AMC 10 Weekly Practice Round 2

Complete problem set with solutions and individual problem pages

Problem 10 Easy

There are two bags of fruit. Take \frac{1}{5} of the fruit out of Bag A and 1103 grams of fruit out of Bag B, and the two bags of fruit will weigh the same. Now, if \frac{1}{2} of the remaining fruit in Bag B is taken out, the weight of the fruit that is still in Bag B will be \frac{1}{3} of the original weight of all the fruit in Bag B. In the first place, there are (   ) grams of fruit in the two bags.

  • A.

    6066.5

  • B.

    6068

  • C.

    6076.5

  • D.

    6079

  • E.

    6076

Answer:A

Let the original amount of fruit in Bag A and Bag B be a grams and b grams, respectively.

From Bag A, \frac{1}{5} was taken out, so the remaining amount is \left(1 - \frac{1}{5}\right)a = \frac{4}{5}a.

From Bag B, 1103 grams were taken out, so the remaining amount is

b - 1103.

Thus, \frac{4}{5}a = b - 1103.

Next, half of the remaining fruit in Bag B was taken out, so the remaining amount is \left(1 - \frac{1}{2}\right)(b - 1103) = \frac{1}{2}(b - 1103).

It is also given that this remaining amount equals \frac{1}{3}b: \frac{1}{2}(b - 1103) = \frac{1}{3}b.

Solving this equation: \frac{1}{2}b - \frac{1103}{2} = \frac{1}{3}b\Rightarrow \frac{1}{6}b = \frac{1103}{2}\Rightarrow b = 3309.

Substitute back: \frac{4}{5}a = b - 1103 = 3309 - 1103 = 2206 \Rightarrow a = \frac{2206 \times 5}{4} = 2757.5.

So the total original weight of the fruit is: a + b = 2757.5 + 3309 = 6066.5 \text{ grams}.

Answer: The original total weight of the fruit was 6066.5 grams.