Question

29: A student starts from his house with a speed
2kmph and reaches his college 3 min late.
Next day he increases his speed by lkmph
and reaches college 3 min earlier the distance
between his house and college is​

Answer 1

Answer:

0.6km

Explanation:

given

u=2×5/18=5/9

h=60min

1min=1/60

3min=1/20

the travel distance from his home to his college is same if he travel slower and faster the distance willnot change

so 2(t+1/20)=3(t-1/20)

2t-3t=-3/20-1/10

-1t=-3-2/20=1/4

d=2(1/4+1/20)=5+1/10

6/10

0.6km is ans