11}{15}$ is greater than $\frac{7}{10}$ .