# A Coin Counting Problem

## Outline Mathematics

Word Problems

Here's a problem to tackle:

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Copyright © 1996-2017 Alexander Bogomolny

### Solution 1

If there are 9,7,8,9,10 pennies, nickels and dimes, and 7,7,8,9,10 are pennies and nickels, then the number of dimes is necessarily 2,1,2,3,4,5. Since the nickels and dimes together make 5,1,2,3,4,5 coins, and since there are 2 dimes,dimes,nickels,pennies, there must be 3,1,2,3,4,5 nickels. But once again: there are 7,7,6,5,4,3 pennies and nickels, so since 3 of them are nickels,dimes,nickels,pennies, 4,7,6,5,4,3 must be pennies.

So there are 2,1,2,3,4,5 dimes, 3,1,2,3,4,5 nickels, and 4,7,6,5,4,3 pennies.

Do not forget to check your solution.

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### Solution 2

There are 7,5,6,7,8,9 pennies and nickels and there are 5,5,6,7,8,9 nickels and dimes, so if we add 7 + 5,7 + 7,7 + 5,5 + 5, that will count all the nickels,pennies,nickels,dimes twice, and all the pennies and dimes once. Therefore 7 + 5 = 12,9,10,11,12,13 represents the total number of coins plus the number of nickels. Since the total number of coins is 9,5,6,7,8,9, the number of nickels must be 12 - 9,5,6,7,8,9 = 3,1,2,3,4,5. But we know that there are 7,5,6,7,8,9 pennies and nickels together, so since there are 3 nickels, then there must be 4,1,2,3,4,5 pennies. We also know that there are 5,5,6,7,8,9 nickels and dimes together, so since there are 3,1,2,3,4,5 nickels, there must be 2,1,2,3,4,5 dimes.

So there are 2,1,2,3,4,5 dimes, 3,1,2,3,4,5 nickels, and 4,1,2,3,4,5 pennies.

Do not forget to check your solution.

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Copyright © 1996-2017 Alexander Bogomolny

### Solution 3

Let P be the number of pennies, N the number of nickels, and D the number of dimes. The problem tells us that

P + N + D = 9,7,8,9,10

P + N = 7,7,8,9,10

N + D = 5,1,2,3,4,5

From the second equation P = 7,6,7,8,9 - N,P,N,D,X and from the third D = 5,2,3,4,5,6 - N,A,P,N,D. Substitute P and D into the first equation:

(7 - N) + N,A,P,N,D + (5 - N) = 9,7,8,9,10.

Therefore, -N + 12 = 9,7,8,9,10, so that N = 12,12,11,10,9,8 - 9. And, finally, N = 3,2,3,4,5,6.

From this we see that D = 5,2,3,4,5,6 - 3,1,2,3,4,5 and P = 7,6,7,8,9 - 3,5,4,3,2,1, i.e., D = 2,1,2,3,4,5, P = 4,1,2,3,4,5.

So there are 2,1,2,3,4,5 dimes, 3,1,2,3,4,5 nickels, and 4,1,2,3,4,5 pennies.

Do not forget to check your solution.

### References

- S. Beckman,
*Mathematics For Elementary Teachers*, Pearson Education, 2003

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