// Numbas version: exam_results_page_options {"name": "David's copy of Find bounds for distance and time spent running, given imprecise measurements", "extensions": ["random_person"], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "David's copy of Find bounds for distance and time spent running, given imprecise measurements", "functions": {}, "preamble": {"css": "", "js": ""}, "statement": "

{person['name']} is a keen runner. {capitalise(pronouns['they'])} run{verbs} at an average speed of {speed}km/h, rounded to the nearest integer.

", "ungrouped_variables": ["speed", "distance", "time", "atime", "person", "pronouns", "verbs"], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "

Multiplication and division of upper and lower bounds.

"}, "variables": {"pronouns": {"name": "pronouns", "templateType": "anything", "definition": "person['pronouns']", "group": "Ungrouped variables", "description": ""}, "person": {"name": "person", "templateType": "anything", "definition": "random_person()", "group": "Ungrouped variables", "description": ""}, "distance": {"name": "distance", "templateType": "anything", "definition": "speed*time", "group": "Ungrouped variables", "description": "

Distance in kms.

"}, "time": {"name": "time", "templateType": "anything", "definition": "random(0.25..1 #0.25)", "group": "Ungrouped variables", "description": "

Time in hours.

"}, "speed": {"name": "speed", "templateType": "anything", "definition": "random(7..11)", "group": "Ungrouped variables", "description": "

Speed in km/h.

"}, "atime": {"name": "atime", "templateType": "anything", "definition": "random(5..30 #5)", "group": "Ungrouped variables", "description": "

Time in minutes

"}, "verbs": {"name": "verbs", "templateType": "anything", "definition": "if(person['gender']='neutral','','s')", "group": "Ungrouped variables", "description": ""}}, "variable_groups": [], "extensions": ["random_person"], "parts": [{"showCorrectAnswer": true, "type": "gapfill", "marks": 0, "stepsPenalty": 0, "variableReplacementStrategy": "originalfirst", "steps": [{"showCorrectAnswer": true, "type": "information", "marks": 0, "variableReplacementStrategy": "originalfirst", "scripts": {}, "variableReplacements": [], "prompt": "

We're not certain about some of the measurements given in this question - we only know the rounded values. This means that the true value could be lower or higher than the given measurement.

\n

Compute upper and lower bounds for {person['name']}'s average speed and the time spent running, then use those to find upper and lower bounds for the distance travelled.

", "showFeedbackIcon": true}], "scripts": {}, "variableReplacements": [], "prompt": "

Suppose {person['name']} ran for {atime} minutes, rounded to the nearest minute.

\n

Using the rounded figures for {pronouns['their']} average speed and time spent running, calculate upper and lower bounds for the distance, $d$, that {person['name']} ran.

\n

\n

[] $\\leq d \\lt$ []km

Now consider {person['name']}'s evening run. {capitalise(pronouns['they'])} covered a distance of {precround(distance,0)}km, rounded to the nearest kilometre.

\n

{capitalise(pronouns['their'])} friend's record time for the evening run is exactly {time*60 +1} minutes.

\n

Can we confidently say that {person['name']} beat {pronouns['their']} friend's record?

\n

First, calculate the lower and upper bounds for {person['name']}'s time, $t$.

\n

\n

[] $\\leq t \\lt$ [] minutes

Therefore, we [] confidently say {pronouns['their']} time was less than {time*60+1} minutes as the [] bound for time is [] the given threshold of {time*60+1} mins.

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upper

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lower

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We're not certain about some of the measurements given in this question - we only know the rounded values. This means that the true value could be lower or higher than the given measurement.

\n

We can find upper and lower bounds for the given measurements. Any values we go on to calculate will also be uncertain and have upper and lower bounds.

\n

\n

To find bounds for a given measurement, we divide the degree of accuracy by 2 and subtract or add this to our estimate to get lower and upper bounds respectively.

\n

For example, $52$ rounded to the nearest integer has a lower bound of $51.5$ and an upper bound of $52.5$.

\n

#### a)

\n

The distance travelled is given by

\n

\$d = \\text{Average speed} \\times \\text{Time taken} \$

\n

We find bounds for speed and time first.

