Question 71997
I have a purse with several coins in it. The coins have values of 1 cent, 
5 cents, 10 cents, and 25 cents. The average value of all coins in the purse
 is 17 cents. If I take out one 1 cent coin, the average value of all coins in
 the purse becomes 18 cents. How many of each coins are there in the purse?
 Can you please help?
:
{{{(1w + 5x + 10y + 25z)/(w + x + y + z)}}} = 17
:
{{{(1(w-1) + 5x + 10y + 25z)/((w-1) + x + y + z)}}} = 18
:
Note the difference in the two equations is 1:
Let t = total value of the coins originally:
Let c = number of coins number of coins originally:
{{{t/c)}}} = 17
 t = 17c
and the removal of 1 cent:
{{{(t-1)/(c-1)}}} = 18
t - 1 = 18(c-1)
t - 1 = 18c - 18
t = 18c - 18 + 1
t = 18c - 17
:
Substitute 17c for t and solve for c
17c = 18c - 17
17c - 18c = -17
-c = -17
c = 17 coins originally
:
Find t: t = 17c
t = 17*17
t = 289 cents, total value originally
:
After removing 1 penny:
288/16 = 18 cents, as the problem states
:
At least we know the total and the number of coins now:
1W + 5x + 10y + 25z = 289
:
From this we can see that there has to be at least 4 pennies, we have:
5x + 10y + 25z = 285; subtracted the 4 cents
:
We know there has to be at least 1 nickel then:
10y + 25z = 280; subtracted the nickel
:
We know there has to be at least 3 dimes then:
25z = 250; subtracted the value of 3 dimes
z = 250/25
z = 10 quarters + 3 dimes + 1 nickel + 4 pennies, that's 18 coins!!
But we found that there should only be 17 coins originally!
:
We can have 11 quarters, 2 nickels and 4 pennies, that 17 coins, however we
don't have any dimes so I am not sure if this is a solution or not
:
Check it anyway:
11 * 25 = 275
 2 * 5 = 10
 4 * 1 = 4
-----------
17 coins=289 cents
:
I'm not sure if there is a solution using dimes, perhaps this will help you anyway.
:
I would appreciate it, if you would let me know what you find out about this problem. ankor@dixie-net.com