SOLUTION: there are pennies, nickels , dimes , and quarters in a piggy bank. if there are 14 coins in the bank and their value is $1.05

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Question 1033082: there are pennies, nickels , dimes , and quarters in a piggy bank. if there are 14 coins in the bank and their value is $1.05


Found 2 solutions by ikleyn, josgarithmetic:
Answer by ikleyn(52805) About Me  (Show Source):
You can put this solution on YOUR website!
.
there are pennies, nickels , dimes , and quarters in a piggy bank. if there are 14 coins in the bank and their value is $1.05
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Although the condition is discontinued by an obvious way, I can suggest that 

1) all necessary data is presented,  and
2) the question is "how many coins of each value are there in the bank?"


Answer. 2 quarters, 3 dimes, 4 nickels and 5 pennies.

Check.  2*25 + 3*10 + 4*5 + 5*1 = 105,
        2    + 3    + 4   + 5    = 14.


Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Four unknown variables and two equations. The easiest restriction is on how many quarters, being some natural number from 1 to 4.

p, n, d, q the count of each coin.
0.01p%2B0.05n%2B0.1d%2B0.25q=1.05
p%2B5n%2B10d%2B25q=105 from account of the money value

p%2Bn%2Bd%2Bq=14 the account of the coins


Solve each for p and equate.
system%28p=105-5n-10d-25q%2Cp=14-n-d-q%29
105-5n-10d-25q=14-n-d-q
-5n-10d-25q%2Bn%2Bd%2Bq=14-105
-4n-9d-24q=-91
highlight_green%284n%2B9d%2B24q=91%29

CASES FOR THE QUARTERS
q=4
4n%2B9d=91-24%2A4
4n%2B9d=-5
-
q=3
4n%2B9d=91-24%2A3
4n%2B9d=19
-
q=2
4n%2B9d=43
-
q=1
4n%2B9d=66
-
Which of these cases will work to give the right whole number values for n and d?


Look at a graph for case q=1. Treat vertical axis as d, horizontal as n,
looks bad!
graph%28300%2C300%2C-1%2C15%2C-1%2C15%2C-%284%2F9%29x%2B66%2F9%29

The graph n and d for case q=2,
guess is that still not good.
graph%28300%2C300%2C-1%2C12%2C-1%2C12%2C-4x%2F9%2B43%2F9%29

See for case q=3,
graph%28300%2C300%2C-1%2C10%2C-1%2C10%2C-4x%2F9%2B19%2F9%29
Maybe....

A different way could be play with the simplest combination of coins to give $1.05 and split some coins to get a count of 14.
system%28q=4%2Cn=1%2Cc=5%29, c for count of coins but we must have c=14. Splitting the nickel might not help much, at least not yet.

Split one quarter.
system%28q=3%2Cd=2%2Cn=2%2Cc=7%29----still not enough coins c.

Split one more quarter.
system%28q=2%2Cd=4%2Cn=3%2Cc=9%29----we want to be able to get an even number of coins.

Maybe split one nickel from that.
system%28q=2%2Cd=4%2Cn=2%2Cp=5%2Cc=13%29-----we want ONE more coin, so maybe split one dime.

system%28Try+then+q=2%2Cd=3%2Cn=4%2Cp=5%2Cc=14%29-----THIS must be the combination.
2 quarters
3 dimes
4 nickels
5 pennies
--That is 14 coins.