Chapter 5, Problem 65A

Interpretation Introduction

Interpretation : The car’s mileage rating in kilometers per liter is to be calculated.

Concept Introduction :

The conversion factor is a numeric ratio which used to prove a measurement in one unit as another unit which is always equal to one.

Answer to Problem 65A

The rating in $\text{km/L}$ are $\text{9}\text{.1 km/L}$ and $\text{6}\text{.1 km/L}$ for highway sue and city driving respectively.

Explanation of Solution

Given information :

The car’s mileage rating for highway use is, $48\raisebox{1ex}{$mi$}\!\left/ \!\raisebox{-1ex}{$gal$}\right.$

The car’s mileage rating for city driving is, $\text{38}\raisebox{1ex}{$\text{mi}$}\!\left/ \!\raisebox{-1ex}{$\text{gal}$}\right.$

Here according to some standard conversion,

  $\text{1}\text{mile=}\text{1}\text{.6 km}$

And,

  $\text{1}\text{gallon}\text{=3}\text{.8L}$

To convert the $\text{48}\raisebox{1ex}{$\text{mi}$}\!\left/ \!\raisebox{-1ex}{$\text{gal}$}\right.$ in kilometers per liter for highway use such that,

  $\begin{array}{l}\text{48}\text{mile}\text{=}\text{48×1}\text{.6 km}=\text{72 km}\\ \text{and}\\ \text{1}\text{gallon}\text{=3}\text{.8 L}\end{array}$

So, the rating is,

  $\frac{\text{72 km}}{\text{3}\text{.8 L}}\text{=9}\text{.1 km}$

Similarly, to convert $\text{38}\raisebox{1ex}{$\text{mi}$}\!\left/ \!\raisebox{-1ex}{$\text{gal}$}\right.$ into kilometer per liter such that,

  $\begin{array}{c}\text{38}\text{mile}\text{=}\text{38×1}\text{.6 km}\\ \text{=60}\text{.8 km}\\ \text{or,}\\ \frac{\text{60}\text{.8 km}}{\text{3}\text{.8L}}\text{=6}\text{.1 km/L}\end{array}$

Conclusion

The car mileage rating for highway use and city driving in kilometer per liter are $\text{9}\text{.1 km/L}$ and $\text{6}\text{.1 km/L}$ respectively.