Acollection of nickels, dimes and quarters totals $6.00. if there are 52 coins altogether and twice as many dimes as nickels, how many of each kind of coin are there?

a.
q = 5
d = 60
n = 30

c.
q = 15
d = 28
n = 6

b.
q = 35
d = 10
n = 5

d.
q = 10
d = 28
n = 14

D=2n:  there are twice as many dimes as nickels.

n+d+q=52:  a total of 52 coins, using d from above in this gives you:

n+2n+q=52  combine like terms on left side

3n+q=52  subtract 3n from both sides

q=52-3n

Then you are told that the coins have a total value of $6 which is equal to 600 cents.  So we can say:

25q+10d+5n=600  divide all terms by 5

5q+2d+n=120, and from earlier we saw q=52-3n and d=2n so you have:

5(52-3n)+2(2n)+n=120  expanding...

260-15n+4n+n=120  combining like terms

260-10n=120  subtract 260 from both sides

-10n=-140  divide both sides by -10

n=14, since q=52-3n and d=2n

q=(52-3(14))=52-42=10 

d=2(14)=28

So there are 10 quarters, 28 dimes, and 14 nickels or as your choices put it:

q=10, d=28, n=14  (the correct answer is d.)

What am i doing here. add?