SOLUTION: kevin and randy muise have a jar containing 82 coins, all of which ae either quarters or nickles. The total value of the coins In the jar is 13.30. How many of each type of coin do
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Question 997434: kevin and randy muise have a jar containing 82 coins, all of which ae either quarters or nickles. The total value of the coins In the jar is 13.30. How many of each type of coin do they have?
The jar contains how many quarters
How many nickles? Found 2 solutions by addingup, ikleyn:Answer by addingup(3677) (Show Source):
You can put this solution on YOUR website! q+n= 82. Subtract n on both sides of the equal sign and we have:
q= 82-n
0.25q+0.10n= 13.30 Now, in this formula, substitute the value of q:
0.25(82-n)+0.10n= 13.30
20.50-0.25n+0.10n= 13.30 Add n on left and subtract 20.50 on both sides
-0.15n= -7.20 Now divide both sides by -0.15
n= -7.20/-0.15 OK, remember that -/- = +:
n= 48 They have 48 nickels and:
82-48= 34quarters.
Proof:
48*0.10= 4.80
34*0.25= 8.50
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Total.....13.30 We have the correct answer
You can put this solution on YOUR website! .
Kevin and Randy have a jar containing 82 coins, all of which are either quarters or nickles. The total value of the coins in the jar is 13.30. How many of each type of coin do they have?
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You can solve this problem even without using equations.
Let us suppose for a moment that all 34 coins are 10-cent.
Then their value is 82*10 = $8.20. It is less than $13.30 in $5.10 = 510 cents.
It is clear that the difference is due to presence of 25-cent coins that we intently counted as 10-cent coins.
It is also clear that the number of these 25-cent coins is = = 34 to compensate the difference.
So, the answer is: there are 34 of 25-cent coins and 82-34 = 48 of 10-cent coins.
