SOLUTION: One night a waiter recieved $11.30 in tips-all coins. He had 3 times as many dimes as nickels, and 10 more quarters than dimes. Find the number of each type of coin. I dont tr

Algebra ->  Customizable Word Problem Solvers  -> Coins -> SOLUTION: One night a waiter recieved $11.30 in tips-all coins. He had 3 times as many dimes as nickels, and 10 more quarters than dimes. Find the number of each type of coin. I dont tr      Log On

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Question 524851: One night a waiter recieved $11.30 in tips-all coins. He had 3 times as many dimes as nickels, and 10 more quarters than dimes. Find the number of each type of coin.

I dont truly understand how to find the formula for this -__- plzz help!!!!

Found 2 solutions by mananth, josmiceli:
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
nickels = n
dimes = 3n
quarters = 3n+10

$11.30 = 1130 cents
5n+10*3n+25(3n+10)=1130
5n+30n+75n+250=1130
110n=1130-250
110n=880
/110
n=8 nickels
dimes = 3n = 24 dimes
quarters = 3n+10= 34
Check
34*25 + 24*10+8*5 = 1130

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let d = number of dimes
Let n = number of nickels
Let q = number of quarters
given:
(1) +d+=+3n+
(2) +q+=+d+%2B+10+
(3) +10d+%2B+5n+%2B+25q+=+1130+ ( in cents )
This is 3 equations with 3 unknowns,
so it's solvable
-------------------
(1) +n+=+d%2F3+
Substitute (1) and (2) into (3)
(3) +10d+%2B+5%2A%28d%2F3%29+%2B+25%2A%28d%2B10%29+=+1130+
(3) +10d+%2B+%285d%29%2F3+%2B+25d+%2B+250+=+1130+
Multiply both sides by 3
(3) +30d+%2B+5d+%2B+75d+%2B+750+=+3390+
(3) +110d+=+3390+-+750+
(3) +110d+=+2640+
(3) +d+=+24+
and, since
(1) +n+=+d%2F3+
(1) +n+=+24%2F3+
(1) +n+=+8+
also,
(2) +q+=+d+%2B+10+
(2) +q+=+24+%2B+10+
(2) +q+=+34+
There are 24 dimes, 8 nickels, and 34 quarters
check:
(3) +10d+%2B+5n+%2B+25q+=+1130+
(3) +10%2A24+%2B+5%2A8+%2B+25%2A34+=+1130+
(3) +240+%2B+40+%2B+850+=+1130+
(3) +1130+=+1130+
OK