SOLUTION: A collection of 40 coins is worth $2.90. The collection consists only of nickels and dimes. What is the total value of the dimes? I am in eighth grade and in quarantine, I have a

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Question 1171536: A collection of 40 coins is worth $2.90. The collection consists only of nickels and dimes. What is the total value of the dimes?
I am in eighth grade and in quarantine, I have a test on Friday that my grade relies on to go from 70 to an 80, if you would like to help me with this problem that would be great, thank you!

Found 3 solutions by josgarithmetic, ikleyn, MathTherapy:
Answer by josgarithmetic(39630)   (Show Source): You can put this solution on YOUR website!
d dimes
n nickels
OR 40-d of the nickels

, counting in cents


but your question was how much money in DIMES;
CENTS
OR
$1.80

Answer by ikleyn(52898)   (Show Source): You can put this solution on YOUR website!
.

Let x be the number of nickels and

let y be the number of dimes.


Then you have these two equations

    x +   y = 40   coins      (1)    (counting for the coins)   and

   5x + 10y = 290  cents      (2)    (counting total money)


How these equations are obtained, should be totally clear.



Now, I do not know, which methods of solution of such equations you know.

So, I will show you the simplest way, using the substitution method.



From equation (1), express  x = 40-y  and substitute it into the second equation, replacing x there.  You will get

    5*(40-y) + 10y = 290.


This equation is for one unknown only, which is y.  Simplify it

    200 - 5y + 10y = 290

    -5y + 10y      = 290 - 200

       5y          = 90

        y          = 90/5 = 18.


Hence, from equation (1)

     x = 40 - 18 = 22.


ANSWER.  22 nickels and 18 dimes.


CHECK.  22*5 + 18*10 = 110 + 180 = 290 cents,  total money.   ! Correct !

Solved.

. . . . . . . . . . .

            I solved the problem for you,  using the system of two equations and the substitution method.

            There is other approach, shorter than that.  I will show it to you, too.


Let x be the number of nickels.

then the number of dimes is (40-x), OBVIOUSLY.


Now you can write the total money equation in this form


    5x + 10(40-x) = 290    cents.


Simplify it and find x


    5x + 400 - 10x = 290

    5x - 10x       = 290 - 400

      -5x          = -110

        x          =  =  = 22.


So, the number of nickels is 22;  the dimes are the rest  40-22 = 18 pieces.


Thus finally you get the same answer.

. . . . . . . . . . .

            Now you know these two approaches.

            One approach starts from two equations and reduce them then to one single equation.

            The other approach starts from one equation and solves it.

            Which approach you select on your test, depends on how you learn the subject in the school.


I completed my post and my teaching.

You may post me, whether you do understand everything in my post and/or ask your questions.

Wish you to be successful at the test.

You may inform me about your results there.

----------------

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 MathTherapy(10557)   (Show Source): You can put this solution on YOUR website!

A collection of 40 coins is worth $2.90. The collection consists only of nickels and dimes. What is the total value of the dimes?
I am in eighth grade and in quarantine, I have a test on Friday that my grade relies on to go from 70 to an 80, if you would like to help me with this problem that would be great, thank you!
Let number of dimes be D
Then number of nickels = 40 - D
We then get: .1D + .05(40 - D) = 2.9
.1D + 2 - 05D = 2.9
.1D - .05D = 2.9 - 2
.05D = .9
Number of dimes, or
With number of dimes being 18, value of dimes =
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