SOLUTION: Jordan has pennies, nickels, dimes, and quarters. He has a total of $14.18. He has one more quarters than pennies and three fewer dimes than nickels. He also has ten more quarters
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Question 1190239: Jordan has pennies, nickels, dimes, and quarters. He has a total of $14.18. He has one more quarters than pennies and three fewer dimes than nickels. He also has ten more quarters than nickels and thirteen more quarters than dimes. How many of each coin does he have?
Found 2 solutions by josgarithmetic, ankor@dixie-net.com:
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
pennies nickels dimes quarters
p, n, d, q
Notice the first four equations, about the coin counts, have q in them.
substitute most of those into the money equation:
----simplify and solve for q.
.
.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Jordan has pennies, nickels, dimes, and quarters.
p = no. of pennies
n = nickels
d = dimes
q = quarters
write an equation for each statement
He has a total of $14.18.
.01p + .05n + .10d + .25q = 14.18
get rid of all these decimals, mult by 100
p + 5n + 10d + 25q + 1418
:
we will try to get all variables in terms q
He has one more quarters than pennies
q = p + 1
p = q - 1
and three fewer dimes than nickels.
d = n - 3
He also has ten more quarters than nickels
q = n + 10
n = q - 10
and thirteen more quarters than dimes.
q = d + 13
d = q - 13
in the first equation (simplified),replace as follows
(q-1)+5(q-10)+10(q-13)+25q = 1418
q + 1 + 5q - 50 + 10q - 130 + 25q = 1418
q + 5q + 10q + 25q - 1 - 50 - 130 = 1418
41q - 181 = 1418
41q = 1418 + 181
41q = 1599
q = 1599/41
q = 39 quarters
then
p = 39 - 1 = 38 pennies
n = 39 - 10 = 29 nickels
d = 39 - 13 = 26 dimes
:
See if this checks out
.01(38) + .05(29) + .10(26) + .25(39) =
.38 + 1.45 + 2.60 + 9.75 = $14.18
:
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