Chapter 9.6, Problem 18P

To determine

To find: the distance from the marina to the destiny.

Answer to Problem 18P

The distance away from the marina is then $\text{d=48miles}\text{.}$

Explanation of Solution

Given:

A motorboat rate going against the current is $\text{20}\text{mi}/\text{h}$ and the Rate of coming back with the current is $\text{30}\text{mi}/\text{h}$ .

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 $\text{t}$ be the time going against the current. Since there is $\text{4}$ hour supply of gasoline, the time going with the current is $\text{4-t}$ .if the rate against the current is $\text{20}$$\text{mi}/\text{h}$ and the rate with the current is $\text{30}$$\text{mi}/\text{h}$ letting $\text{d}$ be the distance then gives:

Using formula:

  $\text{distance=rate×time}$ .

  $\text{d=20t}$ ………………. $\left(\text{i}\right)$

Similarly,

  $\text{d=30}\left(\text{4-t}\right)$ ………….. $\left(\text{ii}\right)$

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

  $\text{20t=30}\left(\text{4-t}\right)$ .

  $\text{20t=120-30t}$ .

Adding $\text{30t}$ to both side:

  $\text{50t=120}$ .

Dividing both side $\text{50}$ :

  $\text{t=2}\text{.4}$ .

Hence, the distance away from the marina is then $\text{d=20}\left(\text{2}\text{.4}\right)\text{=48miles}\text{.}$