SOLUTION: a box contains 77 coins, only dimes and nickels. The amount of money in the box is $5.25. How many dimes and how many nickles are in the box?

Question 1127229: a box contains 77 coins, only dimes and nickels. The amount of money in the box is $5.25. How many dimes and how many nickles are in the box?
Found 2 solutions by stanbon, greenestamps:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a box contains 77 coins, only dimes and nickels. The amount of money in the box is $5.25. How many dimes and how many nickles are in the box?
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Quantity Equation: n + d = 77 coins
Value Equation:: 5n + 10d = 525 cents
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Solve for n and d::
5n + 5d = 5*77
5n + 10d= 525
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Subtract and solve for "d
5d = 525-385
5d = 140
d = 28 (# of dimes)
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Solve for "n"::
n = 77-d = 49 (# of nickels)
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Cheers,
Stan H.
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Answer by greenestamps(13203) (Show Source): You can put this solution on YOUR website!

Here is a common sense way to solve problems like this, without the formal algebra.

If all 77 coins were nickels, the total value would be 77*5 = 385 cents or $3.85.
The actual total value is $5.25, which is $1.40 (140 cents) more than $3.85.
Replacing 1 nickel with 1 dime raises the total value by 5 cents (while keeping the total number of coins the same).
The number of dimes needed to replace nickels to make up the other 140 cents is 140/5 = 28.

So there are 28 dimes and 77-28 = 49 nickels.
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