SOLUTION: Fred has 5 times as many dimes as nickels in his piggy bank. He also has twice as many quarters as dimes. He has $9.15 in total a) Form an equation to model this situation b

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: Fred has 5 times as many dimes as nickels in his piggy bank. He also has twice as many quarters as dimes. He has $9.15 in total a) Form an equation to model this situation b      Log On

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Question 1106176: Fred has 5 times as many dimes as nickels in his piggy bank. He also has twice as many quarters as dimes. He has $9.15 in total
a) Form an equation to model this situation
b) Solve your equation to find how many coin of each type he has.
Found 3 solutions by Alan3354, josgarithmetic, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Fred has 5 times as many dimes as nickels in his piggy bank. He also has twice as many quarters as dimes. He has $9.15 in total
a) Form an equation to model this situation
I would use d for dimes, n for nickels and q for quarters.
========================
b) Solve your equation to find how many coin of each type he has.
What did you get?

Answer by josgarithmetic(39799) About Me  (Show Source):
You can put this solution on YOUR website!
d dime, n nickels, q quarters

system%28d=5n%2Cq%2Fd=2%2C10d%2B5n%2B25q=915%29

Simplify the cents equation and use the first two equations to form a single equation for the cents count equation all in one single variable d. Solve for d, first and use it to find the counts of n and q.

Answer by ikleyn(53763) About Me  (Show Source):
You can put this solution on YOUR website!
.
The strategy is to choose a single  major  unknown variable;
        then to express all other variables via the major one;
        then to build an equation,  and to solve it accurately.
The final step is to calculate all other unknowns,  write the answer and to check it. 


Let N be the number of nickels in the bank (MAJOR unknown variable).

Then the number of dimes is 5N;

     the number of quarters is 2*dimes = 2*(5N) = 10N.



The "value" equation is

nickels + dimes     + quarters = 915   cents,   or


5N      +   10*(5N) + 25*(10N) = 915,

5N + 50N + 250N = 915,

305N = 915  ====>  N = 915%2F305 = 3.


Thus there are 3 nickels in the bank.

Then the number of dimes    is 5*3 = 15  and

     the number of quarters is 2*15 = 30.


Check.  3*5 + 10*15 + 25*30 = 915 cents.   ! Correct !


Answer.  3 nickels, 15 dimes and 30 quarters.

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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.

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.