Question 722728: Can you please answer this with steps: John has a collection of pennies, nickels, and dimes that have a total value of $4.42. He has 4 more dimes than pennies, and 6 more nickels than dimes. How many coins of each type dioes he have? Thank you
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! let p = no. of pennies
let n = no. of nickels
let d = no. of dimes
:
Write an equation for each statement:
:
"John has a collection of pennies, nickels, and dimes that have a total value of $4.42.
.01p + .05n + .10d = 4.42
:
" He has 4 more dimes than pennies,"
d = p + 4
or
p = d - 4
:
"and 6 more nickels than dimes."
n = d + 6
:
In the total$ equation replace p with (d-4); replace n with (d+6)
.01(d-4) + .05(d+6) + .10d = 4.42
.01d - .04 + .05d + .30 + .10d = 4.42
combine like terms
.01d + .05d + .10d -.04 + .30 = 4.42
.16d + .26 = 4.42
.16d = 4.42 - .26
.16d = 4.16
d = 4.16/.16
d = 26 dimes
then
n = 26 + 6
n = 32 nickels
and
p = 26 - 4
p = 22 pennies
:
:
You confirm this by finding the the total value using these numbers
.01(22) + .05(32) + .10(26) =
:
Was this enough steps for you to comprehend this problem? C
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