Question 1064611: A coin bank for decoration contained 1570. There were twice as many 5 php coins as 10 php coins and 14 less 1 php coins as 5 php coins. How many of each kind were there?
Found 2 solutions by Fombitz, ikleyn:
Answer by Fombitz(32388)   (Show Source):
Answer by ikleyn(52879)  (Show Source):
You can put this solution on YOUR website! .
A coin bank for decoration contained 1570. There were twice as many 5 php coins as 10 php coins and 14 less 1 php coins as 5 php coins.
How many of each kind were there?
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Let T be the number of 10 php coins.
Then the number of 5 php coins is 2T, and the number of 1 php coins is (2T-14), according to the condition.
The value equation is
10T + 5*(2T) + (2T - 14) = 1570.
10T + 10T + 2T - 14 = 1570,
22T = 1570 + 14,
22T = 1584 ---> T =  = 72.
72 of 10-php coins.
Now you calculate the rest.
The lesson to learn from this solution: This problem is for ONE unknown.
Not for 3 unknowns and even not for 2. Only for one.
Therefore, the solution MUST be explained adequately.
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As long I live, NEVER saw the coin problem for 3 unknowns.
Coin problems are intended, as a rule, for young students, who even don't know about systems of equations.
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