SOLUTION: A jar has 50 coins: all pennies, nickles, and dimes, with a total value of $3.50. There are twice as many nickles as as pennies. How many are there of each coin? p=pennies

Question 791824: A jar has 50 coins: all pennies, nickles, and dimes, with a total value of $3.50. There are twice as many nickles as as pennies. How many are there of each coin?
p=pennies .01 .01p
2p=nickles .05 .05(2p)
d=dimes .10 .10(50-3p)=5-.30p

p+2p+d=50
d=50-3p
.01p+.05(2p)+5-.3p=3.5
.01p+.1p+5-.3p=3.5
-.19p+5=3.5
5-3.5=.19p
1.5=.19p
email is its.me.pinkey3@gmail.com
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
VARIABLES
p = count pennies
n = count nickels
d = count dimes

TRANSLATE THE WORDED DESCRIPTIONS
.
.
(Twice as many nickels as pennies; ratio of nickels to pennies is 2:1).

MANAGE EQUATIONS TO MAKE SYSTEM EASIER TO SOLVE
Clear the fractional nature of the money sum, multiplying both sides by 100, giving .
Multiply the ratio equation left&right by p, giving .

ALGEBRAIC SOLUTION STEPS
The system is now this:
____________________________________
.
.
.
____________________________________

Solution steps can use the less advanced method of substitution, starting with n. Replace n in the first two equations with n=2p. This gives two equations in two unknowns:
and
and
Various ways to go, but try solving the count equation for d, and substitute this into the money sum equation:
, put into... 11p+10*(50-3p)=350

MISTAKE? (No. See "Second Thought")

(some commentary removed)

SECOND THOUGHT: My answer is like your answer. The mistake is not in our work, but in the problem description.

CORRECTION TO QUESTION: Amount of money should have been $3.10; with this, problem was solved by original poster.
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