Question

use a benchmark to compare fractions — instruction — level d

1. use the benchmark \\(\frac{1}{2}\\) to compare \\(\frac{2}{9}\\) and \\(\frac{9}{10}\\).

(there is a number line with 0, \\(\frac{1}{2}\\), 1 marked, and \\(\frac{2}{9}\\) and \\(\frac{9}{10}\\) plotted on it)

1. now, compare \\(\frac{2}{9}\\) and \\(\frac{9}{10}\\).

\\(\frac{2}{9}\\) ? \\(\frac{9}{10}\\) (with options <, =, >)

Explanation:

Step1: Compare $\frac{2}{9}$ with $\frac{1}{2}$

To compare $\frac{2}{9}$ and $\frac{1}{2}$, we can cross - multiply. $2\times2 = 4$ and $9\times1=9$. Since $4\lt9$, $\frac{2}{9}\lt\frac{1}{2}$.

Step2: Compare $\frac{9}{10}$ with $\frac{1}{2}$

Cross - multiply $\frac{9}{10}$ and $\frac{1}{2}$. $9\times2 = 18$ and $10\times1 = 10$. Since $18\gt10$, $\frac{9}{10}\gt\frac{1}{2}$.

Step3: Compare $\frac{2}{9}$ and $\frac{9}{10}$

If $\frac{2}{9}\lt\frac{1}{2}$ and $\frac{9}{10}\gt\frac{1}{2}$, then $\frac{2}{9}\lt\frac{9}{10}$.

Answer:

$\frac{2}{9}\lt\frac{9}{10}$