application problem

Question

A jar contains twice as many dimes as quarters. There are four more dimes than nickels. There is $5.30 total. How many dimes? I have to write an equation and solve.

Answer 1

 let there are x quarters in the jar

hence therefore, the numbers of dimes are 2x and that of nickels are (2x-4)

Converting the amount attributed by each denominations into cents, the jar has 20x cents in dimes, 25x cents in quarters and 5(2x-4) or (10x-20) cents in nickels. All these denominations amounts to $5.30 0r 530 cents

Therefore, 20x+25x+10x-20=530

or,                        55x-20=530...... adjusting operators

or,                             55x=550 ...... adding 20 to both sides

or,                                 x=10 .........dividing both sides by 55

 

Thus, 2x=20 and 2x-4=20-4=16

 Therefore, there are 10 quarters, 20 dimes and 16 nickels in the jar.

Answer 2

To solve this problem, we can set up an equation for the sum of all the coins and the total amount as follows:

 

(.1*dimes) + (.25*quarters) + (.05*nickels) = $5.30

 

Next we need to set variables for the coins.  We know that there are twice as many dimes as quarters.  If we set the number of quarters to x, we can say that there are 2x dimes.  We also know that there are 4 more dimes than nickels, so we can say that nickels equal 2x-4.

 

Quarters = x

Dimes = 2x

Nickels = 2x-4.

 

Now we can plug these equations into the first one and solve for x:

 

(.1*2x) + (.25*x) + (.05*(2x-4)) = $5.30

                  .2x + .25x + .1x - .2 = $5.30

                                         .55x  = $5.50

x  =  10

By plugging this value of x in for the coin equations, we find that there are 20 dimes, 10 quarters and 16 nickels.

 

We can check this answer by plugging these numbers in the first equation:

 

(.1*20) + (.25*10) + (.05*16) =

      2    +      2.5    +     0.8    =  $5.30