AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 8 Easy

At a school, the number of students in Grade 7 and Grade 8 is the same, and the number of students in Grade 9 is \frac{4}{5} of that in Grade 8.

It is known that the number of boys in Grade 7 is equal to the number of girls in Grade 8, and that the number of boys in Grade 9 accounts for \frac{1}{4} of the total number of boys across all three grades.

What fraction of the total number of students across the three grades are girls?

  • A.

    \frac{9}{19}

  • B.

    \frac{10}{19}

  • C.

    \frac{11}{19}

  • D.

    \frac{11}{21}

  • E.

    \frac{10}{21}

Answer:D

Let the number of boys in Grade 7 be m and the number of girls be n.

Then, in Grade 8, there are n boys and m girls.

The total number of students in Grade 9 is \frac{4}{5}(m + n).

So, the total number of students across all three grades is: m + n + n + m + \frac{4}{5}(m + n) = \frac{14}{5}(m + n)

Let the number of boys in Grade 9 be x.

According to the problem: x = \frac{1}{4}(m + n + x)

Solving this gives: x = \frac{1}{3}(m + n)

So, the number of girls in Grade 9 is: \frac{4}{5}(m + n) - \frac{1}{3}(m + n) = \frac{7}{15}(m + n)

Thus, the total number of girls across all grades is:

n + m + \frac{7}{15}(m + n) = \frac{22}{15}(m + n)

The fraction of girls among all students is: \frac{\frac{22}{15}(m + n)}{\frac{14}{5}(m + n)} = \frac{11}{21}

Answer: C.

Related Problems
Problem 6 AMC 10 Weekly Practice Round 2
Problem 7 AMC 10 Weekly Practice Round 2
Problem 9 AMC 10 Weekly Practice Round 2
Problem 10 AMC 10 Weekly Practice Round 2
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