SOLUTION: A box contains 65 coins, only dimes and nickels. The amount of money in the box is $5.40, how many dimes and nickels are in the box?

Question 1133457: A box contains 65 coins, only dimes and nickels. The amount of money in the box is $5.40, how many dimes and nickels are in the box?
Found 2 solutions by ikleyn, ankor@dixie-net.com:
Answer by ikleyn(52799)   (Show Source): You can put this solution on YOUR website!
.

You can solve this problem mentally.

Let's assume for a minute that all 34 coins are nickels - then the total would be 5*65 = 325 cents, making the shortage of 540 - 325 = 215 cents. Hence, we should replace some number of nickels by dimes to compensate the difference. How much replacement we should do ? - Obviously, = = 43 replacements diminishing the difference of 215 cents by 5 = 10-5 cents at any single replacement. So, the answer is: the original collection has 43 dimes and the rest 65-43 = 22 coins are nickels. Check. 43*10 + 22*5 = 540 cents. ! Correct !

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It is a very standard coin word 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
    - 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 ankor@dixie-net.com(22740)   (Show Source): You can put this solution on YOUR website!
A box contains 65 coins, only dimes and nickels.
d + n = 65
or for substitution the form is
n = (65-d)
The amount of money in the box is $5.40,
.10d + .05d = 5.40
replace n with (65-d)
.10d + .05(65-d) = 5.40
.10d + 3.25 - .05d = 5.40
.10d - .05d = 5.40 - 3.25
.05d = 2.15
d = 2.15/.05
d = 43 dimes
and
65 - 43 = 22 nickels
;
:
See if that checks out, find the actual amt of each
43(.10) = 4.30
22(.05) = 1.10
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Total $ = 5.40

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