SOLUTION: a newspaper carrier has $4.90 in change. He has five more quarters than dimes but two times as many nickles as quarters. How many coins of each type does he have? How many dimes?

Question 1108131: a newspaper carrier has $4.90 in change. He has five more quarters than dimes but two times as many nickles as quarters. How many coins of each type does he have? How many dimes?

Found 3 solutions by josgarithmetic, algebrahouse.com, ikleyn:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
Literally translate the description; use obvious coin count variables, n, d, q.

First, simplify the money count equation; and then use the next two equations to write ONE single equation in the ONE variable q. Solve for q.,
...
Answer by algebrahouse.com(1659)   (Show Source): You can put this solution on YOUR website!
x = number of dimes
x + 5 = number of quarters {he has five more quarters than dimes}
2(x + 5) = number of nickels {he has two times as many nickels as quarters}

0.1x + 0.25(x + 5) + 0.05(2x + 10) = 4.9 {value of coin times number of coins equals total value}
0.1x + 0.25x + 1.25 + 0.1x + 0.5 = 4.9 {used distributive property}
0.45x + 1.75 = 4.9 {combined like terms}
0.45x = 3.15 {subtracted 1.75 from each side}
x = 7 {divided each side by 0.45}
x + 5 = 12 {substituted 7, in for x, into (x + 5)}
2(x + 5) = 24 {substituted 7, in for x, into 2(x + 5)}

7 dimes
12 quarters
24 nickels

For more help like this, visit: www.algebrahouse.com

Answer by ikleyn(52876)   (Show Source): You can put this solution on YOUR website!
There is entire bunch of lessons on coin problems
    - Coin problems
    - More Coin problems
    - Solving coin problems without using equations
    - Kevin and Randy Muise have a jar containing coins
    - Typical coin problems from the archive
    - Three methods for solving standard (typical) coin word problems
    - More complicated coin problems (*)
    - Solving coin problems mentally by grouping without using equations
    - Santa Claus helps solving coin problem
    - OVERVIEW of lessons on coin word problems
in this site.

You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.

Read them and become an expert in solution of coin problems.
In particular, the problem of this post is a TWIN to the problem 2 of the lesson marked (*) in the list.

Also, you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic "Coin problems".

Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.

RELATED QUESTIONS
A newspaper carrier has $4.70 in change. He has seven more quarters than dimes but two... (answered by CubeyThePenguin)
a newspaper carrier has 12.30 in change. He has two more quarters than dimes but three... (answered by ewatrrr)
A newspaper carrier has $4.90 in change. He has five more quarters than dimes but five... (answered by CubeyThePenguin)
a newspaper carrier has $3.35 in change. He has seven more quarters than dimes but two... (answered by LinnW)
a newspaper carrier has $6.30 in change. he has three times more quarters than dimes but (answered by stanbon)
A newspaper carrier has $2.20 in change. He has two more quarters than dimes but five... (answered by ewatrrr)
A newspaper carrier has $4.60 in change. He has two more quarters than dimes but five... (answered by ikleyn)
A newspaper carrier has $4.60 in change. He has two more quarters than dimes but five... (answered by ikleyn,josgarithmetic)
newspaper carrier $6.85 in change 3 more quarters than dimes but four times as many... (answered by josgarithmetic)