AMC 8 Daily Practice Round 8

Complete problem set with solutions and individual problem pages

Problem 24 Medium

In a kindergarten, there are three classes. Class A has 4 more students than Class B, and Class B has 4 more students than Class C. When distributing dates to the children: Each child in Class A receives 3 fewer dates than each child in Class B; each child in Class B receives 5 fewer dates than each child in Class C. As a result: Class A, in total, receives 3 more dates than Class B; Class B, in total, receives 5 more dates than Class C. How many dates were distributed to the three classes in total?

  • A.

    670

  • B.

    673

  • C.

    676

  • D.

    680

  • E.

    682

Answer:B

Let Class C have x children. From the given conditions:

Each child in Class A receives 3 + 5 = 8 fewer dates than each child in Class C.

Thus, x children in Class A receive 8x fewer dates than x children in Class C.

Since Class A in total receives 8 more dates than Class C, and Class A has 8 more children than Class C, these extra 8 children together receive (8x + 8) dates.

Therefore, each child in Class A receives \frac{8x + 8}{8} = x + 1 dates.

Similarly, x children in Class B receive 5x fewer dates than x children in Class C.

Thus, each child in Class B receives \frac{5x + 5}{4} dates.

We then have the equation:

x + 1 + 3 = \frac{5x + 5}{4}

\frac{5}{4}x - x = 4 - \frac{5}{4}

Solving gives x = 11.

Therefore:

Class A has 11 + 8 = 19 children, each receiving 11 + 1 = 12 dates.

Class B has 11 + 4 = 15 children, each receiving 12 + 3 = 15 dates.

Class C has 11 children, each receiving 15 + 5 = 20 dates.

So, the total number of dates distributed is:

12 \times 19 + 15 \times 15 + 20 \times 11 = 673

A total of 673 dates were distributed to the three classes.