A boy has twice as many nickels as pennies and five less dimes than nickels,. If he has $4.46 in all, how many of each coin does he have ?

Question

Please help! Thanks!

Answer 1

Let's first set up some variables

 

N=#Nickels

P=#Pennies

D=#Dimes

 

And then the relationship we are given in the first sentence

N = 2P or P = N/2

D = N-5

 

Then from the second sentence

.01P + .05N + .10D = 4.46

 

Using the relationships above we can restate this equation in terms of only nickels by using substitution.

 

.01(N/2) + .05N + .10(N-5) = 4.46

 

Let's multiply the entire equation by 100, both sides, to get rid of the decimal.

N/2 + 5N + 10(N-5) = 446

 

N/2 + 5N + 10N - 50 = 446

 

Rearranging and combining terms

15.5N = 496

N = 32

Then

P = N/2 = 32/2 = 16

D = N-5 = 32-5 = 27

 

He has

27 dimes

32 nickels

and

16 pennies

 

Let's double check

 

16(.01) + 32(.05) + 27(.10) = .16+1.6+2.7 = $4.46