Results 1 to 9 of 9

Math Help - Coins in a jar word problem

  1. #1
    rex
    rex is offline
    Newbie
    Joined
    Jul 2010
    Posts
    6

    Coins in a jar word problem

    I just joined today, thank you for making this forum. I will be coming here often, again thank you all so much.

    Here is the problem.
    The problem starts with coins in a jar and I must come up with the number of coins of each denomination. The number of quarters is unknown, number of dimes unknown, number of nickels unknown, and number of pennies unknown.

    Here on the known parts of the equation.
    - There are a total of 471 coins worth $43.13.

    - Two Susan B. Anthony dollars, a bunch of quarters, a whole lot of dimes, a good number of nickels and a fair quantity of pennies.
    - The number of nickels is equal to the number of quarters plus the number of dollars.....{so I figure this, n = q + 2}
    The number of pennies is only one more than the number of dimes. .. {so I figure this, p + 1 = d}

    A. Generate and solve the linear equations for this problem.
    B. How many coins, of each type (Susan B Anthony dollars, quarters, dimes, nickels, and pennies) are in the jar?


    Thank you all in advance. I will make a second post right after this one showing what I have so far. Please tell me what I do and do not have correct.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    rex
    rex is offline
    Newbie
    Joined
    Jul 2010
    Posts
    6
    pennies, p = d - 1
    dimes, d = p + 1
    nickels, n = q + 2
    quarters, q = n - 2
    and the number of dollar coins is known, there are 2 susan b anthony coins.

    So I think from this I make my main equation which is this,
    p + d + n + q + 2 = 471

    Once I have the number of coins for each unknown I will have their worth, the total worth of all the coins in the jar is $43.13.
    so now I believe I use this equation,
    p + d + n + q + 2.00 = $43.13

    I know I should just now put all my values listed above into the larger equation but am running into trouble. Please some guidance everybody, thanks again in advance.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    634
    Hello, Rex!

    Welcome aboard!


    There are a total of 471 coins worth $43.13 in the jar: two Susan B. Anthony dollars,
    some quarters, some dimes, some nickels and some pennies.

    The number of nickels is equal to the number of quarters plus the number of dollars.
    The number of pennies is one more than the number of dimes.

    A. Generate and solve the linear equations for this problem.

    B. How many coins of each type are in the jar?

    The jar contains:

    . . \begin{array}{cc}2 & \text{dollars} \\ Q & \text{quarters} \\ D & \text{dimes} \\ Q+2 & \text{nickels} \\ D+1 & \text{pennies} \end{array}



    We have this chart:

    . . \begin{array}{c|cc|ccc|}& \text{coins} & &&  \text{value} \\ \cline{2-3} \cline{4-5}<br />
& 2 & \text{dollars} && 200 \\<br />
& Q & \text{quarters} && 25Q \\<br />
& D & \text{dimes} && 10D \\<br />
& Q+2 & \text{nickels} && 5(Q+2) \\<br />
& D+1 & \text{pennies} && D+1 \\ \cline {2-3} \cline{4-5}<br />
\text{Total:} & 2Q+2D+5 & & & 30Q + 11D + 211 \end{array}


    We have equations: . \begin{array}{ccc}2Q + 2D + 5 &=& 471 \\ 30Q + 11D + 211 &=& 4313 \end{array}


    Solve the system: . \begin{array}{ccc} Q + D &=& 233 \\ 3Q + 11D &=& 4102 \end{array}


    . . and get: . \begin{Bmatrix} 2\text{ dollars} \\ 81 \text{ quarters} \\ 152\text{ dimes} \\ 83\text{ nickels} \\ 153\text{ pennies} \end{Bmatrix}

    Follow Math Help Forum on Facebook and Google+

  4. #4
    rex
    rex is offline
    Newbie
    Joined
    Jul 2010
    Posts
    6
    Thank you so much Soroban.
    I was complicating it more and using four variables, you showed me that I really only had to worry about two. Thanks again.

