Chapter 9.6, Problem 20P

To determine

To find: the rate at which she can paddle in the still water.

Answer to Problem 20P

The rate at which she can paddle in sill water is $\text{6}\text{km}/\text{h}$ .

Explanation of Solution

Given:

The rate of current in Susanna river is $4\text{km}/\text{h}$ .

A canoeist can paddle $\text{5km}$ downstream and $\text{1km}$ upstream in the same amount.

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 $4\text{km}/\text{h}$ ,then the rate downstream is $\text{r+4}$ and the rate upstream is $\text{r-4}$ .

Using formula:

  $\text{distance=rate×time}$ :

  $\text{5=}\left(\text{r+4}\right)\text{t}$ .

Dividing both side by $\text{r+4}$ :

  $\text{t=}\frac{5}{\text{r+4}}$ ………………. $\left(\text{i}\right)$

Similarly,

  $\text{1=}\left(\text{r-4}\right)\text{t}$ .

Dividing both side by $\text{r-4}$ :

  $\text{t=}\frac{1}{\text{r-4}}$ ……………….. $\left(\text{ii}\right)$

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

  $\frac{5}{\text{r+4}}\text{=}\frac{1}{\text{r-4}}$ .

  $5\left(\text{r-4}\right)\text{=1}\left(\text{r+4}\right)$ .

  $\text{r+4=5r-20}$ .

  $\text{r=6}$ .

Hence, the rate at which she can paddle in sill water is $\text{6}\text{km}/\text{h}$ .