Chapter 9.6, Problem 19P

To determine

To find: the rate of the boat in still water.

Answer to Problem 19P

The rate of boat in still water is $15\text{km}/\text{h}$ .

Explanation of Solution

Given:

A motorboat goes $\text{36}$ km downstream and $\text{24}$ km upstream in the same amount of time.

Concept used:

Speed and the rate have same meaning here.

  $\text{Distance=rate}\times \text{time}$ .

In the elimination method:

Either add or subtract equation to get an equation in one variable.

When the coefficients in one variable are opposites then add the equations to eliminate the variable and when the coefficients of one variable are equal and subtract the equation to eliminate a variable.

Calculation:

Let $\text{r}$ be the rate in still water and $\text{t}$ be the time takes for both directions.

If the current is $\text{3}\text{km}/\text{h}$ ,then the rate downstream is $\text{r+3}$ and the rate upstream is $\text{r-3}$ .

Using formula:

  $\text{distance=rate×time}$ :

  $\text{36=}\left(\text{r+3}\right)\text{t}$ .

Dividing both side by $\text{r+3}$ :

  $\text{t=}\frac{\text{36}}{\text{r+3}}$ ………………. $\left(\text{i}\right)$

Similarly,

  $\text{24=}\left(\text{r-3}\right)\text{t}$ .

Dividing both side by $\text{r-3}$ :

  $\text{t=}\frac{\text{24}}{\text{r-3}}$ ……………….. $\left(\text{ii}\right)$

Equating the equation $\left(\text{i}\right)$ and $\left(\text{ii}\right)$ :

  $\frac{\text{36}}{\text{r+3}}=\frac{24}{\text{r-3}}$ .

  $\text{36}\left(\text{r-3}\right)\text{=24}\left(\text{r+3}\right)$ .

  $\text{24r+72=36r-108}$ .

  $\text{r=15}$ .

Hence, the rate of boat in still water is $15\text{km}/\text{h}$ .