SOLUTION: A man has 20 coins in his pocket consisting of 5 peso, 10 peso and 1 peso coins. There are an equal equation of ones and fives among them. He has 88 pesos in all. How many coins of
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Question 969658: A man has 20 coins in his pocket consisting of 5 peso, 10 peso and 1 peso coins. There are an equal equation of ones and fives among them. He has 88 pesos in all. How many coins of each denomination are there? Found 2 solutions by CubeyThePenguin, greenestamps:Answer by CubeyThePenguin(3113) (Show Source):
A solution using three variables for the numbers of 1-, 5-, and 10-peso coins is good algebra; with three variables you will need three equations. But setting up the problem using a single variable will make solving the problem easier, since there will be only one equation.
x = number of 1-peso coins
x = number of 5-peso coins
20-2x = number of 10-peso coins
The total value is 88 pesos:
ANSWER: 8 each 1-peso and 5-peso coins; 20-16=4 10-peso coins
If a formal algebraic solution is not required, a quick solution can be obtained using logical reasoning and some simple mental arithmetic.
Since the numbers of 1- and 5-peso coins are the same, consider them as groups of 2 coins with a value of 6 pesos.
The total value of the 10-peso coins is a multiple of 10; since the total value of all the coins is 88 pesos, the total value of the 1- and 5-peso coins must have units digit 8.
To get a units digit of 8, the number of 6-peso pairs of coins can be either 3 or 8. (3*6=18; 8*6=48).
3 each of 1- and 5-peso coins worth a total of 18 pesos means 7 10-peso coins to make the other 70 pesos; 3+3+7=13 is not the right total number of coins.
8 each of the 1- and 5-peso coins worth a total of 48 pesos means 4 10-peso coins to make the other 40 pesos; 8+8+4=20 is the right total number of coins.
ANSWER: 8 each 1- and 5-peso coins; 4 10-peso coins.
