Expert: Soroban - 5/1/2007

Question
You have 20 coins, 4 piles. The 1st pile is 4 more than the 2nd pile. The 2nd pile is 1 less than the 3rd pile and the 4th pile is twice as much as the 1st pile. How many coins are in each pile?

Answer
Hello, Mandy!

There must be a typo in the problem.
  The answers are not whole numbers.
I'll change the 20 to 28.

You have 28 coins, 4 piles.
The 1st pile is 4 more than the 2nd pile.
The 2nd pile is 1 less than the 3rd pile.
The 4th pile is twice the 1st pile.

Let's call the pile A, B, C, D (in that order).

Let x = number in pile C:  C = x

Pile B is has one less:  B = x - 1

Pile A has 4 more than B:  A = x + 3

Pile D has twice A:  D = 2(x + 3)


We are told that:  A + B + C + D  =  28

There is our equation!  (x + 3) + (x - 1) + x + 2(x + 3)  =  28

   and we have:  5x +8  =  28  �¨  5x = 20  �¨  x = 4


Therefore:
  A has 7 coins
  B has 3 coins
  C has 4 coins
  D has 14 coins
.