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A coin bank contains only nickels and dimes. The total value of the coins in the bank is $2.45.
If the nickels were dimes and the dimes were nickels, the total value of the coins would be $2.80.
Find the number of nickels in the bank.
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Let n be the number of nickels and d be the number of dimes.
Then
5n + 10d = 245 cents (1) ("The total value of the coins in the bank is $2.45. ")
10n + 5d = 280 cents (2) ("If the nickels were dimes and the dimes were nickels, the total value of the coins would be $2.50")
Cancel the factor 5 in both sides of both equations. You will get
n + 2d = 49 (3)
2n + d = 56 (4)
From (3), express n = 49-2d and substitute it into (4). You will get
2*(49-2d) + d = 56.
Simplify and solve for d:
98 - 4d + d = 56, or
-3d = 56 - 98 ---> -3d = -42 ---> d = 14.
Thus the number of dimes is 14.
Then the number of nickels, from (3), is n = 49-2d = 49 - 2*14 = 21.
Check it on your own.
Answer. 14 dimes and 21 nickels.
There is entire bunch of the lessons on coin problems
- Coin problems
- More Coin problems
- Solving coin problems without using equations
- Kevin and Randy Muise have a jar containing coins
- Typical coin problems from the archive
- More complicated coin problems
in this site.
Read them and become an expert in solution of coin problems.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
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It contains many other solved word problems, as well as many other interesting and useful things.