SOLUTION: A coin bank contains 25 coins in nickles, dimes and quarters. There are four times as many dimes as quarters. The value of the coins is $2.05. How many dimes are in the bank?
Algebra ->
Customizable Word Problem Solvers
-> Coins
-> SOLUTION: A coin bank contains 25 coins in nickles, dimes and quarters. There are four times as many dimes as quarters. The value of the coins is $2.05. How many dimes are in the bank?
Log On
Question 201379: A coin bank contains 25 coins in nickles, dimes and quarters. There are four times as many dimes as quarters. The value of the coins is $2.05. How many dimes are in the bank? Found 2 solutions by ptaylor, Theo:Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=number of quarters
Then 4x =number of dimes
And y=number of nickels
Now we are told that:
x+4x+y=25 or
5x+y=25--------------------eq1
also (we'll deal in pennies)
25x+10*4x+5y=205
65x+5y=205-------------------------eq2
multiply eq1 by 5 and then subtract it from eq2
(65x+5y=205)-(25x+5y=125) and we get
40x=80
x=2---------------number of quarters
4x=4*2=8 number of dimes
There must be 25-10=15 nickels
CK
2*25+8*10+15*5=205
50+80+75=205
205=205
We can also work this problem using only one unknown:
Let x=number of quarters
Then 4x =number of dimes
And 25-5x=number of nickels
So we have:
25x+10*4x+5(25-5x)=205 get rid of parens
25x+40x+125-25x=205 subtract 125 from each side
25x+40x+125-125-25x=205-125 collect like terms
40x=80---same as before
Hope this helps---ptaylor
You can put this solution on YOUR website! I believe your answer is going to be
15 nickels (75 cents)
8 dimes (80 cents)
2 quarters (50 cents)
that adds up to be $2.05
here's how I did it.
let n = number of nickels
let d = number of dimes
let q = number of quarters
n + d + q = 25 (number of coins in total)
number of dimes is 4 times the number of quarters (n = 4d)
total value of coins is $2.05
$.05 * n + $.10 * d + $.25 * q = $2.05
take away the dollar sign and multiply both sides of the equation by 100 and you get
5n + 10d + 25q = 205
since d = 4q you can substitute to get
5n + 40q + 25q = 205
since n + d + q = 25 you can substitute 4q for d again to get
n + 5q = 25
solve for n to get
n = 25 - 5q
multiply both sides of the equation by 5 to get
5n = 125 - 25q
substitute in the equation 5n + 40q + 25q = 205 to get
125 - 25q + 40q + 25q = 205
simplify the left side of the equation to get
125 + 40q = 205
solve for q to get
40q = 80
q = 2
since d = 4q, then d = 8
since n + d + q = 25, then n = 15
your answer is
n = 15
d = 8
q = 2
as shown above.
