Question 694642: Stella is looking for the coins that fell through the hole in her pocket. Together the 12 coins are worth $1.75. She remembers that there are four different kinds of coins. She had 3 times as many of one kind as another kind. what coins could Stella have lost?
Answer by ankor@dixie-net.com(22740)   (Show Source):
You can put this solution on YOUR website! Stella is looking for the coins that fell through the hole in her pocket.
Together the 12 coins are worth $1.75.
4 unknowns here, have to make some logical assumptions
:
p + n + d + q = 12
and
.01p + .05n + .10d + .25q = 1.75
:
She had 3 times as many of one kind as another kind.
Assume
q = 3n
:
Assume 5 pennies, (has to be a multiple of 5, & 10 would only leave 2 more coins)
Subtract 5 cents and write the equation:
n + d + q = 7
and
.05n + .10d + .25q = 1.70
replace q with (3n)
.05n + .10d + .25(3n) = 1.70
.05n + .10d + .75n = 1.70
.80n + .10d = 1.70
mult by 10
8n + 1d = 17
d = -8n + 17
Assume n = 1, only value that seems reasonable
d = -8 + 17
d = 9 dimes, 1 nickel, then 3 quarters
:
Let's say, 5 pennies, 1 nickel, 9 dimes, 3 quarters
:
.05 + .05 + .90 + .75 = 1.75
:
Hold everything, I forgot about the 12 coins, this is not right
n + d + q = 7
How about 1 nickel 3 dimes, 4 quarters, but that would only give you 1.35
I don't think this can be done with 12 coins if we use pennies.
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I asked my Grandson Donovan about it and he said use a half dollar as one
of the coins,
I came up with
6 nickels, 2 dimes, 3 quarters, 1 half dollar, 12 coins
.30 + .20 + .75 + .50 = 1.75
and
n = 3q
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