SOLUTION: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but 3 times as many quarters as dimes. The total value of the coins was $11.40. H

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but 3 times as many quarters as dimes. The total value of the coins was $11.40. H      Log On

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Question 201312: Sam found a number of nickels, dimes, and quarters in his room. He found 6 more dimes than nickels but 3 times as many quarters as dimes. The total value of the coins was $11.40. How many coins of each type did Sam find?
6y=11.40
y=1.90
3z=11.40
z=3.80
3.80 + 1.90=5.70
11.40-5.70=5.70

Found 2 solutions by ptaylor, josmiceli:
Answer by ptaylor(2198) About Me  (Show Source):
You can put this solution on YOUR website!

Let x=number of nickels
Then x+6=number of dimes
And 3(x+6)=number of quarters
Now lets deal in pennies
5x+10(x+6)+25*3(x+6)=1140 get rid of parens and simplify
5x+10x+60+75x+450=1140 collect like terms
90x+510=1140 subtract 510 from each side
90x=1140-510=630 divide each side by 90
x=7-------------number of nickels
x+6=7+6=13-------------number of dimes
3(x+6)=3*13=39-----------number of quarters
CK
5*7+10*13+39*25=1140
35+130+975=1140
1140=1140
Hope this helps---ptaylor

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let n = number of nickels
Let d = number of dimes
Let q = number of quarters
given:
d+=+n+%2B+6
q+=+3d
5n+%2B+10d+%2B+25q+=+1140 (in pennies)
---------------------------
d+=+n+%2B+6
n+=+d+-+6
Then, by substituting,
5%2A%28d+-+6%29+%2B+10d+%2B+25%2A%283d%29+=+1140
5d+-+30+%2B+10d+%2B+75d+=+1140
90d+=+1170
d+=+13
And, since
n+=+d+-+6
n+=+13+-+6
n+=+7
Also,
q+=+3d
q+=+3%2A13
q+=+39
Sam found 7 nickels, 13 dimes and 39 quarters
check answer:
5n+%2B+10d+%2B+25q+=+1140
5%2A7+%2B+10%2A13+%2B+25%2A39+=+1140
35+%2B+130+%2B+975+=+1140
1140+=+1140
OK