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A child’s piggy bank contains 35 coins and a total of $6.25. The bank contains quarters, dimes and nickels.
There are twice as many dimes as there are nickles. How many of each coin is in the bank?
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Let N be the number of nickels.
Then the number of dimes is 2N, according to the condition.
Hence, the number of quarters is (35-N-2N) = (35-3N).
The nickels contribute 5N cents towards the total.
The dimes contribute 10*(2N) = 20N cents towards the total.
The quarters contribute 25*(35-3N) cents towards the total.
So, your balance equation is
5N + 20N + 25*(35-3N) = 625 cents. (<<<---=== you counted all the cents)
Simplify and solve for N step by step.
25N + 875 - 75N = 625 ====> -50N = 625 - 875 = -250 ====> N = = 5.
Answer. 5 nickels, 10 dimes and 35-5-10 = 20 quarters.
Check. Can you check it on your own ??
The lesson to learn from this solution:
This problem is FOR ONE UNKNOWN.
Therefore, I made all efforts from my side to solve it as the problem in ONE UNKNOWN.
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There is entire bunch of lessons on coin problems
- Coin problems
- More Coin problems
- Solving coin problems without using equations
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- Typical coin problems from the archive
- Three methods for solving standard (typical) coin word problems
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- OVERVIEW of lessons on coin word problems
in this site.
You will find there the lessons for all levels - from introductory to advanced,
and for all methods used - from one equation to two equations and even without equations.
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
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic "Coin problems".
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