Chapter 9.6, Problem 21P

To determine

To find: the rate or speed of currents.

Answer to Problem 21P

The rate of the current is $\text{540}\text{km}/\text{h}$ .

Explanation of Solution

Given:

The steam boat river queen travels at the rate of $\text{30}\text{km}/\text{h}$ in still water.

It can travel $\text{45km}$ upstream and $\text{63km}$ 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 $\text{c}$ be the rate of the current and $\text{t}$ be the time takes for both directions.

If the rate in still water is $\text{3}\text{km}/\text{h}$ ,then the rate downstream is $\text{30+c}$ and the rate upstream is $\text{30-c}$ .

Using formula:

  $\text{distance=rate×time}$ :

  $\text{63=}\left(\text{30+c}\right)\text{t}$ .

Dividing both side by $\text{30+c}$ :

  $\text{t=}\frac{\text{63}}{\text{30+c}}$ ………………. $\left(\text{i}\right)$

Similarly,

  $\text{45=}\left(\text{30-c}\right)\text{t}$ .

Dividing both side by $\text{30-c}$ :

  $\text{t=}\frac{\text{45}}{\text{30-c}}$ ……………….. $\left(\text{ii}\right)$

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

  $\frac{\text{63}}{\text{30+c}}\text{=}\frac{\text{45}}{\text{30-c}}$ .

  $\text{45}\left(\text{30+c}\right)\text{=63}\left(\text{30-c}\right)$ .

  $\text{1350+45c=1800-63c}$ .

  $\text{108c=540}$ .

Hence, the rate of the current is $\text{540}\text{km}/\text{h}$ .