Question

A soccer coach riding his bike reaches his office in xx hours. If he travels at 24 km/h, he reaches his office 5 minutes late. If he travels 30 km/h, he reaches his office 4 minutes early. How far is his office from his house?

Answer 1

Answer:

18 km

Explanation:

Let 'd' be the distance between his house and office.

Normal time taken to reach the office = 'x' hours.

If speed is 24 km/h, time is increased by 5 minutes.

If speed is 30 km/h, time is reduced by 4 minutes.

We know that,

Time taken = Distance traveled ÷ Speed

So, when speed is 24 km/hr, time is increased by 5 minutes.

$1\ min = \frac{1}{60}\ h\\5\ min =\frac{5}{60}=\frac{1}{12}\ h$

So, time is $x+\frac{1}{12}$

Therefore,

$x+\frac{1}{12}=\frac{d}{24}\\\\x=\frac{d}{24}-\frac{1}{12}\\\\x=\frac{1}{12}(\frac{d}{2}-1)---------1$

Now, when speed is 30 km/h, time is reduced by 4 minutes or $\frac{4}{60}=\frac{1}{15}\ hours$

So, time now is $x-\frac{1}{15}$

Again using the time formula, we have

$x-\frac{1}{15}=\frac{d}{30}\\\\x=\frac{d}{30}+\frac{1}{15}\\\\x=\frac{1}{15}(\frac{d}{2}+1)-------------2$

Equations (1) and (2) are equal. So,

$\frac{1}{12}(\frac{d}{2}-1)=\frac{1}{15}(\frac{d}{2}+1)\\\\\frac{15}{12}(\frac{d}{2}-1)=\frac{d}{2}+1\\\\\frac{5d}{8}-\frac{5}{4}=\frac{d}{2}+1\\\\\frac{5d}{8}-\frac{d}{2}=1+\frac{5}{4}\\\\\frac{5d-4d}{8}=\frac{4+5}{4}\\\\\frac{d}{8}=\frac{9}{4}\\\\d=\frac{9\times 8}{4}=\frac{72}{4}=18\ km$

Therefore, the office is 18 km from his house.