2017 AMC 8

Complete problem set with solutions and individual problem pages

Problem 9 Easy

All of Marcy's marbles are blue, red, green, or yellow. One third of her marbles are blue, one fourth of them are red, and six of them are green. What is the smallest number of yellow marbles that Marcy could have?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    4

  • E.

    5

Answer:D

Solution 1

The 6 green marbles and yellow marbles form 1 - \frac{1}{3} - \frac{1}{4} = \frac{5}{12} of the total marbles. Now, suppose the total number of marbles is x. We know the number of yellow marbles is \frac{5}{12}x - 6 and a positive integer. Therefore, 12 must divide x. Trying the smallest multiples of 12 for x, we see that when x = 12, we get there are -1 yellow marbles, which is impossible. However when x = 24, there are \frac{5}{12} \cdot 24 - 6 = \boxed{\textbf{(D) }4} yellow marbles, which must be the smallest possible.

 

Solution 2

Since \frac{1}{3} of the marbles are blue and \frac{1}{4} are red, it is clear that the total number of marbles must be divisible by 12. If there are 12 marbles, then 4 are blue, 3 are red, and 6 are green, meaning that there are -1 yellow marbles. This is impossible. Trying the next multiple of 12, 24, we find that 8 are green, 6 are red, and 6 are green, meaning that the minimum number of yellow marbles is \boxed{\textbf{(D) }4}.