\n

Lower bound for speed: $\\var{speed} - 0.5 = \\var{speed - 0.5} \\text{ km/h}$

\n

Upper bound for speed: $\\var{speed} + 0.5 = \\var{speed + 0.5} \\text{ km/h}$

\n

First, note that the speed is given in km/h and we want to find the distance in km. We will convert the given time into hours.

\n

Lower bound for time taken:

\n

\\begin{align} \\var{atime} - 0.5 &= \\var{atime - 0.5} \\text{ min} \\\0.5em] &= \\frac{\\var{atime - 0.5}}{60} \\text{ h} \\end{align} \n Upper bound for time taken: \n \\begin{align} \\var{atime} + 0.5 &= \\var{atime + 0.5} \\text{ min} \\\\[0.5em] &= \\frac{\\var{atime + 0.5}}{60} \\text{ h} \\end{align} \n Since we're multiplying the speed and time together, the lower bound for distance is the slowest speed multiplied by the shortest time: \n \\begin{align} \\text{Lower bound} &= \\text{lower bound for speed} \\times \\text{lower bound for time}\\\\ &= \\var{speed - 0.5} \\times \\frac{\\var{(atime - 0.5)}}{60} \\\\ &= \\var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \\text{ km} \\quad \\text{(rounded to 2 decimal places).} \\end{align} \n The upper bound for distance is the fastest speed multiplied by the longest time: \n \\begin{align} \\text{Upper bound} &= \\text{upper bound for speed} \\times \\text{upper bound for time} \\\\ &= \\var{speed + 0.5} \\times \\frac{\\var{atime + 0.5}}{60} \\\\ &= \\var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \\text{ km} \\quad \\text{(rounded to 2 decimal places).} \\end{align} \n Hence, \n \\[\\var{precround((speed-0.5)*(atime - 0.5)/60, 2 )} \\leq d \\lt \\var{precround((speed+0.5)*(atime + 0.5)/60, 2 )} \\text{.}\

\n

\n

#### b)

\n

We're told the speed and the distance travelled, so the time taken is given by

\n

\$t = \\frac{\\text{Distance travelled}}{\\text{Average speed}} \$

\n

We found upper and lower bounds for {person['name']}'s average speed above.

\n

The distance of the evening run is given to the nearest kilometre, so we can compute bounds as follows:

\n

Lower bound for distance: $\\var{distance} - 0.5 = \\var{distance - 0.5} \\mathrm{km}$

\n

Upper bound for distance: $\\var{distance} + 0.5 = \\var{distance + 0.5} \\mathrm{km}$

\n

\n

The upper bound for the time taken is the longest distance divided by the slowest speed:

\n

\\begin{align}
\\text{Upper bound} &= \\text{upper bound for distance} \\div \\text{lower bound for speed} \\\\
&= \\var{distance + 0.5} \\div \\var{speed - 0.5} \\\\
&= \\var{(distance + 0.5)/(speed - 0.5)} \\text{ hours.}
\\end{align}

\n

\n

\\begin{align}
\\var{(distance + 0.5)/(speed - 0.5)} \\text{ hours} &= \\var{(distance + 0.5)/(speed - 0.5)}\\times 60  \\text{ min} \\\\
&= \\var{precround((distance + 0.5)/(speed - 0.5)*60, 2)} \\text{ minutes} \\quad \\text{(rounded to 2 decimal places)}
\\end{align}

\n

\n

The lower bound for time is the shortest distance divided by the fastest speed:

\n

\\begin{align}
\\text{Lower bound} &= \\text{lower bound for distance} \\div \\text{upper bound for speed} \\\\
&= \\var{distance - 0.5} \\div \\var{speed + 0.5} \\\\
&= \\var{(distance - 0.5)/(speed + 0.5)}  \\text{ hours.}
\\end{align}

\n

Converting into minutes, to two decimal places:

\n

\\begin{align}
\\var{(distance - 0.5)/(speed + 0.5)} \\text{ hours} &= \\var{(distance - 0.5)/(speed + 0.5)}\\times 60  \\text{ min} \\\\
&= \\var{precround((distance - 0.5)/(speed + 0.5)*60, 2)} \\text{ min.}
\\end{align}

\n

Therefore, we cannot confidently say {person['name']}'s time was less than {time*60 +1} minutes as the upper bound for {pronouns['their']} time, $\\var{precround((distance + 0.5)/(speed - 0.5)*60, 2)}$ minutes, is above this threshold.

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