100 coins made up of pennies, dimes and quarters worth $5.00

One point of view is, once you have abstracted the "real-world" quantities (number of coins, number of cents) into equations with named variables, you can go at it with algebra and never mind what the next equation "means" in "reality."

You may have to take this view some day for some problem. For this problem, in the equation $9d+24q=400,$ you know that every coin contributes at least one cent to the total value; the $9$ represents the number of extra cents that each dime contributes, and the $24$ represents the number of extra cents that each quarter contributes.

If the $100$ coins were all pennies we would be $400$ cents short of five dollars, so the $d$ dimes and $q$ quarters need to contribute a total of $400$ "extra" cents: $$9d+24q=400.$$


The variables $c,d$ and $q$ are unitless. Their domain is $\{0,1,2, \dots\}$

  • The number of pennies is $c$ coin.
  • The number of dimes is $d$ coin.
  • The number of quarters is $q$ coin.

$$c \ \text{coin} + d \ \text{coin} + q \ \text{coin} = 100 \ \text{coin} \implies c + d + q = 100$$

  • The value of $c$ pennies is $c \ \text{coin} \times \dfrac{1 \ \text{cent}}{\text{coin}} = c \ \text{cent}$

  • The value of $d \ $ dimes is $d \ \text{coin} \times \dfrac{10 \ \text{cent}}{\text{coin}} = 10d \ \text{cent}$

  • The value of $q \ $ quarters is $q \ \text{coin} \times \dfrac{25 \ \text{cent}}{\text{coin}} = 25q \ \text{cent}$

$$c \ \text{cent} + 10d \ \text{cent} + 25q \ \text{cent} = 500 \ \text{cent} \implies c + 10d + 25q = 500$$

Tags:

Elementary Number Theory

Algebra Precalculus

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