Paul has three times as many dimes as nickels and twice as many quarters as dimes. In all she has $5.55. How many coins of each type does she have?

Question

Paul has three times as many dimes as nickels and twice as many quarters as dimes. In all she has $5.55. How many coins of each type does she have?

Answer 1

We know that a certain number of nickels, dimes, and quarters add up to $5.55.  The key seems to be the number of nickles.  Let's make that number X, and the amount would be .05X (since a nickle is $.05).  Since the dimes are 3 times that number, and the amount of each dime is .10, we have .30X.  Now, for quarters we have 6 times the number of nickles, and each amount is .25.  So we need to use 1.5X for quarters. 

 

Here's the formula put together:

.05X + .3X + 1.5X = 5.55

 

From there-

1.85X = 5.55

X = 3

 

So, there would be 3 nickles, 9 dimes, and 18 quarters.

 

You can double check this with (3 x .05)+(9 x .10) + (18 x .25)

Answer 2

This is a system of equations questions. In order to solve you need to set up three equations for your three unknowns. Lets start by setting up the variables.

 

d=number of dimes

n=number of nickels

q=number of quarters

 

Now we will set up our equations.

We know that we have 3 times as many dimes as nickels. Meaning that 3 times nickels equals dimes. To set this up as an equation it will be:

d=3n

 

We know that we have 2 times as many quarters as dimes. Meaning 2 times the number of dimes equals quarters.

q=2d

 

We also know that in total we have $5.55 worth of dimes, nickels, and quarters. Remembering that dimes are worth $.10, nickels are worth $.05, and quarters are worth $.25 we set it up as follows.

.10d+.05n+.25q=5.55

This says that the number of dimes multiplied by how much they are worth plus the number of nickels we have times their worth plus the number of quarters we have times their worth equals our total amount of $5.55

 

Now to Solve the three equations.

 

d=3n              (1)

q=2d              (2)

.10d + .05n + .25q = 5.55              (3)

 

Using the substitution method first plug equation 1 into 2

 

q=2(3n)

q=6n               (4)

 

Now plug (1) and (4) into (3)

 

.10(3n) + .05n + .25(6n) =5.55 (now solve for n)

.30n + .05n + 1.5n =5.55

1.85n            =5.55

n  =  5.55/1.85

n=3

 

Now plug n into equation (1)

d=3(3)

d=9

 

Now plug d into equation (2)

q=2(9)

q=18

 

So you have 3 nickels, 9 dimes, and 18 quarters. 

 

You can of course plug back into equation 3 to verify.

.10(9) + .05(3) + .25(18) =

0.90 + 0.15 + 4.50 = $5.55       (Confirmed)