A collection 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.)