SOLUTION: Kevin and randy Muise have a jar containing 53 coins, all of which are eathir quarters or nickels. The total value of the coins in the jar is $6.05. How many of each type of coin d

Question 1138969: Kevin and randy Muise have a jar containing 53 coins, all of which are eathir quarters or nickels. The total value of the coins in the jar is $6.05. How many of each type of coin do they have?
Found 3 solutions by ikleyn, rothauserc, greenestamps:
Answer by ikleyn(52914)   (Show Source): You can put this solution on YOUR website!
.

Let Q be the number of quarters - then the number of nickels is (53-Q). The "money" equation is 5*(53-Q) + 25Q = 605, or 5*53 - 5Q + 25Q = 605 20Q = 605 - 5*53 Q = = 17. ANSWER. 17 quarters and 53 - 17 = 36 nickels. CHECK. 17*25 + 36*5 = 605 cents. ! Correct !

Solved.

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There are MANY WAYS to solve this problem.

On coin problems, see the lessons
    - 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 (*)
    - Non-typical coin problems
    - 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.

A convenient place to quickly observe these lessons from a "bird flight height" (a top view) is the last lesson in the list.

Read them and become an expert in solution of coin problems.

Pay special attention to lessons marked (*) in the list.

The lesson (**) is specially devoted to these two personages, Kevin and Randy Muise.

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 rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
let x be the number of quarters and y be the number of nickels
:
1) x + y = 53
:
2) 0.25x + 0.05y = 6.05
:
solve equation 1 for y
:
y = 53 - x
:
substitute for y in equation 2
:
0.25x + 0.05(53 - x) = 6.05
:
0.25x + 2.65 - 0.05x = 6.05
:
0.20x = 3.40
:
x = 17
:
y = 53 - 17 = 36
:
********************************************
there are 17 quarters and 36 nickels
:
check answer with equation 2
:
0.25(17) +0.05(36) = 6.05
:
4.25 + 1.80 = 6.05
:
6.05 = 6.05
:
answer checks
*******************************************
:

Answer by greenestamps(13215)   (Show Source): You can put this solution on YOUR website!

If you don't need a formal algebraic solution to a problem like this, here is a quick way to get the answer with logical reasoning and some mental arithmetic.

(1) If all 53 coins were nickels; the total value would be 53*$0.05 = $2.65. That's $3.40 short of the actual total of $6.05.

(2) The difference between the value of each quarter and each nickel is 20 cents.

(3) So to make up those additional $3.40 (340 cents), the number of quarters must be 340/20 = 17.

ANSWER: 17 quarters and (53-17) = 36 nickels
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