Fabio has 22 coins, some of which are dimes and the rest are quarters. Altogether, Fabio's coins are worth more than $3.40.

Question

At least how many of the coins are quarters? At most how many are dimes?

 

Answer 1

He only has 2 types of coins: dimes and quarters.  Assign variables for dimes and quarters.

 

Let x = number of dimes

Let y = number of quarters

 

Set up an equation and inequalitie to represent the situation.

 

 

Since he has exactly 22 coins, we can use an equation.

 

x + y = 22

 

 

Since his total is worth more than $3.40, we can use an inequality.

 

0.10x + 0.25y > 3.40

 

 

We use the substitution method to solve the number of each coin he has.  Substitute the equation into the inequality so that the inequality is in terms of x.

 

0.10x + 0.25(22 - x) > 3.40

 

 

Now we can solve the inequality.

 

 

0.10x + 5.5 - 0.25x > 3.40

-0.15x + 5.5 > 3.40

-0.15x > -2.1

       x > 14

 

When we divide by a negative coefficient in an inequality, we flip the sign.

 

x < 14

 

 

Fabian has at most 13 dimes.

 

 

Next, we can substitute the value of x into the equation, since an equation gives you the exact and least value you can have.

 

y = 22 - 13

y = 9

 

 

Fabian has 13 dimes and 9 quarters.