Question

use a benchmark to compare fractions — quiz — level d

1. use the benchmark \\(\frac{1}{2}\\) to compare \\(\frac{5}{8}\\) and \\(\frac{2}{7}\\).

first, compare each fraction to \\(\frac{1}{2}\\).
\\(\frac{5}{8}\\) is? \\(\frac{1}{2}\\),
\\(\frac{2}{7}\\) is? \\(\frac{1}{2}\\)

Explanation:

Step1: Compare $\frac{5}{8}$ and $\frac{1}{2}$

To compare $\frac{5}{8}$ and $\frac{1}{2}$, we can find a common denominator. The common denominator of 8 and 2 is 8. So, $\frac{1}{2}=\frac{4}{8}$. Now, compare $\frac{5}{8}$ and $\frac{4}{8}$. Since $5 > 4$, we have $\frac{5}{8}>\frac{4}{8}$, which means $\frac{5}{8}>\frac{1}{2}$.

Step2: Compare $\frac{2}{7}$ and $\frac{1}{2}$

To compare $\frac{2}{7}$ and $\frac{1}{2}$, we find a common denominator. The common denominator of 7 and 2 is 14. So, $\frac{1}{2}=\frac{7}{14}$ and $\frac{2}{7}=\frac{4}{14}$. Now, compare $\frac{4}{14}$ and $\frac{7}{14}$. Since $4 < 7$, we have $\frac{4}{14}<\frac{7}{14}$, which means $\frac{2}{7}<\frac{1}{2}$.

Answer:

$\frac{5}{8}$ is greater than $\frac{1}{2}$, $\frac{2}{7}$ is less than $\frac{1}{2}$ (or in boxed form for the comparison symbols: $\frac{5}{8}>\frac{1}{2}$, $\frac{2}{7}<\frac{1}{2}$)