Jorge has $2.10 worth of change in quarters and dimes. If he has 8 times as many dimes as quarters how many of each type does he have?

Question

Jorge has $2.10 worth of change in quarters and dimes. If he has 8 times as many dimes as quarters how many of each type does he have?

Answer 1

Although this problem is fairly easy to figure out by trial and error, we can use our knowledge of algebra to help us solve it easily as well.

 

Set up a system of linear equations relating the number of quarters Q and the number of dimes D using the information given in the problem.

First, think about how many cents there are, and how many of those cents come respectively from the Q quarters and the D dimes:

(210 cents) = (25 cents per quarter)*(Q quarters) + (10 cents per dime)*(D dimes)

 

Then, think about how many dimes there are in relation to the number of quarters:

(8 dimes per quarter)*(Q quarters) = (D dimes)

 

This second equation is a way to write the number of dimes in terms of the number of quarters:

D = 8Q

 

This can then be plugged in for the value of D in the first equation to give an equation with only one variable:

210 = 25*Q + 10*(8Q)

 

We then solve for Q:

210 = 25*Q + 80*Q

210 = 105*Q

2 = Q.

 

This tells us how many quarters we have-- 2 quarters!  We then can reason about the number of dimes by using the fact that we have 8 times as many dimes as quarters (the second equation):

D = 8*Q

D = 8*2

D = 16.

 

Indeed, sixteen dimes and two quarters is exactly $2.10, where the number of dimes is eight times the number of quarters.

Answer 2

Let the number of quarters = x

Let the number of dimes = 8x

 

 

If he combines the quarters and dimes, the total comes out to be $2.10

 

0.25x + 0.10(8x) = 2.10

 

0.25x + 0.80x = 2.10

 

1.05x = 2.10

 

    x = 2

 

Substitute this value of x into the number of quarters and dimes.

 

Jorge has 2 quarters and 16 dimes.