Question

A student walks from his home to school at a speed of 4 km/h and reaches 15 minutes late. If he increases his speed to 6 km/h, he reaches 5 minutes early. What is the distance (in km) from his home to school?

• A. 4 km
• B. 6 km
• C. 8 km
• D. 10 km

Answer 1

Answer:

A. 4 km

Let the distance from home to school be (d) km.

Time taken at 4 km/h:

$\frac{d}{4}\text{ hours}$

Time taken at 6 km/h:

$\frac{d}{6}\text{ hours}$

The difference in arrival times is:

• 15 minutes late to 5 minutes early = 20 minutes

• $(20) minutes (=\frac{1}{3}) hour$

Thus,

$\frac{d}{4}-\frac{d}{6}=\frac{1}{3}$
$\frac{3d-2d}{12}=\frac{1}{3}$

$\frac{d}{12}=\frac{1}{3}$

d=4

Verification

• At 4 km/h:$Time (=\frac{4}{4}=1)$hour.

• At 6 km/h: $Time (=\frac{4}{6}=\frac{2}{3}) hour$= 40 minutes.

Difference (=20) minutes, matching the given condition.

Answer: $(\boxed{4\text{ km}})$