# Math

A bag contains twice as many pennies as nickels and four more dimes than quarters. Find all possibilities for the number of each coin if their total value is \$2.01. I have pennies = 4n nickels = n dimes = 4 + q quarters = q but I don't know where to go from there?

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1. "twice as many pennies as nickels"
number of nickels ----- x
number of pennies --- 2x
"four more dimes than quarters"
number of quarters --- y
number of dimes ---- y+4

5x + 2x + 25y + 10(y+4) = 201
7x + 35y = 161
divide by 7
x + 5y = 23
y = (23-x)/5

since y must be a whole number, 23-x must be a multiple of 5
x y, there are only 4 cases
3 4
8 3
13 10
18 1

check:
case 1: x=3, y = 4
nickels = 3, pennies= 6, quarters = 4, dimes = 8
value = 15+ 6 + 100 + 80 = 201 , yup!!

case 2: x = 8, y = 3
nickels = 8 , pennies = 16, quarters = 3, dimes = 7
value = 40 + 16 + 75 + 70 = 201 , yup!!

I will leave it up to you to verify the other two cases
and then to state a conclusion

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posted by Reiny
2. you forgot to add up all the values.
p+5n+10d+25q = 201
Now, as you say, you also have
p = 2n (twice as many, not 4x as many)
d = 4+q

Now, there may not be a unique solution, since you have only three equations for four unknowns. But let's start substituting and see where we get:
p+5n+10d+25q = 201
2n+5n+10d+25q = 201
7n + 10(4+q)+25q = 201
7n + 40 + 10q + 25q = 201
7n+35q = 161
n+5q = 23
Clearly, q <= 4, so check the possibilities:
q=4, n=3
q=3, n=8
q=2, n=13
q=1, n=18
q=0, n=23

Check the combinations:
(p,n,d,q) = (6,3,8,4): 6+3*5+8*10+4*25 = 6+15+80+100 = 201
and you can check the others...

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posted by Steve

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