SOLUTION: 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 on

Algebra.Com
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

RELATED QUESTIONS

A coin purse contains 5 cents, 10 cents and 25 cents coins. The number of 10 cents-coins... (answered by ikleyn)
A has P5.35 in 5 cents,10 cents and 25 cents. The number of 10 cents coins is one more... (answered by josgarithmetic)
she has 1.10, she has three times as many 5-cents coins as 1 cent coins, and the number... (answered by greenestamps,josgarithmetic)
Eric has a box of coins. The box currently contains 40 coins, consisting of 5 cents, 10... (answered by Alan3354)
David has a total of 4.10 pesos in 5 cent,10 cents and 25 cents coins.He has twice as... (answered by stanbon)
A purse contains a collection of quarters, dimes, nickels, and pennies. The average value (answered by josmiceli)
A purse contains a collection of quarters, dimes, nickels, and pennies. The average value (answered by scott8148)
Dan has 5.35 pesos in 5 cents, 10 cents and 25 cents. The number of 25 centavo coins is... (answered by stanbon)
Paula can have pennies, nickels, dimes, and quarters in her purse. The average value of... (answered by josmiceli)