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 coin are there in the purse? Can you please help?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! 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?
:
 = 17
:
 = 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:
 = 17
t = 17c
and the removal of 1 cent:
 = 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
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