How many of each kind of coin are there?

Question

A collection of coins has a total value of 64 cents. There are two more nickels than dimes and three
times as many pennies as dimes. 

Answer 1

the number of each coin times their value is the total amount, so

 

P = number of pennies

N = number of nickels

D = number of dimes

 

so we can write our total cents equation:

 

P(1) + N(5) + D(10) = 64      ! numbers are in cents

 

N = D+2       !  #nickels  = #dimes + 2

P = 3D         !  #pennies = 3 X #dimes

 

now just substitute N and P into our total cents equation (all in terms of D now)

 

(3D)(1) + (D+2)(5) + D(10) = 64

18D + 10 = 64

18D = 54           

 

D(#dimes) = 3      (54/18

so from N = D+2   N(#nickels)  = 3+2 = 5

from P = 3D          P(#pennies) = 3(3) = 9

 

so to check, 

3 dimes = 30 cents

5 nickels = 25 cents

9 pennies = 9 cents    or  30+25+9 = 64 cents !

 

Answer 2

Anthony, let's start with the basics. We're obviously going to have at least 4 pennies, but 4 is not 3 times anything. Pennies are going to be in multiples of 3.

 

6 pennies won't work because we can't make 58 cents out of nickles and dimes.

 

How about 9 pennies? Then we have 55 cents to make with dimes and nickles. According to the problem, we'd have 3 dimes and 5 nickles.

 

3 dimes is 30 cents, 5 nickles is 25 cents. 30 + 25 + 9 = 64. Bam.

Answer 3

Hi Anthony,

 

Let N = number of nickels and D = number of dimes and P = number of pennies

 

Now let's translate the English into math:

 

2 more nickels than dimes--> N = D + 2  (for example, if there are 5 dimes, then there are 7 nickels)

3x as many pennies as dimes--> P = 3D (for example, if there are 4 dimes, then there are 12 pennies)

The total value is 64 cents--> 5N + 10D + P = 64

 

Now substitute 3D for P in the last equation--> 5N + 10D + 3D = 64

 

Now substitute D+2 for N in the last equation--> 5(D+2) + 10D + 3D = 64

 

Now we can solve for D--> 5D + 10 + 10D + 3D = 64

 

18D + 10 = 64

 

18D = 54

 

D = 3

 

Then N = 5 and P = 9

 

Plug these values back into the problem to make sure they work.

 

Have fun.

 

Dan