SOLUTION: Kevin and randy must have a jar containing 36 coins, all of which are either quarter or nickel. The total value of the coins in the jar is $4.60. How many of each type of coin do t

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Kevin and randy must have a jar containing 36 coins, all of which are either quarter or nickel. The total value of the coins in the jar is $4.60. How many of each type of coin do t      Log On


   

Question 1192482: Kevin and randy must have a jar containing 36 coins, all of which are either quarter or nickel. The total value of the coins in the jar is $4.60. How many of each type of coin do they have?
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52915) About Me  (Show Source):
You can put this solution on YOUR website!
.
Kevin and randy highlight%28cross%28must%29%29 have a jar containing 36 coins, all of which are either quarter or nickel.
The total value of the coins in the jar is $4.60. How many of each type of coin do they have?
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x quarters and (36-x) nickels.


The total money equation is

    25x + 5*(36-x) = 460  cents.


Simplify and find x

    25x + 5*36 - 5x = 460

    25x - 5x        = 460 - 5*36

       20x          =    280

         x          =    280/20 = 14.


ANSWER.  14 quarters and the rest 36-14 = 22 coins are nickels.


CHECK.   14*25 + 22*5 = 460  total cents.   ! Precisely correct !

Solved.

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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
    - Advanced 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.

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(13216) About Me  (Show Source):
You can put this solution on YOUR website!


The other tutor showed a good typical formal algebraic solution.

You can get some good mental exercise by solving the problem informally using logical reasoning and simple mental arithmetic.

(1) 36 coins all nickels would have a value of $1.80
(2) The actual total was $4.60
(3) The difference is $2.80
(4) The value of each quarter is $0.20 more than the value of each quarter
(5) The number of quarters is $2.80/$0.20 = 14
(6) The number of nickels is 36-14=22

ANSWER: 14 quarters and 22 nickels

CHECK: 14($0.25)+22($0.05) = $3.50+$1.10 = $4.60