Chapter 9.5, Problem 5CYP

To determine

To find the value of $t_2$

Answer to Problem 5CYP

the value of $t_2$ is $-80$

Explanation of Solution

Given information:

p varies directly as r and inversely as t

the value of $t_1=20{\text{ when p}}_1=4{\text{ and r}}_1=2$

Calculation:

The values of $t_2$ can be obtained as

First set up a correct proportion for the information given

  $\begin{array}{l}{p}_1=\frac{k{r}_1}{t_1}{\text{ and p}}_2=\frac{k{r}_2}{t_2}\text{ [since, }r\text{ varies directly as }p\text{,so r goes in the numerator and}\\ h\text{ varies inversely as }p\text{,so h goes in the denominator]}\\ k=\frac{p_1{t}_1}{r_1}\text{ and }k=\frac{p_2{t}_2}{r_2}\text{ [solve for }k\text{]}\\ \frac{p_1{t}_1}{r_1}=\frac{p_2{t}_2}{r_2}\text{ [set the two proportion equal to each other]}\end{array}$

Put the values of $p_1=4$ , $r_1=2$ , $r_2=10$ , $t_1=20\text{ and }{p}_2=-5$ in above equation

  $\begin{array}{l}\frac{\left(4\right)\left(20\right)}{2}=\frac{\left(-5\right)\left(t_2\right)}{10}\\ \left(20\right)\left(10\right)\left(4\right)=\left(-5\right)\left(2\right)\left(t_2\right)\left[\text{cross multiply}\right]\\ 800=-10t_2\left[\text{simplify}\right]\\ t_2=-80\left[\text{divide each side by 10}\right]\end{array}$

Hence,

the value of $t_2$ is $-80$