Q. 160 - Intermediate Algebra

Question

Take a handful of two types of coins, and write a problem similar to Example $4.25$ relating the total number of coins and their total value. Set up a system of equations to describe your situation and then solve it.

Step-by-Step Solution

Answer

The problem is "Peter has a pocketful of quarters and dimes. The total value of the coins is $$3.80$. The number of quarter is $4$ less than twice the number of dimes. How many quarter and how many dimes does Peter have? "

The solution is that the number of dimes and quarter is $8$ and $12$.

Step 1. Given information

We are given an reference example $4.25$ and we have to setup a system of equations and solve them.

Step 2. Formation of Problem and system of equations

The problem will be: Peter has a pocketful of quarters and dimes. The total value of the coins is $$3.80$. The number of quarter is $4$ less than twice the number of dimes. How many quarter and how many dimes does Peter have?

Let $x$ and $y$ be the number of quarter and dimes,

So, the system of equation will be,

$$\begin{gathered} x=2y-4 -\left(1\right) \\ 0.25x+0.10y=3.80 -\left(2\right) \end{gathered}$$

Step 3. Solving the system of equations

Putting the value of $x$ in second equation, we get

$$\begin{gathered} 0.25x+0.10y=3.80 \\ 0.25(2y-4)+0.10y=3.80 \\ 0.50y-1+0.10y=3.80 \\ 0.60y=4.80 \\ y=\frac{4.80}{0.60} \\ y=8 \end{gathered}$$

Putting the value of $y$ in first equation, we get

$$\begin{gathered} x=2y-4 \\ x=2\left(8\right)-4 \\ x=16-4 \\ x=12 \end{gathered}$$

Hence the number of dimes is $8$ and number of quarter is $12$. 

Step 4. Checking the solution

Putting the value of $x,y$ in the second equation, we get

$$\begin{gathered} 0.25x+0.10y=3.80 \\ 0.25\left(12\right)+0.10\left(8\right)=3.80 \\ 3+0.80=3.80 \\ 3.80=3.80 \end{gathered}$$

This is true, hence the solution is correct.