SOLUTION: A money box contains only 10-cent and
20-cent coins. There are 25 coins with a
total value of $3.50.
How many coins of each?
I worked out the answer by guessing easily enough
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20-cent coins. There are 25 coins with a
total value of $3.50.
How many coins of each?
I worked out the answer by guessing easily enough
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Question 975494: A money box contains only 10-cent and
20-cent coins. There are 25 coins with a
total value of $3.50.
How many coins of each?
I worked out the answer by guessing easily enough, but now I want to know if there is an algebraic approach to this? Found 2 solutions by solver91311, KMST:Answer by solver91311(24713) (Show Source):
Let represent the number of 10 cent coins and represent the number of 20 cent coins. The total number of coins is 25, so:
The 10 cent coins, being worth 10 cents each, must have a value of cents if there are of them. Likewise, the value of 20 cent coins is cents. The sum of these two values is 350 cents.
Would work just as well, but decimals are always messier than whole numbers, and neatness helps avoid errors.
So now all you need to do is solve the 2X2 linear system for and . This is a fairly straight-forward 2X2 system and either the substitution or the elimination method should work well enough.
John
My calculator said it, I believe it, that settles it
You can put this solution on YOUR website! = number of 10-cent coins. = number of 20-cent coins. = value of 10-cent coins (in $). = value of 20-cent coins (in $). = value of 10-cent coins plus 20-cent coins (in $). <---> (multiplying both sides of the equal sign times 10)
You have two linear equations and two variables, forming a system of linear equations.
You could solve the system by substitution,
solving one equation for one of the variables,
for example ---> ,
and then substituting the expression found for that variable into the other equation to find the value of the other variable: --->--->--->---> ,
anfd finally, substituting the value found into the solved first equation to find the value of the other variable: --->---> .
