35. eileen ran at a speed of 4.8 km/h to complete a 4-km run. sally ran at a speed of 6 km/h to complete the same distance. how much longer did eileen take to complete her run than sally? express your answer in minutes.

36. peter set off from town a at 10 am, driving at an average speed of 84 km/h. he reached town b at 2 pm. if william set off 1 hour 25 minutes earlier than peter and took the same route at an average speed of 70 km/h, at what time would william reach town b?

37. tracy took $2 \frac{1}{3}$ hours to travel $\frac{4}{9}$ of the distance between town x and town y at an average speed of 72 km/h. what was jessicas average speed if she took $4 \frac{1}{5}$ hours to travel from town x to town y?

Question

35. eileen ran at a speed of 4.8 km/h to complete a 4-km run. sally ran at a speed of 6 km/h to complete the same distance. how much longer did eileen take to complete her run than sally? express your answer in minutes.

36. peter set off from town a at 10 am, driving at an average speed of 84 km/h. he reached town b at 2 pm. if william set off 1 hour 25 minutes earlier than peter and took the same route at an average speed of 70 km/h, at what time would william reach town b?

37. tracy took $2 \frac{1}{3}$ hours to travel $\frac{4}{9}$ of the distance between town x and town y at an average speed of 72 km/h. what was jessicas average speed if she took $4 \frac{1}{5}$ hours to travel from town x to town y?

Accepted Answer

{ "quality": "clear", "question_count": 3, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Average Speed", "Time Distance Speed", "Unit Conversion" ], "new_concepts": [], "current_concepts": [ "Average Speed", "Time Distance Speed", "Unit Conversion" ] } ## Solve Question 35 Using the *Time Distance Speed* and *Unit Conversion* knowledge points $$ \begin{aligned} &t_{\text{Eileen}} = \frac{4\text{ km}}{4.8\text{ km/h}} = \frac{5}{6}\text{ hours} = \frac{5}{6} \times 60\text{ min} = 50\text{ minutes}\ &t_{\text{Sally}} = \frac{4\text{ km}}{6\text{ km/h}} = \frac{2}{3}\text{ hours} = \frac{2}{3} \times 60\text{ min} = 40\text{ minutes}\ &\Delta t = 50\text{ min} - 40\text{ min} = 10\text{ minutes} \end{aligned} $$ ## Solve Question 36 Using the *Time Distance Speed* and *Average Speed* knowledge points $$ \begin{aligned} &t_{\text{Peter}} = 4\text{ hours}\quad (\text{from 10 am to 2 pm})\ &d = 84\text{ km/h} \times 4\text{ h} = 336\text{ km}\ &t_{\text{William}} = \frac{336\text{ km}}{70\text{ km/h}} = 4.8\text{ hours} = 4\text{ hours } 48\text{ minutes}\ &\text{Start}_{\text{William}} = 10\text{ am} - 1\text{ hour } 25\text{ minutes} = 8:35\text{ am}\ &\text{Arrival}_{\text{William}} = 8:35\text{ am} + 4\text{ hours } 48\text{ minutes} = 1:23\text{ pm} \end{aligned} $$ ## Solve Question 37 Using the *Time Distance Speed* and *Average Speed* knowledge points $$ \begin{aligned} &d_{\text{partial}} = 72\text{ km/h} \times 2\frac{1}{3}\text{ h} = 72 \times \frac{7}{3} = 168\text{ km}\ &d_{\text{total}} = 168\text{ km} \div \frac{4}{9} = 168 \times \frac{9}{4} = 378\text{ km}\ &v_{\text{Jessica}} = \frac{378\text{ km}}{4\frac{1}{5}\text{ h}} = \frac{378}{\frac{21}{5}} = 378 \times \frac{5}{21} = 90\text{ km/h} \end{aligned} $$ ## Question 35 10 minutes ## Question 36 1:23 pm ## Question 37 90 km/h { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Algebra", "Time Distance Speed" ] }

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