SOLUTION: A box contains $7.15 in nickels, dimes, and quarters. There are 41 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coi

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Question 1113748: A box contains $7.15 in nickels, dimes, and quarters. There are 41 coins in all, and the sum of the numbers of nickels and dimes is 3 less than the number of quarters. How many coins of each kind are there?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.
(1)  nickels + dimes + quarters = 41      ( <<<---=== given )

     N       + D     + Q        = 41      ( the same )


(2)  N       + D   = Q - 3                ( <<<---=== given )


========>  Q + (Q-3) = 41

          2Q - 3 = 41  ====>  2Q = 41 + 3 = 44  ====>  Q = 22


So, we found number of quarters.  It is 22.

Next, you can reduce the problem from 3 unknowns to only two of them:


    we have 41-22 = 19 nickels and dimes, that are worth  7.15 - 22*0.25 = 1.65 dollars.


 N +   D =  19      (coins)    (1)
5N + 10D = 165      (cents)    (2)


Simplify

 N +   D =  19                 (1')
 N +  2D =  33                 (2')


Subtract eq(1') from eq(2').


      D = 33-19 = 14.


Answer.  14 dimes,  22 quarters  and  19-14 = 5 nickels.


Check.   5*5 + 14*10 + 22*25 = 715 cents.   ! Correct !

Solved.

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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.



Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!


Tutor ikleyn's solution is a perfectly good solution using formal algebra.

Here is a solution that is basically the same, but using logical reasoning instead of the formal algebra.

There are 41 coins in all; the number of nickels and dimes together is 3 less than the number of quarters. So temporarily remove the 3 "extra" quarters, leaving a total of 38 coins, of which half are quarters. That means 19 quarters, and a total of 19 nickels and dimes. Then put back the 3 quarters; the number of quarters in the box is 19+3 = 22.

The 22 quarters have a value of $5.50; so the total value of the 19 dimes and nickels is $7.15-$5.50 = $1.65.

If all 19 of the remaining coins were dimes, the total value would be $1.90; but it is $1.65, which is 25 cents less than $1.90. Replacing a dime with a nickel keeps the same total number of coins but reduces the total value by 5 cents. Since all dimes gives us 25 cents more than we want, the number of times we need to replace a dime with a nickel is 25/5 = 5.

So the number of nickels is 5, and the number if dimes is 19-5 = 14.

Answer: 22 quarters, 14 dimes, 5 nickels.

Check: 22(25)+14(10)+5(5) = 550+140+25 = 715
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