David gave mark dimes and quarters for 18 nickels. if david gave mark 6 coins, how many of each coon did mark receive?

Herbert

Let n=the number of nickels and q=the number of quarters.

From the problem statement, we know two things:

The total number of coins is 57; This means that n+q=57

The total value of the coins is $8.85; Since each nickel is worth .05 and each quarter is worth .25, this means that .05n+.25q=8.85

We now have two equations with two unknowns.

We can solve for n in the first equation, n=57-q

Plugging this in for n in the second equation, we get .05(57-q)+.25q=8.85

Multiplying through and simplifying:  .05*57-.05q+.25q=8.85

                                                    2.85   +    .20q    =8.85

                                                                    .20q   =8.85-2.85

                                                                    .20q   =6

                                                                        q   = 6/.20

                                                                        q   = 30 (This is the number of quarters)

Since n+q=57; n=57-q

                      n=57-30

                      n=27 (this is the number of nickels)

We can now double check the results by taking the number of quarters (30) times the value of each quarter (.25) and adding this to the the number of nickels (27) times the value of each nickel (.05):

30*.25+27*.05 = 7.50+1.35 = 8.85, which is the value stated in the problem.

Let me know if you have any questions.

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