Joe has a collection of nickels and dimes that is worth $5.65. ?

Question

If the number of dimes were doubled and the number of nickels were increased by 8, the value of the coins would be $10.45. How many dimes does he have?

I have tried to solve this question, but I keep coming up over on my dollar amounts. Please help!

Answer 1

5n +10d = 565...........(1)

5n+40 +20d = 1045......This becomes 5n+20d = 1005........(2).

Subtract eqn. (1) from eqn. (2)

10d = 440

d = 44

n = 25

Answer 2

Let n be the number of nickels and d be the number of dimes. You can make a system of two equations:

0.05n + 0.1d = 5.65

0.05(n + 8) + 0.1(2d) = 10.45 or 0.05n + 0.2d = 10.05 when simplifed

Solve through subtraction method:

(0.05n + 0.2d = 10.05) - (0.05n + 0.1d = 5.65) = (0.1d = 4.4)

0.1d = 4.4

d = 44

0.05n + 0.2d = 10.05

0.05n + 0.2(44) = 10.05

0.05n + 8.8 = 10.05

0.05n = 1.25

n = 25

ANSWER: There are 25 nickels and 44 dimes.

Answer 3

5*N + 10*D = 565

Divide by 5: N + 2*D = 113

5*(N+8) + 10*(2*D) =1045

5*N + 40 + 20*D = 1045

N + 8 + 4*D = 209

N + 4*D = 201

Subtract: N + 4*D = 201

            - (N + 2*D = 113)

                     2*D = 88

                        D = 44

So, going back to the original - 5*N + 10*D = 565

                    If D = 44 then      5*N + (10*44) = 565

                                             5*N + 440 = 565

                                             5*N = 125

                                                N = 25

Checking: 5*(N + 8) + 10*(2*D) = 1045

If N = 25, then (N + 8) = 33 and 5*(33) = 165

So 10*(2*D) = 20*D = 880

                          D = 44