How to Solve Coin Problems Part 2

I have already introduced how to solve coin problems, so in this post, we solve more problems about it. Coin problems involve problems consisting coins, bills, and of course, any object of value. We have already solved 2 problems in the previous post, so we continue with the third problem.

Problem 3

A box contains 32 bills consisting of Php20 and Php50. The total amount of money in the box is 1000 pesos. How many bills of each kind are there?

Note: For my readers from other countries, Php means Philippine pesos.


This problem is similar to problem 2, so we will not be solving it in details.

Let x be the number of 50-peso bills. Since there are 32 bills, then the number of 20-peso bills is 32 - x Now, if we multiply the number of 50-peso bills by 50 pesos and multiply the number of 20-peso bills by 20 pesos, we have

50x and 20(32-x)


If we add them, the total is 1000 pesos. That is,

50x + 20(32-x) = 1000.

Simplifying and solving for x, we have

50x + 640 - 20x = 1000

30x + 640 = 1000

Subtracting both sides by 640, we have

30x = 360.

Dividing both sides by 30, we have

x = 12.

Therefore, the number of 50-peso bills is 12. Now, since there are 32 bills, the number of 20-peso bills is 32-12 = 20.


12(Php50) + 20(Php20) = Php600 + Php400 = Php1000

Problem 4

In a charity musical show, there are the same number of tickets sold worth $20, $50, and $100. The total cost of the tickets is $5100.


Let x be the number of tickets sold.  Since there are the same number of tickets, if we multiply the number of ticket to each of the price, we have

20x, 50x and 100x

which are the total cost of each kind. If we add them all together, then, it is the total cost of all the tickets which is $5100. That is,

20x + 50x + 100x = 5100.

Simplifying, we have

170x = 5100

Dividing both sides by 170, we have

x = 30.

This means that there are 30 tickets of each price that were sold.


$20(30) + $50(30) + $100(30) = $600 + $1500 + $3000 = $5100.

In the next part of this post, we will solve two more problems to end this series.

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