Help! Math problem!

Home Forums Decaffeinated Coffee Help! Math problem!

Viewing 6 posts - 1 through 6 (of 6 total)
  • Author
    Posts
  • #601178
    supergirl613
    Member

    K, so I have this math problem that I have absolutely have NO clue how to do it. It’s an inequalities problem I think.

    Jo has 37 coins(nickels, dimes and quarters) equaling up to $5.50. She has four more quarters than nickels.

    How many dimes does she have?

    Can someone please help me?? I’m really stuck!!!!

    #835743

    # P T

    nickels x 5 5x

    dimes 37-(x+x+4) 10 10(37-x+x+4)

    quarters x+4 25 25(x+4)

    5x+10(37-x+x+4)+25(x+4)=5.50

    I hope that helps…that’s about all I remember 🙂

    I don’t know why its not staying in chart form, sorry!

    #835744
    Lifeisgreat
    Member

    Let x=number of nickels

    Then x+4=number of quarters

    So the number of dimes would be:

    37-x-(x+4)=

    37-2x-4=

    33-2x —- is the number of dimes

    (lets deal in pennies)

    5x+25(x+4)+10(33-2x)=550

    so 5x+25x+100+330-20x=550

    collect like terms

    10x+430=550

    subtract 430 from each side

    10x+430-430=550-430=

    10x=120

    divide each side by 10

    x=12


    number of nickels

    x+4=12+4=16


    number of quarters

    33-2x=33-24=9


    number of dimes

    Check


    5*12=60

    25*16=400

    9*10=90

    60+400+90=550

    550=550

    #835745
    supergirl613
    Member

    Please someone help!!!

    #835746
    Razzle
    Member

    Just google the question and a whole list of explanations come up. the answer is 9 dimes.

    #835747

    Here you go:

    Set the number of nickels as “n”, the number of dimes as “d”, and the number of quarters as “q”.

    There are a total of 37 coins. Therefore:

    n+d+q=37

    You also know that the total in dollar amounts is equal to $5.50. Therefore:

    .05n+.10d+.25q=5.50

    In the first equation, we can set q = n+4 (as there are 4 more quarters than nickels). Resetting the first equation:

    n+(n+4)+d=37

    Solve for “d” in terms of “n”:

    d=33-2n

    Now we set the second equation (using q=n+4):

    .05n+.10d+.25(n+4)=5.50

    Now you can plug the “d=33-2n” value to the above equation to solve for “n”:

    .05n+.10(33-2n)+.25(n+4)=5.50

    Do all the annoying math, and the final answer is n=12. That gives you the number of nickels. Finally:

    #n=12

    #q=16

    #d=37-28=9

    and that’s your answer! Double check your work and make sure the math is right:

    .05(12)+.1(9)+.25(16)=5.5

    Hope that answers your question! Just keep in mind the moral of the equation: if you have multiple variables (like you do here), you need to set up equations to solve for a variable in terms of a main variable – which in this case I made “n” as our main variable.

Viewing 6 posts - 1 through 6 (of 6 total)
  • You must be logged in to reply to this topic.