SOLUTION: Ramon has P4.80 in 25 cents and 10 cents. He has a total of 30 coins. How many 25 cents coins and how many 10 cents coins does he have?

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Question 1121214: Ramon has P4.80 in 25 cents and 10 cents. He has a total of 30 coins. How many 25 cents coins and how many 10 cents coins does he have?
Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.
1. One-equation setup and solution

Let x be the number of dimes (10-cents coins).

Then the number of quarters (25-cents coins) is 30-x.


The "value" equation is


    10x + 25(30-x) = 480   cents.     (1)


Simplify and solve for x 


    10x + 750 - 25x = 480

    750 - 480 = 25x - 10x

    270 = 15x

    x = 270/15 = 18.


Answer.  18 dimes and (30-18) = 12 quarters.


Check.   18*10 + 12*25 = 480 cents.     ! Correct !

Solved.


2. Two equations setup and solution

     x +   y =  30          (2)    (counting coins)
    5x + 25y = 480          (3)    (counting cents)


From equation (2), express y = 30-x  and then substitute it into equation (4). You will get 


    5x + 25*(30-x) = 480    (4)


Equation (4) is identical to equation (1) of the 1-st solution.


So, in this way you get the same answer.

Solved.


3. Mental/logic solution without using equations

Assume for a minute that all 30 coins are 10-cents dimes - then the total would be 300 cents, making the shortage of 480-300 = 180 cents.


So, we need to replace some dimes by quarters to compensate the difference.


At each replacement, we diminish the difference by 25-10 = 15 cents, so 180%2F15 = 12 replacements are required.


Thus, the collection must have 12 quarters, and the rest 30-12 = 18 are dimes.


Solved (3 times by 3 different methods).


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Congratulations !   You are now familiar with 3 basic methods for solving typical coin problems.

I suggest that algebraic methods will be your basic methods for such problems,
and the logical analysis method will allow you to solve the problems MENTALLY without using equations.
I will be happy if it will make your horizon wider.

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To see other similar solved coin problems, look in the lessons
    - Coin problems
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    - Typical coin problems from the archive
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in this site.

To see how the logical method works for other similar problems, look into the lessons
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in this site.