AMC 10 Daily Practice Round 4

Complete problem set with solutions and individual problem pages

Problem 20 Medium

In a candy distribution scenario, three individuals, A, B, and C, each have a positive integer number of candies. If A gives B 20 candies, B's candy count will be twice the sum of A and C's candy counts. If A gives C 30 candies, C's candy count will be three times the sum of A and B's candy counts. How many candies do A, B, and C have in total?

  • A.

    48

  • B.

    60

  • C.

    80

  • D.

    90

  • E.

    100

Answer:A

Let x, y, and z represent the numbers of candies for A, B, and C, respectively. From the given conditions, we have the following system of equations:

\begin{cases} y+20 = 2(x-20+z) \quad (1) \\ z+30 = 3(x-30+y) \quad (2) \end{cases}

Solving and simplifying, we get:

\begin{cases} 2x+2z-y = 60 \quad (3) \\ 3x+3y-z = 120 \quad (4) \end{cases}

From equation (3), we have y=2x+2z-60. Substituting this into equation (4), we get:

3x+3(2x+2z-60)-z = 120

Hence, 9x+5z=300. Since x and z are positive integers, we have:

9 \leq 9x \leq 300

Therefore, 1 \leq x \leq 33\frac{1}{3}. After checking, we find that x=30 satisfies the equation, and this implies that 5z=30. Thus, z=6 and y=2 \times 30 + 2 \times 6 - 60 = 12.

So, the number of candies for A, B, and C respectively are:

A: 30 candies B: 12 candies C: 6 candies

The total number of candies for A, B, and C is 30+12+6 = 48.

Therefore, the answer is \boxed{\textbf{(A) } 48}.

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