SOLUTION: luis has 27 coins in nickles and dimes. In all he has $1.90. How many of each coin does he have?

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Question 1180542: luis has 27 coins in nickles and dimes. In all he has $1.90. How many of each coin does he have?
Found 2 solutions by ikleyn, Edwin McCravy:
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

x dimes and 27-x nickels


The money equation is


    10x + 5(27-x) = 190  cents


Simplify and solve


    10x + 5*27 - 5x = 190


        5x          = 190 - 5*27


         x          = %28190+-+5%2A27%29%2F5 = 11.


ANSWER.  11 dimes and 27-11 = 16 nickels.

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 Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
Without using algebra, just basic math:

If all 27 had been nickels, he would have had only $1.35. But he had $1.90.
So the extra 55 cents must have come from the dimes contributing an extra 
5 cents each.  So there must have been 55/5 or 11 dimes.  That means he had
 27-11 or 16 nickels.

With algebra.

system%28N%2BD=27%2C0.05N%2B0.10D=1.90%29

Solve that system and get the same answer.

Edwin