Triplets Peter, Reeta and Nikita have two ways for getting home from school each day: cycle on a tandem bike or walk. The bike can carry either one or two riders at a time. Regardless of the number of people pedalling, cycling speed is 5 times walking speed. The triplets always leave school at the same time and always use the same path between school and home, whether walking or cycling. The school is 5km from home and their walking speed is 4 kilometres per hour. A On Monday, Nikita and Peter cycle and Reeta walks. On reaching the point four-fifths of the way home the bike gets a puncture, so Nikita and Peter walk the rest of the way home. How far from school is Reeta when the cyclists arrive home?

You can represent the unknown amounts by the use of variables. Follow whatever the description is and convert it one by one mathematically. For example if it is asked to increase some item by 4 , then you can add 4 in that item to increase it by 4. If something is for example, doubled, then you can multiply that thing by 2 and so on methods can be used to convert description to mathematical expressions.

How to find the speed of an object?

If the object is going linearly, and at constant speed, then the speed of that object is given by the distance it traveled to the time it took to travel that distance.

If the object traveled D distance in T units time, then that object's speed is

[tex]Speed = S = \dfrac{\: Distance \: traveled}{\: Time \: taken} = \dfrac{D}{T} \: \rm unit \: length/unit \: time[/tex]

Given that:

Distance from school to home = 5 km
Walking speed = 4 km / hours

Cycling speed = 5 times walking speed = 20 km / hour
They all go and come together to and from school/home.
On monday: Nikita and Peter are coming by cycle, and Reeta walks.

Cycle punctures at four fifth of the way home = [tex]\dfrac{4}{5} \times 5= 4\: \rm km[/tex] from school (as they're coming towards home, so went from school)
After puncture, cyclists walk to home

To find: Disance of Reeta from school when cyclist reach home.

Suppose at time 0 hours, all three people departed from school (on that monday).

After 't' hours, suppose the cycle gets punctured.

Then, as the cycle was going by 20 km / hour speed, so in 't' hours, it must have covered d kilometers (suppose),

then we get:
[tex]S = \dfrac{D}{T}\\\\20 = \dfrac{d}{t}\\\\d = 20 t \: \rm km[/tex]

This distance is measured from school. But we know that this distance is 4 km, so we get:

[tex]20t = 4\\t = \dfrac{4}{20} = 0.2 \: \rm hours[/tex]

The remaining 1 km (as home is 5 km away from school and 4 km is already traveled) is walked by Cyclists. And walking speed is 5 km / hour, so let they take T hours to travel that 1 km walking, then we get:
[tex]S = D/T\\\\5 = \dfrac{1}{T} \\\\T = 0.2 \: \rm hours[/tex]

So, total time cyclists took to reach home from school is: [tex]0.2 + 0.2 = 0.4 \: \rm hours[/tex]

Reeta is walking that whole 5 km.

The time the cyclist reached home, Reeta had walked for 0.4 hours as they had started at the same time, and it took cyclists 0.4 hours to reach home.

Thus, we have:

Time taken 0.4 hours, speed of Reeta = walking speed= 5 km/hour, then we get:

[tex]S = D/T\\D = ST\\D = 5 \times 0.4 = 2 \: \rm km[/tex]

So Reeta was 2 km away from school when cyclists reached home on that monday.

Thus, the distance between Reeta and the school when the cyclists arrive the home for this case is found being 2 km

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