Hi Emma!
I'm going to represent the coin value in terms of cents
to avoid decimals.
One equation involving the value of the coins is:
5N + 10D + 25Q = 20,000 ($200 is 20000 cents)
We have another restriction on number of coins:
N + D + Q = 1000
Simplify the first equation by dividing through by 5:
5N + 10D + 25Q = 20,000
---- ---- ---- ---------
5 5 5 5
N + 2D + 5Q = 4000
Now, subtract the two systems to eliminate N.
N + 2D + 5Q = 4000
-- (N + D + Q = 1000)
------------------------------
D + 4Q = 3000
Solving this for D, you have:
D = 3000-4Q
Factor out a 4 to simplify
D = 4(750 - Q)
The number of quarters will be the controlling
factor to the number of different possibilities.
Q can't be higher than 750, because then you
would have a negative number of dimes (not possible).
With 750 quarters you will have 0 dimes (plug into the
above equation to see this). Since N + D + Q = 1000,
then you will have 250 nickels. The upper limit for Q
is 750.
Need to find a lower limit constraint for Q.
N + D + Q = 1000
N + 3000-4Q + Q = 1000 D = 3000-4Q (from above)
N + 3000 - 3Q = 1000
-3000 +3Q -3000 + 3Q (solve for N)
---------------------------------------
N = -2000 + 3Q
The number of nickels, N, must also
be >=0.
.
N >= 0
-2000 + 3Q >= 0
+2000 +2000 (add 2000 to both sides)
-------------------------------
3Q >= 2000
--- ------- divide both sides by 3
3 3
The result for Q here is a decimal that
rounds up to 667.
The lower limit for Q is 667. This amounts to
332 dimes and 1 nickel (plug in above to get the
number of other coins).
The range of Q values vary between 667 and 750.
So the number of possibilities of coins, ranging
between 667 and 750 for Q, is 84.
