SOLUTION: The drawer of a cash register contains 30 coins: pennies, nickels, dimes, and quarters. The total value of the coins $3.31. The total number of pennies and nickels combined is the

The unknown quantities are P, N, D and Q.

We are given

    P + N + D + Q = 30     (1)

and

    P + N = D + Q.         (2)


From it, we momentartily conclude that

    P + N = D + Q = 15     (3)


We also are given that

    25Q = 5*(10D)      (4)     (The total value of the quarters is five times total value of the dimes. )

It implies, after reducing/canceling common factors

      Q = 2D.


Then from (3), by substituting (4) there, we get

    D + 2D = 15  --->   3D = 15  --->   D = 5   --->    Q = 10.


So, we just know that the number of dimes is 5 and the number of quarters is 10.

Then the total value of dimes and quarters is  5*10 + 10*25 = 50 + 250 = 300 cents = 3 dollars.


Hence, the total value of pennies and nickels is  $3.31 - $3 = $0.31 = 31 cents.

Thus for P and N we have these two equations

    P +  N = 15    (coins)    (5)

    P + 5N = 31.   (cents)    (6)


Subtract (5) from (6) to get 

        4N = 31 - 15 = 16.


Hence, N = 16/4 = 4;  P = 15 - N = 15 - 4 = 11.


ANSWER.  11 pennies;  4 nickels;  5 dimes and 10 quarters.

The drawer of a cash register contains 30 coins: pennies, nickels, dimes, and quarters. The total value of the coins $3.31. The total number of pennies and nickels combined is the same as the total number of dimes and quarters combined. The total value of the quarters is five times total value of the dimes. How many coins of each type are in the drawer ?

Let number of pennies, nickels, dimes and quarters be P, N, D, and Q, respectively
Then we get:                     P + N + D + Q = 30 ----- eq (i)
             P + N = D + Q ====< P + N - D - Q = 0 ------ eq (ii)
              .25Q = 5(.1D)___.25Q = .5D___.5Q = D ------ eq (iii)
                      .01P + .05N + .1D + .25Q = 3.31 --- eq (iv)
                                       2D + 2Q = 30 ----- Subtracting eq (ii) from eq (i)
                                         D + Q = 15 ----- eq (v)
                                       .5Q + Q = 15 ----- Substituting .5Q for D in eq (v)
                                          1.5Q = 15
                 Number of quarters, or 

                                        D + 10 = 15 ----- Substituting 10 for Q in eq (v)
                    Number of dimes, or 

                 .01P + .05N + .1(5) + .25(10) = 3.31 ----- Substituting 5 for D, and 10 for Q in eq (iv)
                        .01P + .05N + .5 + 2.5 = 3.31
                                   .01P + .05N = 3.31 - 3
                                   .01P + .05N = .31 ---- eq (vi)

       P + N = D + Q, and D + Q = 15, so P + N = 15 ------- eq (vii) 
                                   .01P + .01N = .15 ------ Multiplying eq (vii) by .01 ---- eq (viii)
                                          .04N = .16 ------ Subtracting eq (viii) from eq (vi) 
                  Number of nickels, or 

                                         P + 4 = 15 ---- Substituting 4 for N in eq (vii)
                  Number of pennies, or