Q 4.50 - Intermediate Algebra

Question

Translate to a system of equations and solve:

Priam has a collection of nickels and quarters, with a total value of $\$7.30$. The number of nickels is six less than

three times the number of quarters. How many nickels and how many quarters does he have?

Step-by-Step Solution

Answer

The number of nickels$=51$

The number of quarters$=19$

Step 1. Given information

• The total worth of quarters and dimes is $\$7.30$
• It is given that the number of nickels is six less than three times the number of quarters.

Step 2. Form the equations

We know that the value of each nickel is $\$0.05$ and the value of each quarter is $\$0.25$.

Let the number of nickels be $n$ and the number of quarters be $q$.

According to the question, we get:

$0.05n+0.25q=7.30\_\_\_\_\left(1\right)$

Now, It is given that the number of nickels is six less than three times the number of quarters.

So,

$n=3q-6\_\_\_\_\_\left(2\right)$

Step 3. Solve the equations by substitution method

Substituting the equation $\left(2\right)$ in equation $\left(1\right)$, we get:

$$\begin{gathered} 0.05n+0.25q=7.30 \\ 0.05(3q-6)+0.25q=7.30 \\ 0.15q-0.30+0.25q=7.30 \\ 0.40q=7.30+0.30 \\ 0.40q=7.60 \end{gathered}$$

Dividing both sides by $0.40$, we get:

$$\begin{gathered} \frac{0.40q}{0.40}=\frac{7.60}{0.40} \\ q=19 \end{gathered}$$

Step 4. Find the value of n

Substituting the value of $q=19$in the equation $\left(2\right)$, we get:

$$\begin{gathered} n=3q-6 \\ n=3\times 19-6 \\ n=57-6 \\ n=51 \end{gathered}$$

So, Priam has $51$ nickels and $19$ quraters.