Chapter 5.5, Problem 87STP

To determine

To check: which of the given option is the value of k in the given condition.

Answer to Problem 87STP

D. $k=\frac{13}{60}$

Explanation of Solution

Given:

The given condition is:

The positive difference between $k\text{and}\frac{1}{12}$ is same as the difference between $\frac{1}{3}\text{and}\frac{1}{5}$ .

The given options are:

  $\frac{1}{60}$

  $\frac{1}{20}$

  $\frac{1}{15}$

  $\frac{13}{60}$ .

Calculation:

According to the question, the positive difference between $k\text{and}\frac{1}{12}$ is same as the difference between $\frac{1}{3}\text{and}\frac{1}{5}$ is:

  $\begin{array}{l}\left|k-\frac{1}{12}\right|=\left|\frac{1}{3}-\frac{1}{5}\right|\\ \Rightarrow \left|k-\frac{1}{12}\right|=\left|\frac{5-3}{15}\right|\\ k-\frac{1}{12}=\pm \frac{2}{15}\\ k=\frac{1}{12}\pm \frac{2}{15}\end{array}$

  $\begin{array}{l}\therefore k_1=\frac{1}{12}-\frac{2}{15}\text{ and}{k}_2=\frac{1}{12}+\frac{2}{15}\\ k_1=\frac{5-8}{60}=\frac{-3}{60}\left\{\text{neglet the -value }\right\}\\ \text{and}{k}_2=\frac{1}{12}+\frac{2}{15}=\frac{5+8}{60}=\frac{13}{60}\end{array}$

Thus,

  $k=\frac{13}{60}$

Hence, the given option is D.