Chapter 9.3, Problem 19P

To determine

To find: The required numbers.

Answer to Problem 19P

The larger number is $24$ and the smaller number is $8$ .

Explanation of Solution

Given:

The first number is 16 more than the other number.

The smaller number when subtracted from two third of the other number the result is one fourth of the sum of the two numbers.

Calculation:

Consider the numbers are $x$ and $y$ , where $x$ is the smaller number.

Since, one number is 16 more than the other number than,

  $y=16+x$

Consider smaller number when subtracted from two third of the other number the result is one fourth of the sum of the two numbers. Then,

  $\frac{2}{3}y-x=\frac{1}{4}\left(x+y\right)$

Solve further as,

  $\begin{array}{l}\frac{2}{3}\left(16+x\right)-x=\frac{1}{4}\left(x+16+x\right)\\ 8x+128-12x=6x+48\\ 80=10x\\ x=8\end{array}$

Then, the larger number is,

  $\begin{array}{c}y=16+8\\ =24\end{array}$