Chapter 1.9, Problem 11P

To determine

To calculate:

The kinds of the coin are there

Answer to Problem 11P

The total number of the dimes is $34$ and quarters are $6$ .

Explanation of Solution

Given information:

Number of the dimes: $x$

Number of the quarters: $40-x$

Total value: $4.9$

Calculation:

Let the number of the dimes be $x$ . The number of the quarters be $40-x$ .

The equation relating the value of the type of coin and the total value:

  $A$ : Number of the dimes

  $B$ : Number of the quarters

  $T$ : Total value

  $\frac{1}{10}\left(A\right)+\frac{1}{4}\left(B\right)=T\mathrm{...}\left(1\right)$

Put the value of $A$ and $B$ in equation (1):

  $\frac{1}{10}\left(x\right)+\frac{1}{4}\left(40-x\right)=4.9$

Use the distributive property:

  $\begin{array}{l}\frac{1}{10}x+\frac{40}{4}-\frac{x}{40}=4.9\\ \frac{1}{10}x+10-\frac{x}{40}=4.9\\ \end{array}$

Then subtract $10$ from both sides:

  $\begin{array}{l}\frac{1}{10}x-\frac{x}{4}+10-10=4.9-10\\ \frac{1}{10}x-\frac{x}{4}=-5.1\mathrm{...}\left(2\right)\end{array}$

The lower common denominator for equation (2) is $40$ .

  $\begin{array}{l}\frac{1}{10}\left(40\right)-\frac{x}{4}\left(40\right)=\left(-5.1\right)40\\ 4x-10x=-204\\ -6x=-204\end{array}$

Divide both sides by $-6$ :

  $\begin{array}{l}\frac{-6x}{-6}=\frac{-204}{-6}\\ x=34\end{array}$

Substitute the value of $x$ in $40-x$ :

  $\begin{array}{l}=40-x\\ =40-34\\ =6\end{array}$

The number of dimes is $34$ and number of the quarters is $6$ .

The total money gives $\$4.90$ :

  $\begin{array}{l}\therefore \frac{34}{10}+\frac{6}{4}=4.9\\ \therefore 3.4+1.5=4.9\\ \therefore 4.9=4.9\end{array}$

Hence, the total number of the dimes is $34$ and quarter is $6$ .