The concept of rate in maths and algebra is an important one. Detailed solutions to the questions on finding rate are presented.
The unit rate is the distance divided by the time: \[ \dfrac{300 \ \text{kilometers}}{5 \ \text{hours}} = \left(\dfrac{300}{5}\right) \ (\text{km/hour}) = 60 \ \text{km/hour}. \] Note: This unit rate is also called speed.
\[ \dfrac{10 \ \text{dollars}}{4 \ \text{minutes}} = \left(\dfrac{10}{4}\right) \ (\text{dollars/minute}) = 2.5 \ \text{dollars/minute}. \]
\[ \dfrac{18 \ \text{pages}}{9 \ \text{minutes}} = \left(\dfrac{18}{9}\right) \ (\text{pages/minute}) = 2 \ \text{pages/minute}. \]
\[ \dfrac{240 \ \text{miles}}{12 \ \text{gallons}} = \left(\dfrac{240}{12}\right) \ (\text{miles/gallon}) = 20 \ \text{miles/gallon}. \]
\[ \dfrac{45 \ \text{liters}}{5 \ \text{minutes}} = \left(\dfrac{45}{5}\right) \ (\text{liters/minute}) = 9 \ \text{liters/minute}. \]
\[ \dfrac{16 \ \text{dollars}}{4 \ \text{kilograms}} = \left(\dfrac{16}{4}\right) \ (\text{dollars/kilogram}) = 4 \ \text{dollars/kilogram}. \]
Find the unit rate for each object: \[ \text{Object A: } \dfrac{15}{5} = 3 \ \text{cm/second}, \qquad \text{Object B: } \dfrac{24}{8} = 3 \ \text{cm/second}. \] Both objects move at the same speed.
\[ \text{Car A: } \dfrac{240}{12} = 20 \ \text{miles/gallon}, \qquad \text{Car B: } \dfrac{550}{25} = 22 \ \text{miles/gallon}. \] Car B travels further per gallon.
Since \(1 \ \text{hour} = 60 \ \text{minutes}\), \[ 60 \ \dfrac{\text{km}}{\text{hour}} = \dfrac{60}{60} \ \dfrac{\text{km}}{\text{minute}} = 1 \ \text{km/minute}. \]
Since \(1 \ \text{km} = 1000 \ \text{m}\) and \(1 \ \text{hour} = 3600 \ \text{seconds}\), \[ 72 \ \dfrac{\text{km}}{\text{hour}} = \dfrac{72 \times 1000}{3600} \ \dfrac{\text{m}}{\text{second}} = 20 \ \text{m/second}. \]