AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 11 Medium

There are two groups of numbers. The average of the first group is 12.8, and the average of the second group is 10.2. The overall average of the two groups combined is 12.02. What is the ratio of the number of elements in the second group to that in the first group?

  • A.

    \frac{3}{7}

  • B.

    \frac{7}{3}

  • C.

    \frac{2}{7}

  • D.

    \frac{7}{2}

  • E.

    \frac{2}{3}

Answer:A

There are a numbers in the first group and b numbers in the second group.

According to the problem,

(12.8a + 10.2b) \div (a + b) = 12.02.

Simplifying gives:

0.78a = 1.82b,

which leads to:

b = \frac{3}{7}a.

Therefore, the number of elements in the second group is \frac{3}{7} of the number in the first group.

Related Problems
Problem 9 AMC 10 Weekly Practice Round 2
Problem 10 AMC 10 Weekly Practice Round 2
Problem 12 AMC 10 Weekly Practice Round 2
Problem 13 AMC 10 Weekly Practice Round 2
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