Chapter 6.2, Problem 25PPS

a.

To determine

To find: Time in nearest minute.

Answer to Problem 25PPS

  $8:21:00$ = $501$ min

  $9:26:16$ = $566$ min

  $8:18:37$ = $499$ min

  $9:15:54$ = $556$ min

Explanation of Solution

Given information:

In year $2000$ ,

winning time for men $8:21:00$

winning time for women $9:26:16$

In year $2012$ ,

winning time for men $8:18:37$

winning time for women $9:15:54$

  $8:21:00$ represents $8$ hours $21$ minutes $0$ sec = $\left(8\times 60+21+\frac{0}{60}\right)=\left(480+21+0\right)=501$ min

  $9:26:16$ represents $9$ hours $26$ min $16$ sec = $\left(9\times 60+26+\frac{16}{60}\right)=\left(540+26+0.26\right)\simeq 566$ min

  $8:18:37$ represents $8$ hours $18$ min $37$ sec = $\left(8\times 60+18+\frac{37}{60}\right)=\left(480+18+0.61\right)\simeq 499$ min

  $9:15:54$ represents $9$ hours $15$ min $54$ sec = $\left(9\times 60+15+\frac{54}{60}\right)=\left(540+15+0.9\right)\simeq 556$ min

So, the men’s times would be $501$ min in $2000$ and $499$ min in $2012$ .

The women’s times would be $566$ min in $2000$ and $556$ min in $2012$ .

b.

To determine

To find: Equation to represent men’s and women’s winning times $y$ in any year $x$ .

Answer to Problem 25PPS

men’s winning times, $y=-0.16x+501$

women’s winning times, $y=-0.83x+566$

Explanation of Solution

Given information:

In year $2000$ ,

winning time for men $8:21:00$

winning time for women $9:26:16$

In year $2012$ ,

winning time for men $8:18:37$

winning time for women $9:15:54$

Considering $x=0$ for year $2000$ .

For men’s times,

from given information $\left(x_1,y_1\right)=\left(0,501\right)$ and $\left(x_2,y_2\right)=\left(12,499\right)$

where $y$ represents time and $x$ represents years.

From two-point slope form of a line,

  $\left(y-y_1\right)=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$

  $\begin{array}{l}\Rightarrow \left(y-501\right)=\left(\frac{499-501}{12-0}\right)\left(x-0\right)\\ \Rightarrow \left(y-501\right)=\left(\frac{-2}{12}\right)x\\ \Rightarrow \left(y-501\right)=-0.16x\\ \Rightarrow y=-0.16x+501\end{array}$

For women’s times,

from given information $\left(x_1,y_1\right)=\left(0,566\right)$ and $\left(x_2,y_2\right)=\left(12,556\right)$

where $y$ represents time and $x$ represents years.

From two-point slope form of a line,

  $\left(y-y_1\right)=\left(\frac{y_2-y_1}{x_2-x_1}\right)\left(x-x_1\right)$

  $\begin{array}{l}\Rightarrow \left(y-566\right)=\left(\frac{556-566}{12-0}\right)\left(x-0\right)\\ \Rightarrow \left(y-566\right)=\left(\frac{-10}{12}\right)x\\ \Rightarrow \left(y-566\right)=-0.83x\\ \Rightarrow y=-0.83x+566\end{array}$

c.

To determine

To find: The year in which men’s and women’s winning times to be same.

Answer to Problem 25PPS

  $97.01$ years

Explanation of Solution

Given information:

In year $2000$ ,

winning time for men $8:21:00$

winning time for women $9:26:16$

In year $2012$ ,

winning time for men $8:18:37$

winning time for women $9:15:54$

From given information,

men’s winning times = women’s winning times

  $\begin{array}{l}\Rightarrow -0.16x+501=-0.83x+566\\ \Rightarrow -0.16x+0.83x=-501+566\\ \Rightarrow 0.67x=65\\ \Rightarrow x=97.01\end{array}$

So therefore, after nearly $97.01$ years after $2000$ , the winning times for men’s and women’s will be same.