    Your chart was very helpful to me. How do I go from the equation, which I now know how to build, to finding each coin type?
    I go from; 2Q + 2D + 5 = 471
    to 1; Q = -D -233
    to 2; 2(-D -233) + 2D = 471 and this gives me no answer cause I'm left with nothing.

    Where am I messing up? I still don't really understand how to go from equation to coins.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Quote Originally Posted by rex View Post
    I go from; 2Q + 2D + 5 = 471
    to 1; Q = -D -233
    Rex, are you learning on your own, or do you attend math classes?
    That should be Q = 233 - D.

    Have a peek here:
    SOLVING EQUATIONS
    Follow Math Help Forum on Facebook and Google+

  6. #6
    rex
    rex is offline
    Newbie
    Joined
    Jul 2010
    Posts
    6
    I have not been in a math class for some long time.
    But I do see my obvious error now. Thanks Wilmer.

    Still, I don't quite understand how I go from Q = 233 - D to all the numbers of coins.
    I see that the equations work to get it. But my seeing the whole process is still off.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    rex
    rex is offline
    Newbie
    Joined
    Jul 2010
    Posts
    6
    2Q+2D+5=471
    Q=-D+233
    D=-Q+233

    30Q+11D+211=4313
    Q = (-11D + 4102) / 30
    D = (-2 (15Q - 2051)) / 11

    This is what I come up with still.
    How do you go from the equations and systems you have, to finding the number of coins and the values?
    I am pretty sure I am doing algebra correctly here. Given the equations that is all I can come up with. So do I split the 233 in two values? One is +2 than the other? I am still fully confused on the problem and would really appreciate more explanation on this problem. Thank you.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Dec 2007
    From
    Ottawa, Canada
    Posts
    3,102
    Thanks
    68
    Ok Rex, time to get you out of your misery!

    Forget about the 2 $1 coins: we have 469 coins totalling $41.13

    Always a good idea to number equations; easier to refer to; we have:
    Q + D + N + P = 469 [1]
    N = Q + 2 [2] : The number of nickels is equal to the number of quarters plus 2
    P = D + 1 [3] : The number of pennies is only one more than the number of dimes

    Next step is substitute [2] and [3] in [1]:
    Q + D + Q+2 + D+1 = 469
    2Q + 2D = 466
    Q + D = 233
    D = 233 - Q [4]

    Next step is getting P in terms of Q; substitute [4] in [3]:
    P = 233-Q + 1
    P = 234 - Q [5]

    Now that we have D, N and P in terms of Q, we do the "value" work:
    .25Q + .10D + .05N + .01P = 41.13 ; multiply by 100 to get rid of decimals:
    25Q + 10D + 5N + 1P = 4113 [6]

    Now, substitute [4], [2] and [5] in [6]:
    25Q + 10(233 - Q) + 5(Q + 2) + 1(234 - Q) = 4113
    25Q + 2330 - 10Q + 5Q + 10 + 234 - Q = 4113
    19Q = 1539
    Q = 1539/19 = 81 : so we have 81 quarters.

    [2]: N = 81 + 2 = 83
    [4]: D = 233 - 81 = 152
    [5]: P = 234 - 81 = 153

    Hope that was clear...
    Follow Math Help Forum on Facebook and Google+

  9. #9
    rex
    rex is offline
    Newbie
    Joined
    Jul 2010
    Posts
    6
    omg, thank you so much Wilmer.

    That i get a whole lot easier.
    I'm still going to do it on paper and force the learning into my brain.

    THANK YOU!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Money/Coins Problem
    Posted in the Math Puzzles Forum
    Replies: 4
    Last Post: October 8th 2011, 10:29 AM
  2. Replies: 2
    Last Post: March 12th 2011, 11:15 AM
  3. coins/inequality word problem
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 9th 2010, 12:22 PM
  4. word problem with coins
    Posted in the Algebra Forum
    Replies: 1
    Last Post: November 1st 2008, 01:56 PM
  5. rolling dices and coins problem
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: September 4th 2007, 09:37 AM

Search Tags


/mathhelpforum @mathhelpforum