Indian Olympiad

[outer heart]

Jiya leaves school at the same time everyday if she cycles at 20 km/h, she arrives home at 4:30 pm If she cycles at 10 km/h, she arrives home at 5:15 pm At what speed must she ride to arrive home at 5:00 pm?

[stable bolt]

The answer is 12 kmph, give me a minute to explain.

So Jiya covers the same distance from home to school everyday. Let the distance be x.

I'll take the time taken in 20kmph trip as t1 and the time taken in the 10 kmph as t2

We know that the $Speed=\frac{Distance}{Time}$

[gilded relic]

diaas_(yt)

[stable bolt]

So, $20kmph=\frac{x}{t1}$ and $10kmph=\frac{x}{t2}$

[gilded relic]

diaas_(yt)

[stable bolt]

(I won't be writing units from now on btw)

So we can rewrite these statements as:

$t1=\frac{x}{20}$ and $t2=\frac{x}{10}$

[gilded relic]

diaas_(yt)

[stable bolt]

t2 is greated than t1, so Let's find t2-t1

$t2-t1=\frac{x}{10}-\frac{x}{20}$

[gilded relic]

diaas_(yt)

[stable bolt]

Here's the thing, we know t2-t1 value as It's just the time between 4:30 PM and 5:15 PM. This is because Jiya leaves at the same time everyday.

The time between these two times happens to be 3/4th of an hour

Solving for x in this equation: $\frac34=x(\frac1{10}-\frac1{20})$

[gilded relic]

diaas_(yt)

[stable bolt]

We get x=15 km

Now we can find either t1 or t2 induvidually, let me do that for t2.

$t2=\frac{15}{10} = 1.5 hours$

[gilded relic]

diaas_(yt)

[stable bolt]

As Jiya arrived at 5:15 PM and she left school 1.5 hours ago, she left school at 3:45 PM

To arrive at 5PM she has to travel 15 km in 1.25 hours so the speed will be

$Speed=\frac{15}{1.25}=12kmph$

[gilded relic]

diaas_(yt)

[stable bolt]

There you go