Question

Jack walked at the rate of 4 km in 45 minutes and John walked at the rate of 4.5 km
in 45 minutes. How many minutes will each walk if they begin walking at the same
time from the opposite ends of a 15 m long road to meet at the mall point
located half way through​

Answer 1

Final Answer:

When Jack walking at the rate of 4 km in 45 minutes and John walking at the rate of 4.5 km in 45 minutes, begin walking at the same time from the opposite ends of a 15 m long road to meet at the mall point located half way through, they take 5.06 seconds and 4.50 seconds respectively.

Given:

Jack walked at the rate of 4 km in 45 minutes and John walked at the rate of 4.5 km in 45 minutes. They begin walking at the same time from the opposite ends of a 15 m long road to meet at the mall point located half way through​.

To Find:

How many minutes will each of Jack and John walk if they begin walking at the same time from the opposite ends of a 15 m long road to meet at the mall point located half way through​.

Explanation:

Speed is calculated as the distance divided by the time taken to traverse the said distance.

Thus, Distance is equal to the product of Speed and Time.

Step 1 of 3

Since Jack and John begin walking at the same time from the opposite ends of a 15 m long road to meet at the mall point, which is located half way through​, it implies that both Jack and John  will cover half of the distance of 15m long road, that is, a distance of 7.5m or 0.0075km.

Step 2 of 3

To cover the distance of 4km, Jack takes the time $=45\;mins$.

It follows that to cover the distance of 1km, Jack takes the time $=\frac{45}{4} \;mins$.

So, it is evident that to cover the distance of 7.5mor 0.0075km, Jack takes the time

$=\frac{(45 \times 0.0075)}{4} \;mins\\\\=\frac{(45 \times 0.0075 \times 60)}{4} \;secs\\=5.06\;secs$

Step 3 of 3

To cover the distance of 4.5km, John takes the time $=45\;mins$.

It follows that to cover the distance of 1km, John takes the time $=\frac{45}{4.5} \;mins$.

So, it is evident that to cover the distance of 7.5mor 0.0075km, John takes the time

$=\frac{(45 \times 0.0075)}{4.5} \;mins\\\\=\frac{(45 \times 0.0075 \times 60)}{4.5} \;secs\\=4.50\;secs$

Therefore, when Jack walking at the rate of 4 km in 45 minutes and John walking at the rate of 4.5 km in 45 minutes, begin walking at the same time from the opposite ends of a 15 m long road to meet at the mall point located half way through, they take 5.06 seconds and 4.50 seconds respectively.

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