collection of 41 coins (nickles,dimes,quarters) worth $4.75. There are twice as many quarters as dimes. How many coins of each type?

Question

I have done so many of these but for some reason I cant figure a couple out, please help.

Answer 1

Hi Caleb,

 

Let's begin by seeing which one of the unknowns can be used to reference the other two unknowns.  Looks like the smallest quantity to be expressed is the number of dimes:

    x (number of dimes)

 

Now we'll express the other two unknowns in terms of the number of dimes:

    2x (number of quarters)

    41 - x - 2x (number of nickels)

 

Now we can write an equation to express the given total amount of change in terms of our unknown expressions for the numbers of dimes, quarters, and nickels:

     (0.10)(x) + (0.25)(2x) + (0.05)(41 - x - 2x) =
     4.75

 

Now we'll solve our equation for x (number of dimes):

    0.10x + 0.50x + 2.05 - 0.05x - 0.10x = 4.75

    0.45x + 2.05 = 4.75

    0.45x = 4.75 - 2.05

    0.45x = 2.70

          x = 2.70/0.45

          x = 6 (number of dimes)

 

Finally we'll determine the numbers of quarters and nickels using our expressions for these numbers in terms of x (number of dimes):

  2x = 2 x 6 = 12 (number of quarters)

  41 - x - 2x = 41 - 6 - 12 = 23 (number of nickels)

 

We can verify that our answers are correct by plugging these values into our equation to see if all the coin amounts add up to the given total amount:

    (0.10)(x) + (0.25)(2x) + (0.05)(41 - x - 2x) =
    4.75

    (0.10)(6) + (0.25)(12) + (0.05)(23) = 4.75

    0.60 + 3.00 + 1.15 = 4.75

    3.60 + 1.15 = 4.75

    4.75 = 4.75 (answers are correct) 

 

The trick to solving this problem was choosing the right unknown to use for expressing the other two unknowns.

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan