Summer Problem Solving Marathon Solution #19

Let the amount of money paid by the four kids be a, b, c, and d, respectively. Then we are given a + b + c + d = 60.

The first kid paid half as much as the other three combined. So a = (b + c + d)/2. Substituting this into the original equation, we have:

(b + c + d)/2 + b + c + d = 60

Or:

(b + c + d)3/2 = 60

b + c + d = 40.

Hence a = 20.

Next we are told that the second kid paid one-third as much as the other kids combined. So b = (a + c + d)/3 = (20 + c + d)/3. Substituting this into b + c + d = 40, we have:

(20 + c + d)/3 + c + d = 40

Or (multiplying both sides by 3):

20 + c + d + 3c + 3d = 120

4c + 4d = 100

c + d = 25.

Hence b = 15.

Next, we are told that the third kid paid one-fourth as much as the other kids combined. So c = (a + b + d)/4 = (35 + d)/4. Substituting this into c + d = 25, we have:

(35 + d)/4 + d = 25

35 + d + 4d = 100

35 + 5d = 100

5d = 65

d = 13

So the amount of money paid by the fourth kid is $13.