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Lesson 1: Reason using benchmarks to compare two fractions on the number line. — Practice Questions

  1. 1. Which fraction is closer to the benchmark 1?

    • A. $\frac{4}{5}$
    • B. $\frac{9}{10}$
    • C. They are the same distance
    • D. Cannot be determined
  2. 2. When using the benchmark 1 to compare fractions, the distance of the fraction $\frac{12}{13}$ from 1 is ___.

  3. 3. Using the benchmark 1, which statement correctly compares $\frac{5}{6}$ and $\frac{7}{8}$?

    • A. $\frac{5}{6} > \frac{7}{8}$
    • B. $\frac{5}{6} < \frac{7}{8}$
    • C. $\frac{5}{6} = \frac{7}{8}$
    • D. The fractions are on opposite sides of 1
  4. 4. To compare $\frac{5}{12}$ and $\frac{1}{2}$, we find their difference. The distance between $\frac{5}{12}$ and the benchmark $\frac{1}{2}$ is ___.

  5. 5. Both $\frac{4}{7}$ and $\frac{5}{9}$ are greater than $\frac{1}{2}$. Which comparison is correct?

    • A. $\frac{4}{7} > \frac{5}{9}$
    • B. $\frac{4}{7} < \frac{5}{9}$
    • C. $\frac{4}{7} = \frac{5}{9}$
    • D. They are equidistant from $\frac{1}{2}$
  6. 6. Which fraction is larger, $\frac{2}{5}$ or $\frac{3}{7}$? Both are less than $\frac{1}{2}$.

    • A. $\frac{2}{5}$
    • B. $\frac{3}{7}$
    • C. They are equal
    • D. Cannot be compared
  7. 7. When comparing $\frac{8}{9}$ and $\frac{15}{16}$ using the benchmark 1, the fraction with the larger denominator, $\frac{15}{16}$, has a smaller distance to 1. That distance is ___.

  8. 8. Which symbol correctly compares the fractions $\frac{6}{7}$ and $\frac{10}{11}$?

    • A. >
    • B. <
    • C. =
    • D. ||
  9. 9. To compare $\frac{7}{12}$ and $\frac{3}{5}$ using $\frac{1}{2}$ as a benchmark, we find their distances. The distance of $\frac{7}{12}$ from $\frac{1}{2}$ is $\frac{1}{12}$. The distance of $\frac{3}{5}$ from $\frac{1}{2}$ is ___.

  10. 10. On a number line from 0 to 1, where would you place the fraction $\frac{4}{10}$?

    • A. Between 0 and $\frac{1}{2}$
    • B. Between $\frac{1}{2}$ and 1
    • C. Exactly at $\frac{1}{2}$
    • D. Exactly at 1