A student takes 1.25 hours to travel from home to school at a speed of 4 km/h. By what percentage should he increase his speed to reduce the time by 25% to cover the same distance from school to home?

$33\frac{1}{3}$%

Speed of the student = 4 km/hr

Distance = Speed × Time

Distance between home and school = 4 × 1.25 = 5 km

According to question ,

New time = 1.25 × \(\frac{3}{4}\) =\(\frac{15}{16}\) hr

New speed = \(\frac{5× 16 }{15}\)

= \(\frac{ 16 }{3}\) km/h

Now ,

Ratio of original speed to new speed = 4 : 16/3 = 3 : 4

Let original speed = 3 and New speed = 4

Speed increased by = 4 - 3 = 1

∴ Speed increased by (in %) = \(\frac{1 }{3}\) × 100 = 33\(\frac{1 }{3}\)%