Question

A collection of coins has value of 64 cents. There are two more nickels than dimes and three times as many pennies as dimes in this collection. How many of each kind of coin are in the collection?

Answer 1

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d = number of dimes
n = number of nickels
p = number of pennies
The problem says there are two more nickels than dimes and three times as many pennies as dimes:
d+2(0.05)= n
3d= p
--------------------------
0.10d+0.05*n+0.001p= 0.64
Substitute by plugging into this equation the equivalent values above:
0.10*d + 0.05*d+2(0.05) + 0.01*3d= 0.64
Now solve for d:
0.10*d + 0.05*d+2(0.05) + 0.01*3d= 0.64
0.18d+0.10=0.64
0.18d= 0.54
d= 3 There are 3 dimes, and the problem says
n= d+2= 5 nickels and
p= 3d= 3*3= 9 pennies
3(0.10)+ 5(0.05)+9(0.01)= 0.64
0.30+0.25= 0.09= 0.64
0.64= 0.64 We have the correct answer

Answer 2

heya here is your answers

d = number of dimes

n = number of nickels

p = number of pennies

the problem says there are two more nickels than dimes and three times as
many pennies as dimes:
d+2(0.05)= n
3d= p
--------------------------
0.10d+0.05*n+0.001p= 0.64

Substitute by plugging into this equation the equivalent values above:

0.10*d + 0.05*d+2(0.05) + 0.01*3d= 0.64

Now solve for d:

0.10*d + 0.05*d+2(0.05) + 0.01*3d= 0.64

0.18d+0.10=0.64

0.18d= 0.54

d= 3 There are 3 dimes, and the problem says

n= d+2= 5 nickels and

p= 3d= 3*3= 9 pennies

3(0.10)+ 5(0.05)+9(0.01)= 0.64

0.30+0.25= 0.09= 0.64

i hope it helps you

:₩