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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asked by Kelley
  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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